Development of Beam Equations Development of

CIVL 7/8117

Chapter 4 - Development of Beam Equations - Part 1

Development of Beam Equations In this section, we will develop the stiffness matrix for a beam element, the most common of all structural elements. The beam element is considered to be straight and to have constant cross-sectional area.

Development of Beam Equations We will derive the beam element stiffness matrix by using the principles of simple beam theory. The degrees of freedom associated with a node of a beam element are a transverse displacement and a rotation.

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Development of Beam Equations We will discuss procedures for handling distributed loading and concentrated nodal loading. We will include the nodal shear forces and bending moments and the resulting shear force and bending moment diagrams as part of the total solution.

Development of Beam Equations We will develop the beam bending element equations using the potential energy approach. Finally, the Galerkin residual method is applied to derive the beam element equations

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Beam Stiffness Consider the beam element shown below.

The beam is of length L with axial local coordinate x and transverse local coordinate y. The local transverse nodal displacements are given by vi and the rotations by ϕi. The local nodal forces are given by fiy and the bending moments by mi.

Beam Stiffness At all nodes, the following sign conventions are used: 1. Moments are positive in the counterclockwise direction. 2. Rotations are positive in the counterclockwise direction. 3. Forces are positive in the positive y direction. 4. Displacements are positive in the positive y direction.

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Beam Stiffness The differential equation governing simple linear-elastic beam behavior can be derived as follows. Consider the beam shown below.

Beam Stiffness The differential equation governing simple linear-elastic beam behavior can be derived as follows. Consider the beam shown below. w ( x )dx  dx   2   

Write the equations of equilibrium for the differential element:

M

F

y

right side

 dx   0  (M  dM )  M  Vdx  w ( x )dx    2 

 0  V  (V  dV )  w ( x )dx

dx 2  0

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Beam Stiffness From force and moment equilibrium of a differential beam element, we get:

M F

y

right side

0

0





 wdx  dV  0

w 

 Vdx  dM  0 or

or V  w 

dM dx

dV dx

d  dM  dx  dx 

Beam Stiffness The curvature  of the beam is related to the moment by:



1





M EI

where  is the radius of the deflected curve, v is the transverse displacement function in the y direction, E is the modulus of elasticity, and I is the principle moment of inertia about y direction, as shown below.

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Beam Stiffness The curvature,  for small slopes  



d 2v dx 2

Therefore:

d 2v M  dx 2 EI



dv is given as: dx

M  EI

d 2v dx 2

Substituting the moment expression into the moment-load equations gives: d 2  d 2v   EI   w  x  dx 2  dx 2  For constant values of EI, the above equation reduces to:

 d 4v  EI  4   w  x   dx 

Beam Stiffness Step 1 - Select Element Type

We will consider the linear-elastic beam element shown below.

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Beam Stiffness Step 2 - Select a Displacement Function

Assume the transverse displacement function v is:

v  a1x 3  a2 x 2  a3 x  a4 The number of coefficients in the displacement function ai is equal to the total number of degrees of freedom associated with the element (displacement and rotation at each node). The boundary conditions are:

v ( x  0)  v1

v ( x  L)  v 2

dv dx

dv dx

 1 x 0

 2 x L


Applying the boundary conditions and solving for the unknown coefficients gives:

v (0)  v1  a4

v (L )  v 2  a1L3  a2L2  a3L  a4

dv (0)  1  a3 dx

dv (L )  2  3a1L2  2a2L  a3 dx

Solving these equations for a1, a2, a3, and a4 gives: 1 2  v   3 v1  v 2   2 1  2   x 3 L L  1  3     2 v1  v 2    21  2   x 2  1x  v1 L  L 

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In matrix form the above equations are: v  [N ] d  v1    d    1  v 2  2  where

[N ]  N1 N2 N3 N4 



N1 

1 2 x 3  3 x 2L  L3 L3

N3 

1 2 x 3  3 x 2 L 3 L









N2 

1 3 x L  2 x 2L2  xL3 L3

N4 

1 3 x L  x 2L2 3 L








N1, N2, N3, and N4 are called the interpolation functions for a beam element. 1.000

N1

1.000

0.800

0.800

0.600

0.600

0.400

0.400

0.200

0.200

0.000 0.00

1.00

-0.200

1.000

0.000 0.00

N3

1.000 0.800

0.600

0.600

0.400

0.400

0.200

0.200

-0.200

1.00

-0.200

0.800

0.000 0.00

N2

1.00

0.000 0.00 -0.200

N4

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Beam Stiffness Step 3 - Define the Strain/Displacement and Stress/Strain Relationships

The stress-displacement relationship is:  x  x, y   where u is the axial displacement function.

du dx

We can relate the axial displacement to the transverse displacement by considering the beam element shown below:


u  y

dv dx

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One of the basic assumptions in simple beam theory is that planes remain planar after deformation, therefore:

 x  x, y  

 d 2v  du  y  2  dx  dx 

Moments and shears are related to the transverse displacement as:  d 2v  m  x   EI  2   dx 

 d 3v  V  x   EI  3   dx 

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations

Use beam theory sign convention for shear force and bending moment. m+

V+

m+

V+

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Using beam theory sign convention for shear force and bending moment we obtain the following equations: d 3v f1y  V  EI 3 dx

 x 0

f2 y  V  EI

d 3v dx 3

m1  m  EI

d v dx 2

m2  m  EI

d 2v dx 2

EI 12v1  6L1  12v 2  6L2  L3 

EI  12v1  6L1  12v 2  6L2  L3



EI 6Lv1  4L21  6Lv 2  2L22 3 L

x L

2



x 0

 x L

EI 6Lv1  2L21  6Lv 2  4L22 3 L








In matrix form the above equations are:  f1y   12 m    1  EI  6L   3  f2 y  L  12   6L m2 

6L 4L2 6L

12 6L 12

2L2

6L

6L  v1  2L2   1    6L  v 2   4L2  2 

where the stiffness matrix is:  12  EI 6L k 3  L  12   6L

6L

12

2

4L 6L

6L 12

2L2

6L

6L  2L2   6L   4L2 

 f1y   v1  m     1  1  k   f2 y  v 2  m2  2 

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Beam stiffness based on Timoshenko Beam Theory

The total deflection of the beam at a point x consists of two parts, one caused by bending and one by shear force. The slope of the deflected curve at a point x is: dv   x   x dx



The relationship between bending moment and bending deformation is: d  x  M  x   EI dx

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The relationship between shear force and shear deformation is:

V  x   ks AG   x  where ksA is the shear area.



You can review the details in your book, but by including the effects of shear deformations into the relationship between forces and nodal displacements a modified elemental stiffness can be developed.

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 12  EI  6L k 3 L 1     12   6L

6L  4    L2 6L  2    L2

12 6L 12 6L

6L   2    L2  6L   4    L2 



12EI ks AGL2

Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions Consider a beam modeled by two beam elements (do not include shear deformations):

Assume the EI to be constant throughout the beam. A force of 1,000 lb and moment of 1,000 lb-ft are applied to the midpoint of the beam.

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Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions

The beam element stiffness matrices are: 1

v1

k (1)

 12  EI 6L  3  L  12   6L v

k (2)

6L

12

2

4L 6L

6L 12

2L2

6L



2

 12  EI 6L  3  L  12   6L

v2

2

v

2

6L  2L2   6L   4L2  

3

6L 4L2

12 6L

6L 2L2

12 6L

3

6L  2L2   6L   4L2 


In this example, the local coordinates coincide with the global coordinates of the whole beam (therefore there is no transformation required for this problem). The total stiffness matrix can be assembled as:  F1y  12 6L 6L  12 M   6L 4L2 6L 2L2  1  F2 y  EI  12 6L 12  12 6L  6L M   3  2 2 2  2  L  6L 2L 6L  6L 4L  4L F3 y   0 12 6L 0    0 6L 2L2  0 M3 

Element 1

  v1      1  v 2    2L2  2  12 6L  v 3    6L 4L2  3  Element 2 0 0 12 6L

0 0 6L

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The boundary conditions are:

v1  1  v 3  0

 F1y  6L 6L 12  12 M   6L 4L2 2L2 6L  1  F2 y  EI  12 6L 12  12 6L  6L M   3  2 2 2  2  L  6L 2L 6L  6L 4L  4L F3 y   0 0 12 6L    0 6L 2L2 M3   0

0   v01    0   01   6L  v 2    2L2  2  12 6L  v03    6L 4L2  3  0 0 12 6L


By applying the boundary conditions the beam equations reduce to: 6L  v 2   1,000 lb  24 0   EI   2 2  1,000 lb ft   3  0 8L 2L  2    L 6L 2L2 4L2    0     3

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Beam Stiffness Step 6 - Solve for the Unknown Degrees of Freedom

Solving the above equations gives: v2  

875L3  375L2 lb 12EI

2  

125L2  625L rad 4EI

3  

125L2  125L rad EI

Step 7 - Solve for the Element Strains and Stresses v1    2 2 d v   d N    m  x   EI  2   EI  2   1   dx  v 2   dx  2 

The second derivative of N is linear; therefore m(x) is linear.

Beam Stiffness Step 6 - Solve for the Unknown Degrees of Freedom

Solving the above equations gives: v2  

875L3  375L2 lb 12EI

2  

125L2  625L rad 4EI

3  

125L2  125L rad EI

Step 7 - Solve for the Element Strains and Stresses v1     d 3v   d 3N     V  x   EI  3   EI  2   1   dx   dx  v 2  2 

The third derivative of N is a constant; therefore V(x) is constant.

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Beam Stiffness Example 1 - Beam Problem

Consider the beam shown below. Assume that EI is constant and the length is 2L (no shear deformation).

The beam element stiffness matrices are: v

k (1)

1

 12  EI 6L  3  L  12   6L

1

v

2

6L 4L2

12 6L

6L 2L2

12 6L

2

6L  2L2   6L   4L2 

v

k (2)

2

 12  EI 6L  3  L  12   6L

2

v

6L

12

4L 6L 2L2

6L 12 6L

2

3

3

6L  2L2   6L   4L2 


The local coordinates coincide with the global coordinates of the whole beam (therefore there is no transformation required for this problem). The total stiffness matrix can be assembled as:  F1y  6L 12 6L 0 0   v1   12 M     6L 4L2 6L 2L2 0 0   1   1   F2 y  EI  12 6L 24 12 6L  v 2  0 M   3    2 2 0 8L 6L 2L2  2   2  L  6L 2L F3 y   0 12 6L 12 6L  v 3  0      0 6L 2L2 6L 4L2  3   0 M3  Element 1 Element 2

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The boundary conditions are:

v 2  v 3  3  0

 F1y  6L 12 6L 0 0  v1   12 M     6L 4L2 6L 2L2 0 0   1   1   F2 y  EI  12 6L 24 12 6L  v02  0 M   3    2 0 8L2 6L 2L2  2   2  L  6L 2L F3 y   0 12 6L 12 6L  v03  0      0 6L 2L2 6L 4L2  03  M3   0

Beam Stiffness By applying the boundary conditions the beam equations reduce to: 6L 6L  v1   P   12   EI   2 2  0   3  6L 4L 2L  1   0  L  6L 2L2 8L2        2 Solving the above equations gives:

 7L   3  v1      PL2   1    3    4EI    2    1 

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The positive signs for the rotations indicate that both are in the counterclockwise direction. The negative sign on the displacement indicates a deformation in the -y direction.  F1y  6L 12 6L 0 0   7 L 3   12 M   6L 4L2 6L 2L2 0 0  3   1    F2 y  P  12 6L 24 12 6L   0  0 M      2 0 8L2 6L 2L2   1   2  4L  6L 2L F3 y   0 12 6L 12 6L   0  0      0 6L 2L2 6L 4L2   0  M3   0


The local nodal forces for element 1: 6L 12 6L    7L 3   P   f1y   12       2 2   m1  P  6L 4L 6L 2L   3   0        f2 y  4L  12 6L 12 6L   0   P   2 2  m2   6L 2L 6L 4L   1  PL 

The local nodal forces for element 2: 6L 12 6L  0   1.5P   f2 y   12 m   6L 4L2 6L 2L2   1  PL   2 P            f3 y  4L  12 6L 12 6L  0   1.5P   2 2  m3   6L 2L 6L 4L  0   0.5PL 

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The free-body diagrams for the each element are shown below.

Combining the elements gives the forces and moments for the original beam.


Therefore, the shear force and bending moment diagrams are:

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Beam Stiffness Example 2 - Beam Problem Consider the beam shown below. Assume E = 30 x 106 psi and I = 500 in4 are constant throughout the beam. Use four elements of equal length to model the beam.

The beam element stiffness matrices are: vi

k( i )

 12  EI  6L  3 L  12   6L

i

v ( i 1) ( i 1)

6L 4L2

12 6L

6L 2L2

12 6L

6L  2L2   6L   4L2 

Beam Stiffness Example 2 - Beam Problem Using the direct stiffness method, the four beam element stiffness matrices are superimposed to produce the global stiffness matrix. Element 1

Element 2

Element 3 Element 4

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Beam Stiffness Example 2 - Beam Problem The boundary conditions for this problem are:

v1  1  v 3  v 5  5  0

Beam Stiffness Example 2 - Beam Problem The boundary conditions for this problem are:

v1  1  v 3  v 5  5  0 After applying the boundary conditions the global beam equations reduce to: 24 0  0 8L2 EI  6L 2L2 3 L  0 0  0 0

0  v 2  10,000 lb   0 0  2       0 8L2 6L 2L2  3      v  10,000 lb  0  4 6L 24     2 0 2L 0 8L2  4    6L 2L2

0 0

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Beam Stiffness Example 2 - Beam Problem Substituting L = 120 in, E = 30 x 106 psi, and I = 500 in4 into the above equations and solving for the unknowns gives:

2  3  4  0

v 2  v 4  0.048 in

The global forces and moments can be determined as: F1y 

M1 

5 kips

25 kips·ft

F2 y  10 kips F3 y  10 kips

M2  0 M3  0

F4 y  10 kips F5 y  5 kips

M4  0 M5  25 kips·ft

Beam Stiffness Example 2 - Beam Problem The local nodal forces for element 1:

 f1y   12 m    1  EI  6L    3  f2 y  L  12   6L m2 

6L

12

4L2

6L

6L 2L2

12 6L

6L   0   5 kips  2L2   0   25 k·ft     6L  0.048  5 kips   4L2   0   25 k·ft 

The local nodal forces for element 2:  f2 y   12 m    2  EI  6L    3  f3 y  L  12   6L m3 

6L

12

2

4L

6L

6L 2L2

12 6L

6L  0.048   5 kips  2L2   0  25 k·ft     6L   0   5 kips   4L2   0  25 k·ft 

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Beam Stiffness Example 2 - Beam Problem The local nodal forces for element 3:

 f3 y   12 m    3  EI  6L   3  f4 y  L  12   6L m4 

6L

12

2

4L

6L

6L 2L2

12 6L

6L   0   5 kips  2L2   0   25 k·ft     6L  0.048  5 kips   4L2   0   25 k·ft 

The local nodal forces for element 4:  f4 y   12 m    4  EI  6L   3  f5 y  L  12   6L m5 

6L

12

4L2

6L

6L 2L2

12 6L

6L  0.048   5 kips  2L2   0  25 k·ft     6L   0   5 kips   4L2   0  25 k·ft 

Beam Stiffness Example 2 - Beam Problem Note: Due to symmetry about the vertical plane at node 3, we could have worked just half the beam, as shown below.

Line of symmetry

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Beam Stiffness Example 3 - Beam Problem Consider the beam shown below. Assume E = 210 GPa and I = 2 x 10-4 m4 are constant throughout the beam and the spring constant k = 200 kN/m. Use two beam elements of equal length and one spring element to model the structure.

Beam Stiffness Example 3 - Beam Problem The beam element stiffness matrices are: 1

v1

k (1)

 12  EI 6L  3  L  12   6L

v2

6L

12

4L2 6L 2L2

6L 12 6L

2

v2

6L  2L2   6L   4L2 

k (2)

 12  EI 6L  3  L  12   6L

2

6L

12

4L2

6L

6L 2L2

12 6L

The spring element stiffness matrix is: v3

k

(3)

 k   k

v3

v4

k  k 



k

(3)

3

v3

v4

 k 0 k  0 0 0    k 0 k 

3

6L  2L2   6L   4L2 

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Beam Stiffness Example 3 - Beam Problem Using the direct stiffness method and superposition gives the global beam equations. Element 1

Element 2

 F1y  6L 12 6L 0 0 0   v1   12 M   6L 4L2 6L 2L2 0 0 0   1   1   F2 y   12 6L 24 0 6L 0  v 2  12   EI    2 0 8L2 2L2 0  2  6L  M2   3  6L 2L L F   0 0 12 6L 12  k ' 6L k '  v 3   3y     2 2 0 0 6 2 6 4 0  L L L L  M3    3  F   0 0 0 0 k ' k '  v 4   0  4y 

k' 

kL3 EI

Element 3

Beam Stiffness Example 3 - Beam Problem The boundary conditions for this problem are: v1  1  v 2  v 4  0  F1y  6L 12 6L 0 0 0   v01   12 M   6L 4L2 6L 2L2 0 0 0   01   1   F2 y   12 6L 24 12 0 6L 0  v02    EI    2 6L 0 8L2 2L2 0  2   M2   3  6L 2L L F   0 12 6L 12  k ' 6L k '  v 3  0  3y     2 2 L L L L  0 0 6 2 6 4 0  M3    3  F    v04  k k  0 0 0 0 ' 0 '   4y 

k' 

kL3 EI

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After applying the boundary conditions the global beam equations reduce to:  8L2 6L 2L2  2   0  M2        EI  F3 y   3  6L 12  k ' 6L  v 3   P   M  L  2L2 6L 4L2  3   0   3  Solving the above equations gives:

 3PL2  1     EI  12  7k '   2   3 1    7PL    v 3          EI  12  7k '    3  9PL2  1      EI  12  7k '  

k' 

kL3 EI

Beam Stiffness Example 3 - Beam Problem Substituting L = 3 m, E = 210 GPa, I = 2 x 10-4 m4, and k = 200 kN/m in the above equations gives: v 3  0.0174 m 2  0.00249 rad

3  0.00747 rad

Substituting the solution back into the global equations gives:  F1y   69.9 kN  M     1  69.7 kN  m  F2 y   116.4 kN      0  M2     F   50 kN   3y    0  M3    F   3.5 kN    4y  

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Beam Stiffness Distributed Loadings Beam members can support distributed loading as well as concentrated nodal loading. Therefore, we must be able to account for distributed loading. Consider the fixed-fixed beam subjected to a uniformly distributed loading w shown the figure below.

The reactions, determined from structural analysis theory, are called fixed-end reactions.

Beam Stiffness Distributed Loadings In general, fixed-end reactions are those reactions at the ends of an element if the ends of the element are assumed to be fixed (displacements and rotations are zero).

Therefore, guided by the results from structural analysis for the case of a uniformly distributed load, we replace the load by concentrated nodal forces and moments tending to have the same effect on the beam as the actual distributed load.

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Beam Stiffness Distributed Loadings The figures below illustrates the idea of equivalent nodal loads for a general beam. We can replace the effects of a uniform load by a set of nodal forces and moments.

Beam Stiffness Work Equivalence Method This method is based on the concept that the work done by the distributed load is equal to the work done by the discrete nodal loads. The work done by the distributed load is: L

Wdistributed   w  x  v  x  dx 0

where v(x) is the transverse displacement. The work done by the discrete nodal forces is:

Wnodes  m11  m22  f1y v1  f2 y v 2 The nodal forces can be determined by setting Wdistributed = Wnodes for arbitrary displacements and rotations.

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Beam Stiffness Example 4 - Load Replacement Consider the beam, shown below, and determine the equivalent nodal forces for the given distributed load.

Using the work equivalence method or: L

 w  x  v  x  dx  m 

1 1

Wdistributed  Wnodes

 m22  f1y v1  f2 y v 2

0

Beam Stiffness Example 4 - Load Replacement

Evaluating the left-hand-side of the above expression with: w  x   w 1 2  v ( x )   3 v1  v 2   2 1  2   x 3 L L  3 1      2 v1  v 2    21  2   x 2  1x  v1 L  L 

gives: L

 w v  x  dx  0



Lw L2w v1  v 2   1  2   Lw v 2  v1  2 4

L2w L2w  21  2   1  wLv1 3 2

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Using a set of arbitrary nodal displacements, such as: v1  v 2  2  0

1  1

The resulting nodal equivalent force or moment is: L

m11  m22  f1y v1  f2 y v 2   w  x  v  x  dx 0

 wL2 2 2 L2  wL2 m1     Lw  w 3 2  12  4


Using a set of arbitrary nodal displacements, such as: v1  v 2  1  0

2  1

The resulting nodal equivalent force or moment is: L

m11  m22  f1y v1  f2 y v 2   w  x  v  x  dx 0

 wL2 wL2  wL2 m2      4 3   12

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Setting the nodal rotations equal zero except for the nodal displacements gives: LW Lw f1y    Lw  Lw   2 2 f2 y 

LW Lw  Lw   2 2

Summarizing, the equivalent nodal forces and moments are:

End of Chapter 4a

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