Structural Analysis

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1 STRUCTURAL ANALYSIS UNIT 1

CONTENTS: 1.1

Introduction

1.2

Differential equation of the deflection curve

1.3

Slopes and deflections by direct integration method

1.4

Cantilever beams

1.4.1

Cantilever beam with u.d.1

1.4.2

Cantilever beam with concentrated load at free end

1.4.3

Cantilever beam with variable loading

1.4.4

Cantilever beam with a moment at free end

1.5.0

Macaulay’s Method

1.5.1

Cantilever beam with complex loading Self assessment questions Numerical Examples Reference books Answers to S.A.Q.

AIMS: 1.

To obtain the relation between the moment and curvature.

2.

To obtain the slope and deflection equations for cantilever beams with different types of loadings.

3.

To introduce Macaulay’s method by singularity functions.

OBJECTIVES: After going through this lesson you should be able to 1.

obtain the slope and deflection equations for cantilever beams with any type of loading

2.

know the application of singularity functions.

1.1

INTRODUCTION: In the design of a beam the design engineer has to limit the maximum fibre stresses

and also the maximum deflections. When a beam is subjected to loads, the axis of the beam deflects from its original straight position. Design codes specify the limits on the maximum deflection. In buildings the floor beams and slabs should not deflect excessively to avoid adverse psychological effect on the occupants and also to minimise distress in brittle finishes, ceiling’s partitions etc.

Further, knowledge of slopes and deflections is essential for

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analysing statically indeterminate beams. In this chapter the expressions for the deflection curve (elastic curve) are developed. 1.2

DIFFERENTIAL EQUATION OF THE DEFLECTION CURVE: Consider a beam AB simply supported at A and B which is straight before the

application of the load. The shape A p B represents the deflected shape of the axis of the beam under loading. This curve is known as the deflection curve or elastic curve.

Fig 1.1 To derive the expression for the deflection curve or elastic curve the co-ordinate axis are taken as shown in Fig. 1.1. Take another point q in the neighbourhood of p at a distance dx. Let M be the bending moment and R be the radius of curvature at p. We know the relation between the moment M the radius of curvature R and flexural rigidity of the beam EI. It is given as 1 M = R EI

(1.1)

We know from analytical geometry that in Cartesian co-ordinates, the curvature at any point is defined as 1 d2 y dy = (1 + ( ) 2 )3/2 2 R dx dx

(1.2)

where y is the displacement in the y direction in the x - y plane. However if the deflections are small, the slope

dy dy is small and hence the term ( ) 2 can be neglected. The dx dx

equation 1.2 can then be re-written as 1 d2 y = R dx 2

(1.3)

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Refer Fig 1.2 points p and q are a distance dx apart. tangent at p with the x-axis and d angle

is the angle made by the

is the angle between the normal at p and q. Here the

decreases and x increases. This means that for positive bending moment which

produces concavity upwards, d

is negative for a positive increase dx in x.

There

1 d2 y M E 1 = = - 2 Using the relation , the relation between M and the curvature can be R dx I R R obtained as 2

M = - E I

d y dx 2

(1.4)

Fig 1.2 The equation 1.4 is the differential equation of the deflection curve which must be integrated in each case to obtain the equation of the deflection (or elastic) curve 1.3 SLOPES AND DEFLECTIONS BY DIRECT INTEGRATION METHOD: Rewriting the equation 1.4 as : E I

d2 y dx 2

= -M

On first integration, we get, EI

dy = - M dx + C1 dx

(1.5)

Integrating again, we obtain EIY = -

[ M dx] dx + C1 x + C2

The equations 1.5 and 1.6 give the slopes and deflections respectively at a given point x. The constants of integration C1 and C2 can be determined from the boundary conditions for slopes and deflections. Here the equation 1.4 has been integrated twice, hence this

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method is known as the double integration method. This direct method of integration is helpful in solving simple problems. 1.4

CANTILEVER BEAMS: The student might be familiar with various types of beams. A cantilever beam is one

which is fixed at one end and free at the other end (Fig 1.3)

Fig 1.3 In Fig 1.3, the beam is fixed at A and free at B. At A, the beam can neither deflect nor rotate and hence both slope and deflection are zero at fixed end. Cantilevers have wide applications in actual practice.. For example cantilever sun shades in buildings, cantilever porticos, and cantilever balconies in theatres are widely constructed. In the sections that follow, we derive the deflection equations for cantilever beams under different loading conditions. 1.4.1 Cantilever beam with uniformly distributed load: Consider a cantilever beam A B, fixed at A and free at the end B (Fig 1.4).

Fig 1.4 The U.d.1 acting on the beam is w/m. Taking the free and ‘B’ as origin, the bending moment Mx at a distance x will be : Mx = -

Wx 2 2

(a)

From equation 1.4, we get EI

d2y wx 2 = dx 2 2

The first integration gives

(b)

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5 dy Wx 3 EI + C1 = dx 6

(c)

The constant of integration C1 is found from the condition that the slope at the built in end A is zero, i.e.,

dy = 0 for x = l . Substituting these values in (c), we find dx

Wl3 C1 = , 6

(d)

The second integration gives Wx 4 Wl 3 - ( ) x + C2 EIY = 24 6

(e)

The constant C2 is obtained from the condition that the deflection is also zero at the builting and A. Thus by substituting x =l and Y = 0, we obtain Wl 4 C2 = 8

(f)

Substituting C1 and C2 in the equations (c) and (e), we obtain: EI

dy Wx 3 Wl 3 = dx 6 6

EIY=

(1.7)

Wx 4 Wl 4 Wl 4 = ( )x + 24 6 8

Substituting x = 0 in the relations 1.7 and 1.8, we obtain the slope

(1.8) B

and YB at free

end B. B

=

Wl3 Wl 4 and YB = 6 EI 8 EI

1.4.2 Cantilever beam with a concentrated load at the free end

Fig 1.5 The cantilever beam A B carries a concentrated load P at the free end B. The moment at x is equal to : Mx = - px

(a)

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6 d2y EI = Px dx 2

(b)

First integration gives: EI

dy Px 2 = + C1 dx 2

The constant C1 is determined from the condition that

(c) dy = 0 at x = l dx

substitution these in equation (1), we get Pl 2 C1 = 2

(d)

Integrating (c), we get EIY =

Pl 2 Px3 - ( ) x + C2 6 2

(e)

The constant C2 is obtained from the condition that the deflection is zero at x = l on substitution C2 =

Pl3 3

Substituting the values of C1 and C2 in equations © and (e), we obtain: EI

dy Px 2 Pl 2 = dx 2 2

EIY

Px3 Pl 2 Pl 3 - ( )x + 6 2 3

(1.9) (1.10)

Substituting x = 0 in the equations (1.9) and (1.10), we find the slope and deflection at the free end as: B

=

Pl 2 Pl3 and YB = 2 EI 3EI

1.4.3 Cantilever beam with variable loading: Consider a cantilever beam carrying a triangular load with zero intensity at free end and an intensity of w/m at the fixed end.

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Fig 1.6 Taking B as origin, the intensity of loading Wx at the x is equal to Wx =

Wx l

Wx 2 2 total load acting upto x is equal to and its c.g. lies at x from B. Thus 2l 3 Wx 2 2 M = -( )× x 2l 3

(a)

from equation 1.4, we have d2y wx 3 EI = dx 3l

(b)

first integration gives EI

dy Wx 4 = + C1 dx 12l

(c)

The constant C1 is obtained from the condition that at x = l, C1 =

dy = m0 dx

Wl3 12

second integration gives EIY

Wx 5 Wl 3 - ( ) x + C2 60l 12

C2 is determined from the condition that at x = l. y = 0 C2 =

Wl 4 15

(d)

The

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substituting these values of C1 and C2 in equations (c) and (d) we get EI

dy Wx 4 Wl3 = dx 12 12

EIY

(1.11)

Wx 5 Wl 3 Wl 2 - ( ) x + 60l 12 15

(1.12)

1.4.4 Cantilever beam with a moment at the free end:

Fig 1.7 The cantilever beam is acted upon by a clock wise moment M at the free end. The moment at x is equal to : Mx = - M EI

(a)

d2y =M dx 2

(b)

Integrating: EI

dy = Mx + C1 dx

from the condition

(c)

dy = 0, at x = l, we get dx

C1 = - Ml so further integration; we obtain, Mx 2 EIY = - M l + C2 2

(d)

substitution values of C1 and C2 Equations (c) and (d) we obtain EI

dy = Mx = Ml dx

EIY B=-

Mx 2 Ml 2 - Ml x+ 2 2 Ml Ml 2 and YB = EI 2 EI

(1.13) (1.14)

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MACAULAY’S METHOD: In the earlier section we have considered cantilever beams with simple loading. In all

these cases the moment is represented by a single function over the entire span. In many loading conditions, the moment may have different expressions in different zones over the span.

Fig.1.8 See the Fig 1.8. Here the expressions for the bending moments in the various zones BC, CD, DE and EA have different forms. Hence if we use double integration method it is some what complicated and time consuming. To avoid this difficulty, to solve this kind of problem, a versatile technique known as Macaulay’s method can be adopted.

Before

developing this method, we have to know about singularity functions based on which this method was developed. Singularity functions: fn(x) = [x - a]n is a singularity function. The function has the property: [x - a]n = 0

for 0 < x < a

(x - a)n

for a < x <

where n

0 (i.e., n = 0, 1, 2, .............)

The function in the brackets is non existent until x reaches a. For x beyond a, the expression becomes an ordinary binomial. For n = 0 and x > a, the function is unity. If n has a negative value, for the first two negative values the functions are always zero except when x = a. Here they are infinite but they are integrated as follows: x 0

[ x - a]-2 dx = [x - a]-1

(1.15)

x 0

[ x - a]-1 dx = [x - a]0

(1.16)

For positive values of n the function is integrated as follows: x 0

[ x - a]n dx = [x - an) +1/n+1 for n > 0

1.5.1 Cantilever beam with complex loading:

(1.17)

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Macaulay devised a method of continuous expression for bending moment and integrating in such a way that constants of integration are valid for all sections of the beam even though the law of bending moment varies from section to section. Now we shall discuss the application Macaulay’s method.

Fig 1.9 See the Fig 1.9 P1 and P2 are concentrated loads. M is the couple acting at a distance (a + b) from the free end. Taking B as origin, the moments in the various zones are given as: Between B and C : Mx = -P1 x

(1.18 a)

Between C and D : Mx = -P1 x - P2 (x - a)

(b)

Between D and A : Mx = -P1 x - P2 (x - a) - M ( x - a - b)0

(c)

According to the property of the singularity functions, negative terms do not exist. The three expressions (1.18 a), (b) and (c) can thus be conveniently combined into a single expression with vertical lines separating them as : Mx = -P1 x | - P2 (x - a) | - M (x - a - b)0

(d)

In the expression (d), P1 x exist every where on the span, P2 (x - a) exists for X > a, and M(x - a - b)0 exist for x > (a + b). Using the equation 1.4. EI

d2y = P1 x | + P2 (x - a) | + M (x - a - b)0 dx

(e)

Integrating we get EI

dy = P1 x 2/2| + P2 (x - a)2/2 | + M (x - a - b) | C1 dx

The constant C1 is found from the condition that at x = l, are deleted in the expressions for convenience hereafter.

(f)

dy = 0. The vertical lines dx

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P1l 2 P2 (l a) 2 0 = + + M (l- a - b) + C1 2 2

(g)

Solving for C1, C1 =

P1l 2 P2 (l a) 2 - M (l- a - b) 2 2

Integrating the expression (f) again; we obtain: EIY=

P ( x a )3 M (x - a - b) 2 P1 x 3 + 2 + + C1 x + C 2 6 6 2

(h)

The constant C2 is obtained from the condition that at x =l, y = 0 0=

P1l3 P (l a )3 M (l - a - b) 2 + 2 + + C1 x + C2 6 6 2

Simplifying: P1l3 P2 (l a ) 2 (l + a ) M (l - a - b) l - M C2 = + + 3 12 2 (l - a - b) 2 substituting C1 and C2 in the equation (f) and (h), we get the expressions for slope and deflection. SELF ASSESSMENT QUESTIONS: 1.

The deflection at the free end of a cantilever beam carrying a concentrated load ‘P’ at the free end is equal to. (a)

2.

Pl 2 Pl3 Pl 4 (b (c) 2 EI 3EI EI

(d) None of the above

For the cantilever beam carrying u.d.1 over the entire span, the maximum deflection occurs: (a) At fixed end

3.

(b) At centre

(c) At free end

(d) None of the above

Write down the expression for the bending moment for the cantilever loaded as shown.

Fig 1.10 4.

The constants of integration C1 and C2 are determined from the conditions:

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(a) Deflection at fixed end is zero

(b) Slope at fixed end is zero

(c) Both slope and deflection are zero at fixed end (d) None of the above 5.

In a cantilever beam, the equation of the deflection curve is (a) Straight line (b) Parabola

(c) Catenary

(d) None of the above

Numerical Examples: In this section, a few numerical examples are worked out to give clear concepts in typical loading cases. Example 1.6.1 : A cantilever beam of span 10m carries an u.d.1. of intensity 10 KN/m over the right hand half of the span. Find the slope and deflection at the free end.

Fig 1.11 Solution: Assume the u.d.1 is acting over the entire span on top and apply negative loading acting upwards at bottom over the left hand half of the span (See Fig.b) Taking ‘B’ as origin 2

Mx = EI

Wx + 2

W(x -

l 2 ) 2

2

d2y Wx 2 l = - W (x - ) 2 2 dx 2 2

(a) (b)

Substituting W = 10 KN/m and l= 10 m dy 5x 3 EI = - 5 (x - 5) 2 dx 3

(c)

Integrating: EI

dy 5x 3 5 (x - 5)3 = + C1 dx 5 3

At x = 10 m,

dy = 0 dx

C1 = -1458.33 Integrating again:

(d)

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5x4 5 EIY = ( x 5) 4 - 1458.33x + C2 12 12

(e)

At x = 10 m, y = 0 C2 = 10677.08 Substituting the values of C1 and C2 in the equations (d) and (e) we get dy 5 = EI dx 3

x 3 - 5 (x - 5)3 - 1458.33 3 3

5 x4 5 EIY = (x - 5) 4 - 1458.33x + 10677.08 12 12 substituting x = 0, we get slope and deflection at the free end. B

= -

YB =

1458.33 EI 10677.08 EI

Example 1.6.2: Find the slope and deflection for the cantilever beam at the free end. Adopt Macaulay’s method. A couple 20 KN/m is acting at 2 m from the free end.

Fig 1.12 Taking ‘B’ as origin Mx = -50 x - 20 (x-2)0 – EI

10 (x - 3) 2 2

d2y 10 ( x - 3) 2 = 50 x + 20 (x-2)0 + dx 2

Integrating: EI

dy 5 = 25 x 2 + 20 (x - 2) + (x-3)3 + C1 dx 3

Substituting

dy = 0 at x = 6 dx

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0 = 25 x 36 + 20 × 4 +

5 x (3)3 + C1 3

C1 = -1025 Integrating again: EIY = (

25 5 ) x 3 + 10 (x-2)2+( ) (x-3)4 + C1 x + C2 3 12

Substituting y = 0 at x = 6 25 5 x 216 + 10 x 16 + x 81 - 1025 x 6 + C2 3 12

0 =

C2 = 4156.25 Substituting C1 and C2 we have: EI

dy 5 = 25 x2 + 20 (x-2) + (x-3)3 - 1025 dx 3

EIY = B

= -

25 3 5 x + 10 (x-2)2 + (x-3)4 - 1025 x + 4156.25 3 12

1025 EI 4156.25 EI

YB = SUMMARY:

The differential equation for the deflection is : EI

d2y = -M dx 2

Integrating the first integration gives slope and the second integration gives deflection.

The constants of integration C1 and C2 are determined from the boundary

condition at the fixed end i.e.,

dy = 0 and y = 0. dx

EXERCISE PROBLEMS 1(a) A uniform cantilever beam of span ‘l’ fixed at A and free at B carries u.d.1 for a length of l/2 from fixed end. Further, it carries a concentrated load P at the free end B. Find the slope and deflection at the free end.

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(b) What upward force is required at B to bring the free end “B” to the same level as the fixed end A. 2.

A cantilever beam of span 5 m carries loading as shown. Find the slope and deflection at the free end.

Fig 1.13 3.

A cantilever of span ‘L’ is subjected to a point load ‘w’ and an external moment ‘M’ at a distance of ‘a’ from its fixed end. Calculate the deflection at the loading point and the maximum deflection in the beam?

4.

A cantilever beam of span 20 m carries an U.D.L. of intensity 20 KN/m over the entire span and also a concentrated load of 10 KN is acting at the centre of the span. Find the slope and deflection at the free end.

5.

A cantilever beam of span 10m carrying a linearly varying load of intensity zero at the free end and 15 KN per unit length at the fixed end. Find the slope and deflection at the free end.

Fig 1.14 6

A Cantilever beam of length ‘l’ carries u.d.1. of intensity of W KN/m over right hand half length from free end. Find the equation of the deflection curve. Hence find the slope and deflection at free end.

Answers to self assessment questions: 1.

(b)

2.

(c)

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3.

Mx = -M - P1(

4.

(c)

5.

(b)

16 x 1 ) 2

Reference Books: 1.

Analysis of structures : Vol.-I By V N Vazirani and M M Ratwani

2.

Strength of Materials : Part - I By S Timshenko D Vass Nostrand company INC New York

3.

Introduction to Mechanics of solids By E P Poppy: Prentice-Hall of India Pvt Ltd., New Delhi

4.

Structural Theory and Analysis By J D Todd. E L B S and Macmillan

5.

Theory and problems of strength of Materials By W A Nash : Schaum’s Outline series McGraw-Hill internal Book company, Singapore. *** STRUCTURAL ANALYSIS UNIT 2

CONTENTS: 2.1

Introduction

2.2

Simply supported beam with a central concentrated load

2.3

Simply supported beam with U.D.L. over entire span

2.4

Simply supported beam with linearly varying load

2.5

Simply supported beam with non-central load Numerical Examples Self Assessment Questions Summary Answers to S.A.Q.

AIMS: 1. To develop deflection equations for simply supported beams and over hanging beams with different loading conditions.

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OBJECTIVES: This unit will enable you to develop the deflection equations for simply supported beams and over hanging beams for any kind of loading. Also the values of slopes and deflections for standard loading cases of simply supported beams are developed. 2.1. INTRODUCTION: Simply supported beams have wide application in the actual construction. Both R.C.C. and steel beams are used to support floors and bridge decks. The design criterion of these structures also limits the magnitude of deflections. In this lesson we discuss about the development of elastic curves for simply supported beams with different loading conditions.

Fig 2.1 See Fig.2.1. The beam is simply supported at the ends A and B. RA and RB are the reactions developed due to external loads acting on the span. These reactions can be readily determined from static equilibrium equations. The curved shape is the deflected shape of the beam after loading. From the deflected shape we can conclude that the deflections are zero at the ends but slopes are present. These geometric boundary conditions on deflection are used to determine the integration constants. 2.2 SIMPLY SUPPORTED BEAM WITH A CENTRAL CONCENTRATED LOAD.

FIG 2.2

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See Fig.2.2. The simply supported beam AB carries a concentrated load P at the centre of the span. Because of symmetry the support reactions at A and B are equal and their P 2

value is equal to

Taking A as origin, the moment at any point x is equal to: P 2

Mx =

x

(a)

Applying the relation 1.4 EI

d 2y dx

2

=

P 2

x

(b)

Integrating: EI

dy - p.x 2 = + C1 dx 4

As the beam is loaded symmetrically, x =

(c) dy dy = 0 at the centre. Substituting = 0 at dx dx

L . 2

PL2 + C1 16

0=

PL2 C1 = 16 EI

dy - Px 2 PL2 = + dx 4 16

(2.1)

Integrating again: EI Y =

Px3 PL2 + x + C2 12 16

(d)

We know that at x = 0, y = 0. Applying this condition we get C2 = 0 Px3 PL2 EIY = + x 12 16 At x = 0, At x = 1,

(2.2)

A

PL2 = 16 EI

(2.3)

B

PL2 = 16 EI

(2.4)

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Maximum deflection occurs at centre ‘C’, substituting x = YC =

PL3 48 EI

L in equation 2.2. 2

(2.5)

2.3 SIMPLY SUPPORTED BEAM WITH UNIFORMLY DISTRIBUTED LOAD

Fig 2.3 Consider a simply supported beam AB carrying an uniformly distributed load of intensity w/m over the entire span. Because the load is symmetric, the reactions are equal and their value is equal to Mx = ( EI

d 2y dx

EI =

2

wL . Taking A as origin. 2

wL wx 2 )x 2 2

= -(

dy = dx

(a)

wL wx 2 )x + 2 2 wL x2 4

wx 3 6

(b) + C1

Because of symmetry, the slope at centre i.e., at x = x=

(c) L dy is zero. Substituting = 0 at 2 dx

L , we get 2

wL3 C1 = 24 EI

(d)

wL 2 wx 3 wL3 dy = x + + 4 6 24 dx

(2.6)

Integrating: EI y =

wx 4 wL 3 + x + 12 24

wL3 x + C2 24

Substituting at x = 0, y = 0, we get C2 = 0

(a)

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20 wx 4 wL3 wL 3 x + + 12 24 24

EI Y =

at x = 0,

A=

at x =L ,

B=

x

wL3 24 EI

(2.7)

(2.8)

wL3 24 EI

(2.9)

Maximum deflection occurs at centre: At x =

L 5 , yc = 384 2

wL4 EI

(2.10)

2.4 SIMPLY SUPPORTED BEAM LINEARLY VARYING LOAD: See Fig 2.4 AB is a simply supported beam carrying a linearly varying load of intensity zero at A and w/unit length at B. If wx = intensity of loading at a distance ‘x’ from A, then

Fig 2.4 wx = RA =

wx L

(a)

wL wL and RB = 6 3

wx = total load up to x = The c.g of wx is at Mx =

wL 6

x -

(b) 1 x 2

wx =

wx 2 2L

x from section x 3

wx 3 6L

(c)

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21 d 2y

EI

dx

2

= -

wx 3 wL x + 6L 12

(d)

Integrating: dy wL 2 wx 4 )x + EI = -( + C1 dx 12 24 L EI Y =

wL 3 x + 36

(e)

wx 5 + C1 x + C2 120L

(f)

Substituting the conditions: at x. = 0, y = 0 at x = L, y = 0 we get C2 = 0 and 7 wL3 360

C1 =

Substituting the values of C1 and C2 in the equation (e) and (f), we obtain: EI

wL 2 wx 4 dy =+ x + 12 24L dx

EI y =

wx 5 wL 3 + x + 36 120L

Maximum deflection occurs where

7 wL3 360

(2.11)

7 360

(2.12)

wL3x

dy = 0. Therefore equating equation (2.11) to dx

zero: 0 =

wx 4 7 wL 2 + wL3 x + 12 24L 360

Solving x = 0.5192 L

(g) (h)

The maximum value of deflection is : ymax

wL4 = 0.006505 EI

2.5 SIMPLY SUPPORTED BEAM WITH NON-CENTRAL CONCENTRATED LOAD:

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FIG 2.5 Taking ‘A’ as origin and applying Macaulay’s method. Pb l

Mx = EI

d 2y dx

2

x -P (x - a) Pb l

=

(a)

x + P (x - a)

(b)

Integrating: EI

Pb 2 P (x - a)2 dy =x + 2l 2 dx

EI y =

(

+ C1

P (x - a)3 Pb 3 )x + + C1 x + C2 6l 6

(c) (d)

When x < a, the expression (x - a) is not considered because the ‘argument’ (x - a) is negative and neglected because the term as a whole is negative as per singularity functions. y = 0 at x = 0, we obtain C2 = 0 Similarly putting y = 0, at x = l P (l - a)3 Pb 2 l + C1l + 6 6

=0

(e)

Keeping (l - a) = b and simplifying C1 =

Pb 6l

(l2 - b2)

On substitution C1 and C2 in expressions (c) and (d), we get EI

Pb 2 dy = x + 2l dx

EI y =

Pb 3 x + 6l

Pb 6l

Pb 6l (l 2

b2 ) +

P (x - a)2 2

(2.13)

b2 ) x +

P (x - a)3 6

(2.14)

(l 2

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If a > b, maximum deflection occurs between A and C. For maximum deflection equating

dy dx

= 0 in the zone AC. The last term of eq. (2.13) is not included as x is in zone

AC. 0=

Pb 2l

x2 +

Pb 6l

(l 2

b2 )

(f)

Solving for x, we get, taking the positive value x=

l2

b2 3

and the maximum deflection y max , ymax =

Pb ( 9 3 l EI)

(l2 - b2)3/2

(g)

Numerical examples: In this section a few numerical examples are worked out to give clear concepts in typical loading cases. Example:2.6.1: A simply supported beam AB of span 6m carries a concentrated load of 120 KN at 2 m from left hand support and a clockwise couple of 600 KN.m is acting at 3m from the right hand support. Find the slopes at supports and deflection at the centre of the span.

Fig 2.6 Taking moments about B: RA × 6 - 120 × 4 + 600 = 0 RA = - 20 KN RA + RB - 120 = 0 RB = 140 KN Taking A as origin: Mx = -20x - 120 (x - 2) + 600 (x - 3)0

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24 d2y EI = 20 x + 120 (x - 2) - 600 (x - 3)0 2 dx

Integrating: EI

dy = 10x2 + 60( x - 2)2 - 600 (x - 3) + C1 dx

EI y =

10 3 x + 20 (x - 2)3 - 300 (x - 3)2 + C1 x + C2 3

(a) (b)

From the conditions: (i) y = 0 at x = 0 (ii) y = 0 at x = 6 we get : C2 = 0 and C1 =

700 6

Substituting C1 and C2 in the equation (a) and (b) we have EI

dy 350 = 10x2 + 60 (x-2)2 - 600 (x-3) + dx 3

EI y =

A

=

B

=

10 3

x3 + 20(x -2)3 - 300 (x - 3)2 +

(c) 350 3

x

(d)

350 3EI 1090 3EI

Deflection at centre : i.e., at x = 3 yC =

460 3EI

2.6.2: A simply supported beam of span 8 m carries a u.d.1 of 20 KN.m over a length of 4 m symmetrically about the centre of span. Derive the deflection equation. Find the deflection at centre and slopes at the ends.

Fig. 2.7

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Fig. 2.8 From statics it can be found that RA = RB = 40 KN Assume the u.d.1 of 10 KN.m acts downward on the part DB also and negative loading of same intensity acts upwards at bottom in DB as shown. This does not alter the slopes and deflections of the original beam. Taking ‘A’ as origin; and applying Macaulay’s method Mx = 40 × EI

d 2y dx

2

20 (x - 2) 2 20 (x - 6) 2 + 2 2

= - 40 x + 10 (x - 2)2 - 10 (x - 6)2

Integrating: EI

dy dx

= -20 x2 +

EI y = -

20 3

x3 +

10 3 10 12

(x-2)3 -

10 3

(x-6)3 + C1

(a)

(x-2)4 -

10 12

(x-6)4 + C1 x + C2

(b)

From the geometric boundary conditions, (i) y = 0 at x = 0 (ii) y = 0 at x = 8 We get C2 = 0 and C1 = 293.33 Substituting C1 and C2 in the equations (a) and (b), we have: EI

dy 10 10 = -20 x2 + (x-2)3 12 3 dx

EI y = -

20 3

x3 +

10 12

(x-2)4 -

Hence A

=

293.3 ; EI

B

=

293.3 EI

10 12

(x-6)3 + 243.3

(c)

(x-6)4 + 243.3x

(d)

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Deflection at Centre i.e., at x = 4m yc =

759.86 EI

Note: The third term will not be considered as it enters if x > 4. 2.6.3: A double overhanging beam with two equal overhangs of ‘2m’ and supported span of ‘6m’ carries uniformly distributed load of 15 KN.m over the entire span. Find the slopes at free ends and at supports. Also find the deflection at centre of the span. From statics it can be found RA = RB = 75 KN Taking ‘c’ as origin: Mx = -

15 2

x2 + 75 (x - 2) + 75 (x - 6)

The third term enter for x > 6 EI

d2y = 7.5 x2 - 75 (x - 2) - 75 (x-6) 2 dx

Fig 2.9 Integrating: EI

dy 7.5 x 3 75 (x - 2) 2 75 (x - 6) 2 = C1 dx 3 6 2

EI y

7.5 x 4 75 (x - 2)3 75 (x - 6)3 + C1 x + C2 12 6 6

From the geometric boundary condition (i) y. = 0 at x = 2 (ii) y = 0 at x = 8 We get C2 = -10 and C1 = 18.75

(a) (b)

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27

Substituting C1 and C2 in the expressions (a) and (b) we get dy 7.5 x 3 7.5 7.5 = (x-2)2 (x-6)2+18.75 - ( 6 3 2 dx

(c)

7.5 x 4 7.5 7.5 EI y = (x-2)3 (x-6)3 + 18.75 x - 10 6 6 12

(d)

EI

Substituting values for ‘x’, we get: C

=

18.75 , EI

YA = A

=

10 EI

38.75 EI

D

18.75 EI

=

10 EI

YD B

=

38.75 EI

Deflection at Centre of span YE =

440.6 EI

Note that when we find YE, the third term in the eq. (d) will not be considered as this will be negative for x > 6. 2.6.4: An overhanging beam ABC supported at A and B (AB = 6m, BC = 2m) carries an u.d.1 of intensity 10 KN.m over the span AB and a concentrated load of 30 KN at the free end C. Find the slopes at A, B and C. Also find the deflection at the centre of span AB and at the free end C.

Fig 2.10 RA = 20 KN RB = 70 KN Assume u.d.1 of 10 KN.m is also acting downwards over BC and simultaneously and a negative loading of same intensity acts upwards at bottom in BC to balance the assumed loading as BC

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Fig 2.11 This will not alter the reactions Taking A as origin: Mx = 20 x - 5 x2 + 70 (x-6) + 5 (x-6)2 EI

d2y = -20 x + 5x2 - 70 (x-6) - 5(x-6)2 2 dx

Integrating: EI

dy 5 5 = -10 x2 + x3 - 35 (x-6)2 - (x-6)3 + C1 dx 3 3

EI y =

10 3 5 4 35 5 x + x (x-6)3 (x-6)4 + C1 x + C2 3 12 3 12

(a) (b)

From the conditions: y =0

at x = 0

y = 0

at

x=6m

we have C2 = 0 and C1 = 30 Substituting C1 and C2 in (a) and (b) we have slope and deflection equations: A

=

30 , EI

B

=

30 and EI

C

=

90 EI

Deflection at centre of span AB i.e., x = 3m. Substituting x = 3 in the equation (b) and neglecting terms with negative expressions in the brackets. Yx=3 =

135 4 EI

Yc = Deflection at free end i.e., at x = 8 m =

8673.33 EI

SELF ASSESSMENT QUESTIONS:

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29

1.

w over m

The maximum deflection in a simply supported beam of span L carrying u.d.1 the entire span is equal to : (a)

2.

wL4 48 EI

(b)

wL3 24 EI

5 wL4 384 EI

(c)

(d)

5 wL3 384 EI

In simply supported beams, carrying and loading the maximum slopes occurs at : (a) at one of the supports

(b) at centre

(c) at quarter span

(d) None of the above 3.

In a simply supported beam, maximum deflection occurs where: (a)

4.

d 2y dx

2

= 0 (b) y = 0

(c) Moment is maximum (d)

dy =0 dx

Two simply supported beams with the same value of EI and spans L1 and L2 carry u.d.1 over the entire span. For the maximum deflections to be equal, the ratio of loads

w1 is w2

equal to : (a) 5.

L2 L1

4

(b) 1

(c)

L2 L1

2

(d)

2

A simply supported beam of span 8 m is carrying a central concentrated load of 10 KN. The deflection at the centre of the span is equal to 10 mm . What will be the deflection if the load is doubled. (Assume even though load is doubled, the stress is with in the elastic limit). (a) 10 mm

(b) 20 mm

(c) 15mm (d) None of the above

SUMMARY: The differential equation for the deflection curve is : EI

d2y = - Mx dx 2

Integrating, the first integration gives slope and the second one give deflection. The constants of integration C1 and C2 are determined from the conditions. (i) y = 0 at x = 0 and (ii)

y = 0 at x = 1 for simply supported beams

For maximum deflections, dy/dx = 0 which is solved for ‘x’ to find the position of maximum deflection. EXERCISE: 1.

A simply supported beam AB of span ‘L’ carries a triangular load, with intensity of w/m at centre and zero at ends. Find the slopes at supports and deflection at centre.

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30

A simply supported beam AB of Span ‘L’ carries a sinosoidal loading of intensity w sin x L

per unit length, with maximum ordinate at the centre of span. Find the slopes at

the supports and deflection at centre. 3. A simply supported beam of span 8m carries u.d.1 of intensity 80 KN.m over the left hand half of the span. Further, it carries a central concentrated load of 50 KN and an anti clockwise couple of 120 KN.m at the right hand supports. Find the slopes at supports and deflection at centre of span.

Fig.2.12 4. An over hanging beam carries loading as shown. Find the slope at free end C and deflection at centre of span AB.

Fig. 2.13 5.

A simply supported beam of span 7 m carries U.D.L of intensity 6 KN/m over the entire span. Further it carries a concentrated load of 25 KN acting at 2 m from the left hand support. Find the slopes at supports and deflection at the centre of the span.

6. An over hanging beam carries loading as shown in figure. Find the deflection at the centre of span PQ and at the free end R.

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Fig. 2.14 7. A beam ABC is loaded as shown. Compute (i) deflection at free end C (ii) maximum deflection and (iii) slope at end A

Fig. 2.15 ***

STRUCTURAL ANALYSIS UNIT 3 CONTENTS: 3.1

Moment area method

3.2

Moment area theorems and derivation

3.3

Examples on cantilever beams

3.4

Examples on symmetrically loaded simply supported beams

3.5

Conjugate beam method

3.6

Example in Conjugate beam method Summary Self assessment questions (SAQ)

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Exercise problems AIMS: 1.

To derive the moment area theorems

2.

To find slopes and deflections using moment - area theorems for typical cantilever and symmetrically loaded simply supported beams.

OBJECTIVES: After going through the lesson you should be able to: 1.

To derive the moment area theorems independently and understand the Principles involved in it.

2.

To understand the simplicity of determining slopes and deflections as compared to the double integration method.

3.

To solve typical problems independently.

3.1 MOMENT AREA METHOD: In the previous two chapters we have developed expressions for the deflection curve. From the deflection equation, we can find the slopes and deflections anywhere on the span. In many cases however we need to find the slopes and deflections at a particular section on the span. By the moment - area method we can directly find the slope and deflection at any specified section using the moment diagram, thus greatly reducing the labour and complexity of deriving the equation for the deflection curve. It is based on two theorems, usually known as the Mohr’s moment - area theorems. 3.2 MOMENT - AREA THEOREMS AND THEIR DERIVATION: Theorem - 1: The angle in radians between the tangents at any two points on a beam or the change of slope between these points) is equal to the area of the

M diagram between them. EI

Theorem - II: The deflection of a point B from the tangent drawn at another point A is equal to the moment of the

M diagram between the two points A and B about B. EI

Derivation: Consider two point P and Q which are at a distance dx apart on AB (see Fig .3.1.). R is the radius of curvature and the arc length PQ is approximated by ‘dx’. Therefore,

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33 R d

= dx

or d

=

dx R

d is the angle between the tangents drawn at P and Q, then the angle between the tangents drawn at A and B (or the change of slope between A and B) is given by: =

d

We know that

=

dx R

M 1 1 M or = = EI R R EI

Thus integrating between B and A, we have = A B

A B

M EI .dx

(3.1)

M M is nothing but the area of EI .dx EI

In the figure, the tangents at P and Q are extended until they cut shown that d , the vertical intercept between the tangents is equal to d

= x. d

= x

M EI

dx

on ATB. It can be

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34

Fig 3.1 Integrating between B and A, we have : A

ATB = =

B

d

=

A B

M . × dx EI

M M diagram between B and A about B, where the × dx is the moment of the EI EI

deflection is required. A AT is the tangent drawn at A. 3.3 EXAMPLES ON CANTILEVER BEAMS: Example 1: Find the slope and deflection at the free end of a Cantilever carrying u.d.1. over the entire span. Since A is fixed

A

= 0. The tangent at A is horizontal. According to theorem I, the

angle between the tangents at A and B and consequently B

= Area of

1 3

wl 2 2EI

M diagram between A and B. EI

l =

wl3 6 EI

B

is given by

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35

Fig. 3.2 B is the position of B after deflection. The deflection taking moment of B

=

B

of the free end B is given by

M diagram about B. EI

= Moment of

M diagram about B EI

w l3 3 x l = 6 EI 4

wl4 8 EI

Example 2: Find the slope and deflection at the free end of a cantilever carrying u.d.1 over half the length from the free end. As in the example 1,

B

is equal to the area of

M diagram between A and B. From B EI

to C, the moment varies parabolically and from C to A, it varies linearly, with ordinates as shown. B

= =

1 wl 2 l wl2 l wl 2 l . . + . + x 3 8 EI 2 8 EI 2 4 EI 2 7 wl3 (clock wise). 48 EI

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36

Fig 3.3 The deflection of the free end B is given by taking the moment of the

M diagram EI

about B. B

=

wl 3 3 48 EI 4

wl3 l + 2 16 EI

wl 4 (Downwards) EI

41 384

=

wl 3 l 2 l l l + + + 2 4 16 EI 2 3 2

Example 3: Find the slope and deflection at the free end of a cantilever beam carrying a l from the end. 2

concentrated load P at

B

B

=

Pl 2EI

l Pl 2 = 2 4 EI

= Deflection of free end. = =

1 2

Pl l 2EI

5 24

2 l l + 3 2 2

Pl 3 EI

Example 4: Find the slope and deflection at the free end of the cantilever shown in Fig 3.5.

Moment of inertia of AC is twice the moment of inertia of BC. B

=

= Area of

M diagram EI

1 Pl l Pl 1 1 Pl l × + . + 2 2EI 2 4EI 2 2 4EI 2

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37 5 Pl 2 = 16 EI

Fig 3.4 B

= Moment of M/EI diagram about B

B

=

Pl 2 l 2 l Pl 2 2 l Pl 2 l l + + + + 8EI 3 2 8 EI 2 4 16 EI 2 3 2 =

3 16

Pl 3 EI

Fig 3.5

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38

3.4 EXAMPLES ON SIMPLY SUPPORTED BEAMS: Example 1 : A simply supported beam (Fig 3.6) carries two symmetrically placed loads as shown. Find the slopes at supports and the maximum deflection. As the loading is symmetric, the deflection is maximum at the centre and the slope Zero. Therefore the slope at A. A

= Area of =

M diagram between A and C EI

1 Pa (l - 2a) Pa × a × + . 2 EI 2 EI

Pal Pa 2 = 2 EI 2 EI = deflection at the centre C, which is the same as the deflection of the point B

C

from the tangent drawn at C

Fig.3.6 = Moment of =

=

M diagram between C and B about B EI

(l - 2a) 1 Pa 2 a+ a 2 2 EI 3

Pa 3 3EI

+

Pa l - 2a a+ 2 EI

Pa 2 ( l - 2a) + 2 EI

Simplifying we get

Pa ( l - 2a) 2 4 EI

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39

Pa 3 3a 4a 3 + 3 3EI l l

=

6a 2 l2

Example 2: A simply supported beam shown (Fig 3.7) has a flexural rigidity EI in the

central portion of length

l flexural rigidity 2EI at the end portions. Find the slope at the 2

supports and deflection at Centre. Use moment area method. As the loading and beam is symmetric, maximum deflection occurs at the Centre. Hence the tangent at Centre is horizontal. A

= Slope at A, which is equal to area of

1 l Pl Pl + = 2 4 16EI 8EI c

l 1 l + 4 2 4

M dia. between A and C. EI

8Pl 7 Pl 2 = EI 128 EI

= Deflection at centre: This is same as the deflection of the point ‘A’ from the

tangent drawn at centre. = Moment of M/EI dia between A and C about A =

l pl 2 l pl 2 l 2 l pl 2 2 l . + + + + 128EI 3 4 32 EI 4 8 64 EI 4 3 4

Simplifying we get: =

Pl 3 5 . 256 EI

Example 3: A simply supported beam of span ‘l’ carries u.d.1. over the entire span. Find

the slopes at supports and deflection at centre of span. As the loading is symmetric, the maximum deflection occurs at centre. Hence the tangent at centre ‘C’ is horizontal. A

= Slope at A, which is equal to the area of

1 2 = × 2 3

M dia. between A and C. EI

wl 2 wl 3 l = 8EI 24 EI

= Deflection at centre. This is same as the deflection of point A from the tangent drawn at Centre C. = Moment of

M dia. between A and C about A EI

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40

=

wl 3 24 EI

=

5 384

×

l 5 × 8 2

wl 4 EI

Fig 3.7

CE405/1

41 Fig 3.8

The above values have already been derived early using double integration method. In the previous examples, we have discussed simply supported beams with symmetric loading. Here the maximum deflection occurs at the centre of the span and hence the slope at centre is zero. Therefore the solutions are direct and simple. But when the loading is unsymmetric the solutions are little involved. Therefore in the following examples, simply supported beams with unsymmetric loading are solved. Example 4: A simply supported beam of span ‘l’ carries a non central load ‘P’ at a distance ‘a’ from the left support. Find the slopes at the ends and also determine the slope and deflection under the load. A

= deflection of the Point A from the tangent at B

B

= slope at B

When

B

is small :

=

A

B

(a)

l

But we know that is equal to the moment of M/EI diagram between A and B about A. A

= =

1 Pab (l + a) . × l× l.EI. 2 3 P ab ( l + a) 6EI

(b)

From the relations (a) and (b), we get B

=

P ab ( l + a) 6 EI

Similarly we can find A

=

(c) A

, and is given as:

P ab ( l + b) 6 EI

(d)

Slope and deflection under the load: C

= deflection of the point C from the tangent at B, where C is the position of C

after deflection = Moment of c

= =

M diagram between B and C about C. EI

1 Pab b ×b× × 2 lEI 3

Pab3 6 EIl

(e)

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42 = angle between the tangents at C and B.

CB

Fig 3.9 Deflected shape = Area of

CB

C

M diagram between B and C EI

Pab2 2 EIl

=

= Slope at C

C

=

B

-

=

Pab ( l + a) Pab 2 6 EI l 2 EI l

CB

Simplifying we get C

=

Pab (a - b) 3 EI l

Yc = deflection at the point ‘c’ from the original straight axis i.e., chord AB.

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43

A

=

b

l

-

c

Substituting the values of

and

A

B

from the relations (b) and (e), we get

Pab 3 Pab 2 ( l + a) Yc = 6 EI l 6EI l Pab 2 ( l + a - b) = 6EI l =

Pa 2 b 2 3EI l

(g)

Similarly the slope and deflection at centre can be determined at centre. Example 5 : A simply supported beam of span 8m carries an u.d.1 of intensity of 10 KN.m over left hand half of the span. Find the slopes at the ends. Also determine the slope and deflections at the centre. From statics, the reactions RA and RB can be found. They are equal to : RA = 30 KN RB = 10 KN Taking A as origin Mx = 30 x -

10 x 2 2

0 < x < 4

Taking B as origin Mx = 10 x

0 < x < 4

MC = 40 KN.m = Moment of

A

Moment of

M diagram between A and B about A. EI

M diagram - about A: It can be seen that from A to C the B.M. diagram EI

is Parabola, and C to B is linear. 8 o

M x .x dx = EI 4 0

=

M x x dx =

4 0

1 EI

M x .x.dx + EI 4 0

1 30 x 3 5. x 4 3 4 EI

30 x 4

0

8 4

M x .x.dx EI

10x 2 x dx 2

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44

= 8

320 EI

Mx. ×. dx : As this a triangle, it can be found directly

4

Fig 3.10 8

Mx. ×. dx =

4

40 1 × 4× 2 EI

4+

4 3

1280 EI

=

Adding (a) and (b) we get A=

But

B

B

=

320 1280 746.6 + = EI EI EI

x 8 =

746.6 93.325 = EI EI

Similarly C

a

A

can be found by taking moment of M/EI diagram about B

= Moment of M/EI diagram between C and B about C

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45

=

1 40 4 × 4× × 2 EI 3

=

106.66 EI

yc = deflection at centre C =

B

=

93.325 106.66 266.64 × 4 = EI EI EI

C

=

B

C

-

= area of

CB

= C

× 4-

CB

M diagram between C and B EI

1 40 80 × 4× = 2 EI EI 93.325 80 13.325 = EI EI EI

=

3.5 CONJUGATE BEAM METHOD: The Conjugate beam method is another versatile technique to determine the slopes and deflections in beams at a given section.

The method is developed based on the

mathematical relations between moment Vs load function and deflection Vs

M functions in EI

a beam. If the deflected shape of the beam is described by the function Y(x), the following general relations exist. y = deflection ordinates of the elastic curve dy = dx

`

= slope of the elastic curve

d dM x Vx = = dx dx EI

d3y = dx3

d Mx dx

=

- Vx EI

d4 y dx 4

d Vx dx

=

- Wx EI

=

where Wx. Vx and Mx are the load intensity, shear and moment at the section under consideration. The validity of these relationships depend on the sign convention used for various quantities. The co-ordinate system that satisfy these relationships are y downwards

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46

and x to the right are positive, shear and moment sign conventions are same as adopted earlier. The slopes and deflections by conjugate beam method are obtained as explained below. If the

M diagram for a given beam and loading is considered to be loading on an EI

imaginary beam known as conjugate beam, the following two principles of conjugate beams can be stated. Theorem 1: The shear at any point on the Conjugate beam is equal (in sign and value) to the slopes at the corresponding point on the real beam. Theorem 2: The moment at any point on the Conjugate beam is equal (in sign and value) to the deflection at the corresponding point on real beam. The supports of the Conjugate beam are such that the shear and moment that are obtained in the Conjugate beam are consistent with slopes and deflections in the real beam. The conjugate beam with

M loading for various beams and loadings are as shown in EI

Fig.3.11. With regard to the real beam given in 3.11 a, we see that at the fixed end, no slope and deflection are possible, while at the free end, both slope and deflection would exist. The conjugate beam is supported in a way that no shear and moment are possible at the left end, while shear and moment are generated at the right end. The loading on the conjugate beam acting upwards corresponding to the negative bending moment in the real beam. Similar reasoning is applied to obtain conjugate beam for other beams.

Fig 3.11

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47

3.6 EXAMPLES ON CONJUGATE BEAM METHOD: 1. A cantilever beam of span ‘l’ carries u.d.1 of w/m over the entire span. Find the slope and deflection at free end by conjugate beam method.

Fig.3.12. Fig 3.12.(a) shows the real beam and (b) shows the bending moment diagram on the real beam. Fig (c) shows the loading on the conjugate beam. The shear and moment at ‘B’ in the conjugate beam gives the slope and deflection at ‘B’ in the original beam. For the conjugate beam the reaction (shear) and moment are given as : Vb = the reaction at ‘B’ = Area of =

M i.e., load diagram on the Conjugate beam EI

1 wl 2 wl 4 ×l× = 3 2 EI 8EI

Similarly, the moment at ‘B’ is equal to the moment due to Mb =

M diagram about ‘B’. EI

wl 3 3 wl 4 × l = 6 EI 4 8EI

The shear and moment at ‘B’ in the conjugate beam corresponds to the slope and deflection at B. B

wl 3 = , 6 EI

B

wl 4 = 8EI

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48

2. A simply supported beam of span ‘l’ carries u.d.1 of intensity w/m over the entire span. Find the slopes at the ends and deflection at centre of span. Adopt conjugate beam method.

Fig.3.13 Fig.3.13(a) shows the real beam and (b) shows the bending moment diagram on the real beam. Fig (c) shows the loading on the conjugate beam i.e.,

M diagram. The conjugate EI

beam here is also simply supported beam which is consistent with support conditions. Now the slopes at ‘A’ and “B’ are given by the shears at A and B in the conjugate beam. Similarly the deflection at ‘c’ in the real beam is given by the moment at ‘c’ in the conjugate beam. From statics: VA = VB = = =

1 M [ area of diagram between A and B] 2 EI

1 2 wl 2 × 2 3l 8 EI A

=

wl 3 24 EI

Similarly the moment at section ‘c’ in the conjugate beam is given as: C

= moment of

M diagram between A and C at ‘c’ in the conjugate beam. EI

wl3 l wl 3 3 1 = × × 24 2 24 8 2 5 wl 4 = 384 EI

CE405/1 Summary:

49 In this chapter the moment area theorems are derived. According to these

theorems, if A and B are two points on deflected curve of a originally straight beam, them : AB

= angle between the tangents at A and B. This is given by the area of

M diagram between A and B. EI BA

moment of

= deflection of the point B from the tangent drawn at A. This is given by the M diagram between B and A about B. EI

A few examples are worked on cantilever beams and simply supported beams with symmetric and unsymmetric loading. For cantilever beams, the slope at fixed end is zero and hence it is simple to find slope and deflection. The slope at any section is given directly by the area of

M diagram between the fixed end and the point under reference. The deflection EI

at a point is given by the moment of

M diagram between the fixed end and the point under EI

reference about the point. For simply supported beams with symmetrical loading, maximum deflection occurs at Centre and hence the slope at Centre is zero. Therefore the slope at the ends is given by the area of

M diagram between support and centre. The deflection at centre from the original EI

straight axis is same as the deflection of the point at end (support) from the tangent drawn at centre. Therefore the deflection is equal to the moment of

M diagram over the half of the EI

span about the end. The determination of slopes and deflections for simply supported beams with unsymmetric loading are not direct as in previous case. Examples 4 and 5 illustrate this. Further the Principles of conjugate beam method are developed and examples are presented to illustrate the method. SELF ASSIGNMENT QUESTIONS: Answer the following using moment area theorems 1.

A cantilever beam of span ‘l’ carries a concentrated load P at free end. Find the slope and deflection at free end.

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2.

The area of

50 M diagram between two points gives ____________________ between the EI

tangents drawn at the points. 3.

The moment of the area of

M diagram between two points about the second point EI

gives ___________________ of the second point from the _________________ drawn at first point. 4. In a uniform cantilever beam subjected to specified loading, the slope and deflection are _______________________ is the moment of inertia is doubled uniformly throughout. EXERCISE PROBLEMS 1.

Find the slope and deflection at the free end of a cantilever carrying U.D.l. W per unit length over the entire span. Use moment area method.

2. Find the slope and deflection at the free end of a cantilever carrying U.D.l., W per unit length. Over the entire span. Moment of inertia of AC is twice of the moment of inertia of CB. Use moment area method.

Fig 3. Find the slope and deflection at the free end of cantilever. Using moment area method.

Fig 4. Find the slope at supports and deflection at the centre of the span for the simply supported beam shown. Use moment area method.

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Fig 5.

Solve the problems 1, 2, 3 and 4 by Conjugate beam method. *** STRUCTURAL ANALYSIS UNIT 4

CONTENTS: 4.1

Statically indeterminate beams.

4.2

Propped cantilever beams.

4.3

Examples in propped cantilever beams including numerical examples. Summary Self Assessment Questions Answers to Self Assessment Questions

AIMS: 1. To introduce the term statically indeterminate beams and clearly distinguish between determinate and indeterminate beams. 2. To introduce the concept of geometry/deformation for the solution of statically indeterminate beams in addition to equilibrium equations. 3. To demonstrate that horizontal reactions will not be developed when the beams are subjected to vertical (gravity) loading only. 4. To solve a few typical examples on propped cantilever beams and to draw B.M. and S.F. diagrams. OBJECTIVES: After going through the lesson, you should be able to : 1. To differentiate between statically determinate and indeterminate beams. 2. To determine the degree of indeterminacy of a given beam. 3. To solve examples on propped cantilever beams independently. 4.1 STATICALLY INDETERMINATE BEAMS:

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Consider a simply supported beam AB shown in Fig.4.1. The beam carries a vertical load P1 and an inclined load P2. This loading develops three reactions, two vertical reaction components RA and RB and one horizontal reaction component HA at A. The hinged support at A can resists both horizontal and vertical reactions.

These three reactions can be

determined from the static equilibrium equations. For forces acting in a plane, there are three equilibrium equations namely. Fx = 0, FY = 0 and

M = 0,

Fig 4.1 From the three equations, the three reactions can be determined. In case if P2 = 0, then HA = 0. Now let us discuss about the propped cantilever beam as shown in Fig.4.2

Fig 4.2 The beam is fixed at A and simply supported at B.

There are four reaction

components including the fixing moment MA. Therefore the number of unknowns are four against the three static equilibrium equations. Hence equilibrium alone are not sufficient to solve the reactions completely. One more condition is required to solve it in addition to the three equations given in Eq.(4). This additional condition is obtained from the deformation or geometry of the beam at support.

The number of reaction components in excess of these required for static

equilibrium are known as redundants. The numbers of redundant indicate the degree of static

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indeterminacy. In this case the degree of static indeterminacy is one. The beams for which the reactions can be determined completely from the equilibrium equations are called statically determinate and if they cannot be determined, are known as statically indeterminate beams. For the solution of indeterminate beams, the deformation conditions also called compatibility conditions are obtained from the slopes and deflections of the beams at the supports. 4.2 PROPPED CANTILEVER BEAMS: Let us take a propped cantilever beam subjected to vertical loading only as shown in Fig 4.3.

Fig 4.3 Here as there is no inclined loading, the horizontal reaction becomes zero. The numbers of reaction components are three only. As the beam is subjected to only vertical loading, two equilibrium equations, namely, FY = 0 and the condition

M = 0 are only considered and

Fx = 0 becomes trivial. In this propped cantilever this reaction RB can be

considered as a redundant reaction. To solve this redundant, the geometric condition at B, namely the deflection will be considered. The various steps involved are as follows. The propped cantilever beam can be looked upon as the superposition of two different but individually determinate cases shown in Fig.4.4.

Fig 4.4

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In Fig (a), the propped cantilever with the given loading is shown. In Fig.(b), it is shown as a cantilever subjected to loading as in the original beam, but without the prop.

1

is

the deflection at B of the cantilever due to the applied loading. In Fig ( c ) it is shown as subjected to the redundant reaction RB alone.

2

is the deflection at B in this case. The

geometric condition i.e., the deflection at B in the propped cantilever beam is zero. Therefore the algebraic sum of the deflections

1

and

2

in Figs. 4.4 (b) and (c) shall be equal to zero.

In other words the downward deflection in case (b) is equal to the upward deflection in case (c). 1

+

Here deflection

= 0

2 1

(4.2)

is known quantity where as the deflection

2

is a linear function of

the redundant reaction RB . The equation 4.2 is known as geometric condition from which RB can be solved. Once RB is determined, RA and MA can be found from statics. 4.3 EXAMPLES IN PROPPED CANTILEVER BEAMS: Example 1: To draw B.M. and S.F. diagrams for a propped cantilever beam subjected to u.d.1 over the entire span.

Fig 4.5(a) Consider a cantilever beam subjected to u.d.1 as given.

1

is the deflection at B.

Fig 4.5 (b) 1

=

wL 8EI

Now take cantilever beam subjected to the redundant reaction RB alone.

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Fig 4.5 (c) 2

=

RB L3 3 EI

Since 1

+ RB =

2

is upwards;

2

wL3 = 8 EI

R B L3 3 EI

= 0

3 wL 8

Fig 4.5 (d) From the equilibrium condition, RA + RB = wL We get RA =

5 wL. 8

Taking moments about A: MA =

3 wL2 - wL2 wL2 = 8 2 8

The fixing moment at A is negative which produces tension on top. Taking B as origin moment at a distance ‘X’ is given by: Mx =

3 wx 2 wL x 8 2

To find the position of maximum positive moment, we equate d Mx =0 dx

(a)

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56 dM x 3 = wL - wx = 0 dx 8

x =

3 L. 8

Substituting the values of x i.e.,

3 L in Eq.(a), the maximum positive moment is 8

obtained. Mmax = =

3 3 w 3 wL L1 8 8 2 8

2

9 wL2 128

For position of point of contraflexure, equating Mx = 0, we get

X =

3 wlx 8

wx 2 = 0. 2

3 L 4

Shear force will be zero, where MX is maximum i.e.,

dM x = 0. dx

The B.M. and S.F. diagrams are shown below:

Fig 4.5 (e) Example 2:

central load.

To draw B.M. and S.F. diagrams for a propped cantilever beams with a non-

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Fig 4.6 (a) Step 1: Determine the deflection

1

at B due to the applied load ‘P’ in the cantilever.

Using the moment area method, from theorem 2, we get

Fig. 4.6 (b) 1

= Moment of =

M dia about B EI

1 Pb ×b × 2 EI

a +

2 b 3

Pb 2 (2L + a) 6EI

=

Fig. 4.6 (c) Step 2: Determine deflection 2 = Deflection

2

=

R B L3 3 EI

2

due to RB.

due to reaction RB. (b)

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Fig 4.6 (d) Step 3: Solve RB from the condition 1

Since

2

+

2

= 0

is upwards

1

+

2

=

b2 R B L3 (2L + a) 6 EI 3EI

= 0

Simplifying we get RB =

b2 2

( 2 L + a) L3

In the particular case when a = b = RB =

(c) L 2

5 P 16

Let us take a numerical example: Given P = 100 KN, L = 8m, b = 6 m Substituting the numerical values in Eq.(c), we get RB = 100 x

36 (2 x 8 + 2) = 63.28 KN (8) 3 2

RA = 100 - 63.28 = 36.72 KN.m. MA = 63.28 x 8 - 100 x 6 = -93.76 KN.m Taking B as origin, Mx = 63.28 x - 100 (x - 2) Moment under concentrated load i.e., x = 2 Mx = 2 = 126.56 KN.m. Moment is zero when 63.28 x - 100 (x - 2) = 0 x =

200 = 5.446 m 36..72

This is the point of contraflexure.

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Fig 4.6 (e) Example 3 : For the propped cantilever beam shown find the support reactions and plot b.m. and s.f. diagram. The beams is loaded as shown: Fig (a) shows the given propped cantilever beam with applied loading as in the given beam. Fig. (b) shows the basic determinate cantilever beam without the prop i.e.., reaction RB and Fig. (c) shows the cantilever without the loading and the prop reaction RB only.

Fig 4.7 1

and

2

be the deflection in Fig (b) and Fig (c) respectively. These deflections are

determined using the moment area theorems.

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Fig (d) shows the

M diagram. EI

Fig 4.7 The c.g. os

M EI

is at

1

= moment of

1

=

2

= RB

3 4

L 2

from ‘c’

M dia about B EI

1 L wL2 3 × 1+ × × 4 3 2 4EI

L 2

+

RB =

7 wL4 192 EI

L3 3EI

2

=

(a) (b)

Solve for RB from the condition. 1

=

7 192

2

is negative as it acts upwards

wL4 L3 - RB = 0 EI 3EI

7 w L. 64

(c) (d)

MA =

7 wL L wL2 w L×L × =64 2 4 64

(e)

MC =

L 7 7 wL2 wL × = 2 128 64

(f)

The general expression for moment is given taking ‘B’ as origin as:

Mx = -

7 wL × 64

L 2

w x

2

2

Let us consider a numerical example : L = 8m

w = 10 KN.m

Substituting the values, we get

(g)

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RB =

61 7 × 10 × 8 = 8.75 KN 64

RA = 40 - 8.75 = 31.25 KN (8) 2 MA = -10 × = -10KN.m 64 MC =

7 × 10 × 64 = 35 KN.m 128

The moment will be zero between C and A Substituting in eq.(g), and equating to zero, we get x, Mx =

7 10 (x - 4)2 × 10 × 8 × x = 0 2 64

Solving X = 7.66 mm or 2.088 m. The value 2.088m is not admissible as this lies between B and C. The b.m. and s.f. diagram are drawn as follows:

Fig 4.7 Example 4: An over hanging propped cantilever beam is loaded as shown. Sketch the b.m. and s.f. diagram.

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1

62

= deflection due to applied loading in the cantilever beam at ‘B’ Fig.(b)

2 =

deflection due to prop RB at B

Fig 4.8 Fig (d) shows the b.m.d. for the cantilever beam with the applied loading and (e) shows the b.m.d. due to prop reaction RB. Using the moment area theorems: 1

2

= Moment of

M dia, between C and A about ‘C’ in Fig (d) EI

=

2 30 1 90 × 6×3 + × × 6× ×6 3 EI EI 2

=

1620 EI

= moment of

(a) M dia. In Fig. (e) between C and A about ‘C’ EI

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=

1 6 RB 2 72 R B × 6× × ×6 = 2 EI 3 EI

(b)

But the net deflection in the given propped cantilever beam is zero. 1

+

2

=

1620 72 R B = 0 EI EI

RB = 22.5 KN RA + RB = 15 KN RA = -7.5 KN Taking ‘C’ on origin Mx = - 15 x + 22.5 ( x - 2) MB = - 30 KN.m., MA = 15 KN.m. Mx = 0 when x = 6.0 m The b.m. and s.f. diagram are drawn as follows.

Fig 4.8(f), (g) Example 5: Propped cantilever beam with spring (elastic) support

Fig 4.9 Consider a propped cantilever beam with a spring at support B. This kind of support is termed as elastic support which means that it undergoes deflection under load. K is the

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spring constant which is defined as the force per unit deflection. Rb is the reaction developed at B. If is the compression of the spring, then RB = K 1

= down ward deflection due to u.d.1 over entire span =

2

wL3 8 EI

(a)

= Upward deflection due to RB RB L3 3EI

But if and

2

(b)

is the compression of the spring, then

1

+

2=

, taking the sign of

into account. wL4 = 8 EI

R B L3 3EI

Substituting RB = K =

wL4 8EI

(c)

in Eq. (c), we get:

K L3 3EI

(d)

Rearranging: KL3 3EI

1+

=

=

wL4 8EI

wL4 KL3 8 EI 1 + 3EI

Simplifying 3 8

=

wL 3EI +K L3

(e)

RB = K RB =

3 K 8

wL 3EI K + 3 L

or RB =

3 8

wL 3EI 1+ K L3

Now let us take a numerical example.

(f)

1

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65

Given L = 8m, K = 400 KN.m, w = 3 KN.m Further it is given that, when a vertical load of 10 KN is applied at B when there is no spring there, the deflection at B is 50 mm. 50mm 10 x L3 L3 = = 0.015 1000 3 EI EI EI = 66.66 KN.m L3

Substituting the value of K and

EI in Eq. (f), we get, L3

EI 66.66 = = 0.1666 3 KL 400

RB =

3 3 x 8 = 6.0 KN . × 8 (1 + 3 x 0.166)

24 - 6.0 = 18 KN Now the bending moments and shear forces can be found as usual. Example 6: A propped cantilever beam carries u.d.1 of 4 KN.m over a length of 6 m and a concentrated load of 20 KN is acting at 2 m from the prop. Span of the beam is 10m. Find the reactions developed. Also draw B.M.D indicating salient points.

Fig 4.10 (a) Here it is convenient to calculate the deflection due to applied load as follows (i) due to u.d.1 and (ii) due to concentrated load separately and adding them, using moment area theorem II. (i) Due to u.d.1 Deflection at B =

3 72 1 × 6× 4 + ×6 4 EI 3

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=

66 1224 EI

(ii) Due to concentrated load Deflection at B = = 1

1 160 × 8 × 2 EI

2 x8 3

2 +

4693.30 EI

= total deflection at B due to applied load =

1224 4693.3 5917.3 + = EI EI EI

Fig 4.10 (b)

Fig 4.10 (c) 2

= Deflection at ‘B’ due to RB RB L3 = 3EI RB (10)3 R = 333.33 B 3EI EI From the condition

1

+

2

= 0, taking the signs into account.

R 5917.3 - 333.33 B = 0 EI EI

which gives RB = 17.75 KN

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RA = 44 - 17.75 = 26.25 KN Taking moments about A MA = 17.75 × 10 - 20 × 8 -

4 × (6) 2 2

= -54.50 KN.m Similarly Mc = 17.75 × 2 = 35.50 KN.m MD = 17.75 × 4 - 20 × 2 = 31.0 KN.m Taking ‘B’ as origin the moment between D and A is given by x( 4 ) 2 Mx = 17.75 × - 20 ( x - 2) - 4 2 To find the position of point of contraflexure, equate Mx = 0 and solving, we get x = 7.41 m.

Fig 4.10 (d) SUMMARY:

In this chapter we have discussed about statically indeterminate beams.

Statically determinate beams are, the beams for which the reactions can be solved directly from the static equilibrium alone. For plane determinate structures the three equilibrium equations i.e.,

Fx = 0, .,

FY = 0, and .,

M = 0 are sufficient to determine the

support reactions. But for indeterminate beams, the reactions are in excess of the equilibrium equations. The excess reaction components than the number of the equilibrium equations are called redundant reactions.

These redundant reactions indicate the degree of static

indeterminacy. To solve completely the reaction components for statically indeterminate beams, in addition to the equilibrium equations, additional conditional conditions known as geometric or compatibility conditions are required. The numbers of geometric conditions are equal to the number of redundant reactions. Generally the geometric conditions are obtained from the slope or deflection at the supports. The degree of static indeterminacy for propped cantilever beam is one. examples are worked out to illustrate the solution of propped cantilevers.

A few

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68 SELF ASSESSEMENT QUESTIONS

1. To find the reaction components for a statically determinate beam _______________ equations are sufficient. 2. Any stable structure has to satisfy ________________ equations. 3. A simply supported beam subjected to inclined loading is statically ________________ beam. 4. In a propped cantilever subjected to vertical loading only the degree of indeterminacy is ____________________. 5. To solve indeterminate beams the number of geometric conditions is equal to the _____________ of redundants. 6. ____________________ reactions are defined as the number of excess reaction components over those required for static equilibrium. 7. In a propped cantilever beam the deflection at the prop is ______________________. EXERCISE PROBLEMS: 1. Draw B.M. and S.F. diagrams for a propped cantilever beam of span 10m subjected to a U.D.L. of intensity 15 KN.m at over the entire span. 2. A propped cantilever beam carried U.D.L. of 6 KN.m over the entire length of the span and a concentrated load of 25 KN acting at a distance of 4 meters from the right hand support. Span of the beam is 10 meters. Find the reactions developed. Draw the B.M.D. and also find the point of contraflexure. 3. A propped cantilever beam with an overhang is loaded as shown. Find the prop reaction. Draw b.m. and s.f. diagrams.

Fig 4.11 4. A propped cantilever beam with a overhang is loaded as shown in Fig. Find the prop reaction. Sketch b.m. and s.f. diagram.

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Fig 4.12 5. Solve the problem (1) if the support sinks by 10 mm. 6. Solve the problem (2) if the prop is an elastic spring with the following data. Stiffness of the spring: 400 KN.m, EI for the beam = 35000 KN.m2. *** STRUCTURAL ANALYSIS UNIT 5 FIXED BEAMS CONTENTS: 5.1

Introduction

5.2

Development of the method of analysis of fixed beams

5.3

Examples in fixed beams

5.4

Numerical examples

5.5

Deflections Examples Self assessment questions Summary

AIMS: 1. To introduce the fixed beams and to explain clearly the static indeterminacy. 2. To develop method of analysis of fixed beams using the moment area theorems. 3. Examples are presented to explain clearly the solution of fixed beams with different loading conditions. OBJECTIVES: The main objectives of this lesson are: 1. The student should be able to distinguish clearly the statically indeterminate beams from determinate beams.

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2. After going through the lesson, the student should be able to analyse fixed beams and to draw B.M. and S.F. diagrams and also to find the maximum values of bending moment and shear force in a span. 3. The student should be able to derive the deflection equation for fixed beams for different loading conditions. 5.1 INTRODUCTION: In the previous chapter we have discussed about statically indeterminate and determinate beams. The number redundant reactions give the degree of static indeterminacy. Now let us consider a fixed beam which is subjected to both vertical (gravity) and inclined loading as shown in fig. 5.1. A fixed support is otherwise known as built-in support. At a fixed support, rotation, vertical deflection and horizontal translations are not permitted. Hence three reactions, namely, horizontal reaction H, vertical reaction R and fixing couple M will be developed at a fixed beam, a total of six reactions, three at each support, will be developed as shown in fig 5.1. Hence the degree of static indeterminacy for a beam in general is three.

Fig 5.1 But usually beams are subjected to vertical (gravity) loading only. In that case the horizontal reactions HA and HB become zero, leaving four reaction components. But the available equations are only two, since the condition

FY = 0 become trivial. Hence the

degree of static indeterminacy is reduced to two. For the solution of statically indeterminate beams in to equilibrium equations, additional conditions known as deformation or geometric conditions are required.

The

geometric conditions are developed from the known values of rotation and deflection at supports in the terms of unknown redundants. From these conditions the unknown redundant reaction components are solved. The student should understand that at any support, either the reaction or the corresponding displacement will be known but not both.

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71

DEVELOPMENT OF THE METHOD OF ANALYSIS FOR FIXED BEAMS: Moment area theorems are most conveniently used to solve the fixed beams.

Consider a fixed beam subjected to vertical loading only as shown in fig 5.2 (a).

The

reaction components are: the vertical reactions RA and RB and the fixed end moments MA and MB, the slope as well as the deflection are zero at both supports.

Fig 5.2 (a) As the tangents at A and B are horizontal the angle between them will be zero. Therefore according to moment area theorem I, the area of

M EI

diagram between A and B is

zero. Denote the area of Ms diagram due to applied loading on the beam as if it is a simply supported beam as As . Similarly denote the area of Mi diagram due to the indeterminate fixed end moments as Ai.

Fig 5.2 (b) Simply supported B.M. diagram

Fig (c) Fixed end moment diagram Applying first theorem, if EI is uniform throughout: A A A = s + i = 0 EI EI EI

(5.1)

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72

where Ai = (MA + MB)

L 2

(MA + MB)

L = - As 2

MA + MB =

2As L

(5.2)

Hence As depends on the type of loading on the beam.

In each case, we have to

determine the Ms diagram. Now apply theorem II, as the tangents drawn at A passes through the support at B, the deflection of the support A from the tangent drawn at B is zero. Therefore the moment of M/EI diagram about A will be zero. Ax EI

= 0 where x is the distance of c.g. of

M diagram from the support A. EI

xs and x i are the distance of the c.g. of As and Ai respectively from support A,

If

then A s xs EI

But A i x i = M A ×

+

A i xi EI

= 0

(5.3)

L L L 2L × + MB × × 2 3 2 3

2

L = (M A + 2 MB) 6 From Eq. (5.3), we have 2

L (MA + 2 M B) 6 or (MA + 2 MB) =

=

- As xs

6 A s xs L2

(5.4)

solving (5.2) and (5.4) we get MA and MB. Once MA and MB are known as RA and RB and consequently moments and shears at any section can be found from simple statics. 5.3

EXAMPLES IN FIXED BEAMS:

Example 1:

A fixed beam with central concentrated load P. To draw B.M. and S.F.

diagrams; EI is uniform, throughout. As the beam is symmetrically loaded, We have MA = MB and RA = RB From statics

FY = 0, we get RA + RB = P

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or RA = RB =

P 2

Fig 5.3 From the relation (5.2) MA + MB = As =

- 2 As L

L PL PL2 × L× = 2 4 8

Keeping MA = MB, 2 MA = - 2 x

P L2 - PL = 8× L 4

MA = MB = -

PL 8

Fig 5.3 Simply supported b.m.d. (M diagram) The signs of the fixed end moments are negative and hence these moments develop tension at the top fibres and compression at the bottom fibres. The free body diagram of the fixed beam with load and reactions is as shown in fig. 5.4.

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Fig 5.4 (a) Taking A as origin, the moment at any location X is given as Mx = -

P Pl x , 2 8

0 < x <

L 2

To find out the point of contraflexure (or inflection) i.e., the point where the moment is zero; equating MA = 0, we get =

P Pl x =0 2 8 x = L4

Similarly we have another point of contraflexure at L/4 from B. The maximum positive amount occurs at centre. Mc =

P L x 2 2

PL = 8

PL 8

The resultant moment diagram is as shown.

Fig 5.4 (b) B.M.D.

The S.F.D. is as shown.

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Fig 5.4 (c)S.F.D. Example 2: A fixed beam with u.d.1 over the entire span. To draw B.M. and S.F. diagrams. EI is uniform throughout.

Fig 5.5 Since the beam is loaded symmetrically, we have RA = RB and MA = MB RA = RB = As =

wL 2

2 wL2 wL3 ×L × = 3 8 12

From the relation (5.2) MA = MB = -2 ×

wL3 12 x L 2

MA = MB = -

wL 12

Simply support B.M. diagram

The free body diagram is as shown in fig. 5.6

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Fig 5.6 Free body diagram Taking A as origin Mx =

wl wx 2 wl 2 x 2 2 12

To find the position of points of contraflexure put Mx = 0 wL wx 2 wL2 = 0 x 2 2 12 solving we get x = 0.212L or 0.788L point of contraflexure is at 0.212L from either end. Maximum positive bending moment occurs at centre. 2

Max. positive B.M. =

L wL wL . 2 2 8

2

-

wL 12

2

=

The B.M. and S.F. diagram are as shown

Fig 5.6 (b) B.M.D.

Fig 5.6 (c) S.F.D.

wL 24

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Example 3: Fixed beam with non central load P. Consider a fixed beam carrying a noncentral load P as shown fig 5.7.

Fig 5.7 (a) Since the loading is not symmetric, the fixed end moments MA and MB are not equal. Similarly RA and RB are not equal. As =

L x 2

Pab L

=

Pab 2 (L + a) 3

xs = distance of C.g. of As, from A =

Fig 5.7 (b) Simply support B.M. diagram using the relations (5.2) and (5.4) we have MA + MB = -2 x (MA + 2 MB) = - 6

Pab = 2L

-

Pab L

Pab (L + a) x 2 3L2

(a)

=

-

Pab (L + a) L2

(b)

Solving (a) and (b), we get MA =

Pab 2 L2

MB =

Pa 2 b L2

(c)

Now RA and RB can be determined from statics. This can be better illustrated by a numerical example. 5.4

NUMERICAL EXAMPLES:

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Example 1: A fixed beam of span 10 m carries a non central load of 120 KN at a distance of 4 m from left hand support. EI is uniform throughout. Draw B.M. and S.F. diagrams.

Fig 5.8 (a) Pab 2 120 x 4 x3 6 = = - 172.8 KN.m. MA = 2 L 100 MA = -

Pa 2b 120 x 16 x 6 = = - 115.20 KN.m. 100 L2

(a) (b)

Fig 5.8 (b) Free body diagram Taking moments about B: RA x 10 - 172.8 - 120 x 6 + 115.20 = 0 RA = 77.76 KN

(c)

RB = 120 - 77.6 = 42.24 KN

(d)

Taking A as origin: Mx = 77.76 x - 172.8, ( 0 < x < 4 m)

(e)

To find the position of point of contraflexure, equate Mx to zero. 77.76 x - 172.8 = 0 solving x = 172.8/77.6 = 2.22 m Similarly moment between C and B is given by Mx = 77.76 x - 172.8 - 120 ( x -4) equating Mx to zero 77.76 x - 172.8 - 120 ( x - 4) = 0

(f)

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79

Solving we get x = 7.27 m The points of contraflexure occurs at: x = 2.22 m and x - 7.27 m from A Maximum positive moment occurs at x = 4 m MC = 77.76 x 4 - 172.8 = 138.24 KN.m

Fig 5.8 (b) & (c) Example 2 : A fixed beam AB of span 8 m carries u.d.1 of intensity 10 KN.m over a length of 4 m from A. Draw the B.M. and S.F. diagram.

Fig 5.9 (a)

Fig 5.9 (b)

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80

Fig 5.9 (c) Simply support b.m. diagram First we have to find Ms diagram i.e., simply supported beam B.M. diagram RB = 10 KN RA = 30 KN As = Area of Ms diagram Mx = moment of x AS =

L o

M x . dx

Here the B.M. diagram is a parabola from A to C, and a straight line from C to B. Therefore the area of Ms diagram between A and C can be found by integration. From C to B it can be found directly from the area of the triangle. Taking ‘A’ as origin: Mx = 30 x - 10 x 2/2 A1 =

4 o

= A1 x1 = = =

M x dx = 30x2 2

0 < x < 4 4 o

5x3 3

( 30 x - 5x 2 ) dx 4

= 133.40

4

4

o

o

4 0

Mx . x d x =

(30 x - 5 x 2 ) . x. dx

(30x2 - 5x3 ) dx

30x 3

3

5 4 x 4

4 o

= 320 A1 = area of MS diagram between A and C A2 = area of the MS diagram between C and B. A1 x1

= moment of A1 about A

A1 x2 = moment of A2 about A

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A2 = area of the triangle =

1 x 4 x 40 = 80 2

A2 x2 = 80 ( 4 +

4 ) = 426.60 3

A = A1 + A2 = 133.40 + 80 = 213.30

AS x = A1 x1 +

A 2 x2 = 320 +

426.60

= 746.60 MA + MB = MA + 2 MB =

2 x 213.30 8 6 x 746.60 64

Solving we get MA = - 36.67 KN.m. MB = - 16.66 KN.m. The free body diagram is as shown:

Fig 5.9 (d) Free body diagram

Taking moments about D RA x 8 - 36.67 - 10 x 4 x 6 + 16.66 = 0 RA =

260 8

= 32.50 KN

RB = 40 - 32.5 = 7.5 KN Moment between A and C is given as : 10x Mx = 32.5 x - 36.67 2

2

M(x4) = 32.5 x 4 - 36.67 - 5 x 16 = 13.33 KN Equating Mx to zero, we find point of contraflexure occurs between A and C. 32.5 x - 36.67 - 5 x2 = 0

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Solving we get x = 1.45 m Similarly we find, the point of contraflexure occurs at 2.21 m from B.

Fig 5.9 (e) B.M.D.

Fig 5.9 (f) S.F.D. Example 3 : A fixed beam AB of span 6 m carries two concentrated loads of 50 KN and 60 KN from A as shown. Find the fixed end moments and draw B.M. and S.F. diagrams.

Fig 5.10 (a) When a concentrated load is acting, the fixed end moments are given as MA = -

Pab2 L2

MB = -

Pa2 b L2

using the above relations, the fixed end moments due to the two concentrated loads are calculated separately and are added ( principle of superposition)

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MA = -

50 x 2 x 16 60 x 3 x 9 36 36

= - 44.14 - 45 = - 89.44 KN.m. MB = -

50 x 4 x 4 60 x 3 x 9 36 36

= - 67.22 KN.m.

Fig 5.10 (b) Free body diagram

Taking moments about B: RA x 6 - 87.44 - 50 x 3 - 60 x 2 + 67.33 = 0 RA =

400.22 6

= 66.70 KN

RB = 110 - 66.70 = 43.30 KN M(x = 2) = 66.70 x 2 - 87.44 = 45.96 KN.m. M(x = 3) = 66.70 x 3 - 87.44 - 50 x 1 = 62 KN.m.

Fig 5.10 (c) B.M.D.

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Fig 5.10 (d) S.F.D. Example 4: A fixed beam AB of span “L” carries a clockwise couple ‘M’ from left hand support ‘A’. Find the fixed end moments. If M = 50 KN.m. L = 8.0 m, a = 2 m sketch b.m.d.

Fig 5.11 (a)

Fig 5.11 (b)

Fig 5.11 (c) Simply supported b.m. diagram

Fig 5.11 (d)

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The fixed beam is as shown in Fig 5.11(a). Now take the simply supported beam with applied couple M as shown in Fig 5.11 (b). The reaction at each support will be M/1 as shown in Fig 5.11(b). The bending moment diagram for the simply supported beam taken in Fig (b) is shown in Fig 5.11 © using the moment area theorems: 2As L

MA + MB =

1 Mb 1 Ma x b x a + 2 L 2 L

= -

2 L

= -

M (b2 - a 2 ) 2 L

MA + 2 MB = -

6As xs L2

= -

6 1 Ma 2 1 Mb b xax a+ xbx a+ 2 L 2 L 3 2 L 3

= -

6 M x 2 6L L

[- 2a 3

]

+ 3ab 2 + b3 )

which can be simplified as (MA + 2MB) = -

M (a + b) (b2 + 2ab - a 2 ) 3 L

M (b 2 + 2ab - 2a 3) 2 L M But (MA + MB) = - 2 (b 2 - a 2 ) L M MB = - 2 b2 + 2ab - 2a 2 - b2 + a 2 L M = - 2 (2ab - a 2 ) L = -

[

= -

]

Ma (2b - a) L2

Further MB can be simplified by adding and deducting ‘b’ for the term in the brackets: MB = -

Ma L

3b -1 L

Similarly MA can be obtained as: MA = -

Mb (2 a - b) L2

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= -

86 Mb L

3a -1 L

Substituting the numerical values for M, L, a, and b, we get MA =

- 50 x 6 3 x 2 1 8 8

= 9.375 KNm.

MB =

- 50 x 6 3 x 6 1 8 8

= - 46.875 KNm.

The b.m.d. is as shown in fig. 5.11 (e)

Fig 5.11 (e) B.M.D. 5.5

DEFLECTIONS: The deflections of the fixed beam can be determined by Macaulay method. In a fixed

beam both slope and deflection are zero at both supports. Therefore we have to substitute dy/dx = 0, and y = 0 at x = 0 or at x = L while determining the integration constants. Moment area theorems also can be used to find slope and deflection at any particular section. Example 1: Find the maximum deflection in fixed beam carrying u.d.1 over entire span. RA = RB =

wL 2

wL2 MA = MB = 12

Fig 5.12 (a)

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Fig 5.12 (b) Free body diagram Taking A as origin: wL wx2 x 2 2

Mx = d2y EI dx 2

wL2 12

wL2 wx2 wL + x 12 2 2

=

Integrating dy EI dx

wL2 12

=

wx 3 wLx2 + + C1 6 4

At x = 0, dy/dx = 0 C1 = 0 Integrating again: EIy

wL2 x 2 24

wx 3 wLx 3 + + C2 24 12

At x = 0, y = 0 C2 = 0 Therefore EIy

wL2 24

+

wx 4 24

-

wL 3 x 12

Because of symmetry, maximum deflections occurs at x = L/2 Substituting x = L/2, we get EIymax =

wL2 wL4 ( L / 2) 2 + 24 384

-

wL 3 L 96

Simplifying we get: EIymax =

wL4 384

Example 2: Find the maximum deflection in a fixed beam carrying a concentrated load P at centre.

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Fig 5.13 (a)

Fig 5.13 (b) Free body diagram RA = RB = MA = MB =

P 2

PL 8

Taking A as origin: Mx =

PL P L - xx 8 2 2

EI

P PL L d2y = - + + x2 2 8 2 dx

EI

dy Px 2 PL P =+ x+ 4 8 2 dx

x-

Px 3 PLx 2 P + + 6 12 16

At x = 0, y = 0 EIy =

-

L x2

+ C1

3

+ C2

C2 = 0

Px 3 PL 2 P + x + 12 16 6

x-

Maximum deflection occurs at x = L/2 y EI max = -

2

C1 = 0

At x = 0, dy/dx = 0 EIy =

L 2

P L3 PL L2 x + 12 8 16 4

L 2

3

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89

PL3 = 192 Numerical example 1: A fixed beam of span 7 m carried two concentrated loads of 60 KN and 60 KN at 2 m and 4 m from left hand support. Find the deflection at the centre of span.

Fig 5.14 (a) MA = -

(60 x 2 x 25) 60 x 4 x 9 49 49

= - 105.32 KN.m.

MB = -

(60 x 4 x 5) 60 x 16 x 3 49 49

=

- 83.27 KN.m.

The free body diagram is as shown

Fig 5.14 (b)

Taking moments about B: RA x 7 - 109.32 - 60 x 5 - 60 x 3 + 83.27 = 0 RA =

502.05 = 71.70 KN 7

RB = 120 - 71.7 = 48.30 KN Taking ‘A’ as origin, at ‘x’ from A, Mx = 71.7 x - 105.32 - 60 (x - 2) - 60 (x - 4) EI

d2y dx 2

= - 71.70 x + 105.32 + 60 ( x - 2) + 60 (x - 4) 2

(x - 2) x dy = - 71.70 + 105.32 x + 60 EI 2 2 dx

2

(x - 4) + 60 2

2

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90

x2 x3 EI y = - 71.70 + 105.32 6 2

(x - 2) 3 (x - 4) 3 + 60 + 60 6 6

Integration constants are zero. Deflection at centre is obtained by Substituting x = 3.5 and eliminating the term having the negative value in the brackets, we have EIy (x = 3.5) = - 71.70 x

(3.5) 3 (3.5) 2 + 105.32 x + 60 (3.5 - 2) 3 6 2

= 335.23 335.23 EI

y x = 3.5 =

5.6

SOLUTION TO 4TH ORDER DIFFERENTIAL EQUATION: The fixed beam can be analysed from the solution of 4th order differential equation.

This solution is feasible with symmetric loading acting over the entire span like u.d.1 or sinsoidal loading. The relation between deflection and load intensity acting on the beam element is obtained as follows. The down ward deflection and down ward load are taken as positive. The relations between curvature and moment and shear and load intensity are 2

d y M = - EI dx 2 V =

(a)

dM dx

dV = dx

(b) (c)

-w

From relation (b) & (c) we have V = - EI

d 3y

(d)

dx 3

d4y w = EI dx 4

(5.5)

Thus the Eq.(5.5) gives the relation between the load intensity and deflection of the beam.

The support moments are obtained by successive integration of the 4th order

differential equation and substituting the boundary conditions for shear, moment, slope and deflection. Example: A fixed beam of span ‘L’ carries u.d.1, of intensity of w/m over the entire span. Find the fixed end moments and maximum deflection.

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Solution:

Fig 5.15 RA, RB be the end reactions and MA and MB are the fixed end moments. Due to symmetric loading RA = RB = wL/2. We know that the boundary conditions for slope and deflection, using the 4th order differential equation. d4y =w dx 4

EI

(a)

Integrating 3

d y EI = - V = wdx + C1 dx 4

(b)

- V = wx + C1

(c)

At x = 0, V = wL/2, substituting C1 = -

wL 2

d3y EI 3 dx

= -

wL + wx 2

(d)

Integrating again d2 y EI 2 dx

=

wL wx2 x + 2 2

+ C2

d 2y At x = 0, EI = -MA dx 2 C2 = -MA On further integration: EI

dy dx

EI y =

= -

wL 2 wx3 x + -M 4 6

- wLx 3 12

+

wx 4 24

-

A

x + C3

MA x2 + C3 x + C4 2

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92

At x = 0, y = 0 x = 0,

C4 = 0

dy = 0 dx

C3 = 0

Substituting there values: EI y = -

wx 4 wLx 3 + 24 12

-

M A x2 2

(e)

At x = L, y ==0, we get MA = -wL2/12 Now taking the moments about ‘B’ wL wL2 L 2 2

- MA + M B = 0 wL2 12

MB = MA =

These values are same as the results obtained from fundamentals. The maximum deflection occurs at x = L/2. Substituting x = L/2 in eq. (e), we get wL4 384

EIYmax = Summary:

In general the degree of static indeterminacy for a fixed beam is three. But in the special case where the loading is vertical, the degree of indeterminacy is reduced to two. For the vertical loading condition, the horizontal reactions becomes zero. Using the moment area theorems, the fixed end moments can be found from the relations namely: (MA + MB) = -

2 As L

(MA + 2 MB) = -

6 As x L

(a) (b)

where the various terms are explained in the lesson. At two points in a fixed beam the moment is zero. These are known as contraflexure or inflection points. For symmetric loading for a prismatic (with uniform cross section) beam, the fixed end moments are equal. The fixed end moments are negative in sign which produces tension on top. Further it is found that the deflections in fixed beams are smaller in magnitude as compared to corresponding simply supported beams.

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EXERCISE PROBLEMS: 1. A fixed beam of span 6 m carries u.d.1 of 10 KN.m over the left hand half of the span. Further, it carries a concentrated load of 50 KN at 2 m from the right support. Draw the B.M.D. and S.F.D. 2. A fixed beam of span 8 m carries u.d.L of 12 KN.m over a length of 4 m symmetrically about the centre. Drawn B.M.D. and S.F.D. 3. A fixed beam of span 6 m carries three concentrated loads of 50 KN, 60 KN and 60 KN at 1m , 2 m, and 4 m from the left hand support. Draw B.M.D. and S.F.D. 4. A fixed beam of span ‘L’ carries a triangular load with an intensity of zero at left hand support and an intensity of w/m at right hand support. Find the fixed end moments.

Fig (Hint: wx = intensity of loading at x Treat the load acting on an element of length dx as a concentrated load) (w/1 x dx) Apply the formulae for concentrated loads. MA =

MB = 5.

l o l o

w / L x dx.x (L - x) 2 w / L x dx.x 2 (L - x)

A fixed beam of span ‘l’ carries a triangular load of intensity zero at ends w/m at centre. Find the fixed end moments: (wx = (2w/L) x

Fig

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*** STRUCTURAL ANALYSIS UNIT 6 CONTENTS: 6.1

Introduction

6.2

Moments : Sinking of Supports

6.3

Effect of rotation of support Numerical Example Summary Exercises problem

AIMS: 1.

To explain the effect of sinking of supports of fixed beams.

2.

To develop formulae for the determination of fixed end moments due to sinking and rotation of supports.

OBJECTIVES: The main objectives of this lesson are: 1. The student should be able to understand that additional fixed end moments will be developed if the supports are not at the same level. 2. The student should be able to determine the fixed end moments due to sinking or rotation of support. 6. EFFECTS OF SUPPORT DISPLACEMENTS IN FIXED BEAMS: 6.1 INTRODUCTION: In simply supported beams or any other statically determinate beams, if the supports are displaced by some amount no moments are developed at the supports. But in the case of fixed beams if the supports are displaced additional moments are developed at fixed ends. Consequently shear forces are also developed.

Such displacements include settlements (or

sinking) and rotation of supports. 6.2 MOMENTS DUE TO SINKING (SETTLEMENT) OF SUPPORTS: Consider a fixed beam with relative settlement of one support with respect to the other. See Fig 6.1.

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Fig 6.1 = sinking of support B, B’ is the displaced position of B. Since the slope between the tangents at A and B is equal to Zero, area of

M between A and B is equal to zero. Hence EI

support moments developed due to sinking are equal in magnitude at both supports.

Fig 6.1 L 2

MA EI

+

MB EI

= 0

MA = -MB Using the second moment area theorem, the moments of M/EI diagram about B is equal to 1 MA L 2 L L ( L + MB × ) =EI 2 3 2 3

2 MA + MB = MA =

- 6EI L2

6EI L2 ; MB =

6EI L2

(6.1)

When a fixed beam undergoes support sinking, in addition to loading, the fixed end moments due to loading and sinking are calculated separately and are added. If the support ‘A’ settles with respect to B, then the sign of the moments gets reversed.

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Fig 6.1 Example: 1 : Find the fixed end moments for a fixed beam of span 6m carrying an u.d.L. of 10 KN.m over entire span. Further, the left support settles by 10 mm N E = 2 × 105 mm 2 , I = 8 × 107 mm 4 Fixed end moments: (i) Due to load w12 10 × 36 = - 30 KN.m MA = = 12 12 MB = - 30 KN.m. (ii) Due to sinking MA =

6EI L2

6 × 2 × 105 × 8000 × 104 × 10 = = - 26.60 KN.m. (6000) 2 × 106 MB = 26.60 KN.m. (iii) Final moments: MA = -30 -26.66 = -56.66 KN.m. MB = -30 + 26.66 = -3.34 KN.m.

Fig Taking moments about B: RA x 6 - 56.66 - 10 × 6 × 3 + 3.34 = 0 RA =

233.32 6

= 38.90 KN

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RB = 60 - 38.90 = 21.1 KN Taking A as origin, Mx = 38.90 x-

10x 2 - 56.66 2

To find the position of points of contraflexure, equate Mx to zero. -5 x2 + 38.90 × - 56.66 = 0 solving we get : x = 5.84 m and x = 1.44 m To find the position of max. positive moment: dM x = 38.90 - 10 x = 0 dx

x = 3.89 m Max B.M. = 38.9 × 3.89 -

10 × (3.89) 2 - 56.66 2

= 19.53 KN.m.

Fig 6.3 EFFECT OF ROTATION OF SUPPORT Consider a fixed beam AB, which undergoes an anti-clock-wise rotation angle between the tangents at A and B will be developed due to this rotation

B.

B

at B. The

Let MA and MB be the fixed end moments

B.

According to moment area theorem I, Area of MA EI

M dia between A and B = EI

L MBL + 2 2 EI

=

MA + MB = (

2 EI ) L

B

B

B

(a)

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98

Fig 6.2 As the tangent at A passes through B, using moment area theorem II, MA EI

L 2

2 MB L+ 3 2 EI

L L 2 3

=0

2MA + MB = 0

(b)

Solving (a) and (b), we get MA =

- 2EI L

MB =

- 4EI L

B

(6.2)

B

Example 2: A beam of span 6m fixed at A and B carries a uniformly distributed load of 20 KN/m. The support B undergoes an anti-clock-wise rotation of end moments at A and B. Take E = 2 × 105 N/mm2 and I = 8 × 107 mm fixed end moments: (i) Due to load MA =

- 20 × 36 = - 60 KN.m. 12

1000

radians. Find the fixed

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99 - 20 × 36 = - 60 KN.m. 12

MB =

(ii) Due to rotation: MA MB

=

- 2 × 2 × 105 × 8000 × 104 × = - 16.75 KN.m. (6000) × 106 1000

=

- 4 × 2 × 105 × 8000 × 104 × = - 33.51 KN.m. (6000) × 106 1000

Final moments: MA = - 60 - 16.75 = -76.75 KN.m. MB = - 60 + 33.51 = - 26.49 KN.m. The B.M. and S.F. diagrams can be drawn in the previous example. Summary: In a fixed beam AB, if the support B settles by an amount

, then the fixed end

moments are equal to : MA = MB =

6 EI L2 6 EI L2

If the support A settles, the signs of the moments get reversed. Similarly if the support B undergoes anti-clock-wise rotation

B,

then the fixed end moments developed are

equal to : MA =

- 2EI L

MB =

4EI L

B

B

Answers to S.A.Q. 1. Fixed end moments

2)

6EI l2

3) -

4EI 2EI . B, . l l

B

EXERCISE PROBLEMS: 1. A fixed beam of span 8 m carries a central concentrated load of 50 KN. Further the right hand support of the beam under goes a settlement of 15mm. Take E = 2 × 105 N/mm2 and I = 8 × 107 mm4. Determine the fixed end moments. Sketch b.m.d.

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100

2. A fixed beam of span 6 m carries a central concentrated load 60 KN. The right hand support under goes an anti-clock-wise rotation of moments at A and B. Take E = 2 × 105

1000

radians. Find the fixed end

N and I = 8 × 107 mm4. 2 mm

*** STRUCTURAL ANALYSIS UNIT 7 CONTENTS: 7.1

Introduction

7.2

Three moment equation - Derivation

7.3

Application of the Three moment equation - Continuous beams

7.4

Numerical examples Summary Exercise problems

AIM: To develop ‘ three moment equation’ for the analysis of continuous beams with supports at the same level OBJECTIVES: 1.

To emphasize that the analysis of continuous beams is different from propped cantilever and fixed beams.

2.

To make the student understand that the three moment equation is developed from the condition of geometry at common support i.e., the tangent remains at 180o .

3.

To enable the student to solve continuous beams independently.

7.1 INTRODUCTION: In the previous chapters the student has been introduced to statically determinate and also simple indeterminate beams like propped cantilever and fixed beams. In this chapter continuous beams, which are also statically indeterminate, will be discussed. A beam having more than one span is classified as a continuous beam. Fig. 7.1 shows a continuous beam with three spans.

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101 Fig 7.1

In the continuous beam, there will be four vertical reactions and one horizontal reaction at A. Thus there will be in all five reactions. But there are three static equilibrium equations. Hence this beam is statically indeterminate to second degree. In problems where the loading is purely vertical, then the horizontal reaction at A will become zero, thus developing only four reactions. Then equation

Fx = 0, becomes trivial, leaving the

degree of indeterminacy as two. Sometimes the continuous beam may be fixed at ends as shown in Fig.7.2 In this problem there will be four reactions and two fixed end moments at A and D. Thus the beam is statically indeterminate to degree four.

Fig 7.2 7.2 THE THREE MOMENT EQUATION - DERIVATION: Consider two adjacent spans AB and BC of a continuous beam as shown in Fig.7.3. The supports A, B and C are at the same level. Let L1 and L2 be the span lengths and I1 and I2 the moments of inertia of AB and BC respectively, let MA, MB and MC be the moments at supports A, B and C. Fig. 7.3 (b) shows the bending moment diagram due to external loading on the spans AB and BC, treating the spans as simply supported independent of each other. Fig. 7.3 (c) shows the moment diagram due to MA, MB and MC. The sign if shown positive, assuming MA, MB and MC produce compression on top fibres and tension in bottom fibres, though these moments will be usually negative.

A1 and A2 are the areas of moment

diagrams and a1 and a2 be the distance of their centroids from A and C respectively (Fig 7.3 (b)). Similarly, in Fig 7.3 (c), the moment diagram is divided into triangles and the distances of their centroids are shown. Now consider the span AB where the moment diagram on AB is broken into two parts. Fig 7.4(c) represents the moment diagram due to applied loading on span AB Fig 7.4(b) represents the moments diagram resulting from the moments at supports. The combined moment diagram of (b) and (c) is shown in (a). For convenience MA and MB and are taken as positive but usually they will be negative.

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Figs. 7.3 (b) and (c) represent the moment diagram on two spans AB and BC which is shown as broken into two parts such as for the span AB as in Fig. 7.4. In the analysis A1 and A2 are known and the moments MA, MB and MC are to be determined.

Fig 7.3 (a), (b), (c) Relation between MA, MB and MC can be obtained from the condition that the beam is continuous at intermediate support B or the tangent at ‘B’ to the elastic curve is common for both spans AB and BC. In other words it can be considered that the joint B as rigid joint and the tangent at B to the elastic curve on each side must remain at 180o to each other. In fig. 7.3(a) a tangent is drawn at B and from the condition of continuity, we have AA ' - CC ' = L1 L2

(7.1)

where AA’ = deflection of point A from the tangent at B. From moment area theorems AA’ = moment of

M EI

diagram between A and B about A.

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=

103 1 M L L M L 2 A1a1 + A 1 1 + B 1 L1 EI1 2 3 2 3

(7.2)

Similarly: CC’ = deflection of point C from the tangent at B.

Fig 7.4 (a), (b), (c) =

1 M L L2 M L 2 A 2 a2 + C 2 + B 2 L2 EI 2 2 3 2 3

(7.3)

Substituting equations (7.2) and 7.3) in equation 7.1 we get: =

1 M L (A1a1 ) + A 1 EI1L1 2

L1 M L 2 + B 1 L1 3 2 3

=

-1 M L L M L 2 A 2 a2 + C 2 × 2 + B 2 L2 EI 2 L2 2 3 2 3

(7.4)

simplifying, M A L1 I1

= -

+ 2 MB

6A1a1 I1L1

-

(L1 + L 2 ) L + MC 2 I1 + I 2 I2

6A2 a2 I2 L2

Equation (7.5) is known as the ‘three moment equation’.

(7.5) This equation 7.5 is

applicable for continuous beams for which the supports are at same level.

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104

APPLICATION OF THE THREE MOMENT EQUATION - CONTINUOUS

BEAM: In the analysis of continuous beams using three moment equation, the support moments are taken as unknowns. Therefore the number of unknown support moments gives the degree of static indeterminacy. Therefore the number three moment equations to be solved is equal to the number of support moments. Hence this method is conveniently used when the unknowns are equal to three or less. But it becomes time consuming if the number is more than three using long hand. For example, consider a continuous beam of three spans as shown in Fig 7.5. In this problem there are four supports.

Fig 7.5 But by inspection we know that MA = MD = 0 since the supports are simple supports which do not offer any resistance for rotation. Hence the other support moments to be found are MB and MC. To solve these two unknowns we have to develop two equations using the ‘three moment’ equation. Once the moments at intermediate supports are solved, the reactions and subsequently the moments and shears can be found any where using principles of statics. To find the maximum positive moments at the centre of the spans, different spans are treated as simple spans subjected to external loading with end moments already determined. Shear and bending moment diagrams can be drawn accordingly. In few cases the continuous beam may have fixed end, then the moment at fixed end is one of the unknowns as shown in Fig. 7.6(a). In this problem, support A is fixed. This beam can be treated as follows:

Fig 7.6

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The condition required for geometry which corresponds to the unknown fixed end moment is that the slope of the tangent at A is zero. This condition can be achieved by adding an imaginary span AoA of any length Lo simply supported at Ao and with an infinitely large moment of inertia at all sections. The three moment equation when applied to spans AoA and AB (Fig 7.7), becomes

Fig 7.7 Mo

Lo

+ 2M A

Lo

+

L 6Aa L1 = + MB I IL I Lo

Noting that Mo = 0 and 2 MA

L L + MB I I

= -

(7.6)

= 0 , equation 7.6 becomes: 6Aa IL

(7.7)

Thus the additional moment equation is developed for a fixed end support in a continuous beam. 7.4 NUMERICAL EXAMPLES: Example I: Analyse the continuous beam shown using the three moment equation. Draw the shear and bending moment diagram.

Fig 7.8 (a) given beam Simply supported span moments, at centre: Span AB : Span BC :

15 × 4 = 15 KN.m. 4

4 × (6) 2 = 18 KN.m. 8

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106

Span CD :

6 × (4) 2 = 12 KN.m. 8

The moment diagram on spans AB, BC and CD by considering each span as simple beam are shown in Fig (b).

Fig 7.8 (b) moment diagram on simple span resulting from the applied loading On inspection we know that MA = 0 and MD = -3 KN.m. (Negative because it produced tension on top fibres).

Fig 7.8 (c) moment diagram on the simple spans due to end moments Span AB: A1 =

1 × 4 × 15 = 30 2

a1 =

=

2m

Span BC: A2 =

2 × 6 × 18 = 72 3

a2 = 3 m Span CD: A3 =

2 × 4 × 12 = 32 3

a3 = 2 m Applying three moment equations for spans AB and BC MA

4 + 2 MB 3I

4 6 6 + MC + 3I 4 I 4I

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= -

107 6 × 30 × 2 6 × 72 × 3 4 × 3I 6 × 4I

Keeping MA = 0 and simplifying we get : 17 3 MB + MC = - 84 3 2

(1)

Similarly, applying for spans BC and CD MB

6 6 4 4 + 2 MC + MD + 4I 4I 2I 2I

= -

6 × 72 × 3 6 × 32 × 2 6× 4 I 4 × 2I

Keeping MD = -3 KN.m. and simplifying we get: 3 MB + 7 MC = - 96 2

(2)

Solving equations (1) and (2) we get: MB = - 11.86 KN.m. MC = - 11.185 KN.m. The reaction can be calculated as follows:

Fig 7.8 (d) Taking moments about B and keeping that MB = 11.186 KN.m. with proper sign.

Fig 7.8 (e) we get RA × 4 - 15 × 2 + 11.86 = 0 Solving:

RA = 4.535 KN

Now taking moments about C, keeping that MC = 11.185 with proper direction.

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Fig 7.8 (f) We get, 4.535 × 10 + RB × 6 - 15 × 8 - 4 × 6 × 3 + 11.185 = 0 Solving :

RB = 22.58 KN

Now taking moments about ‘C’ from right hand side, and keeping MC = 11.185 with proper direction.

Fig 7.8 (g) RD x 4 - 3 × 5 - 6 × 4 × 2 + 11.185 = 0 Solving RD = 12.95 KN Now considering statics for the entire continuous beam i.e.,

FY = 0, we get RC = 25.94 KN.

The bending moment and shear force diagrams are given. In the bending moment diagram, moment diagram due to applied loading/and the support moment diagram are superimposed one over the other. The S.F. diagram is drawn using the same principles as for simple beams

Fig 7.8 (h) bending moment diagram

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As the simply supported span moments due to applied loading are positive and the support moments are negative, the resultant moment will be the algebraic sum of these two. The resultant moment diagram is shown hatched. The resultant moments at centre of spans are calculated as follows: 11.86 2

Span AB:

15 -

Span BC :

18 -

(11.86 + 11.185) 2

Span CD :

12 -

(11.85 + 3) 2

= 9.07 KN.m. = 6.48 KN.m.

= 4.91 KN.m.

The resultant moment is drawn as follows: Positive moments are shown above the base line and negative moments are shown below the base line. The points where the moments changes sign are points of contraflexure.

Fig 7.8 (i) bending moment diagram

The student is advised to adopt this diagram The S.F. diagram is shown as follows:

Fig 7.8 (j) shear force diagram

In the shear force diagram shown, the sum of the ordinates of shears above and below the base line at supports is equal to the corresponding reaction. For example at support B the sum of the ordinates is equal to (10.465 + 12.115) = 22.58 KN which is same as reaction RB. Example - 2 : Analyse the continuous beam shown using three moment equation. Draw the

shear force and bending moment diagram

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Fig 7.9 (a) the given beam Simply supported span moments Span AB =

30 × (8) 2 = 240 KN.m. 8

Span BC =

150 × 6 4

= 225 KN.m.

Fig 7.9 (b) moment diagram on simple spans resulting from the applied loading As the given beam has fixed end at A, assume an imaginary span AAo as explained earlier in the section 7.3, with an infinite moment of inertia.

Fig.7.9 © Span AB: A1 =

2 × 8 × 240 = 1280 3

a1 = 4 m Span BC: A2 = a2 =

1 × 6 × 225 = 675 2 6 + 4 10 = 3 3

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On imaginary span Aao, there is no loading. By inspection MC = 0 . Applying three moment equations for spans AoA and AB: MAo = As

Lo

+ 2 MA

Lo

+

8 8 + MB 2I 2I

6 × 1280 × 4 8 × 2I

Lo

becomes zero and simplifying we get

2 MA .

8 8 + MB 2I 2I

=

-

1920 I

8 MA + 4 MB - 1920

(1)

Similarly appearing for spans AB and BC. MA . = -

8 2I

. 2 MB

6 8 6 + MC + 3I 2I 2I

6 × 1280 × 8 6 × 675 10 × 8 × 21 6 × 3I 3

Keeping MC = 0 and simplifying, we get 4 MA + 12 MB = - 2670

(2)

Solving equations (1) and (2) we get MA = - 154.5 KN.m MB = -171.0 KN.m.

Fig 7.9 (d) moment diagram Span AB: at centre of span = 240 -

171 + 154.5 2

= 77.25 KN.m Span BC: Under the load = 275 -

171 × 4 6

Reaction are calculated as follows:

= 161 KN.m.

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Taking moments about B from right hand side

Fig 7.9 (e) RC x 6 - 150 × 2 + 171 = 0 RC = 21.5 KN.m. Taking moments about A from right hand side. 21.5 × 14 + RB × 8 + 154.5 - 150 × 10 - 30 × 8 × 4 = 0 RB = 250.5 KN.m. From the condition FY = 0 we have RA = 118.0 KN `The shear force diagram is drawn as shown:

Fig 7.9 (f) SF diagram Example - 3: Analyse the continuous beam shown by the use of the three moment equation. Draw the shear and moment diagrams: (EI is constant)

Fig 7.10 (a) Simply supported span moments: Span AB : A1 =

60 × 4 4

= 60 KN.m.

1 × 4 × 60 = 120 2

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a1 = 2 m As the given beam is fixed at both ends A and C, we have to assume imaginary spans AAo and CCo on both sides with infinite moment of inertia.

Fig 7.10 (b) Applying three moment equation to spans AAo and AB. MAo = -

Lo

+ 2 MA

Lo

+

4 4 + MB I I

6 × 120 × 2 4× I

keeping

Lo

= 0 and simplifying :

8 MA + 4 MB = - 360

(1)

For the spans : AB and BC: MA

4 4 6 6 + 2 MB + + MC I I I I

= -

6 × 120 × 2 4× I

Simplifying: 4 MA + 20 MB + 6 MC = - 360

(2)

For the spans: BC and CCo: MB

6 6 Lo L + 2 MC + + M Co o I I

Simplifying: Keeping

Lo

= 0

= 0

6 MB + 12 MC = 0

(3)

Now solving the equations (1), (2) and (3), we get the value for MA, MB and MC. In this example, the three equations can be expressed in a matrix form as follows: 8 4 0

4 0 20 6 6 12

MA MB MC

- 360 = - 360 0

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This can be expressed as : [A] {M} = {C} where: [A]

=

8

4

4 0

20 6 6 12

MA MB MC

[M] =

0

- 360 and [C] = - 360 0

The student is advised to note that the matrix (A) is symmetric. The matrix (A) is same for a given continuous beam irrespective of loading on the beam. Only the matrix C changes with loading. Now [M] = [A]-1 [C] Adj[ A] A

[A]-1 =

[A]

-1

1 = 1440

204

-48

24

-48 24

96 48 -48 144

After carrying out the matrix multiplication, we get MA MB MC

=

-39 -12 6

If the student is not aware of the matrix operations, he is advised to solve the three simultaneous equations by elimination method. The moment diagram is drawn as follows:

Fig 7.10 © Moment diagram Resultant moment diagram can be redrawn as follows:

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Moment at centre of span AB = 60 -

39 + 12 2

= 34.5 KN.m.

Fig 7.10 (d) Net moment diagram Positive moment is drawn above base line and negative moment below the base line. It is brought to the notice that the moment at C is positive and not negative. The reaction can be determined as follows: Taking moments about B from right hand: RC x 6 + 6 + 12 = 0 RC = -3 KN (down wards) Taking moments about A from right hand

Fig 7.10 (e) RC x 10 + RB x 4 + 6 + 39 - 60 x 2 = 0 Substituting RC = -3 and simplifying: RB = 26.25 KN From the condition,

Fy : 0, we get

RA = 36.75 KN

Fig 7.10 (f) The shear force diagram is as follows:

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Fig 7.10 (g) Shear force diagram SELF ASSESSMENT QUESTIONS: 1.

The three moment equatio is developed based on the condition the tangent at common support remains at _______________.

2.

In a continuous beam, if one end is a simple support, the support moment is equal to ________________.

3.

In a continuous beam, if the number of support moments is equal to three, then the number of equations to be solved are ______________.

4.

In a continuous beam if the number of support moments is equal to TWO then the static indeterminacy of the beam is ____________.

5.

If a continuous beam consists of a fixed end, then we have to assume an __________________ span with _________________ moment of inertia.

6. Degree of indeterminacy for the given continuous beam is equal to

Fig 7.11

SUMMARY 1.

The three moment equation is given as :

MA

L1 I1

= - 6

+ 2 MB

A1a1 L1I1

- 6

L L L1 + 2 + MC 3 I3 I2 I1 A2 a2 L 2 I2

2. When a continuous beam is having a fixed end, we have to assume an imaginary span of any length with infinitely large moment of inertia as shown below:

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Fig 7.12 Solutions to S.A.O. 1.

180o

2.

zero

3.

3

4.

2

5.

Imaginary, infinite

6.

3

EXERCISE PROBLEMS: 1. Analyse the continuous beam shown using the three moment equation. Draw the S.F. and B.M diagrams.

Fig 7.13 2. Analyse the continuous beam shown using the three moment equation. Draw the S.F. and B.M. diagrams.

Fig 7.14 3. Analyse the continuous beam using three moment equations. Draw the S.F. and B.M. diagrams. Support B’ settles by 20 mm. E = 200 x 106 KN/m2 , I = 100 x 10 –6 m4.

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4. Analyse the continuous beam shown by the use of the three moment equation. Draw the shear and moment diagrams.

Fig 7.16 *** STRUCTURAL ANALYSIS UNIT 8 AIM: To develop the concepts of moment distribution methods for the analysis of statically indeterminate beams. OBJECTIVES: 1. To introduce another but versatile method of solving indeterminate beams and frames. 2. To introduce to the student, the concepts of moment distribution and explain the various steps involved in the analysis of continuous beams. 8.1 INTRODUCTION: The moment distribution method, developed by Hardy Cross, is very convenient to solve complex indeterminate beams and rigid jointed portal frames. In the unit 7, we have discussed about the three moment equation for the solution of continuous beams only. If the number of support moments increases the number of simultaneous equations to be solved increases. Moment distribution method is a versatile technique to solve continuous beams and frames. Basically this method is an iteration method. 8.2 SIGN CONVENTION: In this method the sign convention adopted is as follows: 1. Moment at the end of a member is considered positive when it is anticlockwise direction or on a joint in the clockwise direction. Consider a beam fixed at ends subjected to loading as shown in Fig.8.1.a. The fixed end moments developed are shown in Fig.8.1.b. The left hand support is subjected to clockwise moment or the left end of the member is subjected to anticlock wise moment. Both of them according to our convention are

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positive. And the moments developed at the right hand support is anticlockwise or the right hand end of the member is subjected to clockwise moment. 2. Translation is taken positive when it is upward in the y direction. 3. Angular rotation

is taken as positive when it is anticlockwise direction.

Fig 8.1 Sign Convention for end moments 8.3 STIFFNESS OF MEMBER: CARRY OVER FACTOR Consider a member AB which is fixed at far end B and hinged at near end A as shown in Fig. 8.2 Apply anticlockwise moment MAB at near end A which produces anticlockwise rotation

B,

both of them are positive. Since the far end B is fixed, a moment is developed

and it is designated as MBA. Now we have to develop relation between MAB, MBA and

A.

As the end B is fixed, slope at B is zero. Using the moment area theorem, from the first theorem,

A.

= area of (Since

M diagram between A and B EI B.

= 0)

Fig 8.2 A.L

= deflection of point B from the tangent at ‘A’.

According to beam convention MAB is negative and MBA is positive.

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Fig 8.3

M Diagram EI

From first theorem, we have A

= -

M AB L M L + BA 2 EI 2 EI

M BA L M AB L 2 EI 2 EI

=

(8.1)

and from second theorem, taking moment about B, we get A

L =

M BA L L M AB L 2 . L × 2 EI 3 2 EI 3

Simplifying: A

=

M Ba L MAB L 2EI 2EI

(8.2)

Solving equation 8.1 and 8.2, we obtain MAB =

EI . L

4

and MBA = The factor 4

A

1 MAB 2

(8.3) (8.4)

EI in the equation 8.3 is known as the stiffness of the member, which L

is defined as the moment required to be applied at A to cause a rotation of one radian at A, of span AB simply supported at A and fixed at B. The ratio

EI is referred to as relative L

stiffness and is denoted by K. KAB = E The factor is equal to +

IAB LAB

(8.5)

1 is known as carry over factor. For prismatic member, carry over factor 2

1 . It is the ratio of moment inducted at B to the moment applied at A. 2

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Now consider a member AB which is hinged at far end B as shown in Fig 8.4.

Fig 8.4 Apply an anticlockwise moment at A, which produced a rotation zero. Using conjugate beam method, the relation between MAB and

A. AB

Moment at B is is obtained as

follows: AB

=

M AB L 2 L × 2 EI 3 L

Rearranging: M AB =

=

3EI L

3 4EI ( ) 4 L

(8.6)

AB

(8.7)

AB

Fig 8.4 (a) The terms

3EI is known as the modified stiffness of the member. The relative L

3 stiffness K’ = K . 4

8.4 DISTRIBUTION FACTOR: In Fig 8.5 (a) it is shown that four members are meeting at a rigid joint E.

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Fig 8.5 The joint is subjected to anticlockwise couple M.

As the joint is rigid, all the

members undergo same rotation . Let KAE, KBE, KCE and KDE be the stiffness of the members AE, BE, CE and DE respectively. Now consider free body diagram of the joint E as shown in Fig. 8.5(b). For equilibrium of joint E, MEA + MEB + MEC + MED - M

= 0

(8.8) The end moments of the members can be written in terms of their stiffness and joint rotation

A.

MEA = 4 KAE , MEB = 4 KBE MEC = 4 KCE ,

MED = 4 KDE

(8.9)

Substituting these values in equation 8.8, we obtain: M = 4

(KAE + KBE + KCE + KDE)

= 4 or where

=

M 4

(8.10)

K

K denotes the sum of the relative stiffness of all the members meeting at the

joint E. Inserting the values of MEA = 4 KAE = Defining

in equation 8.9, for the first term. M 4

KAE M K

K

(8.11)

K as the stiffness of the joint, which is equal to the sum of the stiffness of

the members at the joint, the ration

K AE is known as distribution factor. Distribution factor K

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for a member is the factor by which the applied moment M shall be multiplied to get a moment developed in the member. For any member AB the distribution DFAB is given as: DFAB =

K AB K

(8.12)

At any joint the sum of the distribution factors is equal to unity. 8.5 DEVELOPMENT OF THE METHOD: The principle of the moment distribution method can be explained by solving a simple example. Consider a two span continuous beam as shown in Fig. 8.6. The beam is fixed at ends A and C but free to rotate at B. But assume initially the joint B is also restrained against rotation as shown Fig. 8.7(a). The fixed end moments in each span are calculated as follows: MFAB = Fixed end moment for the member AB at end A MFBA = Fixed end moment for the member AB at end B

Fig 8.6

Fig 8.7 (a)

Fig 8.7 (b), (c)

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124

M

F AB

30 × (8) 2 = = 160 KN.m. 12

M

F BA

30 × (8) 2 = = 160 KN.m. 12

M

F BC

150 × 2 × (4) 2 = 133.33 KN.m. = 36

M

F CB

150 × 4 × (2) 2 = = - 66.66 KN.m. 36

The fixed end moments at the ends of the member AB and BC and also the resisting moments developed to hold the joint against rotation are shown in Fig 8.7(b). At joint A there is clockwise moment of l160 KN.m and at the joint C, there is anticlockwise moment of 66.66 KN.m. Now consider joint B. The joint B is subjected to 160 KN.m. (anticlockwise) and 133.33 KN.m (clockwise). Now release the joint B since the joint is not originally restrained. But algebraic sum of the moments acting on the joint B is not equal to zero. At this joint there is an unbalanced moment of - 26.67 KN.m ( -160 + 133.33). This unbalanced moment will act in anticlockwise direction. To keep the joint against rotation, apply a moment in the opposite direction i.e., a clockwise moment of 26.67 KN.m as shown in Fig. 8.7(c). This moment shall be distributed to the members BA and BC in proportion to their stiffness. The distribution factors are calculated as follows: Stiffness of members: KAB = KBC =

2I = 0.25 I 8 3I = 0.50 I 6

while calculating stiffness, the term E is not used, since it is the same for all the members. At joint B, there are two members and the corresponding distribution factors are: DFBA =

0.25 I (0.25 I + 0.5 I)

= 0.333

DFBC =

0.5 I (0.25 I + 0.5 I)

= 0.667

Note that at any joint the sums of the distribution factors shall be equal to unity. Now the moment 26.67 KN.m. at joint B is distributed to the members BA and BC as follows: M

' BA

= 0.333 × 26.67 = 8.88 KN.m.

M 'BC = 0.667 × 26.67 = 17.78 KN.m.

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Now half of these moments shall be carried over to the further ends A and C. M

'' AB

and

' M 'CB

M

'' AB

Let

be the carry over moments, then =

' M 'CB =

1 × 8.88 = 4.44 KN.m. 2 1 × 17.78 = 8.88 KN.m. 2

The beam is now in its true position under the given loading. The final moments at the ends of the members in its true position are obtained by adding algebraically the fixed end moments and the moments caused by the release of the joint B. MAB = 160 + 4.44 = 164.44 KN.m. MBA = - 160 + 8.88 = -151.12 KN.m. MBC = 133.33 + 17.78 = 151.11 KN.m. MCB = - 66.66 + 8.88 = - 57.78 The student is advised to note that the algebraic sum of the moments acting at the joint B (MBA + MBC = -151.12 + 151.11) is equal to zero, which means that the joint is in equilibrium. Now consider the same continuous beam but the end ‘c’ is simply supported as shown in Fig. 8.8.

Fig 8.8

Initially restrain the joints B and C against rotation. The fixed end moments are the same as in the previous example. But the stiffness of the member BC is different since the far end support ‘C’ is simply supported in the given beam. Therefore we have to use modified stiffness for the member BC. The stiffness of the members are calculated as follows:

Fig 8.9 (a)

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Fig 8.9 (b), (c), (d) KAB =

2I = 0.25 I 8

KBC =

3 3I × = 0.375 I 4 6

The distribution factors at joint B are as follows: DFBA =

0.25 I (0.25 I + 0.375 I)

= 0.4

DFBC =

0.375 I (0.25 I + 0.375 I)

= 0.6

Note that the sum of the distribution factors is equal to unity. At ends there will not be any distribution. The fixed end moments at the ends of the members AB and BC and also the resisting moments developed on the joint to hold the joint against rotation are shown in Fig 8.9(b). The directions of these moments are shown according to their signs. Now release the joint ‘C’ as it is originally simply supported. The joint C is under the action of an anticlockwise moment (-66.66 KN.m). As the support C is simply supported it cannot resist any moment. Therefore apply an equal and opposite moment, thus making the net moment at ‘C’ zero, as shown in Fig 8.9c. Half of this moment i.e.,

1 × 66.66 = 33.33 2

KN.m will be carried over to the joint B. Now the joint B is under the action of the moments as shown Fig 8.9(d). The unbalanced moment at joint B is equal to : 6.66 KN (-160 + 133.33 - 33.33). To keep the joint B against rotation apply a moment of -6.66 KN.m as shown in Fig.8.9 (d). This moment shall be distributed among the members BA and BC as follows: M

' BA

= - 0.4 × 6.66 = - 2.664 KN.m.

M 'BC = - 0.6 × 6.66 = - 3.996 KN.m. Further, half the value of -2.664 KN.m. will be carried over to the further end A and no carry over moment to the support C since it is simply supported. M

'' AB

= -

1 × 2.664 = - 1.332 KN.m. 2

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The final moments at the ends of the members are obtained by adding algebraically the fixed end moments and the moments caused by the release of the joints. MAB = 160 - 1.332 = 158.664 KN.m. MBA = - 160 - 2.664 = -162.664 KN.m. MBC = 133.33 + 33.33 - 3.996 = 162.664 KN.m. MCB = 0 Note that MBA - MBC is equal to zero. The whole procedure can be condensed and performed in a tabular form.

It is

described as follows. In the top three lines, joints, the members corresponding to the joints and distribution factors are recorded.

In fourth line fixed end moments with signs are

recorded. In fifth line joint C is realised first. As the joint C is simply supported the final moment at c shall be zero. Therefore apply a moment of +66.66 KN.m at ‘C’ and half of this value i.e., 33.33 KN.m. is carried over to the farther end B of the member BC. In sixth line, joint B is released. Now the unbalanced moment at B is equal to 6.66 KN.m (-160 + 133.33 + 33.33). We have now to distribute a moment equal and opposite to unbalanced moment i.e., -6.66 KN.m. After distribution, the moments are :-0.4 × 6.66 and -0.6 × 6.66 for members BA and BC respectively. After distribution, half the value

1 (-0.4 × 6.66) is carried 2

over to A, i.e., -1.332 KN.m. This is recorded in the sixth line. The final moments i.e., algebraic sum of all moments are recorded in the last line. Joint

A

B

Member

AB

BA

BC

CB

Distribution factor

-

0.4

0.6

-

Fixed end moments

160

-160

133.33

-66.66

33.33

66.66

Balancing at c and carry

C

over Bal. c. o at B

-1.332

-2.664

-3.996

Final moments

158.668

-162.664

162.664

0

SELF ASSESSMENT QUESTIONS 1.

In the moment distribution method theorem

2.

The term

I is known as ____________ L

4 EI is known as ______________. L

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128

When the far end is hinged the modified stiffness of the members meeting at _____________.

4.

Stiffness of a joint is the sum of the stiffness of the members meeting at ___________.

5.

Distribution factor is defined as the ration of stiffness of member to the stiffness of _______________.

6.

The sum of the distribution factors at a joint is equal to ______________.

7.

For prismatic members the carry over factor is equal to _______________.

8.

The sign of the carry over moment is ______________ as the moment .

9.

At a joint the sign of the distributed moments is _____________ to the sign of the unbalanced moment.

10.

At a joint algebraic sum of the final moments is equal to ___________.

SUMMARY:

Fig 8.10 1.

In a member AB, when the far end B is fixed, the relation between MAB and

A

is given

as: MAB = 4 4EI L

2.

EI L

A

is the stiffness of the member

In a member when the far end is hinged, the relation between MAB and 3EI . L

A

4 EI . L

A

MAB = =

3 4

I L

3.

Relative stiffness L

4.

The distribution factor for a member AB is given DFAB =

K AB K

A

is given as:

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where KAB = stiffness of the member and K = sum of the stiffness of the members meeting at joint i.e., stiffness of joint. Answers to S.A.Q. (1) Stiffness (2) Relative stiffness (3)

3 4

4 EI . L

(4) joint

EXERCISE PROBLEMS 1. Analyse the continuous beam shown using the moment distribution method. Draw the S.F. and B.M. diagrams

Fig 8.11 2. Analyse the continuous beam shown using the moment distribution methods. Draw the S.F. and B.M. diagrams.

Fig 8.12 *** STRUCTURAL ANALYSIS UNIT - 13 CONTENTS: 13.1

Introduction: Analysis of continuos beams

13.2

Numerical Examples

AIMS: To explain the slope deflection method for continuous beams including support settlements by solving typical examples.

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OBJECTIVES: After going through this lesson, the student should able to solve problems on continuous beams by slope - deflection method independently. Slope Deflection method: Analysis continuous beams. 13.1 INTRODUCTION: In unit 12, slope-deflection equations were developed. Equations (12-4 and (12-5) give the basic slope - deflection relationships. In section 12-4, joint equilibrium equations were discussed. A simple example was also presented to explain the principles involved in the slope- deflection method. We shall now discuss the analysis of continuous beams including support settlements. The slope - deflection equations are presented here for ready reference: MAB = MFAB + EI/L (4 MBA = MFBA + EI/L ( 2 Here

= /L, where

A

+2

A

+4

B B

-6 )

(12.4)

-6 )

(12.5)

is the support settlement. When there is a support settlement, it will

be more convenient to rewrite the equations. (12.4) and (12.5) in the modified form as given. MAB = MFAB + EI/L (4 MBA = MFBA + EI/L ( 2

A

+ 2 B) - (6EI/L)

(a)

+ 4 B) - (6EI/L)

(b)

A

A few examples will clarify the method.

13.2 NUMERICAL EXAMPLES: Example. 1:

Fig 13.1(a)

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Analyse the two-span continuous beam ABC with an over hang CD. Shown in Fig 13.1 (a) by the slope deflection method. Assume that all the supports are at the same level. Draw the bending moment diagram. The given beam has two spans AB and BC with an overhang CD support A is fixed and hence

A

= 0, Support B and C are freely supported add therefore

B

and

C

are the two

unknown relations Fixed end moments MFAB = 20 × (6)2 / 12 = 60 KN.m. MFBA = 20 × (6)2 / 12 = 60 KN.m. MFBC = 60 × 4 × (2)2/(6)2 = 26.66 KN.m. MFCB = 60 × 2 × (4)2/(6)2 = 53.33 KN.m. Here there is an overhang CD. The cantilever moment MCD at C is equal to 20 × 1 KN.m., as shown in Fig.13.1 (b). This is positive according to our sign convention

Fig 13.1 (b) MCD = 20 KN.m. Relative stiffness KAB = EI/6 KBC = 1.5EI/6 Keeping EI/6 = I, KAB = 1 and KBC = 1.5. Slope deflection equations Since all the supports are at the same level, the term and BC. Also =

A

A

= /L will be zero for both spans AB

= 0. Substituting

= 0, in the equations (a) and (b) we have MAB = 60+1 (2 B) MBA = -60+1 (4

B)

(c)

MBC = 26.66 + 1.5 (4

B+

MCB = -53.33 + 1.5 (2

B

MCD = 20

2 C)

+ 4 C)

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In this example

and

B

C

are the unknowns we need two equations to solve two unknowns.

Applying joint equilibrium equations at joints B and C, we get the following two equations:At joint B: MBA + MBC = 0 At joint C: MCB + MCD = 0

(d)

Substituting the expression for MBA, MBC, MCB, and MCD from (c) we obtain: MBA + MBC = (-60+4 C) + (26.66 + 60B+3 C) = 0 MCB + MCD = (-53.33+3 B) + (6

C

+ 20) = 0

(e)

Simplifying (e), we have 10 3

B B

+3

=6

C C

= 33.34

= 33.33

(f)

The above two equations can be expressed in matrix from as 10

3

B

3

6

C

=

33.34 33.33

(g)

Expressing symbolically

[k ]{ } = {P} where [ k ] =

10 3 3

6

# B& ' % C(

[ ]= $ and

#33.34& ' %33.33(

{P} = $

The student is advised to note that [ k ] is a symmetric matrix of size 2×2. The size of this matrix will be equal to the number of unknowns { } and {p} are column vectors. The [ k ] matrix is the same for a given continuous beam irrespective of loading. The { P} matrix only will change with loading on the beam. By matrix operations:-

{ } = [k ] 1{p}

(gh)

In this problem

[k ] 1 = B C

=

3 Adj[k ] 1 6 = 51 3 10 [k ]

3 #33.34 & 1 6 $ ' 51 3 10 %33.33(

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133

After carrying matrix multiplication B

=

C

1962 .

(I)

4.574

NOTE: When using relative stiffnesses of their absolute values the relations obtained are not the actual magnitudes but proportional to them. Substituting the numerical values of

B

and

C

in the expressions (C), we get the final moments as: MAB = 60 + 2 × 1.962 = 63.92 KN.m. MBA =- 60 + 4 × 1.962 = - 52.15 KN.m. MBC = 26.66 + 1.5 ( 4 × 1.962 + 2 × 4.574) = 52.15 KN.m. MCB =- 53.33 + 1.5 ( 2 × 1.962 + 4 × 4.57) = - 20 KN.m. Now the student can verify the joint equilibrium equations(d) with the computed numerical values of moments. The bending moment and shear force diagram can be drawn as usual. If the student is not familiar with matrix algebra, he can solve equations (f) for

B

and

C

by

the conventional elimination method. Example 2: Analyse the three span continuos beam ABCD fixed at the ends A and D (Fig 13.2) by the slope deflection method. Support B sinks by 10mm. Take E = 20 × 106 KN/m2 and I = 20 × 10-4 m4. Solution: The beam has three spans AB, BC and CD. the supports A and D are fixed and hence

A

=

D

= 0. As the supports B and C are freely

Fig 13.2 supported, there will be rotations 10mm, the term

B

and

C

which are unknown. As the support B sinks by

= /L exists for the two spans AB and BC, and for the span CD,

since C and D are as same level.

=0

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134

Fixed end moments:

60 × 2 × (4)2

MFAB =

(6)2

60 × (2)2 × 4

MFBA = M FBC =

= 53.37KN.m.

(6)2

= 26.66KN.m.

(a)

40 × (5) 2 = 83.33KN . m. 12 40 × 52 = 83.33KN . m. 12

M FBC =

M FCD = 30 × (6) 2 = 90 KN . m.

MFDC = - 90KN.m.

(b)

Relative stiffnesses: KAB = 2I/6 KBC = I/5 KCD = 2I/6

(c)

Multiplying the stiffeness values by 15/I, we have the relative values as: KAB = 5, KBC = 3, and KCD = 5. (see note under equation 1. in example.1) Slope - deflection equations: = /L for the two spans is computed as follows: for the span AB, the

The term

support B is at lower level than at A, hence

is negative for span AB. For the span BC, the

support C is at higher level than at B, hence is positive for span BC. Span AB,

= - 10/6000 = - 1/600

Span BC,

= 10/5000 = 1/500

The quantities (6EI/L)

are computed.

Span AB =

1 6 × 2 × 20 × 106 × 20 × 10 4 = 133.33 KN.m. 6 600

Span AB =

6 × 20 × 10 6 × 20 × 10 4 1 = 96.66KN .m. 5 500

Substituting

A

=

D

(e)

= 0 and the above quantities, the slope deflection equations are obtained.

Thus we have: MAB = 53.33 + 5 (2 B) + 133.33 KN.m. MBA = -26.66 + 5 (4 B) + 133.33 KN.m.

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MBC = -83.66 + 3 (4

B

+2 C) -96.66 KN.m.

MCD = -83.66 + 3 (2

B

+2 C) -96.66 KN.m.

MCD = 90 + 5 (4 C) MDC = -90 + 5 (2 C)

(f)

Joint equilibrium equations At joint D MBA + MBC = (-26.66 + 20

B

+ 133.33) + (83.33 + 12 B) + 6

C

- 96.66) = 0

At joint C. MCB + MCD = (-83.33 + 6

B

+ 12

C

- 96.66) + (90+20 C) = 0

(g)

Simplifying we have: B+

32 6

B

6

+ 32

C

= - 93.34

C

= 90

32

6

B

6

32

C

=

(h) 93.34 90

[k ]{ } = {p} 32

6

[ ] = [k ] 1[p] = 1/ 988 6 32

93.34 90

=

2.476 3.48

Substituting the above values in equation (f), we get the final moments as: MAB = 161.9 KN.m. MBA = 22.16 KN.m. MBC = 22.16 KN.m. MCB = -153.10 KN.m. MCD = 159.00 KN.m. MDC = 5.20 KN.m. Example 3: Analyse the two span prismatic beam under applied moment by slope deflection method. Draw the bending moment diagram. EI is constant.

Fig 13.3 (a)

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136

As there is no loading on the spans, fixed end moments are equal to zero. Also the supports are at the same level and hence

= 0. Supports A and C are fixed and hence

the joint clockwise couple M is acting and hence positive Relative stiffness: KAB = EI/2L KBC = EI/L Keeping EI/2L = 1, KAB = 1 and KBC = 2. Slope deflection equations: MAB = 1 (2 B) MBA = 1 (4 B) MBC = 2 (4 B) MCB = 2 (2 B) Applying joint equilibrium equations at ‘B’ MBA + MBC + M = 4 B

B

+8

B

+M=0

= -M/12

Substituting the value

B

in equations (a), we get:

MAB = 2

B

= -M/6

MBA = 4

B

= - M/3

MBC = 8

B

= -2/3 M

MCB = 4

B

= - M/3

The free body diagram is as follows.

The corresponding B.M diagram on the tension side is drawn as follows.

Fig 13.3 (b) & (c)

A

=

C

= 0. On

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Example 4: Analyse an overhanging propped cantilever beam with loading as shown. Further a clockwise couple of 100 KN.m. is acting a joint B. Use slope deflection method.

Fig 13.4 (a) EI is constant Fixed end moments 2

20 × (6) MFAB = = 60 KN.m. 12 = -60KN.m.

MFBA A

= 0,

B

is unknown

Relative stiffness: KAB = EI/6; Keep EI/6 = 1,

KAB = 1

Slope deflection equations: MAB = 60 + 1 (2 B) MBA = -60 + 1 (4 B) MBC = 50 KN.m. Joint equilibrium at B: MBA + MBC + 100 = - 60 + 4 B

B

+ 50 + 100 = 0

= -22.5

Substituting the value of

B,

we get the final end moments as:

MBA = 60 + 2 × (-22.5) = 15 KN.m. MBC = - 60 + 4 × (-22.5) = - 150 KN.m. MBC = 50 KN.m.

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138

Free body diagram

B.M.D. is drawn on tension side Fig 13.4 (b)

Exercise Problems: Solve the problems by slope deflection method: 1. Analyse the two-span continuous beam ABC fixed at A. EI is constant for the two spans sketch B.M.D.

Fig 13.5 2. Analyse the two-span continuous beam ABC fixed at A. Support B sinks by 10mm. E = 20 × 106 KN/m2 and I = 20 × 10-4 m4

Fig 13.61 3. Analyse the given two span prismatic beam by slope deflection method. Sketch B.M.D. A clock wise couple 50 K.N.m. is applied B.

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139 Fig 13.7 **** STRUCTURAL ANALYSIS UNIT - 13

CONTENTS: 13.1

Introduction: Analysis of continuos beams

13.2

Numerical Examples

AIMS: To explain the slope deflection method for continuous beams including support settlements by solving typical examples. OBJECTIVES: After going through this lesson, the student should able to solve problems on continuous beams by slope - deflection method independently. Slope Deflection method: Analysis continuous beams. 13.1 INTRODUCTION: In unit 12, slope-deflection equations were developed. Equations (12-4 and (12-5) give the basic slope - deflection relationships. In section 12-4, joint equilibrium equations were discussed. A simple example was also presented to explain the principles involved in the slope- deflection method. We shall now discuss the analysis of continuous beams including support settlements. The slope - deflection equations are presented here for ready reference: MAB = MFAB + EI/L (4 MBA = MFBA + EI/L ( 2 Here

= /L, where

A

+2

A

+4

B B

-6 )

(12.4)

-6 )

(12.5)

is the support settlement. When there is a support settlement, it will

be more convenient to rewrite the equations. (12.4) and (12.5) in the modified form as given. MAB = MFAB + EI/L (4 MBA = MFBA + EI/L ( 2

A

+ 2 B) - (6EI/L)

(a)

+ 4 B) - (6EI/L)

(b)

A

A few examples will clarify the method.

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13.2 NUMERICAL EXAMPLES: Example. 1:

Fig 13.1(a) Analyse the two-span continuous beam ABC with an over hang CD. Shown in Fig 13.1 (a) by the slope deflection method. Assume that all the supports are at the same level. Draw the bending moment diagram. The given beam has two spans AB and BC with an overhang CD support A is fixed and hence

A

= 0, Support B and C are freely supported add therefore

B

and

C

are the two

unknown relations Fixed end moments MFAB = 20 × (6)2 / 12 = 60 KN.m. MFBA = 20 × (6)2 / 12 = 60 KN.m. MFBC = 60 × 4 × (2)2/(6)2 = 26.66 KN.m. MFCB = 60 × 2 × (4)2/(6)2 = 53.33 KN.m. Here there is an overhang CD. The cantilever moment MCD at C is equal to 20 × 1 KN.m., as shown in Fig.13.1 (b). This is positive according to our sign convention

Fig 13.1 (b) MCD = 20 KN.m. Relative stiffness

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141

KAB = EI/6 KBC = 1.5EI/6 Keeping EI/6 = I, KAB = 1 and KBC = 1.5. Slope deflection equations Since all the supports are at the same level, the term and BC. Also =

A

A

= /L will be zero for both spans AB

= 0. Substituting

= 0, in the equations (a) and (b) we have MAB = 60+1 (2 B) MBA = -60+1 (4

B)

(c)

MBC = 26.66 + 1.5 (4

B+

MCB = -53.33 + 1.5 (2

B

2 C)

+ 4 C)

MCD = 20 In this example

and

B

C

are the unknowns we need two equations to solve two unknowns.

Applying joint equilibrium equations at joints B and C, we get the following two equations:At joint B: MBA + MBC = 0 At joint C: MCB + MCD = 0

(d)

Substituting the expression for MBA, MBC, MCB, and MCD from (c) we obtain: MBA + MBC = (-60+4 C) + (26.66 + 60B+3 C) = 0 MCB + MCD = (-53.33+3 B) + (6

C

+ 20) = 0

(e)

Simplifying (e), we have 10 3

B B

+3

=6

C C

= 33.34

= 33.33

(f)

The above two equations can be expressed in matrix from as 10

3

B

3

6

C

=

33.34 33.33

Expressing symbolically

[k ]{ } = {P} where [ k ] =

10 3 3 # B& ' % C(

[ ]= $

6

(g)

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and

142 #33.34& ' %33.33(

{P} = $

The student is advised to note that [ k ] is a symmetric matrix of size 2×2. The size of this matrix will be equal to the number of unknowns { } and {p} are column vectors. The [ k ] matrix is the same for a given continuous beam irrespective of loading. The { P} matrix only will change with loading on the beam. By matrix operations:-

{ } = [k ] 1{p}

(gh)

In this problem

[k ] 1 = B

=

C

3 Adj[k ] 1 6 = 51 3 10 [k ]

3 #33.34 & 1 6 $ ' 51 3 10 %33.33(

After carrying matrix multiplication B C

=

1962 .

(I)

4.574

NOTE: When using relative stiffnesses of their absolute values the relations obtained are not the actual magnitudes but proportional to them. Substituting the numerical values of

B

and

C

in the expressions (C), we get the final moments as: MAB = 60 + 2 × 1.962 = 63.92 KN.m. MBA =- 60 + 4 × 1.962 = - 52.15 KN.m. MBC = 26.66 + 1.5 ( 4 × 1.962 + 2 × 4.574) = 52.15 KN.m. MCB =- 53.33 + 1.5 ( 2 × 1.962 + 4 × 4.57) = - 20 KN.m. Now the student can verify the joint equilibrium equations(d) with the computed numerical values of moments. The bending moment and shear force diagram can be drawn as usual. If the student is not familiar with matrix algebra, he can solve equations (f) for

B

and

C

by

the conventional elimination method. Example 2: Analyse the three span continuos beam ABCD fixed at the ends A and D (Fig 13.2) by the slope deflection method. Support B sinks by 10mm. Take E = 20 × 106 KN/m2 and I = 20 × 10-4 m4.

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143

Solution: The beam has three spans AB, BC and CD. the supports A and D are fixed and hence

A

=

D

= 0. As the supports B and C are freely

Fig 13.2 supported, there will be rotations 10mm, the term

B

and

C

which are unknown. As the support B sinks by

= /L exists for the two spans AB and BC, and for the span CD,

=0

since C and D are as same level. Fixed end moments:

60 × 2 × (4)2

MFAB =

(6)2

60 × (2)2 × 4

MFBA = M FBC =

= 53.37KN.m.

(6)2

= 26.66KN.m.

(a)

40 × (5) 2 = 83.33KN . m. 12

M FBC =

40 × 52 = 83.33KN . m. 12

M FCD = 30 × (6) 2 = 90 KN . m.

MFDC = - 90KN.m.

(b)

Relative stiffnesses: KAB = 2I/6 KBC = I/5 KCD = 2I/6 Multiplying the stiffeness values by 15/I, we have the relative values as: KAB = 5, KBC = 3, and KCD = 5. (see note under equation 1. in example.1) Slope - deflection equations:

(c)

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144

The term

= /L for the two spans is computed as follows: for the span AB, the

support B is at lower level than at A, hence

is negative for span AB. For the span BC, the

support C is at higher level than at B, hence is positive for span BC. Span AB,

= - 10/6000 = - 1/600

Span BC,

= 10/5000 = 1/500

The quantities (6EI/L)

are computed.

Span AB =

6 × 2 × 20 × 106 × 20 × 10 4 1 = 133.33KN.m. 6 600

Span AB =

6 × 20 × 10 6 × 20 × 10 4 1 = 96.66KN.m. 500 5

Substituting

A

=

D

(e)

= 0 and the above quantities, the slope deflection equations are obtained.

Thus we have: MAB = 53.33 + 5 (2 B) + 133.33 KN.m. MBA = -26.66 + 5 (4 B) + 133.33 KN.m. MBC = -83.66 + 3 (4

B

+2 C) -96.66 KN.m.

MCD = -83.66 + 3 (2

B

+2 C) -96.66 KN.m.

MCD = 90 + 5 (4 C) MDC = -90 + 5 (2 C)

(f)

Joint equilibrium equations At joint D MBA + MBC = (-26.66 + 20

B

+ 133.33) + (83.33 + 12 B) + 6

C

- 96.66) = 0

At joint C. MCB + MCD = (-83.33 + 6

B

+ 12

C

- 96.66) + (90+20 C) = 0

(g)

Simplifying we have: 32 6

B+ B

6

+ 32

C

= - 93.34

C

= 90

32

6

B

6

32

C

=

(h) 93.34 90

[k ]{ } = {p} 32

6

[ ] = [k ] 1[p] = 1/ 988 6 32

93.34 90

=

2.476 3.48

Substituting the above values in equation (f), we get the final moments as:

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145

MAB = 161.9 KN.m. MBA = 22.16 KN.m. MBC = 22.16 KN.m. MCB = -153.10 KN.m. MCD = 159.00 KN.m. MDC = 5.20 KN.m. Example 3: Analyse the two span prismatic beam under applied moment by slope deflection method. Draw the bending moment diagram. EI is constant.

Fig 13.3 (a) As there is no loading on the spans, fixed end moments are equal to zero. Also the supports are at the same level and hence

= 0. Supports A and C are fixed and hence

the joint clockwise couple M is acting and hence positive Relative stiffness: KAB = EI/2L KBC = EI/L Keeping EI/2L = 1, KAB = 1 and KBC = 2. Slope deflection equations: MAB = 1 (2 B) MBA = 1 (4 B) MBC = 2 (4 B) MCB = 2 (2 B) Applying joint equilibrium equations at ‘B’ MBA + MBC + M = 4 B

B

+8

B

+M=0

= -M/12

Substituting the value

B

in equations (a), we get:

MAB = 2

B

= -M/6

MBA = 4

B

= - M/3

MBC = 8

B

= -2/3 M

A

=

C

= 0. On

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146

MCB = 4

B

= - M/3

The free body diagram is as follows.

The corresponding B.M diagram on the tension side is drawn as follows.

Fig 13.3 (b) & (c) Example 4: Analyse an overhanging propped cantilever beam with loading as shown. Further a clockwise couple of 100 KN.m. is acting a joint B. Use slope deflection method.

Fig 13.4 (a) EI is constant Fixed end moments 2

MFAB =

20 × (6) = 60 KN.m. 12 = -60KN.m.

MFBA A

= 0,

B

is unknown

Relative stiffness: KAB = EI/6; Keep EI/6 = 1, Slope deflection equations: MAB = 60 + 1 (2 B)

KAB = 1

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147

MBA = -60 + 1 (4 B) MBC = 50 KN.m. Joint equilibrium at B: MBA + MBC + 100 = - 60 + 4 B

B

+ 50 + 100 = 0

= -22.5

Substituting the value of

B,

we get the final end moments as:

MBA = 60 + 2 × (-22.5) = 15 KN.m. MBC = - 60 + 4 × (-22.5) = - 150 KN.m. MBC = 50 KN.m.

Free body diagram

B.M.D. is drawn on tension side Fig 13.4 (b)

Exercise Problems: Solve the problems by slope deflection method: 1. Analyse the two-span continuous beam ABC fixed at A. EI is constant for the two spans sketch B.M.D.

Fig 13.5 2. Analyse the two-span continuous beam ABC fixed at A. Support B sinks by 10mm. E = 20 × 106 KN/m2 and I = 20 × 10-4 m4

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Fig 13.61 3. Analyse the given two span prismatic beam by slope deflection method. Sketch B.M.D. A clock wise couple 50 K.N.m. is applied B.

Fig 13.7 **** STRUCTURAL ANALYSIS UNIT - 14 CONTENTS: 14.1

Introduction

14.2

Numerical examples - symmetric frames

14.3

Analysis of unsymmetric frames.

14.4

Numerical examples.

AIMS: 1. To introduce to the student symmetric and unsymmetric frames. 2. To clearly explain the difference in the analysis of symmetric and unsymmetric frames. OBJECTIVES: After going through the lesson the student should be able to solve symmetric and unsymmetric frames independently. 14.1 INTRODUCTION: In unit 10, moment distribution method was developed for the analysis of frames. Now we shall develop the slope-deflection method. see figs. 14.1 (a) to (f)

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149

All these frames are single - bay single - storey frames. Figs. (a) and (b) show symmetrical frame under symmetrical loading. Fig.(c) shows an unsymmetric frame with unequal columns. Fig (d), shows a symmetric frame with unsymmetric loading. Fig. (c) shows a symmetric frame with lateral load. Fig (f) shows an unsymmetric frame with lateral load.

(d)

(e)

(f)

Fig 14.1 14.2 SYMMETRIC FRAMES UNDER SYMMETRIC LOADING: In this section we shall discuss about the analysis of symmetric frames under symmetric loading. For these frames, there will be joint rotations but there will be no lateral displacement or ‘sway’. These are some times called ‘non-sway’ frames. Fig. 14.2 (a), shows a symmetric frame under symmetric loading. the frame is symmetric about the vertical axis passing through the centre of the beam BC. Fig (b), shows the corresponding deflected shape in dotted form. From the symmetric deformed shape, we can observe that the rotations and moments at B and C are equal in magnitude but opposite in sign. Thus in

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150

Fig 14.2 symmetric frames, the number of unknown rotations are halved, thus considerably reducing the numerical work. Fig. (c) shows a symmetric frame where the axis of symmetry passes through the centre column. The corresponding deformed shape is shown in Fig. (d). In this frame the slope at ‘C’ is zero. Also, the rotations and moments at B and E are equal but opposite in sign. Thus it is sufficient to analyse a substitute frame (Fig. (e)) with end C fixed.

Fig 14.2 (e) A great amount of work is thus reduced. A few examples will clarify the analysis. Symmetric frames under symmetric load do not undergo lateral displacement or sway. 14.2

NUMERICAL EXAMPLES: SYMMETRIC FRAMES UNDER SYMMETRIC LOADS:

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151

Example.1: Find the moments in the members of the frame shown in Fig 14.3 EI is constant

Fig 14.3 This frame is symmetric with symmetric loading. Here the supports A and D are fixed and hence

A

=

D

= 0. The two unknown rotations are

Fixed end moments 2

MFDC =

30 × (8) = 160KN.m. 12 2

MFCD =

30 × (8) = 160KN.m. 12

Relative stiffnesses KAB : KCD = I/4; KBC = I/8 Keeping I/8 = 1; KAB = KCD = 2, KBC = 1 Slope deflection equations:

Here the term Substituting

= /L will not be present. A

=

D

= 0 and

MAB = MFAB + EI/L (4

A

= 0: we have

+ 2 B) - 6(EI/L)

= 2(2 B) MBA = 2 (4 B) MBC = 160 + 1 (4

D

+ 2 C)

MCB = - 160 + 1 (2

B

+ 4 C)

But because of symmetry,

C

=-

B:

for joint equilibrium: MBA + MBC = 8

B

+ 160 + 4

B

+2

C

=0

D

and

C.

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152 12

Substituting 10

B

B

+2

C

= -160

C

=-

B,

equation (a) becomes:

= - 160

B

Substituting the value of

= -16

B:

MAB = 4

B

= - 64 KN.m.

MBA = 8

B

= - 128 KN.m.

MBC = 160 + 1 (4

B

+ 2 C)

= 160 - 32 = 128 KN.m. MCD = - 128 KN.m., MCD = 128 KN.m. MDC = 64 KN.m. Example: 2 Find the moments of the members of the frame shown in Fig 14.4

Fig 14..4 The frame is single-bay two-storey frame. Since it is symmetric, there will be two unknown rotations

B

and

C.

Fixed end moments: MFDE =

30 × 2 × (4)2 (6)2

+

30 × 4 × (2)2 (6)2

MFDB = - 40KN.m. MFCB = 20 × 36/12

= 60KN.m.

MFCB

= - 60 KN.m.

Relative stiffness: KAB = I/3 = KBC KBE = KCD = 4I/6 Keeping I/3 = 1, KAB = KBC = 1 KBE = KCD = 2

= 40KN.m.

D

=-

C

and

E

=-

B,

and

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153

Slope deflection equations: =

A

F

=0

MAB = 1 (2 B) MBA = 1 (4 B) MBC = 1 (4

+ 2 C)

B

MBE = 40 + 2 (4 MCB = 1(4

C

+2 E)

C

+ 2 B)

MCD = 60 + 2 (4

C

+2

D)

Joint conditions: (i) At joint B MBA = MBC + MBE = 0 (ii) At joint C: MCB + MCD = 0 Substituting

E

=-

B

(b)

and

D,

4

B

+4

B

+2

4

C

+2

B

+ 60 + 4

C

we have

+ 40 + 4 C

B

=0

=0

(c)

Simplifying: 12 2

B B

+2

+8

= -40

C C

= -60

(d)

writing in matrix form 12

2

B

2

8

C

40

=

60

Solving: B C

=

Substituting for

2.170 6.956 B,

C

and

E

=-

B,

D,

=-

in equations (a), we get MAB = 2

B

= - 4.34 KN.m.

MBA = 4

B

= -8.68 KN.m.

MBC = (4

B

+ 2 C) = - 22.592 KN.m.

MBE = (4

B

+ 2 C) = -22.592 KN.m.

MBE = 31.32 KN.m.

C

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154

MCB = - 32.164 KN.m. MCD = 32.17 The moments: MDC = -MCD

MEB = -MBE

MDE = - MCB

MED = - MBC

MEF = - MBA

MFE = - MAB

Example 3: Find the moments of the members of the frame shown in fig 14.5

Fig 14.5 This is two-bay, two-storey frame. The frame is symmetric about the central column. Comparing with fig.14.2 (c), this frame can be replaced by a substitute frame shown in Fig 14.6, with ends fixed at D and E.

Fig 14.6 Here the unknown are

B

and

C.

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155 =

D

E

=

A

=0

Fixed end moments: MFAB = MFBA= MFBC = MFCB = 0 MFBE = 20 × 36/12 = 60 KN.m. MFEB = - 60 KN.m. MFCD = 30 × 36/12 = 90 KN.m. MFDC = - 90 KN.m. Relative stiffness: KAB = KBC = I/3 KBE = KCD = 2I/6 Keeping I/3 = 1, KAB = KBC = KBE = KCD = 1 Slope deflection equations: MAB = 1 (2 B) MBA = 1 (4 B) MBC = 1 (4

B

+ 2 C)

MDE = 60+1 (4 B) MCB = 1 (4

C

+ 2 B)

MCD = 90+1 (4 C)

(a)

Joint conditions At joint B, MBA + MBC + MBE = 0 C, MCB + MCD = 0 4

4

C

+4

C

+2

B

B

+2

+90 + 4

C

(b) + 60 + 4 C

B

=0

=0

Simplifying 12 2

B

+2

C

= - 60

B

+8

12

2

B

2

8

C

C

= - 90 =

60 90

Solving we get B C

=

3.26 10.43

Substituting the values of

B

and

C,

the final end moments are:

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MAB = 2

B

= - 6.52 KN.m.

MBA = 4

B

= -13.04 KN.m.

MBC = (4

B

+2 C) = -33.9KN.m.

MBE = 60 + 4 MCB = (4

C

B

= 46.96 KN.m.

+ 2 B) = - 48.24 KN.m.

MCD = 90 +4

C

= 48.28 KN.m.

the moments in the other half of the frame will be equal but of opposite sign. 14.3 ANALYSIS OF UNSYMMETRIC FRAMES: In Fig 14.1, the frame shown at (d) is symmetric but the loading is unsymmetric. (e) shows a symmetric frame with lateral load. (c) and (f) show unsymmetric frames. When symmetric frames are subjected to symmetric loading. there is no “sway”. Sway is defined as lateral displacement hence it is not necessary to suffix “lateral” “sway”). But even when the frame is symmetric but subjected to unsymmetric loads or when the frames are subjected to lateral loads, the frame under goes sway. Similarly when the frame is unsymmetric irrespective of type of loading, they may (A non-symmetric frame under a particular loading pattern may not sway) under go lateral sway (Fig 14.1 c). Now look at the fig 14.7 (a). The frame is symmetric

Fig 14.7 but the load is unsymmetric. The frame deforms laterally by an amount

. This lateral

displacement is known as sway. Therefore in this frame, in addition to the joint rotations and

C,

B

there is the lateral sway . thus there are three unknowns. But we have only two joint

equilibrium conditions. therefore we have to develop an additional equation from the

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horizontal equilibrium of the frame as a whole. this equation is called the shear or bent equation. Take the frame shown in Fig 14.8. Here

A

=

D

= 0 and the unknowns are

B,

C

and

Fig 14.8 The joint conditions are : MBA + MBC = 0 MCB + MCD = 0 Now consider the free body diagram of the members AB and CD. HA and HD are the horizontal reactions at A and D. MAB and MBA are the end moments for the member AB.

Fig 14.9 MCD and MDC are the end moments for the member CD. (Fig.b)

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For the member AB, taking moments about B: HA h1 - MAB - P1 a = 0

HA =

(M

AB

(b)

+ M BA ) P1a + h1 h1

(14.1)

In the above expression if P1 is not present, the term

P1a vanishes. Similarly for the member h1

CD, taking moments about C: HD h2 - MDC - MCD - P3b = 0

HD =

(M

DC

(c)

+ M CD ) P3b + h2 h2

(14.2)

It P3 is absent, the term P3b/h2 vanishes Now the equilibrium of the frame in the horizontal direction gives: P1 + P2 - P3 - HA - HD = 0

(14.3)

This equation is known as ‘shear’ or “bent” equation. From the three equations i.e. Equations (a) and (f), we can solve

B C

and .

14.4 NUMERICAL EXAMPLES: Example .1 Analyse the given portal frame by slope deflection method. EI is constant. = sway ( to the right assumed) Since the supports A and B are fixed. A

=

D

= 0. the three unknowns are

B,

C

and the lateral sway .

14.10 Fixed end moments:

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MFAB =

60 × 3 × (3)2 (6)2

= 45KN.m.

MFBA = - 45KN.m.

MFBC =

2

40 × (6) = 120KN.m. 12

MFCB = - 120 KN.m. Since there is no loading on CD. MFCD = MFDC = 0 Relative stiffness: KAB = I/6, KCD = I/4 KBC = I/6 Multiplying by 12/I: we have: KAB = I/6 × 12/I = 2 KBC = I/6 × 12/I = 2 KCD = I/4 × 12/I = 3 Slope - deflection equations: = sway or lateral displacement considering AB and CD as two beams supported at A and B for span AB and supported at C and D for the span CD. Since there is a level difference between A and B and also C and D, the term

= /L

exists for the spans AB and CD. But for the span BC, the term does not exist since B and C are at same level.

Fig 14.11

For span AB and CD, is negative (Note for vertical or inclined members, imagine then rotated till horizontal and then visualise the sign of For AB: for

CD;

= - /6,

= - /4

or

as for a beam) and therefore:

BC = 0

(c)

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Substituting for , the slope deflection equations are obtained from equation 13.4 and 13.5 as: MAB = 45 + 2 (2

B

+ 6 /6)

MBA = -45 + 2 (4

B

+ 6. /6)

MBC = 120 + 2 (4

B

+2 C)

MCD = -120 + 2 (2

B

+ 4 C)

MCD = 3 (4

C

+ 6. /4)

MDC = 3 (2

C

+ 6. /4)

Joint equilibrium equations: Joint B: MBA + MBC = 0 Joint C: MCB + MCD = 0 shear equation: using equations : 14-1-,14-2 and 14-3, we have: HA = (MAB + MBA)/ 6 +

60 × 3 6

HD = (MCD + MDC) / 4 and the shear equation becomes

(M AB + M BA ) + 30 (MCD + MDC ) = 0

60

6

(c)

4

simplifying the equation (c), we get: 2(MAB + MBA) + 3 (MCD + MDC) = 360

(d)

Substituting the relation (a) in equations (b) and (d) we get MBA + MBC = 8

B

+2 +8

MCB + MCD = 4

B

+8

2 (4

B+

B

-2 +8

C

B

+12

+4 C

2 ) + 3 (12

C

- 45 + 120 = 0

+ 4.5 - 120 = 0 C

+ 4.5 + 6

C

(e)

+ 4.5 ) = 360

Simplifying the equation (e) 16 4

B B

24

+4

+ 20 B

C

+ 2 = -75

C+

4.5 = 120

(ii)

C

+ 35 = 360

(iii)

+ 54

Dividing the third equation by 12 and rewriting the above: 16 4

+4

C

+ 2 = -75

+ 20

C

+ 4.5 = 120

B B

(i)

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161 B

+ 4.5 + 35/12 = 30

(f)

Expressing in the matrix relation: 16 4 2 4 20 4.5 2 4.5 35 / 12

B C

- 75 120 30

[K ]{ } = {p}

[K ]

1

1 = 2 × 69.32

9.52

0.666

0.666 10.666 5.5 16

5.50 16 76

After carrying out matrix multiplication 9.52 Symmetric # 75& 13.83 + + 1 -.666 10.666 $120 ' = 1/ 2 12.26 C = 2 × 69.32 5.5 16 76 +% 30 +( 11.14 B

# B& The student should note that the matrix [ K ] is symmetric. After substituting the value $ ' % C(

in the relations (a), we get MAB = 28.48 KN.m. MBA = - 89.18 KN.m. MBC = 89.20 KN.m. MCB = -98.62KN.m. MCD = 61.84 KN.m. The B.M diagram is drawn as follows:

Fig 14.12

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span moments Moment at centre of span AB: = 60 × 6/4 - (89.2 + 28.48)/2 = 31.16KN.m. Moment at centre of span BC: = 40 × (6)2/8 - (89.2 + 98.62)/2 = 86.1 KN.m. To find the shear force in the members, the free body diagrams of the members are most useful: Span AB

Fig 14.13 HA =

(M AB + M BA ) + 60 × 3 6

6

= (29.48 - 89.18)/6 + 30 = 19.90 KN.m. HB = 60 - 19.90 = 40.10 KN. Span CD:

HD =

(MDC + MCD ) 4

= (98.62 + 61.84)/4 = 40.11 KN HC = -40.11 KN. Span BC:

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Fig 14.14 VB + VC = 40 × 6 = 240 KN. Taking moments C: VB × 6 - 89.2 + 98.62 - 40 × 6 × 3 = 0 VB = 118.43 KN VC = 240 - 118.43 = 121.57 KN. The reactions at the support are shown as given below:

Fig 14.15 RA = VB = 118.43 KN. RD = VC = 121.57 KN. Example 2: Analyse the given portal frame by slope deflection method. EI is constant.

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Fig 14.16 In this frame, there is only lateral load of 80 KN at joint B. This frame is the same as in example 1. As there is no transverse load on the members, fixed end moments for all the members vanish. Here the unknowns are the rotations

B

and

C

and the sway .

Relative stiffness: K = KBC = 2, KCD = 3 Slope-deflection equations: MAB = 2 (2

B

+ 6 /6)

MBA = 2 (4

B

+ 6 /6)

MBC = 2 (4

B

+ 2 C)

MCB = 2 (2

B

+ 4 C)

MCD = 3 (4

C

+6 /4)

MDC = 3 (2

C

+ 6 /4)

(a)

Joint equilibrium equations: Joint B: MBA + MBC = 0 Joint C: MCB + MCD = 0 The horizontal shears at A and D are HA =

HD =

(M AB + M BA ) 6

(MCD + MDC ) 4

(b)

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The shear equation: 80 - HA - HD = 0

(M AB + MBA ) + (MCD + M DC ) = 80 6

4

(c)

Substituting the expression for the moments MAB, MBA .......etc., and simplifying the equations (b) and (c) we get. 16

B

+4

C

+2 =0

C

+ 4.5 = 0

4

B

+ 20

2

B

+ 4.5

C

+ (35/12) = 80

Rewriting in the matrix form 16 4 2

4

2

20 4.5 4.5 35 /12

B

0

C

= 0 80

[ k ]{ } = { P} The student should note that the matrix (K) is the same as in the previous example. Therefore (k)-1 is the same. Hence the values: # + $ + %

& 9.52 0 1 + .666 10.666 Symmetric 0 C' = + 2 × 69.32 55 16 76 80 . ( B

on multiplication we get B C =

3.173 9.232 43.85

Substituting the above values in relations (a), we get,

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Fig 14.17 MAB = 75 KN.m. MBA = 62.31 KN.m. MBC = - 62.31 KN.m. MCB = - 86.55 KN.m. MCD = 86.54 KN.m. MDC = 141.93 KN.m. The bending moment diagram is drawn as follows:

Fig 14.18 EXERCISE PROBLEMS: Solve the portal frames by slope deflection method. Sketch b.m.d. Problem 1

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Problem 2

Problem 3

Problem 4

167

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Problem 5

Problem 6

168

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**** STRUCTURAL ANALYSIS UNIT 15 CONTENTS: 15.1

Introduction to Kani’s method

15.2

Development of the method for continuous beams

15.3

Numerical examples.

AIMS: To develop Kani’s method for the analysis of beam and frames and to introduce the student to the advantages of this method over other methods. OBJECTIVES: After going through this unit the student should able to understand the Principles of Kani’s method and in a position to solve beams and frames. 15.1 INTRODUCTION: This method was developed by Gasper Kani of Germany in 1947. this method is very fertile for frames with lateral sway in particular. It has eliminated the two stage analysis of frames with lateral sway in moment distribution method. It has not any special advantage over moment distribution method for the solution of continuous beams and non-sway frames. Also the student learned the slope-deflection method in the earlier units. In the slopedeflection method, number of simultaneous equation are to be solved. Now with the introduction computers, any number of simultaneous equations can be solved. Hence the moment distribution method and Kani’s method has no more advantages except to solve simple frames and beams by long hand. 15.2 BASIC PRINCIPLES: Members with no translation of joints: Consider a member AB in a continuos beam or a frame with transverse loading as shown in Fig 15.1 (a). The deflected shape of the member is shown in Fig 15.1 (b). It is assumed that the joints under go translations, and also the ends A and B will rotate through A

and

B

respectively. Let MAB and MBA be the end moments of the member (beam) AB.

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Fig 15.1 Sign Convention: The sign convention followed in the slope-deflection method is adopted herein, that is: 1. Anti clockwise end moments are positive and 2. Anticlock wise rotations are positive. The actual member end moments developed due to super position of the following four components of deformation. a) The member AB is considered as completely restrained or fixed. The fixed end moments for this condition are expressed as MFAB and MFBA at ends A and B respectively (Fig 15.1(c)) b) The end A only is rotated through an angle

A

inducing a moment 2M1AB at end A and

M1AB at the farther end B with is fixed (Fig. 15.1.d). The moment M1AB is known as rotation moment at end A. c) Now the end ‘A’ is considered as fixed in the deformed position and the end B is rotated through angle

B

which induces a moment 2 M1BA at B and moment M1BA at end A

(Fig 15.1e) the moment M1BA is termed as the rotation moment at end B. d) Now the member AB fixed at both ends A and B, the ends A and B under goes translation A

and

B

respectively ‘ ‘ is the difference (

(Fig.15.1f.) The fixed end moments developed are M”AB = M”BA = 6EI /L2

B

-

A),

between the supports A and B

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When the translation of joints occurs along with rotations, the final end moments can be expressed as the super position of the four moments, that is MAB = MFAB + 2M’AB + M1BA + M”AB ’

MAB = MFBA + M AB +

2M’BA

+M

”

and

BA

(15.1)

The quantity M”AB = M”BA is known as displacement moment of the member AB. For member AB, when we refer to the final moment MAB at A, and A may be refereed as the near end and B as the far end. Similarly when use refer to moment MBA at B, end B may be referred as the near end and end A as the far end. Therefore, the relationship in Equation 15.1, may be stated as the true moment at the near end of a member is the algebraic sum of (a) the fixed end moments at the near end (b) twice the rotation moment of near end and (c) rotation moment of the far end (d) the moment due translation at near end.

Fig 15.2 Equation 15.1 is applicable to all members of the frame. Now we consider the different members meeting at joint A. The member end moments at A for the members meeting at ‘A’ are given by: MAB = MFAB + 2M’AB + M’BA + M”AB MAC = MFAC + 2M’AC + M’CA + M”AC MAD = MFAD + 2M’AD + M’DA + M”AD MAE = MFAE + 2M’AE + M’EA + M”AE For the equilibrium of the joint, the sum of the end moments at ‘A’ must be zero, that is

or where

M AB = 0

(15.3)

MFAB + 2 M1AB + M’BA + M”AB = 0

(15.4)

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MFAB = Algebraic sum of fixed end moments at A of all members meeting at A M’AB = Algebraic sum of rotation moments at A of all the members meeting at A. M’BA = Algebraic sum of the rotation moments of far ends of the members meeting A. M1AB = (-1/2) ( MFAB + M’BA + M”AB)

(15.5)

From the moment - rotation relationship

Fig 15.3 2M’AB = 4EIAB/LAB = 4E KAB

(15.6)

A

where KAB = IAB/LAB, the relative stiffness for member AB Therefore M’AB = 2E KAB.

(15.7)

A

At joint A all the members undergo rotation M’AB = 2E

A

A.

Assuming E is the same for all

KAB.

(15.8)

Dividing equation 15.7 by equation 15.8, we have

M ,AB K = AB ZM AB , ZK AB or

MAB , =

Substituting for

(15.9)

K AB ZM AB , ZK AB

(15.10)

M’AB from equation 15.5

K MAB , = ( 12 ) AB (MFAB + K AB

M ,AB +

M ,AB , )

(15.11)

The ratio (-1/2) KAB/ KAB is known as rotation factor for the member AB at joint A. Denoting the rotation factor as UAB UAB = (-1/2) (KAB./ KAB)

(15.12)

The Equation 15.11 can be written as M’AB = UAB ( MFAB + M’BA + M”AB)

(15.13)

In this equation the summation of fixed end moment MFAB is known quantity for the given joint. If joint translation

is known, the displacement moments can be determined

using the relation 6EI /L2. If

is zero, then the displacement moments is taken as zero.

Further if translation ‘ ‘ is known additional equation is to be used which will be discussed

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later. To start with far end rotation moments M’BA are not known and hence they may be taken as zero. By similar approximation the rotation at other joints are also determined. With the approximate values of rotation moments computed it is possible to again determine a more correct value of the rotation moment at A for member AB using Equation 15.13. The procedure explained above is carried out till the accurate values of the rotation moments are obtained. Then the final end moments are obtained from Equation 15.1. Some points: 1. The sum of the rotation factor is (-1/2) 2. If an end of member is fixed, the rotation at that end zero and consequently rotation moment is zero. 3. If an end of a member is hinged or pinned it convenient to consider it as fixed and take the relative stiffness as (3/4) I/L. Some examples are presented to illustrate the Kani’s method. Example 1: Analyse the given continuous beam using the Kani’s method Sketch b.m.d, EI is same for all the members.

Fig 15.4 The fixed end moments are

MFAB =

30 × 1 × (2)2 (3)2

+

30 × 2 × (1)2 (3) 2

= 20KN.m.

MFBA = -20KN.m. MFBC = 60 × 2 × 4/16 = 30 KN.m. MFCB

= -30 KN.m.

MFCD = 30 × (4)2/12 = 40 KN.m. MFDC

= - 40 KN.m.

Now the relative stiffness and rotation factors are evaluated Joint B

Member BA

Rel.Stiffness K I/3

K (7/12) I

Rotation factor U = (-1/2) K/ K UBA = - .285

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174 BC

I/4

CB

I/4

CD

I/4

UBC = - .214 I/2

UCB = - .25 UCD = - .25

The sum of fixed end moments: at B : - 20 + 30 = 10 KN.m. at C : -30 + 40 = 10 KN.m. The scheme for proceeding with the method of rotation contribution is as shown in Fig. 15.5. The beam line is drawn and joints B and C are marked by two squares one inside the other. The sum of fixed end moments at each joint are entered in the inner squares.

Fig 15.5 Rotation factors (-0.285) and (-.214) for members BA and BC at B and (-.25), and (-.25) for members CB and CD at C are entered in the annular spaces as shown in Fig 15.4. The member fixed end moments are written above the beam line. The rotation moments can now be determined by iteration process as presented below. First take joint B. Apply Equation 15.13. Here the displacement moments M”AB = M”BA = 0. Therefore we have: M’BA = UBA ( MFAB + M’AB) and

M’BC = UBC ( MFBC + M’BC)

in which MFAB = sum of fixed end moments at B = 10 KN.m. M’AB = Sum of rotation moments of far ends of members meeting at joint B. we know M’AB = 0 since end A is fixed M’CB = 0 assumed to start with

(15.14)

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Substituting these values in equation 15.14. M’AB = -0.286 (10 + 0 + 0) = -2.86KN.m. M’BC = - 0.214 (10+0+0) = - 2.14 KN.m. These rotation moments are entered below the beam line. Now consider the joint C. Rotation moments M’CB and M’CD will be determined using equation 15.13. M’CB = -0.25 (10 - 2.14 + 0) = -1.965 KN.m. M’CD = -0.25 (10 - 2.14 + 0) = - 1.965KN.m. These rotation moments are shown in Fig 15.4. in the appropriate places. This completes one cycle. This procedure is repeated starting again from joint B. More accurate rotation moments at joint B can be obtained by taking the approximate values obtained in the first cycle. For example considering joint B. The sum of the rotation moments at the far ends. at A = 0 at C = -1.965 Therefore M’BA = -0.286 (10 - 1.965 + 0) = -2.298 KN.m. M’BC = -0.214 (10 - 1.965 + 0) = -1.719 These values of rotation moments super side the values (-2.86 and -2.14) obtained earlier. Now take joint C. Revised values of rotation moments are determined as explained earlier. Therefore: M’CB = - 0.25 (10 - 1.719 + 0) = -2.07 M’CD = -0.25 (10 - 1.719 + 0) = -2.07 These values replace the previous values of -1.965. The new values are shown below and the previous values are struck. This completes the second cycle. Proceeding again starting from joint B. M’BA = -0.286 ( 10 - 2.07 + ) = -2.27 M’BC = -0.214 (10 - 2.07 + 0) = -1.70 At joint C: M’CB = -0.25 (10 - 1.70 + 0) =-2.08 M’CD = -0.25 (10 - 1.70 + 0) =-2.08 These values replace the previous values. This completes the third cycle. At this stage it is seen that the difference between the 2nd cycle previous values in the present values of the

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third cycle is small. Hence no further iterations are necessary. We can take the values of the 3rd cycle can be taken as final values. Now we have acceptable values of rotation moments so that the final moments can be determined using Equation 15.1. The final moments are calculated as follows: MAB = 20 + 2(0) - 2.27 = 17.70 KN.m. MBA = -20 + 0 -2 × 2.27 = -24.54 KN.m. MBC = 30 - 2 × 1.7 - 2.08 = 24.52 KN.m. MCB = -30 - 1.7 - 2 × 2.08 = -35.86 KN.m. MCD = -40 - 2 × 2.08 + 0 = 35.84 KN.m. MDC = -40 - 2.08 + 2(0) = -42 KN.m. The span moments i.e., at the centre of the span are determined as given below: Span AB : 30 - (17.33 + 24.56)/2 = 8.86 KN.m. Span BC : 60 - (24.56 + 35.84)/2 = 29.80 KN.m. Span CD : 60 - (35.84 + 42.08)/2 = 21.0 KN.m. The support reactions can be determined as explained in the previous chapters. The B.M.D is drawn below.

Fig 15.6 BOOKS: 1. Basic structural Analysis by C.S.Reddy. **** STRUCTURAL ANALYSIS UNIT 16 CONTENTS: 16.1 Introduction 16.2 Kani’s method - continuous beams 16.3 Kani’s method - Symmetrical Frames Exercise Problems

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AIMS: 1. Examples are presented to make the student understand the solution of continuous beams by Kani’s method. 2. Examples are presented to make the student understand the solution of symmetric frames by Kani’s method. 16.1 INTRODUCTION: In the unit 15, Kani’s method is developed the expression for rotation factors, rotation moment and final member end moments are presented. An example is presented to make student understand the Kani’s method. In this unit some more examples are presented in continuous beams with different support conditions, including settlement of supports. Further symmetric frames are also presented. Numerical Examples: continuous beams: Example :1 Determine the end moments of the continuous beam shown in Fig 16.1. The relative values of I of the span are shown along the span. E is constant.

Fig 16.1 The beam has hinge at A and roller supports at D. It will be convenient to consider that: a) The ends are fixed and the relative stiffness of span AB and CD are taken as (3/4) KAD and (3/4)KCD respectively, and b) The fixed end moments MFBA and MFCD are modified as (MFBA - 1/2 MFAB) and (MFCD 1/2 MFDC) respectively. With the above modifications the hinged or roller ends are assumed to be fixed having zero fixed end moments. The fixed end moments are: MFAB = 40 × (3) (1)2 / (4)2 = 7.50 KN.m. MFBA = 40 × (1) (3)2 / (4)2 = -22.50 KN.m. MFBC = -MFCB = 10(8)2/12 = 53.33 KN.m. MFCD = - MFDC = 40 × 6/8 = 30.0 KN.m.

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The moments are adjusted to account for hinged or roller supports at A and D. The modified moments are: MFBA = -22.50 - 1/2 (7.5) = - 26.25 KN.m. MFCD = +30.0 - 1/2 (-30) = 45.00 KN.m. MFAB = MFDC = 0 The rotation factors are: Joint

Member

Rel. Stiffness

K

Rotation factor U =

1 2

K K

K B C

BA

3/4(I/4)

BC

(2I/8)

CB

2I/8

CD

(3/4) (1.5I/6)

(7/16)I

-0.21 -0.29

(7/16)I

-0.29 -0.21

The fixed and moments and the rotation moments are entered in Fig 16.2

Fig 16.2 Cycle1 Joint B: To begin with the rotation moments at four ends are assumed to be zero. Therefore, M’BA = (-.21) (27.08 + 0 + 0) = -5.8 KN.m. M’BC = (-0.29) (27.08) = - 7.74 KN.m. Joint C: M’CB = -0.29 (-8.33 -7.74 + 0) = 4.59 KN.m. M’CD = (-0.21) (-8.33 -7.74 + 0) = 3.44 KN.m. These rotation moments are shown entered in Fig 16.2.

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The rotation moments at A and D are noted as Zero since they have been replaced by fixed ends. Cycle 2: Joint B: More accurate rotations can be obtained from previously computed values of rotation moments. For example at joint B: M’BA = (-0.21) (27.08 + 0 + 4.59) =-6.79 KN.m. M’BC = (-0.29) (27.08 + 0 + 4.59) = -9.05. KN.m. Joint C: M’CB = -0.29 (-8.33 - 9.05 + 0) = 4.97 KN.m. M’CD = -0.21 (-8.33 - 9.05 + 0) = 372 KN.m. The procedure is continued and values upto four cycles are shown entered in Fig 16.2. Now the rotation moments are known and hence the final moments are evaluated MAB = 0 MBA = -26.25 - 2 × 6.87 = - 39.99KN.m. MBC = 53.33 - 2 × 9.17 + 5 = 39.99 KN.m. MCB = -53.33 + 2 × 5.0 - 9.17 =-52.50 KN.m. MCD = 45.00 + 2 × 3.75 + 0 = + 52.50 KN.m. MDC = 0 Since the supports A and D are simply supported the moments at these supports must be zero. Hence moment calculations need be carried out here. Example 2: Analyse the continuous beam by Kani’s method. EI is constant

Fig 16.3 The beam is fixed at A and supported on roller at ‘C’ with an over hang of 2m. The fixed end moments are

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180

MFAB = 10 (8)2/12 = 53.33 KN.m. MFBA = 10(8)2/12 = -53.33 KN.m. MFBC = 50 × 6/8 = 37.5 KN.m. MFCB = - 50 × 6/8 = -37.5 KN.m. MFCD = 30 × 2 = 60 KN.m. Relative stiffness KAB = I/B, KBC = I/6, KCD = 0 Rotation factors are UBA = -0.21, UBC = -0.29, UCB = -0.50 The values are shown entered in Fig 16.4. Proceeding in the usual manner the rotation moments are evaluated. The rotation moments for five cycles are shown.

Fig 16.4 The final end moments are computed as usual MAB = 53.33 + 6.64 + 0 = 59.97 KN.m. MBA = - 53.33 + 2 × 6.64 = - 40.05 KN.m. MBC = 37.5 + 2 × 9.18 - 15.80 = 40.06 KN.m. MCB = -37.5 - 2 × 15.00 + 9.18 = -59.92 KN.m. MCD = 60 KN.m. Beams with support settlements: Now let us take continuous beams with support settlements. In these problems, the fixed end moments due to applied loading and due to sinking of supports are added algebraically and the net moments are considered the final fixed ends moments in further computations.

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Example 3: Analyse the given continuous beam using the Kani’s method. Support ‘B’ sinks by 15mm. Take E = 200 × 106 KN/mm2 and I = 100 × 106 mm4. draw the b.m.d. ‘D’ is simply supported.

Fig 16.5 It will be convenient to (a) consider the end ‘D’ is fixed and the relative stiffness for CD is taken as 3/4 KCD as a fixed end. (b) the fixed end moments MCD is modified as (MFCD - 1/2 MFDC). with the above modifications the roller support at ‘D’ is assumed to be fixed having zero fixed end moments. The fixed end moments are: MFAB = 40 (6)2/12 = 120 KN.m. MFBA = - 120 KN.m. MFBC = 40 × (6)2/12 = 120 KN.m. MFCB =

120 KN.m.

MFCD = (120 × 6/B - 1/2 × 120 ×/8 ) = 130 KN.m. The fixed end moments due to support settlement ‘ ‘ are calculated as follows. When we consider the span AB, the support ‘B’ is at lower level and hence the fixed end moments are positive (see fig 16.6(a))

Fig 16.6 As the support ‘C’ is at higher level the fixed end moments are negative.

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MFAB = MFBA = 0.015 × 6 × 2 × 200 × 106 × 100 × 106/(6)2 = 100KN.m. MFBC = MFCB = -100 KN.m. The fixed end moments due to the known settlement of supports are to be added to the moments due to loading to arrive at net fixed end moments. The total fixed end moments are: MFAB = 120 + 100 = 220 KN.m. MFBA = -120 + 100 = - 20 KN.m. MFBC = 120 - 100 = 20 KN.m. MFCB = -120 -100 = - 220 KN.m. MFCD = 135 KN.m. Relative stiffness: KAB = 2I/6 = I/3 KBC = 2I/6 = I/3 KCD = 3/4 (3I/6) = 3 I/8 Rotation factors are: UBA = - 1/2 (I/3) / (I/3 + I/3) = - 1/4 UBC = -1/2 (I/3) / (2I/3) = -1/4 UCB = -1/2 × I/3/ (I/3+3I/8) = -0.235 UCD = -1/2 × 3I/8/ (I/3+3I/8) = - 0.265

Fig 16.7 The remaining procedure is same as the problems solved earlier. The results are tabulated as given in 3 cycles gave accurate results. The final member end moments are computed as usual: MAB = MFAB + 2M’AB + M’BA MAB = 220 - 5.3 = 214.70 KN.m.

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MBA = -20 - 2 × 5.3 = - 30.60 KN.m. MBC = 20 - 2 × 5.3 + 21.22 = + 30.62 KN.m. MCB = -220 + 2 × 21.22 - 5.3 = 182.86 KN.m. MCD = 135 + 2 × 23.92 = 182.84 KN.m. Now the student is advised to draw the bending moment diagram. 16.2 FRAMES WITH NO LATERAL TRANSLATION OF JOINTS: The frame in which lateral translations are prevented are analysed in the same way as continuous beams. The lateral sway is prevented either due to support conditions or due to symmetry of the frame and loading. The procedure is illustrated by solving the following example. Example 4: A rectangular portal frame has a beam member of ‘6m’ and column members of height 4.5m height. The moments of inertia of the beam and columns are 2I and I respectively. An u.d.1. of 30 KN/m acts on the length of the beam. Analyse the frame by Kani’s method. Sketch b.m.d.

Fig 16.8 Due to symmetry of geometry of the frame and loading the frame does not undergo lateral translation. The fixed end moments are MFBC = 30 (6)2/12 = 90 KN.m.

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MFCB = -90 KN.m. MFAB = MFBA = MFCD = MFDC = 0 Relative stiffness: KAB = KCD = I/4.5, KBC = 2I/6 = I/3 Rotation factors: At joint B: UBA = -1/2 KBA / (KBA + KBC)= -1/2 × I/4.5/ (I/4.5 + I/3) = -0.20 UBC = -1/2 × I/3 (I/4.5 + I/3) = -0.3 At joint C : UCB = -0.3 UCD = -0.2 Now the rotation moments are worked but in the manner as was worked out in the previous examples for continuous beams. These are shown entered in Fig 16.9

Fig 16.9 The final moments are: MAB = -25.56 KN.m. MBA = -2 × 25.56 = - 51.12 KN.m. MBC = 90 - 2 × 38.5 + 38.50 = 51.5 KN.m. MCB = -90 + 2 × 38.5 - 38.5 = - 51.5 KN.m. MCD = 2 × 25.56 = 51.12 KN.m. MDC = 25.56 KN.m. The bending moment diagram is drawn on tension side.

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Fig 16.10 16.3 SYMMETRICAL FRAMES UNDER SYMMETRICAL LOADING: When a frame is symmetric in both geometry and loading conditions computational work is reduced considerably. Two cases of symmetry arises, namely frames in which the axis of symmetry passes through the centre line of the beams and frames with a six of symmetry passing through column base. Case1: Axis of symmetry passes through centre of beams. Let AB be any horizontal member of frame through the centre of which the axis of symmetry passes.

Fig 16.11

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(a) End moments and rotations (b) Rotation at end A due to moment MAB (c) rotation at end A due to moment MBA. Let MAB and MBA be the end moments. Due to symmetry of deformation, MAB and MBA are numerically equal but are opposite in sense. The slope at A =

A

is obtained as a superposition of the rotations due to MAB and -

MBA as shown, in Fig 16.11 ‘b’ and ‘c’ respectively. Therefore, =

A

’ A

+

” A

From the moment rotation relationships , = M AB L , - = M L / 6EI = M L / 6EI A BA AB 3EI

A

Therefore A

=

’ A

+

” A

= MABL/2EI

Let this member be replaced by member AB’ whose end will undergo rotation

A

due to

moment MAB applied at end A, while end B’ is being restrained. The substitute member will have the same value of I as for the original member. For such a beam the force displacement relation is A

= MAB L’/4EI

Fig 16.12 where L’ is the length of the substitute member. Hence for the equality of rotations between original member AB and the substitute member AB’. A

= MAB L/2EI = MAB L’/4EI

or I/L = 2I/L’ or k = 2k’ or k’ = k/2

(16.1)

Thus if k is the relative stiffness of original member AB, this member can be replaced by substitute member AB’ having relative stiffness k/2. With this substitute member, the analysis then needs to be carried out for only one half of the frame considering the line of symmetry as the fixed end. The following examples illustrate the various steps involved in the analysis of symmetric frames. Examples: Analyse the given frame. Use the concept of symmetric frame.

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Fig 16.13 Since the axis of symmetry passes through the middle of the beam BC, only one half of the frame will be considered. The substitute frame is shown in Fig 16.14.

Fig 16.14 (a) Substitute frame (b) Rotation moments Fixed end moments are MFBC = 40 × (6)2/12 = 120 KN.m. MFCB = -120 KN.m. The relative stiffnesses are: KCD = KAB = I/3, KBC = 3I/6 KBC’ = 1/2 KBC = I/4 The rotation factors are UBA = - 4/14, UBC = - 3/14

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The true rotation moments are obtained in the first distribution only. The final moments are worked as follows: MAB = - 34.29 KN.m. MBA = -2 × 34.29 = - 68.58 KN.m. MBC = 120 - 2 × 25.71 = + 68.58 KN.m. From symmetry: MCB = - 68.58 KN.m. MDC = 34.29 KN.m. Example 6: Analyse the frame shown in Fig 16.15 adopting the advantage of symmetry

Fig 16.15 The fixed end moments are

MFBE =

30 × 2 × (4)2 (6)2

+

30 × 4 × (2)2 62

= 40KN.m.

2

20 × (6) MFCD = = 60KN.m. 12 The substitute frame is shown 16.16 (a). For the substitute frame the relative stiffnesses and the rotation factors are worked out. The rotation moments are evaluated at joints B and C. These are shown in Fig 16.16 b. Relative stiffnesses are KAB = KBC = I/3 KBE’= KCD’ = 1/2 × 4I/6 = I/3 Rotation factors:

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Joint B: UBA = UBC = 1/ 2 ×

I/3 = 1/ 6 (I / 3 + I / 3 + I / 3)

Joint C: UCD’ = UCB = -1/2 × I/3 / (I/3 + I/3) = -1/4 The final moments are computes as shown:

Fig 16.16 MAB = -4.35 KN.m. MBA = -2 × 4.35 = -8.70 KN.m. MBE’ = 40 - 2 × 4.35 = 31.30 KN.m. MBC = -2 × 4.35 - 13.91 = -22.61 KN.m. MCB = - 2 × 31.91 - 4.35 = - 32.17 KN.m. MCD’ = 60 - 2× 13.91 = 32.18 KN.m.

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Fig 16.17 The moments on opposite sides are shown. B.M.D is drawn on tension side

Fig 16.18 Span moment CD =

20(6) 8

2

32.15 = 57.85KN.m.

Span moment BE = 30 × 2 - 31.3 = 28.70 KN.m. Exercise Problems:

1. Analyse the continuous beam shown by Kani’s method. Sketch b.m.d.

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Fig 16.19 2. Analyse the continuous beam shown by Kani’s method. Sketch b.m.d

Fig 16.20 3. Analyse the continuous beam given. EI is same for all spans. Sketch b.m.d. Support B sinks by 5mm. I = 25 × 10-6 m4, E = 210 × 106 KN/m2.

Fig16.21 4. Analyse the frame by Kani’s method. Sketch b.m.d.

Fig 16.22 5. Analyse the frame by Kani’s method. Sketch b.m.d.

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Fig 16.23 Books: (1) Basic structural Analysis by C.S.Reddy. **** STRUCTURAL ANALYSIS UNIT 17 CONTENTS: 17.1 Introduction 17.2 Kani’s method - Frames with lateral sway 17.3 Numerical Examples 17.4 Frames with unequal column heights 17.5 Example Exercise problem. 17.1 INTRODUCTION: In the unit 16, Kani’s method for continuous beam and symmetric frames without lateral sway are discussed. This unit deals with discussion about frames with lateral sway which differ from the frames discussed in the unit 16. Considering the sway, the equations are developed. 17.2 FRAMES WITH LATERAL SWAY: Consider a multistoreyed frame shown in Fig 17.1(b). The frame carries both vertical (gravity) loading and lateral loading. The lateral loads may be due to wind or earthquake forces. For simplicity the lateral loads are taken as acting at joints only. Thus these lateral loads do not cause any fixed end moments on the vertical members (columns) but develop

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shear in the vertical members. Fixed end moments are developed due to vertical loads only acting on horizontal members (beams). Let AB represent a vertical member in any storey of a multistroreyed frame (Fig 17.1 (a)) P is the lateral load acting at the joints. MAB and MBA be the end moments at A and B. Let H be the horizontal force developed on column AB. If the hr is the height of the rth storey.

Fig 17.1 then from the equilibrium consideration of member AB. MAB + MBA + H (hr) = 0 or

H=

(M AB + MBA )

(17.1) (17.2)

hr

Consider now a general building frame shown in Fig.17.1 (b). Let AB, CD, EF and GH represent columns in a particular storey. In general

H represent the shear in that storey. If we denote Qr = shear in the rth

storey, we can write: Qr = - (MAB + MBA)/hr

(17.3)

where hr = height of columns of the rth storey MAB = sum of end moments at the upper ends of all the columns in the rth storey MBA = sum of end moments at the lower ends of all the columns in the rth storey. From equation 17.3, we can rewrite Qr hr = - ( MAB +

MBA)

(17.4)

we know the general expression for end moments for the member AB is MAB = MFBA + 2M’AB + M’BA + M”AB and MBA = MFBA +2M’BA + M’AB + M”BA For columns with lateral load acting at joints, the fixed end moments

(17.5)

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MFAB = MFBA = 0. For any prismatic member it may be noted that M”AB = M”BA Therefore MAB + MBA = 3 M’AB + 3 MBA + 2 M”AB From Equation 17.4. - ( MAB + MBA) = Qrhr Qr hr = or

Qr hr / 3 =

r

(3(M’AB + M’BA) + 2M”AB)

# %

, + r $ M,AB + MBA

2 & M,,AB ' ( 3

This gives MAB ,, = 3/ 2{Qr hr / 3 + r (M AB , + MBA , )}

(17.6)

Here Qr hr/3 is known as storey moment. This is positive when Q acts from right to left. We know that the relative lateral displacement

is the same for all the columns in any one

storey. For any column the translation moment is

MAB ,, = 6EI h 2 = 6EI h r r where = /hr Thus the translation moment of a column in a storey is proportional to the relative stiffness K = I/hr Therefore

M”AB/ M”AB = KAB/ KAB

or

M”AB = (KAB/ KAB). M”AB

(17.7)

Substituting for M”AB from equation 17.6 we can write equation. 17.7 M”AB = (KAB/ KAB) (-3/2) {Qrhr/3 + or

M”AB = .AB { Qrhr/3 +

r

r

(M’AB + M’BA)}

(M’AB + M’BA)}

(17.8)

in which .AB = (-3/2) KAB/ KAB is called displacement factor for member AB. It may be noted that in equation 17.18. (M’AB + M’BA) represents the sum of the rotation moments at top and bottom ends of all the columns of the sotrey under consideration. KAB = sum of relative stiffnesses of all the columns of a storey under consideration. Obviously the sum of the translation factors of all the columns of a storey will be equal to (-3/2). Summing the various relationships obtained earlier are rewritten as follows: M’AB = UAB (MFAB +

M’BA + M”AB)

(17.9)

CE405/1 M’BA = UBA (MFBA + M’AB + M”AB = M”BA = .AB (Qrhr/3 +

195 M”BA) ’ r(M AB

+ M’BA)

(17.10)

MAB = MFAB + 2M’BA + M’BA + M”AB MBA = MFAB + M’AB +2M’BA + M”BA

(17.11)

Applying Equations 17.9 and 17.10, the rotation and translation moments are determined by iteration for all storeys in turns. In any cycle there are two steps. In the first, the rotation moments are determined with trail values of displacement moments. In the second step the displacement moments are determined with the trial values of step 1. Thus the cycle completes. This process of iteration continues till acceptable values of rotation and translation moments are known. Thus the final moments are found from equations 17.11. This method is explained with examples. 17.3 NUMERICAL EXAMPLES: Example 1: Analyse the given protal frame by Kani’s methods

Fig 17.2 The fixed end moments are: MFBC = 67.5 × 2 × (4)2/(6)2 = 60.0 KN.m. MFCB = -67.5 × (2)4 × 4 / (6)2 = - 30.0 KN.m. The rotation factors are computed as usual are equal to:

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UBA = UBC = -1/4 and UCB = UCD = -1/4 The translation factor for each column is: .AB = .CD = -3/2 KAB/ KAB =

3/ 2. 2I

(I / 2 + I / 2 )

= 3/ 4

The fixed end moments and rotation factors are entered as shown. Now we can proceed with iteration process. The various computations follow the following order. First the rotation moments are calculated at joints B and C and then displacement moments for column AB and CD.

Fig 17.3 Rotation & Translation moments Here in this problem, there is no lateral load acting at joint B. Hence Qr hr /3 = 0. Cycle1: Rotation Moments Joint B: Assuming all far end rotation and translation moments to be zero, the following first approximations are obtained for the near end rotation moments: M’BA = -1/4 (60) = - 15 KN.m. M’BC = - 15 KN.m. Joint C:

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Sum of fixed end moments

=

-30.0 KN.m.

Rotation moment at B

=

-15.0 KN.m.

Rotation moment at D

=

0.0 Fixed end

Displacements for columns AB,CD = Total

0.0 (Assumed in first trail) -45.0 KN.m.

Therefore M’CB = -1/4 (-45) = +11.26 KN.m. Translation M’CD =

+11.26 KN.m.

Storey1: (There is only one storey in present case) Rotation moments at top of Columns AB

=

- 15.0 KN.m.

Column CD

=

+ 11.26 KN.m.

At bottom of columns

=

0.0 KN.m. -3.74 KN.m .

Therefore: M”AB = -3/4 (-3.74) = 2.80 KN.m. M”CD = -3/4 (-3.74) = 2.80 KN.m. The rotation moments are entered for the beam ends as earlier and columns at their ends. Note that the translation moments are entered along the columns at the mid height of storey. This completes the first cycle. Cycle 2: Joint B: Sum of fixed end moments

= + 60KN.m.

Rotation moment at A

= 0.0 KN.m

Rotation moment at B

= 11.26 KN.m.

Translation moments of column AB

= 2.80 KN.m .

Total

= 74.06 KN.m.

M’BC = -1/4 (74.06) = - 18.52 KN.m. M’BA = -1/4 (74.06) = - 18.52 KN.m. Joint C: Sum of fixed end moments

= -30.0 KN.m.

Rotation moments at B

= -18.52 KN.m.

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Rotation moments at D

= + 0.0 KN.m.

Translation moments of CD

= + 2.8 KN.m .

Total M’CD

-45.72 KN.m .

= -1/4 (-45.72) = 11.44 KN.m.

M’CD = -1/4 (-45.72) = 11.44 KN.m. Storey 1: Rotation moments at top of column AB

=

-18.52

Rotation moments at top of column CD

=

+ 11.44

Total

- 7.08

Translation moments M”AB = M”CD = -3/4 (-7.08) = 5.22 KN.m. This completes the second cycle. In a similar manner the computations were carried out upto fives cycles. The rotation and translation moments in successive cycles are entered. Once the rotation moments and translation moments are known with desired accuracy, the final moments can be computed using the equations 17.5. They are given as follows: MAB = -19.30 + 6.44 = - 12.86 KN.m. MBA = -2 × 19.30 + 6.44 = - 32.12 KN.m. MBC = 60 - 2 × 19.30 + 10.72 = + 31.12 KN.m. MCB = - 30 + 2 × 10.72 - 19.3 = - 27.86 KN.m. MCD = 2 × 10.72 + 6.44 = +27.88 KN.m. MDC = 6.44 + 10.72 = 17.16 KN.m. Examples: Analyse the given frame by Kani’s method

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199 Fig 17.4

Fixed end moments: MFBC = 100 × 2 × (4)2 / (6)2 = 88.88 KN.m. MFCB = - 100 (2)2 × 4/ (6)2 = -44.44 KN.m. Relative stiffness and rotation factors: KAB = 2I/4 = I/2 KBC = 3I/6 = I/2 KCD = 2I/4 = I/2 Rotation factors: UBA = -1/4 and UBC = -1/4 UCB = -1/4 and UCD = -1/4 Translation factors: UAB = UCD = -3/2 (I/2 / (I/2 + I/2)) = -3/4 Storey moment Qrhr/3 = - 50 × 4/3 = - 66.67 KN.m. The storey moment is recorded in a small rectangular block. The fixed end moments, rotation factors are entered as shown.

Fig 17.5 The rotation moments and displacement moments are given by the relation: M’AB = UAB ( MFAB + M’BA + M”AB) M”AB = .AB [(Qrhr/3 +

(M’AB + M’BA)]

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Cycle:1 Joint B: Assuming all far end rotation and translation moments to be zero, the following first approximations are obtained for the near end rotation moments: M’BA = -1.4 (88+0+0) = -22.22 KN.m. M’BC =

= - 22.22 KN.m.

Joint C: Sum of fixed end moments

= -30.0KN.m.

Rotation moment at B

= -22.22KN.m.

Rotation moment at D

= 0.00 (fixed end)

Displacement moments for columns AB and CD

= 0.00 (assumed)

Total

= -52.22 KN.m .

Therefore M’CB = -1/4 (-66.67) = + 16.56 KN.m. M’CD

= 16.56 KN.m.

Translation moments: Storey 1: Storey moment

= - 66.67 KN.m.

Rotation moments at top of Column AB

= -22.22

Column CD

= +16.66 Total

-72.23 KN.m.

Therefore: M”AB = -3/4 (-72.23) = 54.16 KN.m. M”CD

= 54.16 KN.m.

Cycle 2: Joint B: Sum of fixed end moments

= 88.88 KN.m.

Rotation moment at A

= 0.0 KN.m.

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Translation moment at B

= 16.56 KN.m.

Moment for AB

= 54.16 KN.m

Total

= 159.6 KN.m .

Therefore: M’BC = 1/4 (159.6) = -39.9 KN.m. M’BA = -39.9 Joint C: Sum of fixed end moements

= -44.44 KN.m.

Rotation moment at B

= -39.90 KN.m.

Rotation moment at D

= 0.0KN.m.

Translation moment of column CD = 54.16 KN.m. Total

-30.18 Kn.m .

Therefore M’CB = -1/4 (-30.18) = 7.54 KN.m. M’CD

= 7.54 KN.m.

Translation moments: Storey 1: Rotation moment at top of column AB

= -39.90

Rotation moment at top of column CD

= 7.54

Storey moment

= -66.67 Total

-99.00 KN.m.

Translation moments: M”AB = M”CD = -3/4 (-99.0) = 74.25 KN.m. In similar way the student is advised to continue and do some more trials till acceptable values are obtained. Then the final moments may be computed just like in example 1:

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Fig 17.6 Example 3: We shall consider two storey frame subjected to both vertical and lateral loading. The details of the frame and loading are given in frame.

Fig 17.7 Fixed end moments MFCD = 12 × (4)2/12 = 16 KN.m. MFDC = -16KN.m. MFBE = 24 × 1 × (3)2/(4)2 = + 13.50 KN.m. MFEB = 24 × (1)2 × 3/(4)2 = - 4.50 KN.m.

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Rotation factors: Joint

Members

Rel. K

B

BA

I/4

BC

I/4

-1/8

BE

I/2

-1/4

CB

I/4

-1/6

CD

I/2

C

U

K I

-I/8

3/4I

-1/3

Translation factors: In each storey there are only two columns and both of them have the same relative stiffness. Therefore the translation factor for each column. = 1/2 (-3/2) = -3/4 Storey moments: Storey 2; Qz = 8KN Qrhr/3 = = - 8 × 4/3 = - 10.67 KN.m. Storey1: Q1 = (-8) + (-16) = - 24 KN.m. Qrhr/3 = = - 24 × 4/3 = - 32.0 KN.m. The storey moments are recorded in small rectangular blocks at the mid height of each storey to the first column line. The iterations are carried out in the following order. First joints B-ED-C and then storeys 2 and 1. Cycle1: Joint B: The rotation moments and translation moments are initially assumed to be zero. Sum of fixed end Moments at the joint Sum of rotation moments at ends:

= 13.50 KN.m. A

= 0.00 fixed end

C

= 0.00 (assumed)

E

= 0.00 (assumed)

Sum of translation moments of : Column

BC

= 0.00 (assumed)

BA

= 0.00 (assumed) + 13.50 KN.m.

Using Equations 17.9:

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M’BA = M’BC = (-1/8) (13.5) = -1.69 KN.m. M’BE = -1/4 (13.5) = - 3.38 KN.m. Joint C: Sum of fixed end moments at joint

= -4.5 KN.m.

Sum of rotation moments at far ends:

Total Therefore

B

= -3.38 KN.m.

D

= 0.00 (Assumed)

F

= 0.00 (Assumed) -7.88 KN.m.

M’ED = M’EF = (-1/8) (-7.88) = 0.99 KN.m. M’ED = (-1/4) (-7.88) = 1.97 KN.m.

Joint D: Sum of fixed end moments at joint

= -16.0 KN.m.

Sum of rotation moments at far ends: at C

= 0.0 KN.m.

at E

= 0.99 KN.m.

Sum of translation moments of Column : DE Total M’DC

= 0.00 KN.m. 15.01 KN.m .

= -1/3 (-15.01) = 5.0 KN.m.

M’DE = (-1/6) (-15.01) = 2.50 KN.m. Joint C: Sum of joint moments

= 16.0 KN.m.

Sum of rotation moments at far ends B

= -1.69 KN.m.

D

= 5.00KN.m.

Sum of translation moments of column BC Total M’CB = (-1/6) (19.31) = - 3.22 KN.m. M’CD = (-1/3) (19.31) = - 6.44 KN.m.

= 0.00 KN.m. 19.31KN.m.

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Storey 2: Storey moment

= -10.67 KN.m.

Rotation moments at ends of columns BC

= -3.22 KN.m. (Top)

Rotation moments at ends of columns BC

= 1.69 KN.m.(Bottom)

Column DE

= 2.50 (Top)

Column DE

= 0.99 (bottom)

Total

= -12.09 KN.m.

M”BC = M”DE = (-3/4) (-12.04 = 9.07 KN.m. Storey 1: Storey moment

= -32.0 KN.m.

Rotation moment from column ends AB

= -1.69 (top) = 0.00 (bottom)

EF

= 0.99 (top) = 0.00 (bottom )

Total

=-32.70 KN.m .

M”AB = M”EF = (-3/4) (-32.7) = 24.53 KN.m. The rotation moments and translation moments are entered. Cycle 2: Improved values of rotation moments and translation moments can be obtained by taking the values obtained in the first. Cycle. Consider Joint B: Sum of joint moments

= 13.50 KN.m.

Sum of rotation moments at far ends

at Sum of translation moments column

Total

A

= 0.00 KN.m.

C

= 3.22 KN.m.

E

= 1.97 KN.m.

BA

= 24.53 KN.m.

BC

= 9.07 KN.m. 45.85 KN.m .

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M’BC = M’BA = (-1/8) (45.85) = -5.73 KN.m. M’BE = (-1/4) (45.85) = -11.46 KN.m. The computations are carried out at joints E,D,C in that order as in cycle 1. Similarly translation moments are found out for storey 1 and storey 2. Thus it completes the 2nd cycle. In a similar manner the subsequent cycles are carried out taking each time improved rotation and translation moments obtained in the immediate previous cycle. These are entered in the fig. The final moments are computed using the equations 17.5.

Fig 17.8 Final moments: MAB = MFAB + 2M’AB + M’BA + M”AB MBA = MFBA + M’AB + 2M’BA + M”AB Using the above relations, the final end moments are evaluated. MAB = 0+ 0 + (-6.22) + 32 = + 25.70 KN.m. MBA = -2 × 6.22 + 32 = 19.56 KN.m.

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MBC = -2 × 6.22 - 5.42 + 18.46 = 0.61 KN.m. MCB = 2 × (-5.42) - 6.22 + 18.46 = 1.41 KN.m. MBE = 13.15 - 2 × 12.44 - 8.91 = -20.29 KN.m. MCD = 16.00 - 2 × 10.83 + 4.25 = -1.41 KN.m. Similarly the member end moments are calcualted 17.4 FRAMES WITH UNEQUAL COLUMN HEIGHTS - BASES FIXED OR HINGED: Now let us consider a single storey frame with unequal column heights and some of which may be fixed and others hinged at their bases. For simplicity, let us take a single storey, single bay portal frame as in Fig 17.2, but the conclusions drawn from them are quite general. Lateral displacements develope additional moments in the columns and Equations 17.11 are applicable. Moments denoted as M” are known as linear displacement moments and are as follows:

and

M”AB = M”BA = 6E KAB /hAB

(17.12)

For member CD : M”CD = 3EKCD/hCD

(17.13)

M”DC = 0

(17.14)

Now let us replace column CD which is hinged at the base by an equivalent column C’D’ fixed at the base as shown in Fig. 17.2b. For substitute column we take KC’D’ = 3/4 KCD and H C’D’ = 1.5 hCD. If the two columns were to undergo the same lateral displacement at the top, we have MC,, ,D , = 6E KC, D, / hC, D , = 6E(3 4)KCD /1.5hCD = MCD ,,

(17.15)

that is so far as displacements are concerned, a substitute column may be used instead of a hinged one as shown in Fig 17.2(c), provided we ensure that the two frames have the same shear force due to the displacement. For this we introduce factors MAB and MCD for the columns in the substitute frame and equate the shear force due to displacement in the frame that is 2 M AB ,, / hAB + MCD ,, / hCD = MAB 2 M AB ,, / hAB + MCD 2 MC,, , D, / hC, D, which gives MAB = 1, for column fixed at base and

(17.16)

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MCD = (MCD ,, / MC,, ,D , ).(hC, D, / 2hCD )= 3/4

(17.17)

for the hinged column. We are now in a position deal with the substitute frame by choosing storey height hr = hAB and writing

(a) frame with unequal column heights one leg fixed and the other hinged at the base (b) Substitute column (c) frame with substitute column Fig 17.9 CAB = hr/hAB and CC,D , = hr / hC, D, or

C’CD = hr/h’CD

(17.18)

for convenience of rotation. Now summing up the horizontal forces to zero, we have Qrhr + CAB (MAB + MBA) + C’CD (MCD + MDC) = 0

(17.19)

Substituting for moments from Equations 17.11 and noting the fixed end moments = 0, we have

[

]

Qr hr + 3 CAB (MAB , + MBA , ) + CCD , (MC, ,D , + M,D, C , ) +2(MAB CAB MAB ,, + MCDCCD , MD ,, , C , ) = 0

(17.20)

This after transformation gives , MD ,, , C, = 2 / 3(Qhr / 3 + MAB .CAB M ,,AB + MCDCCD

Cik (Mik , + Mki , ))

(17.21)

, / KCD , ) = (K AB / KCD , ).(C AB / CCD , ) we know MAB ,, / MC,, , D, = (KAB / hAB )(hCD

(17.22)

Here i-k in general represents the two ends of a column

or

(

)

2 MAB .CBA M ,,AB / MCDCCD , M ,,D, C , = (MAB / MCD ) C2AB / CCD , ) (17.23) (K AB / KCD

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which gives M AB ,, = C AB . K AB

( M C M ,, (M C K AB

AB

M C,,,D , = CCD , KCD ,

AB

2 AB

AB AB

+ M CDCCD , M C,,,D , ) + M CDCCD , 2 KCD , )

,, + M CDCCD , M C ,D , ) ( M AB CAB M AB

(

2

2

M AB C AB K AB + M CDCCD , KCD ,

)

(17.24)

(17.25)

with the help of equations 17.8 the translation moment for any column i-k may be written as

Mik ,, = ( 3/ 2)Cik Kik /

(

{

Mik ,, = . ik (Qrhr / 3 +

or

where . ik = ( 3 / 2)Cik Kik /

(

)

2 Mik Cik Kik {Qrhr / 3 + Cik (Mik , + M,ki )}

}

Cik (Mik , + Mki , ))

Mik C 2ik Kik

)

(17.26) (17.27)

which are familiar as the translation factors for column i-k. A control on the calculations .ik is given by Mik Cik = -3/2

(17.28)

These derivations are perfectly general in nature and can be extended to different cases as described below. Case 1:

All the columns fixed at base For this condition: Mik = 1 for all columns and .ik = (-3/2) Cik Kik/( C2ik Kik)

(17.29)

.ik = (-3/2) Cik Kik/( C2ik Kik)

(17.30)

and the control is Cik Kik = -3/2 Case 2:

All the columns hinged at base For this condition Mik = 3/4 for all columns and

(17.31)

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.ik = -2 Cik Kik/ C2ik Kik

(17.32)

and the control is Cik .ik = -2 The method can be understood by the examples. Example 4: Analyse the given portal frame using Kani’s method

Substitute frame Fig 17.10 The hinged column is replaced by a column fixed at base having h’AB = 1.5 hAB = 1.5 (4) = 6.0m. and K’AB = (3/4) KAB = 3/4 (1) = 3/4. If we choose frame height hr = hCD = 4m, C’AB = 4/6 = 2/3 and CCD = 4/4 = 1.0 Fixed end moments: The fixed end moments are calculated as usual. However the fixed end moments MFBA is modified as (MFBA - 1/2 MFAB) to take into account hinged condition at support A. Storey moment: Lateral force Qr = (25+ 37.5/4) = 34.375 KN and the storey moment Qrhr/3 = - 34.375 × (4)/3 = - 45.83 KN.m. Rotation factors: Joint B:

Joint C:

UBA = (-1/2)

(3 4)

(1 + 3/ 4 )

= -0.21

UBC = (-1/2)

1 = -0.29 (1 + 3/ 4 )

UCB = (-1/2)

1 = -0.25 (1 + 1)

UCD = (-1/2)

1 = -0.25 (1 + 1)

Translation factors:

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For AB, MAB = 0.75, C’AB = 2/3, K’AB = 3/4 C’AB K’AB = 0.5, MAB C2AB K’AB = 0.25 For CD: MCD = 1.0, CCD = 1, KCD = 1, CCD KCD = 1.0 M2CD C2CD KCD = 1.0 Therefore using equation 17.27

. AB = ( 3/ 2)

05 = 0.6 (0.25 + 1.0)

. CD = ( 3/ 4)

1.0 = 1.2 (0.25 + 1.0)

These values satisfy the check given by equation 17.31. The fixed end moments, storey moment, rotation and translation factors are shown. The iterations have been carried out in the usual manner. To start with the rotation and translation moments are considered to be zero. Cycle1: Rotation moments: Joint B:

M’BC = -0.29 (+22.5+0+0) = - 6.53 KN.m. M’BA = -0.21 (22.5) = -4.73KN.m. Joint C:

M’CB = M’CD = -0.25 (-60 - 6.53 + 0) = + 16.63 KN.m. Translation moments:

M”AB = -0.6 (-45.83 + 2/3 (-4.73) + 16.63) = + 19.41 KN.m. M”CD = -1.2 (-32.30) = + 38.82 KN.m. This completes the first cycle of computations and the values are shown entered in the fig. Cycle 2: Rotation moments: Joint B:

M’BC = -0.29 (+22.5+16.63 + 19.41) = - 16.98 KN.m. M’BA = -0.21 (58.54) = - 12.29 KN.m. Joint C:

M’CB = M’CD = 0.25 (-60-16.98+38.82) = + 9.54 KN.m. Translation moments:

M”AB = -0.6 (-45.83 + 2/3 (-12.29) +9.54) = + 26.69 Kn.m.

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M”CD = -1.2 (44.48) = 53.58 KN.m. This completes the 2nd cycle. In like manner further iterations are carried out each time improving the values of previous cycle. The rotation and translation moments upto five cycles have been shown entered. The final moments are computed as given. MAB = 0 (Hinged support) MBA = -37.5 + 2 × 11.9 +29.64 = -31.66 KN.m. MBC = 60 - 22 × 16.44 + 4.37 = 31.49 Kn.m. MCB = -60 + 2 × 4.37 - 16.44 = -67.70 KN.m. MCD = 2 × 4.37 + 59.27 = 68.01 KN.m. MDC = 4.37 + 59.27 = 63.64 KN.m. Reference: Basic structural analysis by C.S.Reddy Exercise problems: Analyse the given protal frames by Kani’s method. Sketch b.m.d.

Fig 17.11 ****

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213 STRUCTURAL ANALYSIS UNIT 18

AIMS: 1. To derive the strain energy principles, Castigliano’s theorem II and unit load method. 2. To use the above principles for the determination of slopes and deflections in typical problems. OBJECTIVES: After going through the lesson you should be able to: 1. Apply the above principles to new problems 2. Appreciate the simplicity of the above principles for the determination of slopes and deflections for typical problems. 18.1 INTRODUCTION: In the previous chapters, the slopes and deflections for beams are developed using double integration method and moment area theorems. There is another versatile approach i.e., energy method to solve deflections and slopes. This method gives simpler solution in many complicated problems. Methods of computing internal strain energy will be reviewed first. The expressions for the elastic strain energy in axially loaded bars and members subjected to flexure and derived. Then by using the law of conservation of energy and equating the internal strain energy to the external work, the deflection will be obtained. The direct solution of problem by equating the external work to the internal strain energy turns out to be useful in cases only when one force is applied to a member. Therefore a general theorem applicable to elastic systems subjected to any number of loads is derived from deflections and rotations of any element. This theorem is known after Castigliano. 18.2. ELASTIC STRAIN ENERGY: When an element is subjected to external load, the member under goes deformation. Internal stresses and internal strains are induced. When the stresses are within the elastic limit, the stresses and strains obey Hooke’s law i.e., the relation between them is linear. Hence these systems are termed linearly elastic systems. The relation between the externally applied loads and the corresponding displacements of the load are also linear. In mechanics, energy is defined as the capacity to do work and work is the product of a force and the distance in the direction the force moves. In solid deformable bodies, stresses multiplied by their respective areas are forces, and deformations are distances. The product of these two quantities is the internal work done in a body by externally applied forces. The internal work is stored in a body as the internal elastic energy of deformation or the elastic

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strain energy. We will discuss the method of determining the internal strain energy. Consider an infinitesimal element such as shown in fig 18.1 a subjected to normal stress

(a) An element intentions

(b) Stress- strain diagram Fig 18.1

The sides of the element are dx, dy and dz. The area of the element normal to the direction of x is dy dx. On this area /x is the normal stress. The force acting right or left face of the element is /x. dydz where dydz is an infinitesimal area of the element. Due to this force the element undergoes elongation by an amount exdx where ex is the strain in the x direction. If the element is made of a linearly elastic materal, stress is proportional to strain as shown in fig 18.1b. Therefore, if the element is initially free of stress, the force which finally acts on the element increases linearly from zero until it attains its full value. The average force acting on the element while deformation is taking place is /x dydz/2. This average force multiplied by the distance through which it acts is the work done on the element. For a perfectly elastic body no energy is dissipated and the work done on the element is stored as recoverable internal strain energy. Thus the internal elastic strain energy U for an infinitesimal element subjected to uniaxial stress is dU = 1/2 /x. dy.dz . ex dx = 1/2 /xex dx dy dz. (average force x distance) = 1/2 /x ex dv

(18.1)

where dV is the volume of the element. By rewriting the equation 18.1 we obtain the strain energy stored in perfectly elastic body per unit volume of the material or its strain-energy density U0. Thus dU/dV = U0 = /x ex/2

(18.2)

This expression may be graphically interpreted as an area under the inclined line on the stress-strain diagram. Fig 18.1 (b). The corresponding area enclosed by the inclined line and the vertical axis is called the complimentary energy. For linearly elastic materials the two

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areas are equal. Expression 18.2 can be extended to normal stresses /y and /z and to the corresponding linear strains ey and ez. 18.3 STRAIN ENERGY IN AXIALLY LOADED BAR: In this problem when a bar of uniform section is subjected to an axial load P, the internal strain energy can be derived as follows: dU = /x ex/2 dV. = /x ex/2. dx dy dz using ex = /x/E; and substituting in the above we obtain dU = (/x2/2E) dV. But /x = P/A, substituting and integrating over the volume V, U= V

(P 2 / 2A2 E.)dxdydz

P 2 / 2A2 E

= L

dydz dx. A

P 2 / 2AE dx

(18.3)

L

On further integration along the bar length L, gives the total energy in the body. If P,A and E are constant U is given as U = P2L/2AE

(18.4)

18.4 STRAIN ENERGY IN BENDING: See the fig 18.2(a) and (b). Fig (a) shows the cross

section of the rectangular section and (b) shows the variation of stress diagram across the depth. According to well known flexure formula, the bending stress /x at any fibre ‘y’ from N.A is given as: /x = My/I

(18.5)

(a)

(b) Fig 18.2

where /x = bending stress at a distance ‘y’ from neutral axis

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M = bending moment M I = moment of intertia of section. Substituting the relation 18.5 for /x in equation 18.1 and is integrated over the volume V. Nothing that M and I are functions of x only and that by definition y 2dydz = I.

we have,

/x 2

U= V

2E

dV

1/ 2E (My/I)2 dx dy dz

= V

= M 2 / 2EI 2 L

2

y dydz

dx

A

= M 2 / 2EI.dx

(18.6)

L

Equation. 18.6 reduces the volume integral for elastic strain energy of a beam in bending to a single integral over the length of the beam. 18.5 Examples: On strain energy: Example 1: A typical application of the law of conservation of energy can be explained by

referring to a bar subjected to axial pull P gradually applied as shown in fig. 18.3.

Fig 18.3

The bar has elongated by an amount energy is equal to - P

in the direction of P. Here the change in the potential

/2. According to the law of conservation of energy it must appear

elsewhere. In this case it is found in the form of strain energy in the bar. This is given by Wi = P 2dx / 2AE

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where Wi = energy stored in the body or internal work Let We = P /2 where We is the external work. Therefore, according to the law of conservation of energy L

Wi + We = - P

/2 + P 2dx / 2AE = 0 0

The minus sign of the external work is taken to account for the loss potential energy -We + Wi = 0 or We = Wi that is the external work done is equal to internal strain energy. Example 2: To find the deflection at the free end of a cantilever subjected to a concentrated

load P at its free end.

Fig 18.4

Let EI be the flexural rigidity and

be the deflection at the free end.

Taking B as origin Mx = -Px then the integral energy U is given as’

M2 / 2EI dx

U= L

= ( Px)2 / 2EI dx L

= P2 L3/6EI

(a)

The external work We is given as: We = P /2 Equating (a) and (b), we get = PL3/3EI which is same as the value obtained from double integration method.

(b)

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Example 3: To find the deflection at centre of a simply supported beam of span L carrying a central concentrated load P.

Fig 18.5 Consider a simply supported beam carrying load P at its centre. Let

be the deflection at

centre under the load. Then We = P /2

(a)

Taking A as origin, the moment M is given as: Mx = P/2.x

0 < x < L/2

Since the beam is symmetric, the total strain energy U over the length 'L' is equal to twice the value over half of its length. Therefore L

U=

M2x / 2EI dx

0 L/2

M2x / 2EI dx

=2 0

L/2

(P / 2x)2 / 2EI dx

=2 0

After integration and simplifying, we have Wi = U = P2 L3 /96 EI

(b)

Equating (a) and (b) we have 3

PL = 48EI which is same as obtained by double integration method. 18.6 Castigliano's Theorem II: In 1879 Castigliano published two theorems. Theorem II,

stated below is having wide application to determine the deflection

i

in the direction of any

particular load Pi, when any system is subjected to a system of loading. Theorem II states as follows. The partial derivative of the strain energy of a linearly elastic system with respect to

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any selected force acting on the system gives the displacement of the force in the direction of its line of action.. The words force and displacements have a generalized sense and include respectively, couple and angular rotation. The derivation is as follows: Consider a simple beam fig. 18.4a subjected to a system of loading P. The beam undergoes deflections as shown in fig 18.4b.

Fig 18.5 The external work We is equal to internal strain Energy U. given as

U=

n i =1

Pi i / 2

where the summation is taken over for all the loads acting on the beam. Here Pi and

i

be any

particular load and the corresponding deflection in its direction among the system of loads. Now if any one of the loads say Pi is increased: by a differential amount say dPi, the strain energy of the system will change by an amount 0U/0Pi) dPi. The expression for the total strain Energy becomes U' = U + (0U/0Pi). dPi

(a)

which can be equated to W’e. In terms of these loads and the corresponding displacements shown in fig. (a) n

'

Uo = W e = U + The term 1/2 dPi d

i=1

i can

Pid i + 1/ 2dPi d i

be neglected. n

U' = 'We' = U + i =1

Pid i

(b)

Now suppose the sequence of loading is reversed, ie., dPi is applied first and then the system of loads P. See the fig. 18.4 (b) then W’e = U’ = U + dPi.

i

(c)

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Since the order of the application of the loads is immaterial, the total work done or the total strain energy must be same in both cases U+ (0U/0Pi) dPi = U + dPi 0U/0Pi =

i

(18.8)

i

Here Pi can be moment and corresponding

i

can be rotation.

In other cords.

0U / 0Mi =

(18.9)

i

In applying the expressions 18.8 and 18.9, the solution of the problem is not limited by the number of applied Forces. The addition of a ficticious force at a location where no actual force is applied enables one to employ the above equations at any location on a body. On setting the ficticious force equal to zero, one determines the actual displacements in the end. A few examples will illustrate the application of Castigliano's theorem II. 18.7 Examples On Castigliano's theorem II: Example 1: Use Castigliano's theorem to determine the maximum deflection of a simply supported beam subject to a single concentrated load P applied at the mid point.

Fig 18.6 Taking 'A' as origin, the bending moment to the left of the vertical load P is given by M = P/2 x from which 0M/0P = x/2 for 0 < x < L/2 Due to symmetry about mid point, the internal strain energy stored in the half of the beam to the right of the concentrated load is equal to that stored in the half to the left of the load. The total internal strain energy is L

U=

2

L /2

(Px / 2)2 dx / 2EI

M dx / 2EI = 2 0

0

According to the Castigliano's theorem, the deflection under the point of application of the load P is

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= 0U / 0Q = M (0M / 0Q)dx / EI 0 L/2

(Px / 2 )(x / 2)dx / EI = PL3 / 48EI

=2 0

Example 2: Use Castigliano's theorem to determine the deflection and slope at the free end

of a cantilever beam subject to a uniformly distributed load of intensity w/m

Fig 18.7 (i) Deflection at free end:

The problem is somewhat different nature than the previous problem, since we sought the deflection under the point of application of a real concentrated load. Here we require the deflection at the point where there is no concentrated load, so we must temporarily introduce a ficticious load Q at the free end B, represented by dotted line indicating it is not a real force. At the end of the problem, the value of Q will be made equal to zero. Taking B as origin, the bending moment at a section at a distance x from B is given as: M = - (Qx + wx2/2) from which 0M/0Q = -x Here M is negative according to the convention adopted in the previous chapters. The internal strain energy is L

U=

M2 dx / 2EI =

0

L 0

(Qx + Wx2 / 2) dx / 2EI 2

From Castigliano's theorem, the deflection under the point of application of the fictitious load Q is L

= 0U / 0Q = M (0M / 0Q)dx / EI 0 L

=

[(Qx + wx ) / 2 EI ] xdx 2

0

It is brought into the notice of the reader, that the load Q can be set equal to zero at any time after the partial derivative has been taken. Thus

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222 L

= 0

Here

(wx2 / 2)xdx / EI = wL4 / 8EI

is positive which means the deflection is downward, the same direction as the load Q.

(ii) Slope at free end: Since there is no couple at free end, apply an auxilary couple M at free

end as shown. Taking B as origin,

Fig18.8

Moment M at a section at a distance x from free end is Mx = -(M + wx2/2) from which 0Mx/ 0M= -1. The internal strain energy is L

U=

M 2 xdx / 2EI =

0

L 0

(M + Wx 2 / 2)/ 2EI dx

From the Castiglisno's theorem the slope at the free end in the direction of M is equal to partial derivative of Mx with respect to M. L B = 0U / 0M =

Mx 0

L

= 0

(0Mx / 0M ) EL

dx

((M + wx / 2)/ EI )(1)dx 2

Keeping M = 0 and performing integration we have B

= wL3/6EI.

Example 3: A simply supported beam of span 6m carries an U.D.L. of intensity 20 KN/m.

over the entire span. Further a concentrate load of 100 KN is applied at centre. Find the deflection at the centre using Castigliano's theorem. EI is constant.

Fig 18.9

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Solution: When a numerical problem is given like this, the student is advised to apply a fictitious concentrated load Q at the point where the deflection is required, even when there is a real concentrated load already, as in this problem. At the end the load Q can be set equal zero. This avoids lot of confusion.

Fig 18.10 RA = RB = (20 × 6 +100)/2 + Q/2 = Q/2 + 110. Taking 'A' as origin, the moment Mx ia given as: Mx = (Q/2 + 110) x = 20x2/2 for 0 < x < 3 and 0Mx/ 0Q = x/2. The strain energy is given as L

U =

M x 2 / 2 EIdx 0

Since the beam is symmetric, the strain energy U is equal to L 2

Mx2/2EI dx

U=2 0

This deflection at centre is given by L /2

M x .(0M.x / 0Q).dx / EI

= 0U / 0Q = 2 0

Substituting for Mx and Mx/x, we have. = ( 2 / EI )

3

0

((Q / 2 + 110) x

Setting Q = 0 since it is not real load 3

= 2 / EI (110 x 10 x 2 ) x / 2dx. 0 3

= 1 / EI (110 x 2 10 x 3 ) dx 0

)

10 x 2 x / 2dx.

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[

]

= 1 / EI (110 x 3 / 3 10 x 4 / 4) 0 = 787.5 / EI 3

18.8 UNIT LOAD METHOD: This method is other wise known as dummy load method. Consider a simple beam shown in fig 18.5. It is required to find deflection at D. Apply a unit load at D.

Fig 18.11 Then, Mx = moment at any section due to loads P1 and P2 etc. and due to fictious load Q applied at ‘D’ where the deflection is required. m = moment at any section due to unit load applied at ‘D’ where the deflection is required. mQ = moment a section due to load Q, which is equal to the product of ‘m’ and Q. m = moment at a section due to applied loads P1 and P2 etc. excluding Q. Mx = m + mQ and 0Mx/0Q = m setting Q = 0, according to the Castigliano’s theorem, the deflection at ‘D’ is given as L

= 0 U / 0Q = M.

0M / 0Qdx

0

L

= 0

EI

M.mdx EI

(18.10)

Equation 18.10 will be expression for the deflection using unit load method. Similarly if slope is required at a section, it is given as L

=

M . m, / EIdx

(18.11)

0

where m’= moment at any section due to unit couple applied at the point the slope is required. Now we will solve a few examples. 11.9 Examples: On unit load method Example 1: Find the slope and deflection at the free end of a cantilever carrying U.d.l of

intensity w/m over entire length. According to unit load method

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L

=

Mm / EIdx 0 L

=

Mm, / EIdx. 0

To find the deflection at free end, apply a unit load at free end and moment at section at a distance ’x’ from free end is given as:

M = wx2/2 m = -1. x. Substituting the values for Mx and m L B = 0

( wx2 / 2)( 1.x) / EI dx

L

= wx 3 / 2EI dx 0

= wL4/8EI Similarly to find slope at ‘B’ apply a unit clock wise couple at B. M = wx2/2 m’ = -1.

L B =

Mm, / EI dx 0

L

= wx 2 / 2(1/ E)dx 0

= wL4/6EI

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Example 2: A simply supported beam of span L carries U.d.l. of intensity w/m over the entire span. Find the deflection at centre and slope at left hand support.

To find the deflection at centre, apply a unit load at centre. M = wL/2 x - wx2/2 for 0 < x < L/2 m = 1/2 x for 0 < x < L/2

According to unit load method L

= Mm / EI dx 0

Since the beam is symmetrically loaded, L/ 2

=2

Mm / EIdx 0 L/ 2

= 2 / EI 0

(wL / 2x

L/ 2

= 2 / EI 0

(wL / 4.x2

)

wx 2 / 2 ( x / 2)dx

)

wx 3 / 4 dx

= 2/EI (wL/4.x3/3-wx4/16) 0L / 2 Simplifying we get = 5/384 wL4/EI Now to find slope at left support, apply a unit couple in the anti clockwise direction.

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M = wL/2 x - wx2/2 m = 1-x/L L A=

Mm / EI dx 0

L

= 0 L

= 0

(wL / 2x

wx 2 / 2 (1 x / L)dx.

)

(wL / 2x

wx 2 / 2 w / 2.x 2 + wx 3 / 2L dx.

)

After integration & simplifying, we get A

= wL3/24EI

Summary: (1) Strain energy: In this, the deflections are found from the relation

External work = Internal strain energy We = Wi We = Work done by applied load Wi = Internal strain energy = M2/2EI dx in bending (2) Castigliano’s Theorem II: According to this theorem, the partial derivative of strain

energy U with respect to any force/couple gives deflection/slope’ = M0M / 0Qdx = M0M / 0Q10dx

where Q/Q10 is the fictitious force/ couple where the deflection/ slope is required. The fictitious force/couple should be set equal to zero at the end. (3) Unit load method: The deflection/slope is given by = Mm / EI dx

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228 = Mm, / EI dx

where M = moment due to applied load m = moment due to unit load applied at the point where deflection is required. m’= moment due to unit couple applied at the point where slope is required. Exercise Problems: 1. Find the vertical deflection at joint 5, for the given truss. AE is same for all the members.

2. Determine the deflection at joint ‘C’ for the given truss shown. AE is same for all the members.

3. Determine the vertical and horizontal deflections of the joint ‘A’ for the cantilever truss shown. AE is same for all the members. Use unit load method.

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**** STRUCTURAL ANALYSIS UNIT 19 AIMS: 1. To find deflections for the trusses using Castigliano’s theorem II and unit load methods. 2. To solve typical examples. OBJECTIVES: After going through this lesson the student should be able to: 1. Find deflections of trusses using Castigliano’s theorem or unit load method. 19.1 INTRODUCTION: In the units 18 we have developed Castigliano’s theorem and unit load methods to determine the deflections for beams and frames. Now in this unit we shall modify those two methods to apply for the trusses. 19.2 CASTIGLIANO’S THEOREM II - DEFLECTION OF TRUSSES: Consider a simple truss shown in fig 19.1. It is required to find the deflection at joint ‘C’ due to applied load P1 and P2. As there is no load at ‘C’ apply an auxiliary fictitious load Q at ‘C’. Then:

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Fig 19.1 S = forces developed in the members due to load including Q. = change in length of the member i.e., SL/AE Where L and A are the length and cross sectional area of the member. The strain energy U stored in the member is given as: 2

U = S /2 =

S.SL S L = 2AE 2AE

(19.1)

As the truss consists of a number of discrete members, the strain energy for the entire truss is the summation of the strain energy of the individual members. U=

S2L/2AE

(19.2)

According to Castigliano’s theorem, the partial derivative of U with respect to Q gives the displacement of the load (deflection of joint C) in its direction. = 0 / 0Q =

S2L 2AE

SL/AE. (0S/0Q)

(19.3)

Equation 19.3 is the deflection equation for trusses. At the end, set Q = 0. In numerical problems also even if there is load at joint ‘C’, apply an auxiliary load Q at ‘C’. After carrying out the partial derivative, set Q = 0. 19.3 UNIT LOAD METHOD:

Apply a unit load at joint ‘C’ (fig 19.1) in place of Q, with the unit load only acting on the turss, let the forces developed in the turss members be K. Now apply the loads P1, P2...etc. and also the fictitious load Q in place of the unit load. Now the forces developed in the members of the truss are S1 = S + K Q where S = force in the member due to P1 , P2 etc. excluding Q.

(19.4)

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Now inserting S1 in place of S in Equation 19.2, we have: S12L/2AE

U=

(S + KQ)2 L

=

2AE

And the deflection

is given as

= 0U / 0Q =

(S + KQ)LK AE

After setting Q = 0, =

(SL/AE) K

(19.5)

Equation 19.5 gives the deflection of joint by unit load method. In general unit load method is most convenient to find deflections in the turss. However example will be worked using both methods. Example : 1 Determine the vertical deflection of joint C for the given truss shown. AE is same for all the members. Use Castigliano’s theorem II.

Fig 19.2 Apply auxiliary load Q at joint C. In the analysis it is convenient to find forces in the members due to known loading and due to Q separately and then add them together. The student is aware that the method of joints is more convenient compared to the method of section in cases where the forces in all the members are required. The forces are tabulated as given below. Tension and compression are indicated by positive and negative signs respectively.

AB

Member

L/AE

S (KN)

0S/0Q

SL 0S AE 0Q

(1)

(2)

(3)

(4)

(5)

2/AE

-28.90-0.577Q

-0.577

33.35/AE

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AC

2/AE

14.50+0.288Q

+0.288

8.35/AE

BC

2/AE

5.77+0.577Q

+0.577

6.66/AE

BD

2/Ae

-17.30-0.577Q

-0.577

19.96/AE

CD

2/AE

-5.77+0.577Q

+0.577

-6.66/AE

CE

2/AE

20.2+0.288Q

0.288

5.82/AE

DE

2/AE

-40.3000.577Q

-0.577

23.25/AE = 90.73/AE

In the last column (5), the value SL/AE. (0S/0Q) is computed after setting Q = 0 column (3). =

SL/AE. (0S/0Q) = 90.73/ AE

The units in the above expression are in KN and ‘m’. Converting them to N and mm: and taking A = 200 mm2 and E = 2 × 105 N/mm2

=

90.73 × 1000 × 1000 = 2.26mm 200 × 2 × 105

Example 2: Find the horizontal deflection at joint ‘C’ for the given truss shown. AE is same

for all the members. Use Castigliano’s theorem, II.

Fig 19.3 (a)

To find the horizontal deflection at joint ‘C’ an auxiliary load Q is applied at ‘C’ as shown in Fig 19.3 (b). The forces due to Q are evaluated as follows.

Fig 19.3 (b)

Taking moments about A, RD × 18 - Q × 4.5 = 0

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233 RD = (4.5/18) Q = Q/4 = 0.25Q

RA + RD = 0

RA = - 0.25 Q.

The horizontal reaction at A is equal to Q. The joint A is analysed by the method joints: Fig 19.3(c) tan

= 4.5/6 = 36.86°

Fig 19.3(c) Fy = FAB sin

- 0.25 Q = 0

FAB = 0.25 Q / sin Fx = FAB cos

= 0.416 Q

+ FAF - Q = 0

FAF = FAB cos

+Q

= - 0.332 Q + Q = 0.668 Q Similarly the forces in all the members can be evaluated, The results are tabulated as follows, both due to applied loading and due to Q. Member

L/AE

S (KN)

0S/0Q

SL 0S . AE 0Q

AB

7.5/AE

-75+0.416Q

0.416

-234/AE

BC

6.0/AE

-60+0.668Q

0.668

-240.48/AE

CD

7.5/AE

-75-0.416Q

0.416

234/AE

AF

6.0/AE

60+0.668Q

0.688

240.48/AE

FE

6.0/AE

60+0.668Q

0.668

240.48/AE

ED

6.0/AE

60+0.668Q

0.668

240.48/AE

BE

7.5/AE

0.-0.416Q

-0.416

-3.12/AE

CE

4.5/AE

45+0.25Q

0.25

50.625/AE

BF

4.5/AE

45-0

0

0 = 530.33/AE

Taking A = 500 mm2 and E = 2 × 105 N/mm2

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234 =

530.33 × 1000 × 1000 = 5.3mm 500 × 2 × 10 5

Example 3: Find the horizontal and vertical deflection of the joint for the cantilever truss shown. AE is the same for all the members. Use unit load method.

Fig 19.4 (a) According to unit load method, the deflection is given as: =

SL/AE. K

Where S = force in the member due to applied loading K = force in the member due to unit load at the joint. At support E, there is horizontal reaction and at support D, there will be both horizontal and vertical reactions. RD = 20 KN. Taking moments about D: HE × 3-20 × 6 = 0 HE : 40 KN HD = - 40 KN The forces in the members can be found as usual. (i) Vertical deflection. Apply a unit load at joint A. K: Force in the member due to unit load acting vertically. These forces can be solved as done for the 20 KN load. (ii) Horizontal deflection: Apply a unit load at A in the horizontal direction as shown in Fig. 19.4 (b).

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Fig 19.4 (b) ’

K = force in the member due to this unit load. The horizontal reaction at D is: HD = 1. For the above loading the forces in the members are found easily FAB = FBD = 1. The forces in the other members are zero. All the values are tabulated. Values of K and K’ are tabulated in separate columns. Member

L/AE

S (KN)

K

K’

(SL/AE).K

(SL/AE).K ’

AB

3/AE

20.0

1.0

1.0

60/AE

60/AE

AC

3.35/AE

-22.37

-2.24

0

167.86/AE

0

BD

3/AE

20.0

1.0

1.0

60/AE

60/AE

BC

1.5/AE

0

0

0

0

0

CD

3.35/AE

11.18

1.12

0

41.94/AE

0

CE

3.35/AE

-33.56

-3.35

0

376.38/AE

0

DE

3.0/AE

15.0

1.50

0

67.5/AE

0

773.6/AE

120/AE

The vertical and horizontal deflections at A are given as: v

=

(SL/AE) K = 773.6/AE

h

=

(SL/AE) K’ = 120/AE

Giving suitable values A and E, the numerical values can be found. Taking A = 200 mm2, E = 2 × 105 N/mm2 v

= 773.6 × 1000 × 1000/(200 × 2 × 105) = 19.30 mm.

h

=

120 × 1000 × 1000 = 3.0mm 200 × 2 × 10 5

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The student should note that the values S and L are in ‘KN’ and in metres respectively. Therefore to convert them into ‘N’ and ‘mm’ units, they are multiplied by 1000 and 1000 respectively. Example 4: The Howe truss supports three loads as shown in Fig 19.5. Determine the vertical deflection of joint E. Use unit load method.

Fig 19.5 Applying a unit load at ‘E’ the forces in the members are found. The forces due to the applied loading and unit loads are tabulated. Member

L/A

S

K

(mm/mm2)

(KN)

(KN)

AB

30

-25

-1.25

937.5

BC

30

5

1.0

150

BD

18

-15

-.75

202.5

CD

30

-12.5

-.63

236.25

CE

22.5

22.5

1.13

572

DE

-

0

1.0

-

DF

30

-12.5

-.63

236.2

DG

18

-15.0

-.75

202.5

FG

30

5.0

1.0

150

FH

22.5

15

0.75

253.1

GH

30

-25

-1.25

937.5

EF

22.5

22.5

1.13

572

AC

22.5

15

0.75

253.1

From the above table:

(SL/A)K

(SL/A).K = 4702.8, taking the cross sectional area of tension

members as 200 mm2 and compression members as 250 mm2. Given E = 2 × 105 N/mm2

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= (1/ E ) (SL / A),K =

237

4702.8 × 1000 = 23.5mm 2 × 105 EXERCISE PROBLEMS

1. Analyse the given pin-jointed truss. Area of cross section of the members is same for all the members i.e., 15 cm2, and E = 2 × 106 N/mm2.

Fig 19.6

2. Analyse the pin jointed truss. Area of the members is same for all i.e., 10 cm2 and E = 2 × 106 N/mm2.

Fig 19.7

3. Find the stresses of the given pin jointed frame due to temperature rise of 100C. Take A = 10 cm2 and E = 2 × 105 N/mm2.

Fig 19.8

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**** STRUCTURAL ANALYSIS UNIT 20 AIMS: 1. To develop the concept of statically indeterminate trusses. 2. To find the degree of internal indeterminacy of a redundant truss 3. To analyse a given redundant truss for the given loading 4. To analyse a given redundant truss due to temperature change or lack of fit. 5. To analyse externally indeterminate trusses. 20.1 STATICALLY INDETERMINATE TRUSSES: In unit 4, we have discussed about statically indeterminate beams. In this unit we shall discuss about statically indeterminate (redundant) trusses. A truss is statically indeterminate internally or externally or both. The total degree of indeterminacy (redundancy) is the sum of the internal and external indeterminacy. 20.2 INTERNALLY INDETERMINATE TRUSSES: A simple truss is shown in Fig 20.1 a. A truss is made up of individual members joined together to form a series of triangles. The members are either in tension or compression. A truss is just internally stable if it consists of a series of triangles as shown in Fig 20.1 b. The first triangle is made up of three joints and three members. Each successive triangle requires two additional members but only one additional joint. Then if m is the number of members in a truss and j is the number of joints, then (m - 3) = 2 (j - 3) or

m = 2j - 3

(20.1)

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Fig 20.1 The truss shown in (a) is statically determinate since it satisfies the condition (20.1). The truss shown in Fig 20.1(c) is statically indeterminate. According to equation 20.1, the truss requires 9 (nine) members to preserve geometry, but actually it has ten members. This kind of truss which has excess members than required for a stable truss is termed an internally statically indeterminate or redundant truss. The excess members are known as redundant members. The number of redundant members give the degree of internal indeterminacy. Therefore the internal indeterminacy for the truss shown in Fig 20.1 (c) is one. If the members are less than required i.e. if m < 2j - 3, then the truss is internally unstable or deficient. Therefore the degree of internal indeterminacy Dsi for a plane truss is given by the equation. Dsi = m - (2 j - 3)

(20.2)

20.3 ANALYSIS OF INTERNALLY INDETERMINATE TRUSSES: For the analysis of indeterminate trusses (pin-jointed frames). Castigliano’s theorem of least work gives the solution. The theorem states that “ the redundant reactions or forces in a statically indeterminate structure are such as to make the internal-work a minimum”. See the Fig 20.2a. The truss has 10 members against the 9, required for geometric stability. Of the two members 3-4 or 2-5, any one can be removed and still the truss will be stable and determinate. For example remove the member 3-4 (fig .b)

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Fig 20.2 but carrying the given loads p1 and p2. Here the member 3-4 and consequently the force developed in the original truss is a redundant. Let S be the forces developed in the members of the truss in the determinate structure (Fig. b.). Now apply a pair unit loads in the direction of the member 3-4 at joints 3 and 4 as shown in fig (c). In this, the external load will not be acting. Let ‘K’ be forces developed in the members of truss due to unit loads. Let ‘X’ be the force in the redundant members 3-4, then the forces in the members of the truss due to ‘X’ be KX. Then the total forces in the members of the truss due to loading and due to the redundant force X will be the sum of S and KX. Therefore S1 = S + KX

(20.3)

where S1 = forces in the members of the redundant truss. The strain energy stored in the truss is given as:

U=

(S + KX )2 2AE

L+

2

X L0 2AE

(20.4)

There the first term gives the strain energy of the members excluding the redundant member. Since the redundant member 3-4 is also elastic member it also stores strain energy during its deformation and hence one finds the second term X2 L0/2AE in equation 20.4. where L0 is the length of the redundant member. Applying Castigliano’s theorem of least work on equation (20.4) we get

0U = 0X

(S + KX ) AE

KL +

XL0 =0 AE

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=

241

SKL + AE

2

K LX XL0 + =0 AE AE

Solving for X, we have X=

SKL AE 2 K L L0 + AE AE

(20.5)

Equation 20.5 is the basic expression for determination of the force in the redundant member. To sum up, the analysis of internally indeterminate truss is done in two stages. Step 1: Remove the redundant member. Find the forces in the determinate structure due to

the given loading. Let ‘S’ be the force. Step 2: Apply a pair of unit loads in the direction of the redundant member. Find the forces

in the members of the truss. Let the forces be K. Now no applied load acts on the truss. If X is the force in the redundant member, then forces in the members due to X are KX. Here X is the unknown quantity. Step 3: Solve for X from Equation 20.5 Step 4: Once X is found, the forces in the members of the given redundant truss are given by

S + KX. A few examples will clarify the above steps. Tension and compression are indicated by positive and negative signs respectively. 20.4 STRESSES IN A REDUNDANT TRUSS DUE TO TEMPREATURE VARIATION AND LACK OF FIT:

In general structures may be subjected to temperature variation. Also sometimes the length of a member may be too short or too long, what we call fabrication error or lack of fit. In determinate trusses the above factors will not induce any additional stresses. But in a redundant structure, the temperature change or lack of fit induce additional stresses in addition to the stresses due to applied loading. In this section we will discuss about these factors.

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Fig 20.3 See Fig 20.3(a). The truss is indeterminate internally to a degree one. Consider the member BC as a redundant. Let BC be too short by an amount ‘ ‘ than required length. This error in length is called lack of fit. Since it is too short, it is required to be stretched by applying some force X and then fit in position. This causes stresses in the members of the truss. The effect is explained as follows. Remove the member BC as shown in fig 20.3(b) and apply a pair of unit loads at joints B and C in the direction BC. Let ‘K’ be forces developed in the members developed due the loads.

be the deflection of this pair of unit load in its direction. From

Equation 13.5, can be obtained keeping S = K as:

KLK = AE

=

2

K L AE

(20.6)

Then if X is the force developed in the redundant member due to lack of fit, the deflection due to X (fig .c) is equal to : X It is given as: 2

K L AE

X=X

(20.7)

To this value, the elongation of the redundant member due to X is added. This sum is equal to the lack of fit 2

X

or

X =

K L XL0 + = AE AE K 2 L L0 + AE AE

(20.8)

where L0 is the length of the redundant member. If the member is too short, the numerical value of

is positive and if the member is too long, the numerical value of

is negative. X

will be tensile (positive) if the member is too short and it will be compressive (negative) if the member is too long.

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Similarly effect of temperature variation can be derived. Let T be the temperature variation and ‘2‘ be the coefficient of linear expansion. If T is the drop in temperature then of the member due to T will L02 T, where L0 is the length of the redundant

the shortening

member under consideration. Substituting L02 T in place of X=

in equation 20.8, we have

L02 T

(20.9)

K 2 L L0 + AE AE

If T is rise in temperature X=

will be negative and X is given as:

L02T

(20.10)

K 2 L L0 + AE AE

Equation (20.9) and (20.10) will be the force in the redundant member due to temperature variation. Once X is found, the stresses due to X in the other members of the truss are equal to KX. Few examples will clarify the above.

20.5 Examples: Example 1: Analyse the given pin-jointed truss. Area of the cross section for all the members is same i.e., 10.0cm2 E = 20 × 106 N/mm2.

Fig (a) The truss is statically indeterminate to degree one. m = 10, j = 6 Osi = m - (2j-3)

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= 10 - (2 × 6 - 3) = 1 Consider the member 3-4 as redundant member. Step 1: Remove the member 3-4 analyse the determinate truss for the given loading, as shown in Fig. (b). The values of forces ‘s’ are tabulated in column 3.

Fig (b) Step 2: Apply unit loads at joints 3 and 4 in the direction member 3-4 as shown in Fig. (c). The force ‘K’ due to this unit loads are shown in column 4.

Fig 20.4 Member

L A

S

K

SKL A

KX

S+KX

(5)

K 2L A (6)

(1)

(2)

1-2

75

(3) KN -150

(4) KN 0

(7)

(8)

0

0

0

-150.0

1-3

60

+120

0

0

0

0

120.0

2-3

45

+90

-0.6

-2430

16.2

-15.83

74.17

2-4

60

-120

-0.8

5760

38.4

-21.10

-141.10

2-5

75

0

1.0

0

75

26.38

26.28

3-5

60

120.0

-0.8

-5760

38.4

-21.10

98.90

4-5

45

90.0

-0.6

-2430

16.2

-15.83

74.17

4-6

75

-150.0

0

0

0

0

-150.0

5-6

60

120

0

0

0

0

120

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3-4

75

0

1.0

0

75

26.38

26.38

SKL 4860 = AE E

184.2 K 2 L L0 + = AE AE E For redundant member L0/AE is same as K2L/AE since K = 1 in the redundant member. The redundant force X is obtained as: X =

SKL AE K2L AE

+

L0 AE

=

4860 = 26.38 KN. 184.2

Now the columns (7) and (8) are filled. Example 2: Find the forces in the members of a redundant frame shown due to lack of fit, if the member is too long by 0.50mm. Take A = 800mm2 for all the members, and E = 2 × 105 N/mm2.

Fig 20.5 Remove the member AC and a pair of unit loads is applied at A and C as shown in fig. (b). The forces K due to this are tabulated. Member

L A

K

K2L

KX(KN)

AB

5

-0.707

2.5

+2.93

BC

5

-0.707

2.5

+2.93

CD

5

-0.707

2.5

+2.93

AD

5

-0.707

2.5

+2.93

BD

7.07

1.0

7.07

-4.142

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AC

7.07

1.0

Since the member is too long,

7.07

-4.142

is negative. from the table

K 2 L L0 24.14 = + E AE AE Substituting these values in 20.8, we get X =

2

K L L0 + AE AE

=

0.5 24.14 2 × 105

= - 4142.5N or - 4.142 KN. The same example can be solved for temperature variation. For example if there is a drop in temperature of 100C, and taking 2 = 12 × 10-6/0C the redundant force in the member AC is given as follows.

= LT2 = 2 × 4000 × 12 × 10 6 × 10 = 6.78mm X =

K 2 L L0 + AE AE

=

6.78 × 2 × 105 24.14

or = 56172N 56.17 KN Once X is determined, the forces in the members are equal to KX. In examples where there is applied loading and lack of fit simultaneously, the redundant truss should be solved separately for loading and lack of fit. The final stresses will be the sum of the two. 20.6 INTRODUCTION: We have discussed about internally indeterminate trusses. A truss

can be indeterminate internally or externally indeterminate truss. In units 4 and 5, the student has been introduced to statically indeterminate beams. The indeterminacy of a beam depends on the number of reaction components including fixed end moments. Similarly external indeterminacy for a truss depends on the support reactions. See the fig 20.6(a). Here the truss has a hinged support at A and a roller support at B. At the hinged support there will be two reaction components (RA, HA) and there will be one reaction (RB) at the roller support B.

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Thus the total number of reaction components is three. We know that three unknown reactions can be solved from the available three static equilibrium equations: FX = 0, FY = 0 and

M= 0 in a plane. Hence the truss shown in (a) is statically determinate externally.

Consider the truss shown in Fig. (b). Here the supports A and B are hinged. Hence there will be four reaction components, two at A (HA, RA) and two at B (HB, RB).

Fig 20.6 This truss thus has four reaction components, which are in excess of three static equilibrium equations. It is therefore externally indeterminate to a degree one. Similarly the trusses shown in Fig (C) has four reactions two at A (HA, RA), one at B(RB) and one at C (RC). This is also externally indeterminate to a degree one. The excess reaction components over those required for static equilibrium i.e. 3, are known as redundant reactions. The number of redundant reactions give the degree of indeterminacy. 20.2 ANALYSIS OF EXTERNALLY INDETERMINATE TRUSS: For the analysis of externally indeterminate trusses Castigliano’s theorem of least work can be used. But the method of consistent deformation will be most convenient. It is otherwise known as geometric or deformation compatibility. The student is advised to refer to unit 4 for the analysis of propped cantilever beams. The analysis of trusses is also approached in a similar manner. See the fig 20.7 (a). The supports A and B are hinged. At each support there will be

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Fig 20.7 two reactions as shown in fig (a). The truss is statically indeterminate to a degree one, which means, that any one of the two horizontal reactions, say HB can be removed and the truss would still remain stable but become determinate as shown in Fig (b). The support B is no more hinged and it allows horizontal movement. Let R’A, RB’ and HA be the three reactions developed in the determinate truss due to load P. Now, since there is no horizontal reactions at B, it moves by amount say ‘ ‘ to the right (assumed arbitrarily at the positive sense) due to applied loading. The value of

can be found by the unit load method. From equation (19.5)

we have SKL AE

=

(19.5)

where S = forces in the members due to applied load in the determinate truss K = forces in the members due to unit load applied in the horizontal direction to the right, since

is considered to be towards the right.

But in the original truss (fig 20.7 (a), the support is not free to move. Apply a force HB which will prevent this movement as shown in Fig (c). No other applied load is assumed to act on the truss. The magnitude and direction of HB will be such as to keep the truss at B in the original position. The movement due to HB is equal to

B

HB where

B

= movement

(deflection due to unit load applied at B in the horizontal direction towards the right. The deflection B

B

is give as KKL AE

=

which is obtained by replacing S by K in equation 19.5. Then the value of HB can be determined from the geometric condition: +

B

HB = 0

(20.1)

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Once HB is found the stresses in the given truss are determined as S1 = S + KHB

(20.2)

where S1 = Stresses in the members of the given indeterminate truss (Fig a) S = Stresses in the members of the determinate truss due to applied loading (Fig .b) K = Stresses in the members of the truss due to a unit horizontal load applied at B to the determinate truss (Fig b) The reactions HA, RA and RB can be found from statics. A few examples will clarify the above procedure. 20.7 Examples: Example 1: Compute the reactions and forces in the members of the truss shown. AE is same for all the members

(a) m = 5, j = 4

2j-3 = 5.

Since m = 2j-3, the truss is internally determinate Reactions = RA, RB, RC and HC: Consider RB as redundant reaction. The method is explained in step by step.

(b)

(c)

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Step 1: Remove the support B, Find the vertical deflections

B

(

B

is in a directions opposite

to that of RB) in the statically determine truss due to applied loading as in fig (b) as in the unit 19. B

=

SKL AE

Step2: Apply vertical unit load in the upward direction with out applied loading (Fig C). Find the deflection =

B

B

due to unit load

K2 L AE

3. Now find RB from the condition of geometry at support B B

RB +

B

=0

Solution: The reactions RA’, RB’, & HA’ in Fig. (d) are found from statics due to applied loading

(d)

(e)

Fig 20.8

HC’ - 100 = 0 (1)

HC’ = 100 KN.

RA’ + RC’ = 0

(2)

Taking moments about A: 100 × 4 + RC’ × 6 = 0

(3)

RC’ = 66.66 KN ( - ) RA’ = 66.66 KN. The reaction RC’ is negative which means that it acts downwards. Reactions due to unit load are from Fig. (e) RA” = RC” = - ½

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The stresses (Forces) in the members due to applied load (100 KN) and due to unit load are tabulated. L AE

S(KN)

AB

3 AE

50

0.375

56.25 AE

0.4218 AE

45.17

BC

3 AE

50

0.375

56.25 AE

0.4218 AE

45.17

AD

5 AE

-83.3285

0.625

260.4 AE

1.9531 AE

-91.36

BD

4 AE

0

-1.0

0

4 AE

-12.86

CD

5 AE

83.325

0.625

260.4 AE

1.9531 AE

75.28

Member

Tan = 4/3,

Sin

B

=

= 0.6 ;

SKL AE

S+KRB

K 2L AE

SKL 112.5 = AE AE

= 53.13° ;

= 0.8; cos

K(KN)

K 2 L 8.7498 = AE AE

SKL 112.5 = AE AE

K 2 L 8.7498 = B= AE AE

BRB

+

RB = -

B

B

=0

= - 112.5/8.7498

B

RB = -12.86 KN As RB is negative, it acts downwards: Now using statics, reactions can be computed: Hc = 100 KN

(1)

RA + RB + RC = 0 RA - 12.86 + RC = 0 Taking moments about A: 100 × 4 + Rc × 6 + RB × 3 = 0 100 × 4 + Rc × 6 - 12.86 × 3 = 0 Rc = -60.23 KN.

(2)

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(f) Rc also acts downwards. From (2), we find. RA = 73.10 KN. Final stresses are shown in the last column. Example 2: A plane pin jointed frame (truss) is as shown in fig. It is pin jointed at 1,2,3 and 4. The truss carries a horizontal load of 50 KN to the right at joint 2. Analyse the truss. The cross sectional areas in mm2 are given in brackets corresponding to members.

(a)

(b)

Select the determinate truss as shown in Fig. (b) with roller support at 4, carrying the load 50 KN. Let R3’, R4’ and H3’ be the reactions. H3’ = 50 KN Taking moments about 4: R3’ × 3 + 50 × 3 = 0 R13 + R14 = 0 R14 = 50KN

R13 = 50 KN (Acts downward)

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(c) Fig 20.9 To find the horizontal deflection apply unit load, at 4 to the right, as shown in Fig (d). the forces ‘S’ and ‘K’ are tabulated. H4 is the redundant reaction. tan

=

3 ; 3

= 45°

(d) Member

L A

S(KN)

K(KN)

SK L A

K

2

A

L

S+KH4 (KN)

1-2

1.5

0

-1.0

0

1.50

22.63

1-3

1.5

0

-1.0

0

1.50

22.63

1-4

2.83

0

1.414

0

5.66

-32.00

2-3

2.83

70.71

1.414

282.95

5.66

38.71

2-4

1.5

-50

-1.0

75.00

1.50

72.63

357.95

15.82

4

= horizontal displacement at 4, due to 50 KN. SKL 357.95 = AE E

=

K2L = 1582 . AE

S4 =

The geometric condition demands that: 4

+

4

H4 = 0

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solving, H4 = -

357.95 1582 .

= - 22.63 KN (i.e., to the left) Now the last column can be filled. The last column gives the stresses (forces) in the members of the given redundant truss. The support reactions can be determined from statics. while calculating

4,

unit load is taken as acting to the right. But we got negative sign for H14,

From statics; which means that it acts to the left. H3 + 22.63 - 50 = 0

(1)

H3 = 27.37 KN (Acts to the left) R3 + R4 = 0

(2)

Taking moments about 4, 50 × 3 + R3 × 3 = 0;

R3 = - 50 KN (Acts downward)

R4 = 50KN.

(e) Exercise Problems: 1. The Howe truss supports three loads as shown in Fig. Determine the vertical deflection of joint ‘F’, Use unit load method.

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Fig 20.10 Note: Area of tension members 250 mm2. Compression members 300 mm2, E = 2 × 105 N/mm2. 2. Find the forces in the given redundant truss shown in Fig. Take the member ‘CD’ as redundant. Take A = 1200 mm2 and E = 2 × 105 N/mm2 for all the members.

Fig 20.11 3. In the problem 3, if there is a rise in temperature of 100C find the forces in the members due to this rise of temperature. Take A = 1200 mm2 and E = 2 × 105 N/mm2 for all members 2 = 12 × 10-6/0C. 4. In the truss ABCD both ends A and D are provided with hinged supports. It carries loads of 75KN and 50KN, as shown. Treat the horizontal reaction at ‘D’ as redundant. Compute the forces in all the members AE is the same for all members.

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Fig 20.12 5. A plane pin jointed frame (truss) is as shown in fig. It is pin jointed at ‘A, B, C and D’. The truss carries a horizontal load of 100 KN to the right at joint ‘C’. Analyse the truss, cross sectional areas in Cm2 are given in brackets corresponding to members.

Fig 20.13

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