EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 9.4, Problem 83P

Use the method of superposition to solve the following problems and assume that the flexural rigidity El of each beam is constant.

9.83 and 9.84 For the beam shown, determine the reaction at B.

Chapter 9.4, Problem 83P, Use the method of superposition to solve the following problems and assume that the flexural

Fig. P9.83

Expert Solution & Answer
Check Mark
To determine

Find the reaction at point B of the beam using superposition method.

Answer to Problem 83P

The vertical reaction at point B is By=13wL32()_.

The moment at point B is MB=11wL2192(Clockwise)_.

Explanation of Solution

The flexural rigidity of the beam is EI.

Convert the beam into two sections as in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 9.4, Problem 83P , additional homework tip  1

Consider portion AC of the beam:

Show the free-body diagram of the superimposed beam as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 9.4, Problem 83P , additional homework tip  2

Loading I:

The downward reaction Cy is action at point C.

Refer to case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for point load acting in a cantilever beam as follows;

y=PL33EIθ=PL22EI

Find the deflection at point C due to vertical reaction as follows;

(yC)I=Cy(L2)33EI=CyL324EI

Find the slope at point C due to vertical reaction as follows;

(θC)I=Cy(L2)22EI=CyL28EI

Loading II:

The counter-clockwise moment is acting at point C.

Refer to case 3 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for moment acting in a cantilever beam as follows;

y=ML22EIθ=MLEI

Find the deflection at point C due to moment as follows;

(yC)II=MC(L2)22EI=MCL28EI

Find the slope at point C due to moment as follows;

(θC)II=MC(L2)EI=MCL2EI

Find the deflection at point C as follows;

(yC)AC=(yC)I+(yC)II

Substitute CyL324EI for (yC)I and MCL28EI for (yC)II.

(yC)AC=CyL324EI+MCL28EI (1)

Find the slope at point C as follows;

(θC)AC=(θC)I+(θC)II

Substitute CyL28EI for (θC)I and MCL2EI for (θC)II.

(θC)AC=CyL28EI+MCL2EI (2)

Consider portion CB of the beam:

Show the free-body diagram of the superimposed beam as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 9.4, Problem 83P , additional homework tip  3

Loading III:

The upward reaction Cy is action at point C.

Refer to case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for point load acting in a cantilever beam as follows;

y=PL33EIθ=PL22EI

Find the deflection at point C due to vertical reaction as follows;

(yC)III=Cy(L2)33EI=CyL324EI

Find the slope at point C due to vertical reaction as follows;

(θC)III=Cy(L2)22EI=CyL28EI

Loading IV:

The downward uniformly distributed load (udl) is spread throughout the portion BC.

Refer to case 2 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for udl spread throughout the cantilever beam as follows;

y=wL48EIθ=wL36EI

Find the deflection at point C due to udl as follows;

(yC)IV=w(L2)48EI=wL4128EI

Find the slope at point C due to udl as follows;

(θC)IV=w(L2)36EI=wL348EI

Loading IV:

The clockwise moment is acting at point C.

Refer to case 3 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for moment acting in a cantilever beam as follows;

y=ML22EIθ=MLEI

Find the deflection at point C due to moment as follows;

(yC)V=MC(L2)22EI=MCL28EI

Find the slope at point C due to moment as follows;

(θC)V=MC(L2)EI=MCL2EI

Find the deflection at point C as follows;

(yC)CB=(yC)III+(yC)IV+(yC)V

Substitute CyL324EI for (yC)III, wL4128EI for (yC)IV, and MCL28EI for (yC)V.

(yC)CB=CyL324EIwL4128EI+MCL28EI (3)

Find the slope at point C as follows;

(θC)CB=(θC)III+(θC)IV+(θC)V

Substitute CyL28EI for (θC)III, wL348EI for (θC)IV, and MCL2EI for (θC)V.

(θC)CB=CyL28EI+wL348EIMCL2EI (4)

The resultant deflection at point C is equal.

(yC)AC=(yC)CB

Equate the equations (1) and (3).

CyL324EI+MCL28EI=CyL324EIwL4128EI+MCL28EI2CyL324EI=wL4128EICy=3wL32

The resultant slope at point C is equal.

(θC)AC=(θC)CB

Equate the equations (2) and (4).

CyL28EI+MCL2EI=CyL28EI+wL348EIMCL2EI2MCL2EI=wL348EIMC=wL248

Show the free-body diagram of the portion CB as in Figure 4.

EBK MECHANICS OF MATERIALS, Chapter 9.4, Problem 83P , additional homework tip  4

Find the vertical reaction at point B by resolving the vertical component of forces.

Fy=0Cyw(L2)+By=03wL3216wL32+By=0By=13wL32()

Find the moment at point B by taking moment about point B.

MB=0wL2483wL32(L2)MB+w(L2)(L4)=0wL2483wL264MB+wL28=04wL21929wL2192MB+24wL2192=0

MB=11wL2192(Clockwise)

Therefore,

The vertical reaction at point B is By=13wL32()_.

The moment at point B is MB=11wL2192(Clockwise)_.

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Chapter 9 Solutions

EBK MECHANICS OF MATERIALS

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