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 9.3, Problem 60P

9.59 through 9.62 For the beam and loading indicated, determine the magnitude and location of the largest downward deflection.

9.60 Beam and loading of Prob. 9.46.

9.46 For the beam and loading shown, determine (a) the slope at end A, (b) the deflection at point B. Use E = 29 × 106 psi.

Chapter 9.3, Problem 60P, 9.59 through 9.62 For the beam and loading indicated, determine the magnitude and location of the

Fig. P9.46

Expert Solution & Answer
Check Mark
To determine

Find the magnitude and location of the largest downward deflection of the beam.

Answer to Problem 60P

The location of the largest downward deflection is xm=26.4in._.

The largest downward deflection of the beam is ym=0.1520in.()_.

Explanation of Solution

Given information:

The modulus of elasticity of the material is E=29×106psi.

Calculation:

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

Mechanics of Materials, 7th Edition, Chapter 9.3, Problem 60P

Write the singularity equation for load intensity as follows;

dVdx=w(x)=10x240+10x400

Integrate the equation to find the shear force.

V=Ay10x241+10x401

By definition, the change in bending moment with respect to change in distance is shear force.

dMdx=V=Ay10x241+10x401200x400

Integrate the equation to find the bending moment.

M=Ayx5x242+5x402200x401 (1)

Write the second order differential equation as follows;

d2ydx2=M(x)EI

Here, the moment at the corresponding section is M(x), the modulus of elasticity of the material is E, and the moment of inertia of the section is I.

Substitute (Ayx5x242+5x402200x401) for M(x).

d2ydx2=Ayx5x242+5x402200x401EIEId2ydx2=Ayx5x242+5x402200x401

Integrate the equation with respect to x;

EIdydx=Ayx2253x243+53x403100x402+C1 (2)

Integrate the Equation (2) with respect to x.

EIy=Ayx36512x244+512x4041003x403+C1x+C2 (3)

Boundary condition 1:

At the point D; x=48in.;M=0.

Substitute 48 in. for x and 0 for M in Equation (1).

0=Ay(48)548242+548402200484010=48Ay2,880+3201,600Ay=86.667lb

Boundary condition 2:

At the point A; x=0;y=0.

Substitute 86.667 lb for Ay, 0 for x and 0 for y in Equation (3).

EI(0)=86.667(0)365120244+512040410030403+C1(0)+C20=00+00+0+C2C2=0

Boundary condition 3:

At the point D; x=48in.;y=0.

Substitute 86.667 lb for Ay, 48 in. for x, 0 for y, and 0 for C2 in Equation (3).

EI(0)=86.667(48)3651248244+51248404100348403+C1(48)+00=1,597,446.144138,240+1,706.6666717,066.66667+48C1C1=30,080.128lb-in.2

Determine the moment of inertia (I) of the circular cross section using the equation.

I=πd464

Here, the diameter of the circular cross section is d.

Substitute 1.25 in. for d.

I=π×1.25464=0.119842in.4

At point A; x=0;dydx=θA.

Substitute 29×106psi for E, 0.119842in.4 for I, θA for dydx, 86.667 lb for Ay, 0 for x, and 30,080.128lb-in.2 for C1 in Equation (2).

29×106×0.119842×θA=86.667×(0)22530243+530403100040230,080.12829×106×0.119842×θA=00+0030,080.128θA=8.66×103rad

At point B; x=24in.;dydx=θB.

Substitute 29×106psi for E, 0.119842in.4 for I, θB for dydx, 86.667 lb for Ay, 24 in. for x, and 30,080.128lb-in.2 for C1 in Equation (2).

29×106×0.119842×θB={86.667×(24)225324243+53244031002440230,080.128}29×106×0.119842×θB=24,960.0960+0030,080.128θB=1.473×103rad

At point C; x=40in.;dydx=θC.

Substitute 29×106psi for E, 0.119842in.4 for I, θC for dydx, 86.667 lb for Ay, 40 in. for x, and 30,080.128lb-in.2 for C1 in Equation (2).

29×106×0.119842×θC={86.667×(40)225340243+53404031004040230,080.128}29×106×0.119842×θC=69,333.66,826.667+0030,080.128θC=9.33×103rad

At point D; x=48in.;dydx=θD.

Substitute 29×106psi for E, 0.119842in.4 for I, θD for dydx, 86.667 lb for Ay, 48 in. for x, and 30,080.128lb-in.2 for C1 in Equation (2).

29×106×0.119842×θD={86.667×(48)225348243+53484031004840230,080.128}29×106×0.119842×θD=99,840.38423,040+853.3336,40030,080.128θD=11.847×103rad

The slope changes from negative to positive in the section BC.

The maximum deflection occurs where the slope changes sign. i.e., dydx=0.

Substitute 0 for dydx, 86.667 lb for Ay, and 30,080.128lb-in.2 for C1 in Equation (2).

EI(0)=86.667xm2253xm243+53xm403100xm40230,080.1280=43.333xm253(xm24)3+0030,080.1280=43.333xm253(xm24)330,080.128

Solve the Equation;

xm=26.4in..

Therefore, the location of the largest downward deflection is xm=26.4in._.

At largest deflection point; xm=26.4in.;y=ym.

Substitute 29×106psi for E, 0.119842in.4 for I, yB for y, 86.667 lb for Ay, 26.4 in. for x, 30,080.128lb-in.2 for C1, and 0 for C2 in Equation (3).

29×106×0.119842×ym={86.667×26.43651226.4244+51226.4404100326.440330,080.128(26.4)+0}=265,775.102213.824+00794,115.3792ym=0.1520in.=0.1520in.()

Therefore, the largest downward deflection of the beam is ym=0.1520in.()_.

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