Mechanics of Materials Chapter 5

Third Edition

CHAPTER

MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University

Analysis and Design of Beams for Bending

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

Third Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Analysis and Design of Beams for Bending Introduction Shear and Bending Moment Diagrams Sample Problem 5.1 Sample Problem 5.2 Relations Among Load, Shear, and Bending Moment Sample Problem 5.3 Sample Problem 5.5 Design of Prismatic Beams for Bending Sample Problem 5.8


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Introduction • Objective - Analysis and design of beams • Beams - structural members supporting loads at various points along the member • Transverse loadings of beams are classified as concentrated loads or distributed loads • Applied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couple (from the normal stress distribution) • Normal stress is often the critical design criteria σx = −

My I

σm =

Mc M = I S

Requires determination of the location and magnitude of largest bending moment © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Introduction Classification of Beam Supports


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Shear and Bending Moment Diagrams • Determination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending couple. • Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the section. • Sign conventions for shear forces V and V’ and bending couples M and M’


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Sample Problem 5.1 SOLUTION: • Treating the entire beam as a rigid body, determine the reaction forces

For the timber beam and loading shown, draw the shear and bendmoment diagrams and determine the maximum normal stress due to bending.

• Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples • Identify the maximum shear and bending-moment from plots of their distributions. • Apply the elastic flexure formulas to determine the corresponding maximum normal stress.


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Sample Problem 5.1 SOLUTION: • Treating the entire beam as a rigid body, determine the reaction forces from ∑ Fy = 0 = ∑ M B : RB = 40 kN

RD = 14 kN

• Section the beam and apply equilibrium analyses on resulting free-bodies ∑ Fy = 0

− 20 kN − V1 = 0

V1 = −20 kN

∑ M1 = 0

(20 kN )(0 m ) + M1 = 0

M1 = 0

∑ Fy = 0

− 20 kN − V2 = 0

V2 = −20 kN

∑ M2 = 0

(20 kN )(2.5 m ) + M 2 = 0

M 2 = −50 kN ⋅ m

V3 = +26 kN

M 3 = −50 kN ⋅ m

V4 = +26 kN M 4 = +28 kN ⋅ m V5 = −14 kN

M 5 = +28 kN ⋅ m

V6 = −14 kN M 6 = 0 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.1 • Identify the maximum shear and bendingmoment from plots of their distributions. Vm = 26 kN M m = M B = 50 kN ⋅ m

• Apply the elastic flexure formulas to determine the corresponding maximum normal stress. S = 16 b h 2 = 16 (0.080 m )(0.250 m )2 = 833.33 × 10− 6 m3 MB 50 × 103 N ⋅ m σm = = S 833.33 × 10− 6 m3

σ m = 60.0 ×106 Pa


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Sample Problem 5.2 SOLUTION: • Replace the 10 kip load with an equivalent force-couple system at D. Find the reactions at B by considering the beam as a rigid body. • Section the beam at points near the support and load application points. Apply equilibrium analyses on The structure shown is constructed of a resulting free-bodies to determine W10x112 rolled-steel beam. (a) Draw internal shear forces and bending the shear and bending-moment diagrams couples. for the beam and the given loading. (b) determine normal stress in sections just • Apply the elastic flexure formulas to to the right and left of point D. determine the maximum normal stress to the left and right of point D.


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Sample Problem 5.2 SOLUTION: • Replace the 10 kip load with equivalent forcecouple system at D. Find reactions at B. • Section the beam and apply equilibrium analyses on resulting free-bodies. From A to C : ∑ Fy = 0 ∑ M1 = 0

− 3x − V = 0

(3x )(12 x )+ M

From C to D : ∑ Fy = 0 − 24 − V = 0

V = −3 x kips = 0 M = −1.5 x 2 kip ⋅ ft

V = −24 kips

∑ M 2 = 0 24( x − 4) + M = 0 M = (96 − 24 x ) kip ⋅ ft

From D to B : V = −34 kips


M = (226 − 34 x ) kip ⋅ ft

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Sample Problem 5.2 • Apply the elastic flexure formulas to determine the maximum normal stress to the left and right of point D. From Appendix C for a W10x112 rolled steel shape, S = 126 in3 about the X-X axis. To the left of D : M 2016 kip ⋅ in = S 126 in 3 To the right of D :

σ m = 16.0 ksi

M 1776 kip ⋅ in = S 126 in 3

σ m = 14.1 ksi

σm =

σm =


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Relations Among Load, Shear, and Bending Moment • Relationship between load and shear: ∑ Fy = 0 : V − (V + ∆V ) − w ∆x = 0 ∆V = − w ∆x dV = −w dx xD

VD − VC = − ∫ w dx xC

• Relationship between shear and bending moment: ∑ M C′ = 0 :

(M + ∆M ) − M − V ∆x + w∆x ∆x = 0 ∆M = V ∆x − 12 w (∆x )

2

2

dM =0 dx xD

M D − M C = ∫ V dx xC © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.3 SOLUTION: • Taking the entire beam as a free body, determine the reactions at A and D. • Apply the relationship between shear and load to develop the shear diagram. Draw the shear and bending moment diagrams for the beam and loading shown.

• Apply the relationship between bending moment and shear to develop the bending moment diagram.


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Sample Problem 5.3 SOLUTION: • Taking the entire beam as a free body, determine the reactions at A and D. ∑MA = 0 0 = D(24 ft ) − (20 kips )(6 ft ) − (12 kips )(14 ft ) − (12 kips )(28 ft ) D = 26 kips ∑ Fy = 0

0 = Ay − 20 kips − 12 kips + 26 kips − 12 kips Ay = 18 kips

• Apply the relationship between shear and load to develop the shear diagram. dV = −w dx

dV = − w dx

- zero slope between concentrated loads - linear variation over uniform load segment © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.3 • Apply the relationship between bending moment and shear to develop the bending moment diagram. dM =V dx

dM = V dx

- bending moment at A and E is zero - bending moment variation between A, B, C and D is linear - bending moment variation between D and E is quadratic - net change in bending moment is equal to areas under shear distribution segments - total of all bending moment changes across the beam should be zero © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.5 SOLUTION: • Taking the entire beam as a free body, determine the reactions at C. • Apply the relationship between shear and load to develop the shear diagram. Draw the shear and bending moment diagrams for the beam and loading shown.

• Apply the relationship between bending moment and shear to develop the bending moment diagram.


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Sample Problem 5.5 SOLUTION: • Taking the entire beam as a free body, determine the reactions at C. ∑ Fy = 0 = − 12 w0 a + RC a⎞ ⎛ ∑ M C = 0 = 12 w0 a⎜ L − ⎟ + M C 3⎠ ⎝

RC = 12 w0 a

a⎞ ⎛ M C = − 12 w0 a⎜ L − ⎟ 3⎠ ⎝

Results from integration of the load and shear distributions should be equivalent. • Apply the relationship between shear and load to develop the shear diagram. a

2 ⎞⎤ ⎡ ⎛ x ⎛ x⎞ VB − V A = − ∫ w0 ⎜1 − ⎟ dx = − ⎢ w0 ⎜ x − ⎟⎥ ⎜ a⎠ 2a ⎟⎠⎥⎦ ⎢⎣ ⎝ 0 ⎝ 0

a

VB = − 12 w0 a = − ( area under load curve)

- No change in shear between B and C. - Compatible with free body analysis © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.5 • Apply the relationship between bending moment and shear to develop the bending moment diagram. a

⎡ ⎛ x 2 x3 ⎞⎤ ⎛ x 2 ⎞⎟ ⎞⎟ ⎜ ⎜ M B − M A = ∫ − w0 x − dx = ⎢− w0 ⎜ − ⎟⎥ ⎜ 2 6a ⎟ ⎜ ⎜ 2a ⎟⎠ ⎟⎠ ⎢⎣ 0⎝ ⎠⎥⎦ 0 ⎝ ⎝ a⎛

M B = − 13 w0 a 2 L

(

)

M B − M C = ∫ − 12 w0 a dx = − 12 w0 a(L − a ) a

a w0 ⎛ a⎞ M C = − 16 w0 a(3L − a ) = ⎜L− ⎟ 2 ⎝ 3⎠

Results at C are compatible with free-body analysis


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Design of Prismatic Beams for Bending • The largest normal stress is found at the surface where the maximum bending moment occurs. M max c M max σm = = I S

• A safe design requires that the maximum normal stress be less than the allowable stress for the material used. This criteria leads to the determination of the minimum acceptable section modulus. σ m ≤ σ all S min =

M max

σ all

• Among beam section choices which have an acceptable section modulus, the one with the smallest weight per unit length or cross sectional area will be the least expensive and the best choice. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.8 SOLUTION: • Considering the entire beam as a freebody, determine the reactions at A and D. A simply supported steel beam is to carry the distributed and concentrated loads shown. Knowing that the allowable normal stress for the grade of steel to be used is 160 MPa, select the wide-flange shape that should be used.

• Develop the shear diagram for the beam and load distribution. From the diagram, determine the maximum bending moment. • Determine the minimum acceptable beam section modulus. Choose the best standard section which meets this criteria.


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Sample Problem 5.8 • Considering the entire beam as a free-body, determine the reactions at A and D.

∑ M A = 0 = D(5 m ) − (60 kN )(1.5 m ) − (50 kN )(4 m ) D = 58.0 kN ∑ Fy = 0 = Ay + 58.0 kN − 60 kN − 50 kN Ay = 52.0 kN

• Develop the shear diagram and determine the maximum bending moment. V A = Ay = 52.0 kN VB − V A = −(area under load curve) = −60 kN VB = −8 kN

• Maximum bending moment occurs at V = 0 or x = 2.6 m. M max = (area under shear curve, A to E ) = 67.6 kN © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Sample Problem 5.8 • Determine the minimum acceptable beam section modulus. S min =

M max 67.6 kN ⋅ m = σ all 160 MPa

= 422.5 × 10− 6 m3 = 422.5 × 103 mm3

• Choose the best standard section which meets this criteria. Shape

S , mm3

W410 × 38.8

637

W360 × 32.9

474

W310 × 38.7

549

W250 × 44.8

535

W200 × 46.1

448

W 360× 32.9


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Mechanics of Materials Chapter 5

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