Fundamentals Of Structural Analysis:
Fundamentals Of Structural Analysis:
5th Edition
ISBN: 9781260083330
Author: Leet, Kenneth
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 7, Problem 2P
To determine

Find the equations for slope and deflection for the beam by the double integration method and calculate the deflection at B and C.

Expert Solution & Answer
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Answer to Problem 2P

The equation for slope is dydx=1EI(3wLx240wL2x30wx424L)_.

The equation for deflection is y=1EI(wLx340wL2x260wx5120L)_.

The deflection of the beam at B is ΔB=0.000659wL4EI_.

The deflection of the beam at C is ΔC=0.001302wL4EI_.

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 Free Body Diagram of the beam as shown in Figure 1.

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

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about D is equal to 0.

MD=0RA×LwL23012×L×w×13L+wL220=0RAwL30wL6+wL20=0RA=9wL60

Summation of forces along y-direction is equal to 0.

+Fy=0RD+9wL6012wL=0RD=7wL20

Consider a section at a distance x from A.

Sketch the cross section as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 7, Problem 2P , additional homework tip  2

Refer to Figure 1.

Calculate the load intensity at a distance x from A.

wL=wxxwx=wLx

Calculate the moment at section x from B.

M(x)=wL230+9wL60×x12(wLx)x(13x)=3wLx20wL230wx36L

Apply double integration method as shown below.

EId2ydx2=M(x)EId2ydx2=3wLx20wL230wx36L

Integrate with respect to x.

EIdydx=3wL20x22wL230xw6Lx44+C1=3wLx240wL2x30wx424L+C1        (1)

Integrate with respect to x.

EIy=3wL40x33wL230x22w24Lx55+C1x+C2=wLx340wL2x260wx5120L+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)=3wL(0)240wL2(0)30w(0)424L+C1C1=0

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

EI(0)=wL(0)340wL2(0)260w(0)5120L+C1(0)+C2C2=0

Substitute 0 for C1 in Equation (1).

EIdydx=3wLx240wL2x30wx424L+0dydx=1EI(3wLx240wL2x30wx424L)        (3)

Hence, the equation for slope is dydx=1EI(3wLx240wL2x30wx424L)_.

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

EIy=wLx340wL2x260wx5120L+(0)x+(0)y=1EI(wLx340wL2x260wx5120L)        (4)

Hence, the equation for deflection is y=1EI(wLx340wL2x260wx5120L)_.

Calculate the deflection at B as shown below.

Substitute L4 for x in Equation (3).

yB=1EI(wL40(L4)3wL260(L4)2w120L(L4)5)ΔB=wL4120EI(36421611024)   =wL4120EI(4812811024)   =0.000659wL4EI

Hence, the deflection of the beam at B is ΔB=0.000659wL4EI_.

Calculate the deflection at C as shown below.

Substitute L2 for x in Equation (3).

yC=1EI(wL40(L2)3wL260(L2)2w120L(L2)5)ΔC=wL4120EI(3824132)   =wL4120EI(1216132)   =0.001302wL4EI

Therefore, the deflection of the beam at C is ΔC=0.001302wL4EI_.

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