CIVL 7/8117
Chapter 4 - Development of Beam Equations - Part 2
Beam Beam Stiff ness General Formulation
We can account for the distributed loads or concentrated loads acting on beam elements by considering the following formulation for a general structure: F = K d - F0
where F0 are the equivalent equivalent no dal forces , expressed in terms of the global-coordinate components. These forces would yield the same displacements as the original distributed load. If we assume that the global nodal forces are not present (F = 0) then: F0 = K d
Beam Beam Stiff ness General Formulation
We now solve for the displacements, d , given the nodal forces F0. Next, substitute the displacements and the equivalent nodal forces F0 back into the original expression and solve for the global nodal forces. F = K d - F0
This concept can be applied on a local basis to obtain the local nodal forces in individual elements of structures as: f = k d - f0
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Chapter 4 - Development of Beam Equations - Part 2
Beam Beam Stiffness Stiff ness Example Example 5 - Load Replaceme Replacement nt
Consider the beam shown below; determine the equivalent nodal forces for the given distributed load.
The work equivalent nodal forces are shown above. wL 2 6L v 1 12 wL 6L 2L2 1 12 12 6L v 2 wL 6L 4L2 2 2 wL 12
Using the beam stiffness equations: f1y 12 m 1 EI 6L 3 f2 y L 12 m2 6L
6L 2
4L
6L 2L2
2
2
Beam Beam Stiff ness Example Example 5 - Load Replaceme Replacement nt
Apply the boundary boundary conditions: conditions: v1 1 0
6L 12 2 EI 6L 4L L3 12 6L 2 6L 2L
12 6L 12
6L
6L v1
2L2 1 6L v 2
2 4L2
We can solve for the displacements
wL 2 EI 12 2 3 wL L 6L 12
6L v 2 4L2 2
wL4 v 2 8EI 3 2 wL 6EI
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Chapter 4 - Development of Beam Equations - Part 2
Beam Beam Stiffness Stiff ness Example Example 5 - Load Replaceme Replacement nt
Consider the beam shown below; determine the equivalent nodal forces for the given distributed load.
The work equivalent nodal forces are shown above. wL 2 6L v 1 12 wL 6L 2L2 1 12 12 6L v 2 wL 6L 4L2 2 2 wL 12
Using the beam stiffness equations: f1y 12 m 1 EI 6L 3 f2 y L 12 m2 6L
6L 2
4L
6L 2L2
2
2
Beam Beam Stiff ness Example Example 5 - Load Replaceme Replacement nt
Apply the boundary boundary conditions: conditions: v1 1 0
6L 12 2 EI 6L 4L L3 12 6L 2 6L 2L
12 6L 12
6L
6L v1
2L2 1 6L v 2
2 4L2
We can solve for the displacements
wL 2 EI 12 2 3 wL L 6L 12
6L v 2 4L2 2
wL4 v 2 8EI 3 2 wL 6EI
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Chapter 4 - Development of Beam Equations - Part 2
Beam Beam Stiff ness Example Example 5 - Load Replaceme Replacement nt
In this case, the method of equivalent nodal forces gives the exact solution for the displacements and rotations. To obtain the global nodal forces, we will first define the product of Kd to be Fe, where Fe is called the effective global nodal forces . Therefore: wL 2 F e1y 12 6L 12 6L 0 5wL2 e 2 2 M 1 EI 6L 4L 6L 2L 0 12 e 3 wL4 1 2 6 1 2 6 L L L wL F 8 EI 2y 3 2 2 e wL M 2 6L 2L 6L 4L 6EI 2 wL2 12
Beam Beam Stiffness Stiff ness Example Example 5 - Load Replaceme Replacement nt
Using the above expression and the fix-end moments in:
F = K d - F0
wL wL wL 2 2 F1y 5wL2 wL2 wL2 M 2 1 12 12 F w L w L 2y 0 M2 2 2 wL2 wL2 0 12 12
F M
y
0 F1y wL
wL2 L 0 wL 1 2 2
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiffness Example 6 - Cantilever Beam
Consider the beam, shown below, determine the vertical displacement and rotation at the free-end and the nodal forces, including reactions. Assume EI is constant throughout the beam.
We will use one element and replace the concentrated load with the appropriate nodal forces.
Beam Stiff ness Example 6 - Cantilever Beam
The beam stiffness equations are: f1y 12 m 1 EI 6L 3 f2 y L 12 m2 6L
4L2
12 6L
6L
12
2
6L
6L
2L
P 2 6L v1 PL 2L2 1 8 6L v 2 P 2 4L2 2 PL 8
Apply the boundary conditions: v1 1 0
6L 12 6L 4L2 EI L3 12 6L 2 6L 2L
12 6L 12
6L
6L v1
6L v 2 4L2 2 2L2 1
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 6 - Cantilever Beam
The beam stiffness equations become:
P 2 EI 12 3 PL L 6L 8
5PL3 v 2 48EI 2 2 PL 8EI
6L v 2 4L2 2
To obtain the global nodal forces, we begin by evaluating the effective nodal forces.
F e1y 12 6L 12 6L 0 e 2 2 M 1 EI 6L 4L 6L 2L 0 e 3 5PL F 2 y L 12 6L 12 6L 48EI 6L 2L2 6L 4L2 PL M e 2 8 EI 3
2
Beam Stiff ness Example 6 - Cantilever Beam
Using the above expression in the following equation, gives:
F = K d - F0
P P 2 2 P F1y 3PL PL PL M 1 8 8 2 F P P 2y 0 M2 2 2 PL PL 0 8 8 Kd = Fe
F0
F
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 6 - Cantilever Beam
In general, for any structure in which an equivalent nodal force replacement is made, the actual nodal forces acting on the structure are determined by first evaluating the effective nodal forces Fe for the structure and then subtracting off the equivalent nodal forces F0 for the structure. Similarly, for any element of a structure in which equivalent nodal force replacement is made, the actual local nodal forces acting on the element are determined by first evaluating the effective local nodal forces for the element and then subtracting off the equivalent local nodal forces associated only with the element. f = k d - f0
Beam Stiffness Comparison of FE Solution to Exact Solution
We will now compare the finite element solution to the exact classical beam theory solution for the cantilever beam shown below. Both one- and two-element finite element solutions will be presented and compared to the exact solution obtained by the direct double-integration method. Let E = 30 x 10 6 psi, I = 100 in4, L = 100 in, and uniform load w = 20 Ib/in.
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Chapter 4 - Development of Beam Equations - Part 2
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Beam Stiffness Comparison of FE Solution to Exact Solution
To obtain the solution from classical beam theory, we use the double-integration method: M( x ) y EI where the double prime superscript indicates differentiation twice with respect to x and M is expressed as a function of x by using a section of the beam as shown:
F
y
0 V ( x ) wL wx
wx 2 wL2 M 0 M( x ) wLx 2 2
Beam Stiff ness Comparison of FE Solution to Exact Solution
To obtain the solution from classical beam theory, we use the double-integration method: M( x ) y EI w x2 L2 y Lx dx dx 2 EI 2
w x 3 Lx 2 xL2 y C 1 dx EI 6 2 2
y
w
Boundary Conditions
y (0) 0
y (0) 0
x Lx x L C1x C2 6 4 EI 24 4
3
2 2
y
w
x4
EI 24
Lx 3
6
x 2L2
4
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Comparison of FE Solution to Exact Solution
Recall the one-element solution to the cantilever beam is:
wL4 v 2 8EI 3 2 wL 6EI Using the numerical values for this problem we get: 4 20 lb in 100 in 6 4 8 30 10 100 psi in 0.0833 in v 2 3 lb 2 20 in 100 in 0.00111 rad 6 4 6 30 10 psi 100 in
Beam Stiff ness Comparison of FE Solution to Exact Solution
The slope and displacement from the one-element FE solution identically match the beam theory values evaluated at x = L. The reason why these nodal values from the FE solution are correct is that the element nodal forces were calculated on the basis of being energy or work equivalent to the distributed load based on the assumed cubic displacement field within each beam element.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Comparison of FE Solution to Exact Solution
Values of displacement and slope at other locations along the beam for the FE are obtained by using the assumed cubic displacement function. v(x)
1 3
L
2 x
3
3 x 2L v 2
1 3
L
x L x L 3
2 2
2
The value of the displacement at the midlength v(x = 50 in) is: v ( x 50 in ) 0.0278 in
Using beam theory, the displacement at v(x = 50 in) is: v ( x 50 in ) 0.0295 in
Beam Stiff ness Comparison of FE Solution to Exact Solution
In general, the displacements evaluated by the FE method using the cubic function for v are lower than by those of beam theory except at the nodes. This is always true for beams subjected to some form of distributed load that are modeled using the cubic displacement function. The exception to this result is at the nodes, where the beam theory and FE results are identical because of the workequivalence concept used to replace the distributed load by work-equivalent discrete loads at the nodes.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Comparison of FE Solution to Exact Solution
The beam theory solution predicts a quartic (fourth-order) polynomial expression for a beam subjected to uniformly distributed loading, while the FE solution v(x) assumes a cubic (third-order) displacement behavior in each beam all load conditions. The FE solution predicts a stiffer structure than the actual one. This is expected, as the FE model forces the beam into specific modes of displacement and effectively yields a stiffer model than the actual structure. However, as more elements are used in the model, the FE solution converges to the beam theory solution.
Beam Stiff ness Comparison of FE Solution to Exact Solution
For the special case of a beam subjected to only nodal concentrated loads, the beam theory predicts a cubic displacement behavior. The FE solution for displacement matches the beam theory solution for all locations along the beam length, as both v(x) and y(x) are cubic functions.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Comparison of FE Solution to Exact Solution
Under uniformly distributed loading, the beam theory solution predicts a quadratic moment and a linear shear force in the beam. However, the FE solution using the cubic displacement function predicts a linear bending moment and a constant shear force within each beam element used in the model.
Beam Stiff ness Comparison of FE Solution to Exact Solution
We will now determine the bending moment and shear force in the present problem based on the FE method. M ( x ) EIy EI
d 2 Nd dx 2
d N d EI 2
dx 2
M ( x ) EI B d
6 12 x 4 6 x 6 12x 2 6 x 3 2 2 3 2 2 L L L L L L L L
B
6 12 x 4 6x v 1 L L2 1 2 3 L L 6 12 x 2 6x 2 3 v 2 2 2 L L L L
M EI
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Comparison of FE Solution to Exact Solution
We will now determine the bending moment and shear force in the present problem based on the FE method.
Position
MFE
Mexact
X=0
-83,333 lb-in
-100,000 lb-in
X = 50 in
-33,333 lb-in
-25,000 lb-in
X = 100 in
16,667 lb-in
0
Beam Stiffness Comparison of FE Solution to Exact Solution
The plots below show the displacement, bending moment , and shear force over the beam using beam theory and the one-element FE solutions.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiffness Comparison of FE Solution to Exact Solution
The FE solution for displacement matches the beam theory solution at the nodes but predicts smaller displacements (less deflection) at other locations along the beam length.
Beam Stiffness Comparison of FE Solution to Exact Solution
The bending moment is derived by taking two derivatives on the displacement function. It then takes more elements to model the second derivative of the displacement function.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiffness Comparison of FE Solution to Exact Solution
The shear force is derived by taking three derivatives on the displacement function. For the uniformly loaded beam, the shear force is a constant throughout the singIe-element model.
Beam Stiffness Comparison of FE Solution to Exact Solution
To improve the FE solution we need to use more elements in the model (refine the mesh) or use a higher-order element, such as a fifth-order approximation for the displacement function.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiffness Beam Element wit h Nodal Hinge
Consider the beam, shown below, with an internal hinge. An internal hinge causes a discontinuity in the slope of the deflection curve at the hinge and the bending moment is zero at the hinge.
Beam Stiff ness Beam Element wit h Nodal Hinge
For a beam with a hinge on the right end: [K12 ] [K ] 11
f1y 12 m 1 EI 6L 3 f2 y L 12 m2 6L
4L2
12 6L
6L
12
2L2
6L
6L
[K 21 ]
6 L v 1
6L v 2 4L2 2 2L2 1
[K 22 ]
The moment m2 is zero and we can partition the matrix to eliminate the degree of freedom associated with 2. f1 K11d1 K12d 2 f1 K11 K12 d1
d f K K 2 21 22 2
f2 K 21d1 K 22d 2
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Beam Element wit h Nodal Hinge
For a beam with a hinge on the right end:
f1y 12 m 1 EI 6L 3 f2 y L 12 m2 6L
4L2
12 6L
6L
12
2L2
6L
6L
6L v 1 d1 1 v 2L2 1 2
v 1
f 1y f1 m1 f 2y
d 2 2
f2 m2
6L v 2 4L2 2
d 2 K 22 1 f2 K 21d1
f1 K11d1 K12d 2 K11d1 K12 K 22 1 f2 K 21d1 fc kc d1
1
m2 0
fc f1 K 12K 22 f2
kc K11 K12K 22 1K 21
Beam Stiff ness Beam Element wit h Nodal Hinge
We can condense out the degree of freedom by using the partitioning method discussed earlier. kc K11 K12K 22 1K 21
12 EI kc 3 6L L 12
6L 2
4L
6L
12 6L EI 2 1 6L 3 2L 2 6L L 4L 12 6 L
Therefore, the condensed stiffness matrix is:
1 L 3EI kc 3 L L2 L 1 L
1 L 1
2L2
6L
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Beam Element wit h Nodal Hinge
The element force-displacement equations are:
f1y 1 L 1 v 1 3EI 2 m1 3 L L L 1 f L 1 L 1 v 2 2y Expanding the element force-displacement equations and maintaining m2 = 0 gives:
f1y 1 L 1 m 2 1 3EI L L L 3 f2 y L 1 L 1 m2 0 0 0
0 v 1
0 v 2 0 2 0 1
Beam Stiff ness Beam Element wit h Nodal Hinge
Once the displacements d 1 are found, d 2 may be computed: d 2 K 22
1
f2 K 21d 1
2 K 22
1
m2 0 v1 m K 1 2 21 v 2
In this case, d 2 is 2 (the rotation at right side of the element at the hinge.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Beam Element wit h Nodal Hinge
For a beam with a hinge on the left end:
f1y 12 m 1 EI 6L 3 f2 y L 12 m2 6L
4L2
12 6L
6L
12
2L2
6L
6L
6L v 1
6L v 2 4L2 2 2L2 1
Rewriting the equations to move 1 and m1 to the first row:
4L2 m1 f 1y EI 6L 3 f2 y L 6L 2 m2 2L
12
6L 12
12
12
6L
6L
6L
2L2 1 6L v 1
6L v 2 4L2 2
Beam Stiff ness Beam Element wit h Nodal Hinge
For a beam with a hinge on the left end: [K12 ] [K ] 11
4L2 m1 f 1y EI 6L 3 f2 y L 6L 2 m2 2L
12
6L 12
12
12
6L
6L
6L
[K 21 ]
2L2 1 6L v 1
6L v 2 4L2 2 [K 22 ]
The moment m1 is zero and we can partition the matrix to eliminate the degree of freedom associated with 1. f1 K11d1 K12d 2 f1 K11 K12 d1
d f K K 2 21 22 2
f2 K 21d1 K 22d 2
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Beam Element wit h Nodal Hinge
For a beam with a hinge on the right end:
4L2 m1 f 1 EI 6L 3 f2 L 6L 2 m2 2L
12
6L 12
12
12
6L
6L
6L
2L2 1 d1 1 6L v 1 v
1 6L v 2 d2 v 2 2 4L2 2
f1 m1 f 1y f2 f 2y m 2
d1 K111 f1 K12d 2
f2 K 21d1 K 22d 2 K 22d 2 K 21 K111 f1 K 12d 2 fc k cd 2
1
m1 0
fc f2 K 21K11 f1
kc K 22 K 21K111K12
Beam Stiff ness Beam Element wit h Nodal Hinge
We can condense out the degree of freedom by using the partitioning method discussed earlier. kc K 22 K 21K111K12
12 EI kc 3 12 L 6L
12 12
6L
6L
6L EI 1 6L 3 6L 2 6L 6L 2L2 L 2 4L 4L2 2L
Therefore, the condensed stiffness matrix is:
1 1 L 3EI kc 3 1 1 L L L L L2
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Beam Element wit h Nodal Hinge
The element force-displacement equations are:
f1y 1 1 L v 1 3EI f2 y 3 1 1 L v 2 m L L L L2 2 1 Expanding the element force-displacement equations and maintaining m1 = 0 gives:
f1y m 1 3EI 3 f2 y L m2
1 0 L 1
1 v1 0 0 0 1 0 L L v 2 0 L 1 2 0
L
Beam Stiff ness Beam Element wit h Nodal Hinge
Once the displacements d 2 are found, d 1 may be computed: 1
d1 K11
f1 K12d 2
1
1 K11
m1 0 v1 m K v 2 1 12 1
In this case, d 1 is 1 (the rotation at right side of the element at the hinge.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Beam Element wit h Nodal Hinge
For a beam element with a hinge at its left end, the element force-displacement equations are:
f1y 1 1 L v 1 3EI f2 y 3 1 1 L v 2 m L L L L2 2 2 Expanding the element force-displacement equations and maintaining m1= 0 gives:
f1y m 1 3EI 3 f2 y L m2
1 0 1 L
0
1
0
0
0
1
0
L
L v 1
0 1 L v 2
L2 2
Beam Stiffness Example 7 - Beam With Hinge
In the following beam, shown below, determine the vertical displacement and rotation at node 2 and the element forces for the uniform beam with an internal hinge at node 2. Assume EI is constant throughout the beam.
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 7 - Beam With Hinge
The stiffness matrix for element 1 (with hinge on right) is:
f1y 1 L 1 m 2 1 3EI L L L 3 f2 y L 1 L 1 m2 0 0 0 v1
0 v 1
0 v 2 0 2 0 1
v2
1
2
f1y 1 a 1 m 2 1 3EI a a a 3 f2 y a 1 a 1 0 0 0 m2
0 v 1
0 v 2 0 2 0 1
Beam Stiff ness Example 7 - Beam With Hinge
The stiffness matrix for element 2 (with no hinge) is:
f2 12 m 2 EI 6L 3 f3 L 12 m3 6L v2
f2 12 m 2 EI 6b 3 f3 b 12 6b m3
4L2
12 6L
6L
12
2L2
6L
6L
2
v3
6b 4b 2
12 6 b
6b
12
2b 2
6 b
6L v 2
2L2 2
6L v 3 4L2 3 3
6 b v 2
6 b v 3 4b 2 3 2b 2 2
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 7 - Beam With Hinge
The assembled equations are: 3 a3 a f1y 3 3 m a a 1 3 3 f2 y a a EI 0 0 m2 f3 y 0 0 m 3 0 0 3
a3
2
3
2
3
2
12 b3
3 a2
a3
3
6 b2
12 b 3
6 b2
0
0
2
3
0
v 1 0 0 0 1 6 6 12 b b b v 2 6 4 2 b b b 2 12 b6 b6 v 3 b 2 6 4 3 b b b 2
2
2
3
2
2
Beam Stiff ness Example 7 - Beam With Hinge
v1 v3 1 3 0
The boundary conditions are: 3 a3 a f1y 3 3 m a a 1 3 3 f2 y a a EI 0 0 m2 f3 y 0 0 m3 0 0 3
a3
2
3
2
3
2
3 a3
3 a2
12 b 3
6 b2
12 b3 6 b2
0
0
2
3
0
v01 0 0 0 01 6 12 6 v b b b 2 4 2 b6 b b 22 12 b6 b6 v03 b 03 2 6 4 b b b 2
2
2
3
2
2
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 7 - Beam With Hinge
After applying the boundary conditions the global beam equations reduce to:
3 12 a3 b 3 EI 6 b 2
6 b2 4 b
v 2 P 2 0
a 3 b3P 3 3 v 2 3 b a EI a 3 b 2P 2 2 b3 a3 EI
Beam Stiff ness Example 7 - Beam With Hinge
The element force-displacement equations for element 1 are: 0 f1y 1 a 1 3EI 2 0 m1 3 a a a f a 1 a 1 2y 3 3 abP 3 b3 a3 EI
b 3P b 3 a3 f 1y 3 ab P m1 3 3 f b a 2 y b 3P b3 a3
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 7 - Beam With Hinge
The slope 2 on element 1 may be found using the condensed matrix:
m2 0 v 1 d 2 K 22 1 f2 K 21d1 2 element1 K 22 1 m2 K 21 1 v 2 0 a EI 2 6a 0 3 6a 2a 0 2 1 4EI a a b P 3 b a EI 3
3
2 1
3
3
a 2 b 3P 2 b3 a 3 EI
Beam Stiff ness Example 7 - Beam With Hinge
The element force-displacement equations for element 2 are: a bP 3 3 f 2 y 12 6b 12 6b 3 b a EI m 2 2 a3 b 2P 2 EI 6b 4b 6b 2b 3 3 3 f 3 y b 12 6b 12 6b 2 b a EI 2 2 m3 6b 2b 6b 4b 0 0 3
3
a3P b 3 a3 f 2 y 0 m 2 3 f 3 y a P m3 b3 a3 3 ba P b3 a3
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Example 7 - Beam With Hinge
Displacements and rotations on each element.
v2
v1 1 0 1
a3 b3P
a 2b 3P 2 b3 a3 EI
3
2
2
2 1
v 3 3 0
3 b3 a3 EI
2 2
a 3 b 2P 2 b3 a3 EI
Beam Stiff ness Potential Energy Appro ach to Derive Beam Element Equation s
Let’s derive the equations for a beam element using the principle of minimum potential energy. The procedure for applying the principle of minimum potential energy is similar to that used for the bar element. The total potential energy p is defined as the sum of the internal strain energy U and the potential energy of the external forces :
p U
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Chapter 4 - Development of Beam Equations - Part 2
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Beam Stiffness Potential Energy Appro ach to Derive Beam Element Equation s
The differential internal work (strain energy) dU in a onedimensional beam element is: U 21 x x dV
V
For a single beam element, shown below, subjected to both distributed and concentrated nodal forces, the potential energy due to forces (or the work done these forces) is: 2
2
i 1
i 1
Ty v dS Piy v i mi i S
Beam Stiff ness Potential Energy Appro ach to Derive Beam Element Equation s
If the beam element has a constant cross-sectional area A, then the differential volume of the beam is given as: dV dA dx
The differential element where the surface loading acts is given as: dS = b dx (where b is the width of the beam element). Therefore the total potential energy is: p
x A
1 2
x x dA dx Tyvb dx x
2
P v iy
i 1
i
mi i
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Chapter 4 - Development of Beam Equations - Part 2
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Beam Stiff ness Potential Energy Appro ach to Derive Beam Element Equation s d 2v The strain-displacement relationship is: x y 2 dx
We can express the strain in terms of nodal displacements and rotations as:
12 x 6L x y 3 L
6 xL 2L2
6 xL 4L2
12x 6 L
L3
L3
L3
6 xL 4L2
12x 6L
6 xL 2L2
L3
L3
d
x y [B]d 12 x 6L [B ] 3 L
L3
Beam Stiff ness Potential Energy Appro ach to Derive Beam Element Equation s
The stress-strain relationship in one-dimension is:
x [E ] x where E is the modulus of elasticity. Therefore:
x y [E ][B] d
x y [B]d
The total potential energy can be written in matrix form as: p
x A
1 2
x x dA dx bTy v T
T
x
T
dx d
P
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Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Potential Energy Appro ach to Derive Beam Element Equation s
If we define, w bTy as a line load (load per unit length) in the y direction and the substitute the definitions of x and x the total potential energy can be written in matrix form as: L
L
T T T E 2 p y d [B ]T [B ]d dA dx w d [N ]T dx d P 2 0 A 0
Use the following definition for moment of inertia: I y 2dA A
Then the total potential energy expression becomes: L
L
T T T EI p d [B ]T [B ]d dx w d [N ]T dx d P 2 0 0
Beam Stiff ness Potential Energy Appro ach to Derive Beam Element Equation s
Differentiating the total potential energy with respect to the displacement and rotations ( v1, v2, 1 and 2) and equating each term to zero gives: L
L
EI [B ] [B ] dx d w [N ]T dx P 0 0
T
0 L
The nodal forces vector is:
f w [N ]T dx P 0 L
The elemental stiffness matrix is:
k EI [B]T [B] dx 0
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Chapter 4 - Development of Beam Equations - Part 2
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Beam Stiffness Potential Energy Appro ach to Derive Beam Element Equation s
Integrating the previous matrix expression gives: L
k EI [B]T [B] dx 0
12 EI 6L k 3 L 12 6L
4L2
12 6L
6L
12
2L2
6L
6L
6L
2L2
6L 4L2
Beam Stiff ness Galerkin ’s Metho d to Deriv e Beam Element Equations
The governing differential equation for a one-dimensional beam is: d 4v EI 4 w 0 dx We can define the residual R as:
d 4v EI w 0 dx 4 Ni dx 0 L
i 1, 2, 3, and 4
If we apply integration by parts twice to the first term we get: L
0
L
EI v xxxx Ni dx EI v xx N i ,xx dx EI N i v xxx N i ,x v xx 0
where the subscript x indicates a derivative with respect to x
L
0
CIVL 7/8117
Chapter 4 - Development of Beam Equations - Part 2
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Beam Stiff ness Galerkin ’s Metho d to Deriv e Beam Element Equations
Since v [N ]d , then the second derivative of v with respect to x is: v xx
12 x 6L 3 L
6 xL 4L2
12 x 6L
L3
L3
6 xL 2L2
d
L3
or v x x [B ] d Therefore the integration by parts becomes: L
L
N EI [B ] dx d N w dx N V N m d i , xx
i
0
i
i ,x
0
L
0
0
i 1, 2, 3, and 4
Beam Stiff ness Galerkin ’s Metho d to Deriv e Beam Element Equations
The above expression is really four equations (one for each Ni ) and can be written in matrix form as: L
0
L
[B ] EI [B ] dx d [N ]T w dx [N ]xT m [N ]TV T
0
L
0
The interpolation function in the last two terms can be evaluated: [N ]x x 0 [0 1 0 0]
[N ]x x L [0 0 0 1]
[N ] x 0 [1 0 0 0]
[N ] x L [0 0 1 0]
CIVL 7/8117
Chapter 4 - Development of Beam Equations - Part 2
Beam Stiffness Galerkin ’s Metho d to Deriv e Beam Element Equations
Therefore, the last two terms of the matrix form of the Galerkin formulation become (see the figure below): i 1 i 3
V (0)
i 2
m (0)
V (L )
i 4
m(L )
Beam Stiffness Galerkin ’s Metho d to Deriv e Beam Element Equations
When beam elements are assembled, as shown below:
Two shear forces and two moments form adjoining elements contribute to the concentrated force and the concentrated moment at the node common to both elements.
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CIVL 7/8117
Chapter 4 - Development of Beam Equations - Part 2
Beam Stiff ness Problems:
7.
Verify the four beam element equations are contained in the following matrix expression. L
L
EI [B ] [B ] dx d w [N ]T dx P 0 T
0
8.
0
Do problems 4.7, 4.11, 4.18, 4.22, 4.40 and 4.47 on pages 225 - 234 in your textbook “A First Course in the Finite Element Method” by D. Logan.
Beam Stiff ness Problems:
9.
Work problem 4.22 (on page 229 in your textbook “A First Course in the Finite Element Method” by D. Logan) using the SAP2000 analysis system (you may used any available structural analysis software system). Start with two element and increase the number of beam elements until you get good agreement with the exact solution. Plot you results for each discretization to the exact solution. How many elements are enough to accurately model this problem?
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