CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 1/34

Chapter 4b – Development of Beam Equations

Learning Objectives • To introduce the work-equivalence method for replacing distributed loading by a set of discrete loads • To introduce the general formulation for solving beam problems with distributed loading acting on them • To analyze beams with distributed loading acting on them • To compare the finite element solution to an exact solution for a beam • To derive the stiffness matrix for the beam element with nodal hinge • To show how the potential energy method can be used to derive the beam element equations • To apply Galerkin’s residual method for deriving the beam element equations

Beam Stiffness 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=Kd-F0

where F0 are the equivalent nodal 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:

F=Kd0 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 2/34

Beam Stiffness 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=Kd-F0 This concept can be applied on a local basis to obtain the local nodal forces in individual elements of structures as:

f=kd-f0

Beam Stiffness Example 5 - Load Replacement Consider the beam shown below; determine the equivalent nodal forces for the given distributed load.

The work equivalent nodal forces are shown above. wL Using the beam stiffness equations:  2  fv12 6LL 12 6 2 11y  wL 22   m11EI 64LL 62 LL  12    fvLLL3 12 6 12  6 wL 22y   22 2  m2262LL 64 LL  2 wL 12 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 3/34

Beam Stiffness Example 5 - Load Replacement

Apply the boundary conditions: v11 0

12 6LL 12 6 v1 22 EI 64LL 62 LL1  3 v LLL12 6 12  6 2 22 62LL 64 LL2

We can solve for the displacements wL wL4    v  2 EI 12 6L v2 2 8EI    2 32  3 wL LLL64 2 wL  2  12 6EI

Beam Stiffness Example 5 - Load Replacement 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 e  F 12 6LL 12 6 0 2 1y 5wL  22 M e 64LL 62 LL 0  1 EI    12  4   F e LLL3 12 6 12  6 wL wL 2y 8EI  223 e wL 2 M 2 62LL 64 LL 6EI wL2  12 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 4/34

Beam Stiffness Example 5 - Load Replacement Using the above expression and the fix-end moments in: wL wL   wL 22  F 222 1y 5wL wL wL    M1 12 12 2 F=Kd-F0     F wL wL 2y   0 22   M2  wL22 wL 0 12 12 

Fy  0 FwL1y 2  wL L  M1  0 wL  22

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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 5/34

Beam Stiffness Example 6 - Cantilever Beam The beam stiffness equations are: P  2 fv12 6LL 12 6  11y PL 22   m11EI 64LL 62 LL  8    fvLLL3 12 6 12  6 P 22y   22 2  m2262LL 64 LL  PL  8

Apply the boundary conditions: v11 0

12 6LL 12 6 v1 22 EI 64LL 62 LL1  3 v LLL12 6 12  6 2 22 62LL 64 LL2

Beam Stiffness Example 6 - Cantilever Beam The beam stiffness equations become: P 5PL3   2 EI 12 6L v2 v2 48EI   32  2  PL LLL642 2 PL   8 8EI

To obtain the global nodal forces, we begin by evaluating the

effective nodal forces. P e 2 F 12 6LL 12 6 0  1y 3PL e 22  M EI 64LL 62 LL  0 8 1   3  e 3 5PL P F 2y LLL12 6 12  6 48EI    2 e 22PL2 M 62LL 64 LL 8EI PL 2  8 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 6/34

Beam Stiffness Example 6 - Cantilever Beam Using the above expression in the following equation, gives:

P P   P 2 2   F1y 3PL PL PL   M1 8 8 2 F=Kd-F     0 F P P 2y   0 2  M2 2 PL PL  F 0 8 8  e Kd = F F0 F

Beam Stiffness 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=kd-f0 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 7/34

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 106 psi, I = 100 in4, L = 100 in, and uniform load w = 20 Ib/in.

Beam Stiffness Comparison of FE Solution to Exact Solution To obtain the solution from classical beam theory, we use the double-integration method: Mx() 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:

 Fy 0 V() x wL wx wx22 wL  M  0 Mx() wLx  22 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 8/34

Beam Stiffness Comparison of FE Solution to Exact Solution To obtain the solution from classical beam theory, we use the double-integration method: Mx() y   EI wx22 L yLxdxdx  EI 22 wxLxxL322 yCdx1 Boundary Conditions EI  62 2 yy(0) 0 (0) 0 wxLxxL4322 yCxC 12 EI 24 6 4 wxLxxL 4322 y   EI  24 6 4 

Beam Stiffness Comparison of FE Solution to Exact Solution Recall the one-element solution to the cantilever beam is: wL4  v2 8EI  3  2 wL  6EI Using the numerical values for this problem we get:

4 lb 20in  100in  64 v 83010 psi 100 in 0.0833in 2   3    lb 0.00111 rad 2 20in  100in   64 63010 psi 100 in CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 9/34

Beam Stiffness 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.

Beam Stiffness 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. 11 vx() 2 x32 3 xLv  xL322  xL  LL3322 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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 10/34

Beam Stiffness 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 work- equivalence concept used to replace the distributed load by work-equivalent discrete loads at the nodes.

Beam Stiffness 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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 11/34

Beam Stiffness 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.

Beam Stiffness 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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 12/34

Beam Stiffness 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. d 2 Nd d 2Nd Mx() EIy  EI  EI dx2 dx 2 Mx() EIBd 

612x 46xxx 612 26 B 23  2 23  2 LL LL LL LL 612xx 46 MEI23 v11  2 LL LL 612xx 26  23v22  2  LL LL 

Beam Stiffness 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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 13/34

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.

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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 14/34

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.

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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 15/34

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.

Beam Stiffness Beam Element with 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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 16/34

Beam Stiffness Beam Element with Nodal Hinge For a beam with a hinge on the right end:

[]K12 []K11

fv11y 12 6LL 12 6  22  m11EI 64LL 62 LL     fv3 22y LLL12 6 12  6  22  m2262LL 64 LL 

[]K21 []K22

The moment m2 is zero and we can partition the matrix to eliminate the degree of freedom associated with  2 .

fKKd111121   fKdKd1111122    fKKd 221222   fKdKd2211222

Beam Stiffness Beam Element with Nodal Hinge For a beam with a hinge on the right end:

v1 f1y   fv11y 12 6LL 12 6  d11   fm11   22  v  m11EI 64LL 62 LL  2 f2y    fv3 22y LLL12 6 12  6  22 m 62LL 64 LL  22 d22    fm22   

1 dK2222211 fKd

fKdKd Kd K K1 f Kd 111112211 1 12 22 2  21 1

m2  0 fkd 1 1 cc1 ffKKfc 112222 kKKKKc 11 12 22 21 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 17/34

Beam Stiffness Beam Element with Nodal Hinge We can condense out the degree of freedom by using the partitioning method discussed earlier.

1 kKKKKc 11 12 22 21

12 6LL 12 6 EI EI 1 kLLLLLLL64222 6 2 626 c 332  LLL4 12 6LL 12  6  Therefore, the condensed stiffness matrix is: 11L  3EI kLLL2 c L3  11L

Beam Stiffness Beam Element with Nodal Hinge The element force-displacement equations are:

fLv11y 11 3EI  mLLL2  113   L  fLv22y 11 Expanding the element force-displacement equations and

maintaining m2 = 0 gives:

fv11y 110L   2  m113EI LL L0     fv3 22y LL110    m220000  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 18/34

Beam Stiffness Beam Element with Nodal Hinge

Once the displacements d1 are found, d2 may be computed:

 m2  0 v1    dK1 fKd  KmK1  2222211 2222211     v2 

In this case, d2 is 2 (the rotation at right side of the element at the hinge.

Beam Stiffness Beam Element with Nodal Hinge For a beam with a hinge on the left end:

fv11y 12 6LL 12 6  22  m11EI 64LL 62 LL     fv3 22y LLL12 6 12  6  22  m2262LL 64 LL 

Rewriting the equations to move 1 and m1 to the first row:

22 m1142LL66LL    fv11y EI 66LL12 12     fv3  22y L 66LL12 12  22  m2224LL66LL  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 19/34

Beam Stiffness Beam Element with Nodal Hinge For a beam with a hinge on the left end:

[]K12 []K11

22 m1142LL66LL    fv11y EI 66LL12 12     fv3  22y L 66LL12 12  22  m2224LL66LL 

[]K21 []K22

The moment m1 is zero and we can partition the matrix to eliminate the degree of freedom associated with  1 .

fKKd111121   fKdKd1111122    fKKd 221222   fKdKd2211222

Beam Stiffness Beam Element with Nodal Hinge For a beam with a hinge on the right end:

22d     fm m1142LL66LL  11  11    fv12 12  11EI 66LL  v f  1 1y 3    fvL 12 12 dv ff 2266LL  22   22 y 22    2 m2 m2224LL66LL 

1 dK1111122 fKd

fKdKd Kd K K1 f Kd 221122222 2 21 11 1  12 2

m1  0 fkd 1 1 cc2 ffKKfc 221111 kKc 22 KKK 21 11 12 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 20/34

Beam Stiffness Beam Element with Nodal Hinge We can condense out the degree of freedom by using the partitioning method discussed earlier.

1 kKc 22 KKK 21 11 12

12 12 6LL 6 EI EI 1 kLLLLL12 12  6 6 6  6 2 2 c 332  LLL224 664LLL 2 L Therefore, the condensed stiffness matrix is: 11 L 3EI kL11  c 3  L 2 LLL

Beam Stiffness Beam Element with Nodal Hinge The element force-displacement equations are:

fL11y 11 v 3EI  fL11 v 22y 3   L 2  mLLL21  

Expanding the element force-displacement equations and

maintaining m1 = 0 gives:

fv11y 10L  1    m113EI 000 0     fv3 22y LL0 L L    m2210L 1  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 21/34

Beam Stiffness Beam Element with Nodal Hinge

Once the displacements d2 are found, d1 may be computed:  v  m1  0 1 1   dK fKd  KmKv1 1111122 1111122     1 

In this case, d1 is 1 (the rotation at right side of the element at the hinge.

Beam Stiffness Beam Element with Nodal Hinge For a beam element with a hinge at its left end, the element force-displacement equations are:

fL11y 11 v 3EI  fL11 v 22y 3   L 2  mLLL22   Expanding the element force-displacement equations and

maintaining m1= 0 gives:

fv11y 10 1L    m113EI 000 0     fv3 22y LL10 1  2  m22LLL0   CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 22/34

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.

Beam Stiffness Example 7 - Beam With Hinge The stiffness matrix for element 1 (with hinge on right) is:

fv11y 110L   2  m113EI LL L0     fv3 22y LL110    m220000 

vv11 22

fv11y 110a   m aa2  a0  113EI   3  fv22y aa110   m220000 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 23/34

Beam Stiffness Example 7 - Beam With Hinge The stiffness matrix for element 2 (with no hinge) is:

fv2212 6LL 12 6  22  m22EI 64LL 62 LL   fv3 33LLL12 6 12  6   22 m3362LL 64 LL

vv22 3 3

fv2212 6bb 12 6  22  m22EI 64bb 62 bb     fv3 33bbb12 6 12  6  22  m3362bb 64 bb 

Beam Stiffness Example 7 - Beam With Hinge

The assembled equations are: Element 1

33 3 32 3000 fvaa a 1y  1 33 3000  m aa22a  1 1 3312  36  12 6 fv2y aababbb3233232 2  EI  m 00 6642  2 bb22bb 2 fv1266 12  3y 00 3232  3 bbbb  m  3 00 6624 3 bb22bb

Element 2 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 24/34

Beam Stiffness Example 7 - Beam With Hinge

The boundary conditions are: vv1313  0

33 3 32 3000 fvaa a 1y  01 33 3000 m aa22a 0 1  1 333 12 6  12 6 fv2y aaabbbb3233232 v2  EI  m 00 6642  2 bb22bb 2 fv1266 12  0 3y 00 3232  3 bbbb m 660 3 00 24  3 bb22bb

Beam Stiffness Example 7 - Beam With Hinge After applying the boundary conditions the global beam equations reduce to:

abP33 3126   33 33 2 3 baEI ab b v2 P v2  EI      32  642  02 abP bb2 33 2baEI CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 25/34

Beam Stiffness Example 7 - Beam With Hinge The element force-displacement equations for element 1 are:

 bP3 0 ba33 f  fa1y 11 1y 3 3EI 2   ab P maaa1  0  m   a3  1 ba33 fa11     2y f2y 3 abP33 bP   33 33 3baEI ba

Beam Stiffness Example 7 - Beam With Hinge

The slope 2 on element 1 may be found using the condensed matrix:  m  0 v1   2  dK1 fKd  KmK1  2222211 2222211element1      v2    0 aEI 2    0626aa  a 0 2 1 4EI a3  33 abP  33 3 baEI abP23    2 1 2baEI33 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 26/34

Beam Stiffness Example 7 - Beam With Hinge The element force-displacement equations for element 2 are:

3 33 aP abP  33  ba 3 baEI33  f2y 12 6bb 12 6  f2y   2232 0 m2 EI 64bb 62 bb abP  m2       f 3 33 f 3 3y b 12 6bb 12  6 2baEI  3y aP   22  33  m3 62bb 64 bb m3 ba 0  ba3 P 0  ba33

Beam Stiffness Example 7 - Beam With Hinge Displacements and rotations on each element.

abP33 v  0 v  v33 0 11 2 3baEI33 1 3

2 2 abP23 abP32    2  2 1 2baEI33 2 2baEI33 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 27/34

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations 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

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations The differential internal work (strain energy) dU in a one- dimensional beam element is: UdV 1   2 xx 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:

22 TvdS  Pv  m  yiyiii S ii11 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 28/34

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations 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:

2  1  dA dx T vb dx P v m  p 2 xx  y iyi ii xA x i 1

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations dv2 The strain-displacement relationship is:  y x dx 2 We can express the strain in terms of nodal displacements and rotations as: 12xL 6 6 xLL 422 12 xL 6 6 xLL 2  x yd33 3 3 LL L L

 x  yB[] d

12x  6LxLL 6 422 12 xLxLL 6 6 2 []B  33 3 3 LL L L CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 29/34

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations The stress-strain relationship in one-dimension is:

 xx  []E  

where E is the modulus of elasticity. Therefore:

 x  yE[][] B d  x  yB[] d

The total potential energy can be written in matrix form as:

TTT 1 dA dx bT v dx d P pxxy2            xA x

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations

If we define, wbT  y 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:

LL E TTT  yd2 [][] BBddAdxwdNdxdTT[]  P p      00A 2 Use the following definition for moment of inertia: IydA  2 A Then the total potential energy expression becomes:

LL EI TTT  dBBddxwdNdxdP[][]TT[] p     002 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 30/34

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations Differentiating the total potential energy with respect to the

displacement and rotations (v1, v2, 1 and 2) and equating each term to zero gives: LL EI[][] BTT B dx d w [] N dx P 0 00 L The nodal forces vector is: fwNdxP []T  0

L The elemental stiffness matrix is: kEIBBdx [][]T 0

Beam Stiffness Potential Energy Approach to Derive Beam Element Equations Integrating the previous matrix expression gives: L kEIBBdx [][]T 0

12 6LL 12 6 22 EI 64LL 62 LL k   LLL3 12 6 12  6 22 62LL 64 LL CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 31/34

Beam Stiffness Galerkin’s Method to Derive Beam Element Equations

The governing differential equation for a one-dimensional beam is: dv4 EI4  w 0 dx We can define the residual R as: L dv4 EI w N dx0 i  1,2,3,and4 4 i 0 dx If we apply integration by parts twice to the first term we get:

LL L EI v N dx EI  v N dx EI N v N v xxxx i xx i,, xx i xxx i x xx 0 00 where the subscript x indicates a derivative with respect to x

Beam Stiffness Galerkin’s Method to Derive Beam Element Equations Since vNd  []   , then the second derivative of v with respect to x is: 12xL 6 6 xLL 422 12 xL 6 6 xLL 2 vdxx  33 3 3 LL L L or

vBdxx  [] 

Therefore the integration by parts becomes:

LL L NEIBdxdNwdxNVNmdixx,,[]  i i ix  0 0 00 i  1, 2 , 3 , a n d 4 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 32/34

Beam Stiffness Galerkin’s Method to Derive Beam Element Equations

The above expression is really four equations (one for each

Ni) and can be written in matrix form as:

LL L []BEIBdxdTTTT [] [] Nwdx  [] N m  [] NV x 0 00

The interpolation function in the last two terms can be evaluated:

[]Nxxx 0 [0100] []NxL [0001]

[]Nx 0 [1000] [] Nx  L [0010]

Beam Stiffness Galerkin’s Method to Derive Beam Element Equations

Therefore, the last two terms of the matrix form of the Galerkin formulation become (see the figure below):

iV1(0)2 i   m (0)

iVLimL3()4  () CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 33/34

Beam Stiffness Galerkin’s Method to Derive 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.

Beam Stiffness Problems:

7. Verify the four beam element equations are contained in the following matrix expression.

LL EI[][] BTT B dx d w [] N dx P 0 00

8. Do problems 4.10, 4.12, 4.18, 4.22, 4.40 and 4.47 in your textbook. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 2 34/34

Beam Stiffness Problems:

9. Work problem 4.36 in your using the SAP2000.

Attempt to select the lightest standard W section to support the loads for the beam.

The bending stress must not exceed 160 MPa and the allowable deflection must not exceed (L = 6 m)/360

End of Chapter 4b