EBK STRUCTURAL ANALYSIS
EBK STRUCTURAL ANALYSIS
10th Edition
ISBN: 8220106821596
Author: HIBBELER
Publisher: PEARSON
Textbook Question
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Chapter 7, Problem 7.1P

Determine the equation of the elastic curve using the coordinate x, and specify the slope at point A and the deflection at point C. EI is constant.

Chapter 7, Problem 7.1P, Determine the equation of the elastic curve using the coordinate x, and specify the slope at point A

Expert Solution & Answer
Check Mark
To determine

The equation of elastic curve using the coordinate x and to specify the slope at point A and the deflection at point C.

Answer to Problem 7.1P

The equation of elastic curve is shown below.

EIdvdx=wL2×x22w2×x33wL324

EIv=wL4×x33w6×x44wL324x

The slope at point A is, wL324EI.

The deflection at C is, 5wL4384EI.

Explanation of Solution

Calculation:

The following figure shows the free body diagram of the beam.

EBK STRUCTURAL ANALYSIS, Chapter 7, Problem 7.1P

Figure-(1)

Write the Equilibrium Equation for the sum of horizontal forces.

ΣFx=0Ax=0

Here, horizontal reaction at A is Ax.

Write the Equilibrium Equation for the sum of vertical forces.

ΣFy=0Ay+BywL=0

Here, vertical reactions at point A and B are Ay and By.

Due to symmetry of the beam, the reactions at point A and B will be half of the total load acting on the beam.

Ay=wL2By=wL2

Consider the section x-x at distance x from point A as shown below.

Write the Equation for sum of moment about x-x.

ΣMx=0MwL2x+wxx2=0M=wL2xwxx2 ...... (I).

Write the differential equation of the elastic curve as shown below.

EId2vdx2=M ...... (II)

Substitute wL2xwxx2 for M in Equation (II) and integrate it twice with respect to the x.

EIdvdx=wL2×x22w2×x33+C1 ...... (III).

EIv=wL4×x33w6×x44+C1x+C2 ...... (IV).

Here, C1 and C2 are the constant.

Calculate the value of C2.

Apply the boundary conditions at the support points.

At x=0 and v=0.

Substitute 0 for x and v in equation (IV).

EI(0)=wL4×( 0)33w6×( 0)44+C1(0)+C2C2=0

Calculate the value of C1.

At x=L and v=0.

Substitute L for x and 0 for C2 and v in equation (IV).

EI(0)=wL4×( L)33w6×( L)44+C1(L)+0C1=wL324

Calculate the slope equation.

Substitute wL324 for C1, θ for dvdx in equation (III).

θ=1EI(wL2×x22w2×x33wL324)

   ...... (V)

Calculate the deflection equation.

Substitute 0 for C2 and wL324 for C1 in Equation (IV).

v=1EI(wL4×x33w6×x44wL324x) ...... (VI).

Calculate the slope at A.

Substitute 0 for x in Equation (V).

θA=1EI(wL2× ( 0 ) 22w2× ( 0 ) 33w L 324)=wL324EI

Calculate the deflection at C.

Substitute L2 for x in Equation (VI).

vC=1EI(wL4× ( L 2 ) 33w6× ( L 2 ) 44w L 324×L2)=5wL4384EI

Conclusion:

The equation of elastic curve is shown below.

EIdvdx=wL2×x22w2×x33wL324

EIv=wL4×x33w6×x44wL324x

The slope at point A is, wL324EI.

The deflection at C is, 5wL4384EI.

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