Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
5th Edition
ISBN: 9780073398006
Author: Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher: McGraw-Hill Education
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Chapter 7, Problem 1P
To determine

Find the equations for slope and deflection for the beam by the double integration method and compare the deflection at B with the deflection at midspan.

Expert Solution & Answer
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Answer to Problem 1P

The equation for slope of the beam is dydx=16EI(3wL2x3wLx2+wx3)_.

The equation for deflection of the beam is y=16EI(32wL2x2wLx3+wx44)_.

The deflection at midspan is equal to 0.35 times the deflection at B.

Explanation of Solution

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

  • For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
  • For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
  • For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as positive and the counter clockwise moment as negative.

Sketch the deflected shape of the beam as shown in Figure 1.

Fundamentals of Structural Analysis, Chapter 7, Problem 1P , additional homework tip  1

Refer to Figure 1.

Consider a section at a distance x from A.

Sketch the cross section from B as shown in Figure 2.

Fundamentals of Structural Analysis, Chapter 7, Problem 1P , additional homework tip  2

Calculate the moment at the section (Lx) from B.

MLx=0M(x)w×(Lx)×(Lx)2=0M(x)=w(Lx)22

Apply double integration method as shown below.

EId2ydx2=M(x)=w(Lx)22=w2(L22Lx+x2)

Integrate with respect to x.

EIdydx=w2(L2x2Lx22+x33)+C1=w6(3L2x3Lx2+x3)+C1        (1)

Integrate with respect to x.

EIy=w6(3L2x223Lx33+x44)+C1x+C2=w6(32L2x2Lx3+x44)+C1x+C2        (2)

Apply the boundary conditions as shown below.

i) At a distance x=0 the slope dydx=0.

ii) At a distance x=0 the deflection y=0.

Apply boundary condition (i) in Equation (1).

EI(0)=w6(3L2(0)3L(0)2+(0)3)+C1C1=0

Apply boundary condition (ii) in Equation (2).

EI(0)=w6(32L2(0)2L(0)3+(0)44)+C1(0)+C2C2=0

Substitute 0 for C1 in Equation (1).

EIdydx=w6(3L2x3Lx2+x3)+0dydx=16EI(3wL2x3wLx2+wx3)        (3)

Hence, the equation for slope of the beam is dydx=16EI(3wL2x3wLx2+wx3)_.

Substitute 0 for C1 and 0 for C2 in Equation (2).

EIy=w6(32wL2x2wLx3+wx44)+(0)x+(0)y=16EI(32wL2x2wLx3+wx44)        (4)

Hence, the equation for deflection of the beam is y=16EI(32wL2x2wLx3+wx44)_.

Calculate the deflection at B as shown below.

Substitute L for x in Equation (4).

yB=16EI(32wL2×L2wL×L3+w×L44)ΔB=wL46EI(64+14)    =wL48EI

Calculate the deflection at midspan as shown below.

Substitute L2 for x in Equation (4).

yL2=16EI(32wL2×(L2)2wL×(L2)3+w4(L2)4)ΔL2=wL46EI(32×1418+14×116)    =wL46EI(16+164)    =17wL4384EI

Compare the deflection at B with the deflection at midspan as shown below.

ΔL2ΔB=17wL4384EIwL48EI=0.35

Therefore, the deflection at midspan is equal to 0.35 times the deflection at B.

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