CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 1/39

Chapter 4a – Development of Beam Equations

Learning Objectives • To review the basic concepts of beam bending • To derive the stiffness matrix for a beam element • To demonstrate beam analysis using the direct stiffness method • To illustrate the effects of shear deformation in shorter beams • 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

Chapter 4a – Development of Beam Equations

Learning Objectives • 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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 2/39

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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 3/39

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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 4/39

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 on the element level: 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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 5/39

Beam Stiffness At all nodes, the following sign conventions are used on the global level: 1. Bending moments m are positive if they cause the beam to bend concave up. 2. Shear forces V are positive is the cause the beam to rotate clockwise.

Beam Stiffness

(+) Bending Moment

(-) Bending Moment CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 6/39

Beam Stiffness

(+) Shear Force

(-) Shear Force

Beam Stiffness The differential equation governing simple linear-elastic beam behavior can be derived as follows. Consider the beam shown below. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 7/39

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

Write the equations of equilibrium for the differential element:

 dx  Mright side  0 MMdMVdxwxdx     ()  2  0 Fy 0 VVdVwxdx()()  

Beam Stiffness From force and moment equilibrium of a differential beam element, we get: dM M 00Vdx dM or V   right side dx dV F 00wdx dV or w   y dx

ddM w   dx dx CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 8/39

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.

Beam Stiffness dv The curvature,  for small slopes   is given as: dx dv2   dx2 dv22 M dv Therefore: MEI dx22 EI dx

Substituting the moment expression into the moment-load equations gives: ddv22 22EI w x dx dx For constant values of EI, the above equation reduces to: dv4 EI4  w x dx CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 9/39

Beam Stiffness Step 1 - Select Element Type

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

Beam Stiffness Step 2 - Select a Displacement Function

Assume the transverse displacement function v is:

32 vaxaxaxa1234

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:

vx(0) v1 vx() L v2

dv dv  1  2 dx x0 dx xL CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 10/39

Beam Stiffness Step 2 - Select a Displacement Function

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

32 vva(0) 14 vL() v21 aL  aL 2  aL 3  a 4

dv(0) dv() L  a  32aL2  aL  a dx 13 dx 21 23

Solving these equations for a1, a2, a3, and a4 gives: 21 vvv x3 3212 12 LL 31 vv  2  x2   xv  2 12 12 1 1 LL

Beam Stiffness Step 2 - Select a Displacement Function In matrix form the above equations are: vNd [] 

v1  1 dN  [] NNNN v 123 4 2  2 where 11 N23 x323 xL L N  xL3 2 xL 223 xL 12LL33 

11 NxxLNxLxL2332 322  34LL33  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 11/39

Beam Stiffness Step 2 - Select a Displacement Function

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

1.000 1.000 N1 N2

0.800 0.800

0.600 0.600

0.400 0.400

0.200 0.200

0.000 0.000 0.00 1.00 0.00 1.00 -0.200 -0.200

1.000 1.000 N3 N4

0.800 0.800

0.600 0.600

0.400 0.400

0.200 0.200

0.000 0.000 0.00 1.00 0.00 1.00 -0.200 -0.200

Beam Stiffness Step 3 - Define the Strain/Displacement and Stress/Strain Relationships du The stress-displacement relationship is:  xy,  x dx where u is the axial displacement function.

We can relate the axial displacement to the transverse displacement by considering the beam element shown below: CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 12/39

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

dv uy dx

Beam Stiffness Step 3 - Define the Strain/Displacement and Stress/Strain Relationships One of the basic assumptions in simple beam theory is that planes remain planar after deformation, therefore:

du  dv2   x xy,  y  2  dx  dx  Moments and shears are related to the transverse displacement as:

dv2  dv3  mx EI2 Vx EI 3  dx  dx  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 13/39

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Use beam theory sign convention for shear force and bending moment.

M+ M+

+ V+ V

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Using beam theory sign convention for shear force and bending moment we obtain the following equations: dv3 EI fVEI11122y 33 12 v  6 L  12 v  6 L dxx0 L dv3 EI fVEI21122y  33  12 vLvL  6  12  6  dxxL L 2 dv EI 22 mmEI11122 23 6462 LvL   LvL   dxx0 L 2 dv EI 22 mmEI2112223 626 LvL   LvL  4 dxxL L CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 14/39

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations In matrix form the above equations are:

fv11y 12 6LL 12 6  fv11y   22   m11EI 64LL 62 LL m11     k  fv3 fv 22y LLL12 6 12  6  22y   22   m2262LL 64 LL m22  where the stiffness matrix is: 12 6LL 12 6 22 EI 64LL 62 LL k   LLL3 12 6 12  6 22 62LL 64 LL

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations 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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 15/39

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Beam stiffness based on Timoshenko Beam Theory

The relationship between bending moment and bending deformation is: dx  Mx EI dx

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Beam stiffness based on Timoshenko Beam Theory

The relationship between shear force and shear deformation is:

Vx kAGxs  

where ksA is the shear area. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 16/39

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Beam stiffness based on Timoshenko Beam Theory

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.

Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Beam stiffness based on Timoshenko Beam Theory

1266LL 12 22 66LL42LL12EI EI     k  3 2 L 112 66LL 12 kAGLs 22 66LL24LL  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 17/39

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 mid- point of the beam.

Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions The beam element stiffness matrices are:

vv11 2 2 12 6LL 12 6 22 EI 64LL 62 LL k(1)   LLL3 12 6 12  6 22 62LL 64 LL

vv 22 3 3 12 6LL 12 6 22 EI 64LL 62 LL k(2)   LLL3 12 6 12  6 22 62LL 64 LL CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 18/39

Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions 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 6LL 12 6 0 0 v1  22 M 64LL 6 L 2 L 0 0 1 1 F2y EI 12 6LLLL 12  12  6 6 12 6 v2   M 32 222 2 LLL62 6644 LLLL   62 LL 2   F3y 00 126LL 126 v3   00 6LLLL 222 64 M3 3 Element 1 Element 2

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

The boundary conditions are: vv11 3 0

F1y 12 6LL 12 6 0 0 v01  22 M 64LL 6 L 2 L 0 00 1 1 F2y EI 12 6LL 12  12  6 6LL 12 6 v2  32 222 M LLL6 2 664 LLLL 4 62 LL 2     2  F3y 00 126LL 126 v03  00 6LLLL 222 64 M3 3 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 19/39

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

By applying the boundary conditions the beam equations reduce to:

1, 0 0 0lb 2 4 0 6 L  v 2 EI  1,000lbft  0 8 L22 2 L  3  2 L 22  0624LL L 3

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

Solving the above equations gives:

875LL32 375 125 LL 2 625 125 LL2 125 vin rad  rad 2212EI 4EI 3EI

Step 7 - Solve for the Element Strains and Stresses

v1 2 2  dv dN 1 mx EI  EI  2 2 v dx dx 2  2 The second derivative of N is linear; therefore m(x) is linear. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 20/39

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

Solving the above equations gives:

875LL32 375 125 LL 2 625 125LL2  125 vin  rad  rad 2212EI 4EI 3EI

Step 7 - Solve for the Element Strains and Stresses

v1 3 3   dv dN 1 Vx EI 3   EI  dx 2 v   dx 2  2 The third derivative of N is a constant; therefore V(x) is constant.

Beam Stiffness Step 7 - Solve for the Element Strains and Stresses Assume L = 120 in, E = 29x106 psi, and I = 100 in4:

54 vi220.0433 n7.758  10 rad 35.586  10 rad

Element #1: v1 2 2  dv dN 1 mx EI  EI  2 2 v dx dx 2  2 EI mLvLLvL646222  3,875lb ft 11122L3  EI mLvLLvL62622 4 3,562.5lb ft 21122L3  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 21/39

Beam Stiffness Step 7 - Solve for the Element Strains and Stresses Assume L = 120 in, E = 29x106 psi, and I = 100 in4:

54 vi220.0433 n7.758  10 rad 35.586  10 rad

Element #2: v1 2 2  dv dN 1 mx EI  EI  2 2 v dx dx 2  2 EI mLvLLvL646222  2,562.5lb  ft 22233L3  EI mLvLLvL626422   0 32233L3 

Beam Stiffness Step 7 - Solve for the Element Strains and Stresses Assume L = 120 in, E = 29x106 psi, and I = 100 in4:

54 vi220.0433 n7.758  10 rad 35.586  10 rad

Element #1: v1 3 3   dv dN 1 Vx EI 3   EI  dx 2 v   dx 2  2 EI fvLvL12 6 12 6   743.75lb 11122y L3 EI fvLvL12  6  12  6  743.75lb 21y L3 122 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 22/39

Beam Stiffness Step 7 - Solve for the Element Strains and Stresses Assume L = 120 in, E = 29x106 psi, and I = 100 in4:

54 vi220.0433 n7.758  10 rad 35.586  10 rad

Element #2: v1 3 3   dv dN 1 Vx EI 3   EI  dx 2 v   dx 2  2 EI fvLvL12  6 12 6  256.25lb 22233y L3 EI fvLvL12  6  12  6   256.25lb 32233y L3

Beam Stiffness Step 7 - Solve for the Element Strains and Stresses

256.25lb

743.75lb Flb1,000

3,562.5lb ft 2,562.5lb ft

3,875lb ft Mlbft1,000  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 23/39

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:

vv vv 1122 22 3 3 12 6LL 12 6 12 6LL 12 6 22 22 EI 64LL 62 LL EI 64LL 62 LL k(1)  k(2)   L3 12 6LL 12  6 L3 12 6LL 12  6 22 22 62LL 64 LL 62LL 64 LL

Beam Stiffness Example 1 - Beam Problem 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 6LL 12 6 0 0 v1  22  M 64LL 62 LL 0 0 1 1 F2y EI 12 6LL 24 0  12 6 v2   M 32 2 2 2 LLL62 08 L 62 LL2   F3y 00126126LL v3   0062LL22 64 LL M3 3 Element 1 Element 2 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 24/39

Beam Stiffness Example 1 - Beam Problem

The boundary conditions are: vv233 0

F1y 12 6LL 12 6 0 0 v1  22  M 64LL 62 LL 0 0 1 1 F2y EI 12 6LL 24 0  12 6 v02  32 2 2  M LLL62 08 L 62 LL 2  2   F3y 00126126LL v03   0062LL22 64 LL0 M3 3

Beam Stiffness

By applying the boundary conditions the beam equations reduce to: PLL 12 6 6 v1 EI  0642 LLL22  3 1 L 22 0628 LLL2

7L  Solving the above equations gives: 3 v1  PL2     3  1 4EI    2  1  CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 25/39

Beam Stiffness Example 1 - Beam Problem 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.

7L F1y 12 6LL 12 6 0 0 3  M 64LL22 62 LL 0 0 3 1  F2y P 12 6LL 24 0 12 6 0   M 222 2 46LL 2 L 0 8 L 6 LL 2 1 F3y 001261260LL    0062LL22 64 LL 0 M3 

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

7L f1y 12 6LL 12 6 3 P  22 m1 P 64LL 62 LL 3  0     f 2y 4126126LL  L 0  P   22   m2 62LL 64 LL 1 PL The local nodal forces for element 2:

f2y 12 6LL 12 6 0 1.5 P  22 m2 P 64LL 621 LL  PL    f 3y 41261260LL  L  1.5 P   22  m3 62LL 640 LL 0.5 PL CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 26/39

Beam Stiffness Example 1 - Beam Problem The free-body diagrams for the each element are shown below.

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

Beam Stiffness Example 1 - Beam Problem Therefore, the shear force and bending moment diagrams are: CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 27/39

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:

vvii(1)(1) i i 12 6LL 12 6 22 EI 64LL 62 LL k()i   L3 12 6LL 12  6 22 62LL 64 LL

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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 28/39

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

vvv11 3 5 50

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

vvv11 3 5 50 After applying the boundary conditions the global beam equations reduce to:

24 0 6L 0 0 v2 10,000 lb 08LL22 2 0 0 0 2  EI 22 2  3 62LL 8 L 62 LL3  0 L  00 6L 240v 10,000 lb 4  22 002LL 084 0 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 29/39

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:

vv240.048 in 2340

The global forces and moments can be determined as:

Fkips1y  5 Mkipsft1  25 ꞏ

Fkips2y 10 M2  0

Fkips3y  10 M3  0

Fkips4y 10 M4  0

Fkips5y  5 Mkipsft5 25 ꞏ

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

f1y 12 6LL 12 6 0 5 kips  22  m1 EI 64LL 62 LL  0 25kft ꞏ      f 3 2y LLL12 6 12  6  0.048 5 kips 22   m2 62LL 64 LL 0 25kft ꞏ The local nodal forces for element 2:

f2y 12 6LL 12 6 0.0485 kips  22 m2 EI 64LL 62 LL 0 25kft ꞏ     f 3 3y LLL12 6 12  6 0 5 kips  22  m3 62LL 64 LL 025kft ꞏ CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 30/39

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

f3y 12 6LL 12 6 0 5 kips  22  m3 EI 64LL 62 LL  0 25kft ꞏ      f 3 4y LLL12 6 12  6  0.048 5 kips 22   m4 62LL 64 LL 0 25kft ꞏ The local nodal forces for element 4:

f4y 12 6LL 12 6 0.0485 kips  22 m4 EI 64LL 62 LL 0 25kft ꞏ     f 3 5y LLL12 6 12  6 0 5 kips  22  m5 62LL 64 LL 025kft ꞏ

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 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 31/39

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:

vv11 22 vv22 3 3 12 6LL 12 6 12 6LL 12 6 22 22 EI 64LL 62 LL EI 64LL 62 LL k(1)  k(2)   L3 12 6LL 12  6 L3 12 6LL 12  6 22 22 62LL 64 LL 62LL 64 LL The spring element stiffness matrix is:

vv334 vv 34 kk0  kk (3) k(3) 000 k    kk kk0 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 32/39

Beam Stiffness Example 3 - Beam Problem Using the direct stiffness method and superposition gives the global beam equations. Element 2 Element 1

F1y 12 6LL 12 6 0 0 0 v1  22 M 64LL 62 LL 0 00 1 1 F 12 6LL 24 0  12 6 0 v 2y 2 3 EI 22 2 kL M  62LL 08 L 6 L 2 L 0 k '  2 3  2 L  EI F3y 0012612'6LkLk  'v3   M 22 3 0062LL 6 L 4 L 0 3  F   4y 0000kk '0 'v 4 Element 3

Beam Stiffness Example 3 - Beam Problem

The boundary conditions for this problem are: vvv11 2  4 0

F1y 12 6LL 12 6 0 0 0 v01  22 M 64LL 62 LL 0 000 1 1 F 12 6LL 24 0  12 6 0 v0 2y 2 3 EI 22 2 kL M  62LL 08 L 6 L 2 L 0 k '  2 3 2 L  EI F3y 0012612'6LkLk  'v3   M 22 3 0062LL 6 L 4 L 03  F  4y 0000kk '0 'v04 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 33/39

Beam Stiffness Example 3 - Beam Problem

After applying the boundary conditions the global beam equations reduce to: 22 MLLL22862  0 EI   FLkLvP612'6    33y L3  MLLL26422  0 33 

Solving the above 31PL2    equations gives: EI12 7 k ' 2 71PL33 kL vk3   '   EI12 7 k '  EI 3 91PL2    EI12 7 k '

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:

vm3 0.0174

2 0.00249 rad

3 0.00747 rad Substituting the solution back into the global equations gives:

F1y 12 6LL 12 6 0 0 0 0  M 64LL22 62 LL 0 00 0 1  F2y 12 6LL 24 0  12 6 0 0 EI 22 2 M  6LL 2 0 8 L 6 L 2 L 0 0.00249 rad 2 3  L  F3y 0 0 12 6LkLk 12  ' 6 '  0.0174 m   M 22 3 0 0 6LL 2 6 L 4 L 0 0.00747 rad F  4y 0000kk ' 0 '0 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 34/39

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:

vm3 0.0174

2 0.00249 rad

3 0.00747 rad Substituting the solution back into the global equations gives:

F1y 69.9 kN  M 69.7 kN m 1  F2y 116.4 kN  M2  0  F 50 kN  3y  M3 0    F4y 3.5 kN

Beam Stiffness Example 3 - Beam Problem

The variation of shear force and bending moment is: 69.9kN

46.5 kN

69.7kNm

139.5kNm CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 35/39

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. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 36/39

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 Wwxvxdx distributed    0 where v(x) is the transverse displacement. The work done by the discrete nodal forces is:

Wmmfvfvnodes 11 2 2  1y 1 2 y 2

The nodal forces can be determined by setting

Wdistributed = Wnodes for arbitrary displacements and rotations. CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 37/39

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: WWdistributed nodes

L wxvxdxm m  fv fv    11 2 2 1yy 1 2 2 0

Beam Stiffness Example 4 - Load Replacement Evaluating the left-hand-side of the above expression with: wx w 21 vx() v v  x3 3212 12 LL 31 vv  2 x2   xv  2 12 12 1 1 LL gives: L Lw L2 w wv x dx v v Lwv v   12 12 21 0 24 Lw22 Lw 2 wLv 3212 1 1 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 38/39

Beam Stiffness Example 4 - Load Replacement Using a set of arbitrary nodal displacements, such as:

vv122 011 

The resulting nodal equivalent force or moment is:

L mm fvfvwxvxdx 11 2 2 1yy 1 2 2    0

222 wL2 2 L wL mL1  ww   43 2 12

Beam Stiffness Example 4 - Load Replacement Using a set of arbitrary nodal displacements, such as:

vv121 01 2 

The resulting nodal equivalent force or moment is:

L mm fvfvwxvxdx 11 2 2 1yy 1 2 2    0

wL22 wL wL 2 m2    43 12 CIVL 7/8117 Chapter 4 - Development of Beam Equations - Part 1 39/39

Beam Stiffness Example 4 - Load Replacement Setting the nodal rotations equal zero except for the nodal displacements gives: LW Lw fL wL w  1y 22 LW Lw fLw 2y 22 Summarizing, the equivalent nodal forces and moments are:

End of Chapter 4a