Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 9.4, Problem 90P

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

9.90 Before the load P was applied, a gap, δ0 = 0.5 mm, existed between the cantilever beam AC and the support at B. Knowing that E = 200 GPa, determine the magnitude of P for which the deflection at C is 1 mm.

Chapter 9.4, Problem 90P, Use the method of superposition to solve the following problems and assume that the flexural

Fig. P9.90

Expert Solution & Answer
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To determine

Find the magnitude of load P for the given condition using superposition method.

Answer to Problem 90P

The magnitude of load P in the beam is P=5.63kN()_.

Explanation of Solution

Given information:

The gap at the point B is δ0=0.5mm.

The modulus of elasticity of the material is E=200GPa.

The size of the square cross section is a=60mm.

The deflection at point C is δC=1mm.

Calculation:

Find the moment of inertia of the square cross section (I) using the relation.

I=a412

Here, the size of the square cross section is a.

Substitute 60 mm for a.

I=60412=1.08×106mm4×(1m1,000mm)4=1.08×106m4

Consider the portion AB of the beam.

The load P at point C will be converted into a load and moment at point B.

Show the free-body diagram of the superimposed beam AB as in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 9.4, Problem 90P , additional homework tip  1

Loading I:

The downward load P is acting at point B of the beam.

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

Write the deflection equation for concentrated load acting in a cantilever beam as follows;

y=PL33EI

Find the deflection at point B due to load P at point B as follows;

(yB)I=PL33EI

Loading II:

The upward reaction By is acting at point B of the beam.

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

Write the deflection equation for concentrated load acting in a cantilever beam as follows;

y=PL33EI

Find the deflection at point B due to reaction at point B as follows;

(yB)II=ByL33EI

Loading III:

The clockwise moment is acting at point B of the beam.

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

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

y=ML22EI

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

(yB)III=PaL22EI

Find the resultant deflection at point B as follows.

yB=(yB)I+(yB)II+(yB)III

Substitute δ0 for yB, PL33EI for (yB)I, ByL33EI for (yB)II, and PaL22EI for (yB)III.

δ0=PL33EIByL33EI+PaL22EI=1EI(PL33ByL33+PaL22)

Substitute 0.5 mm for δ0, 0.5 m for L, 0.2 m for a, 200 GPa for E, and 1.08×106m4 for I.

0.5mm×1m103mm=1200GPa×109Pa1GPa×1.08×106(P(0.5)33By(0.5)33+P(0.2)(0.5)22)0.5103=1200×109×1.08×106(0.041667P0.041667By+0.025P)108=0.06667P0.041667By (1)

Consider the beam ABC.

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

Mechanics of Materials, 7th Edition, Chapter 9.4, Problem 90P , additional homework tip  2

Loading IV:

The downward load P is acting at point C of the beam.

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

Write the deflection equation for concentrated load acting in a cantilever beam as follows;

y=PL33EI

Find the deflection at point C due to load P at point C as follows;

(yC)IV=P(L+a)33EI

Loading V:

The upward reaction By is acting at point B of the beam.

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

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

θ=PL22EIy=PL33EI

Find the deflection at point B due to reaction at point B as follows;

(yB)V=ByL33EI

Find the slope at point B due to reaction at point B as follows;

(θB)V=ByL22EI

The portion BC remains straight.

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

(yC)V=(yB)V+a(θB)V

Substitute ByL33EI for (yB)V and ByL22EI for (θB)V.

(yC)V=ByL33EIaByL22EI

Find the resultant deflection at point C as follows.

yC=(yC)IV+(yC)V

Substitute δC for yC, P(L+a)33EI for (yC)IV , and (ByL33EIaByL22EI) for (yC)V.

δC=P(L+a)33EIByL33EIaByL22EI=1EI(P(L+a)33ByL33aByL22)

Substitute 1 mm for δC, 0.5 m for L, 0.2 m for a, 200 GPa for E, and 1.08×106m4 for I.

1mm×1m103mm=1200GPa×109Pa1GPa×1.08×106(P(0.5+0.2)33By(0.5)33(0.2)By(0.5)22)216=0.11433P0.041667By0.025By216=0.11433P0.06667By (2)

Solve the Equation (1) and (2).

By=6.42×103NP=5.63×103N×1kN103N=5.63kN()

Therefore, the magnitude of load P in the beam is P=5.63kN()_.

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