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 9.2, Problem 30P

9.29 and 9.30 Determine the reaction at the roller support and the deflection at point C.

Chapter 9.2, Problem 30P, 9.29 and 9.30 Determine the reaction at the roller support and the deflection at point C. Fig. P9.30

Fig. P9.30

Expert Solution & Answer
Check Mark
To determine

Find the reaction at the roller support and the deflection at point C of the beam.

Answer to Problem 30P

The reaction at the roller support B is By=17wL64()_.

The deflection at point C of the beam is yC=wL41,024EI()_.

Explanation of Solution

Consider a section at a distance x from left end A of the section AC.

Show the free-body diagram of the section AC as in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 9.2, Problem 30P , additional homework tip  1

Determine the moment at the section by taking moment about the section.

M=0M+MAAy(x)wx×x2=0M=Ayx+wx22MA

Write the second order differential equation as follows;

d2ydx2=M(x)EI (1)

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 (Ayx+wx22MA) for M(x) in Equation (1).

d2ydx2=Ayx+wx22MAEIEId2ydx2=Ayx+wx22MA

Integrate the equation with respect to x;

EIdydx=Ayx22+wx36MAx+C1 (2)

Integrate the Equation (2) with respect to x.

EIy=Ayx36+wx424MAx22+C1x+C2 (3)

Show the free-body diagram of the section BC as in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 9.2, Problem 30P , additional homework tip  2

Determine the moment at the section by taking moment about the section.

M=0M+MAAy(x)wL2(xL4)+w2(xL2)2=0M=MA+Ayx+wL2(xL4)w2(xL2)2

Substitute (MA+Ayx+wL2(xL4)w2(xL2)2) for M(x) in Equation (1).

d2ydx2=MA+Ay(x)+wL2(xL4)w2(xL2)2EIEId2ydx2=MA+Ay(x)+wL2(xL4)w2(xL2)2

Integrate the equation with respect to x;

EIdydx=MAx+Ayx22+wL4(xL4)2w6(xL2)3+C3 (4)

Integrate the Equation (4) with respect to x.

EIy=MAx22+Ayx36+wL12(xL4)3w24(xL2)4+C3x+C4 (5)

Boundary condition 1:

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

Substitute 0 for x and 0 for y in Equation (3).

EI(0)=Ay(0)36+w(0)424MA(0)22+C1(0)+C20=0+00+0+C2C2=0

Boundary condition 2:

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

Substitute 0 for x and 0 for dydx in Equation (2).

EI(0)=Ay(0)22+w(0)36MA(0)+C10=0+00+C1C1=0

Boundary condition 3:

At the point C; x=L2;dydx=dydx.

Equate Equation (2) and (4).

Ayx22+wx36MAx+C1=MAx+Ayx22+wL4(xL4)2w6(xL2)3+C3wx36+C1=wL4(xL4)2w6(xL2)3+C3

Substitute L2  and x and 0 for C1.

w(L2)36+0=wL4(L2L4)2w6(L2L2)3+C3wL348=wL3640+C3C3=4wL31923wL3192=wL3192

Boundary condition 4:

At the point C; x=L2;y=y.

Equate Equation (3) and (5).

Ayx36+wx424MAx22+C1x+C2=MAx22+Ayx36+wL12(xL4)3w24(xL2)4+C3x+C4wx424+C1x+C2=wL12(xL4)3w24(xL2)4+C3x+C4

Substitute L2  for x, 0 for C1, 0 for C2, and wL3192 for C3.

w(L2)424+(0)(L2)+0=wL12(L2L4)3w24(L2L2)4+wL3192(L2)+C4wL4384=wL47680+wL4384+C4C4=wL4768

Boundary condition 5:

At the point B; x=L;y=0.

Substitute L for x, 0 for y, wL3192 for C3, and wL4768 for C4 in Equation (5).

EI(0)=MAL22+AyL36+wL12(LL4)3w24(LL2)4+wL3192(L)wL47680=MAL22+AyL36+27wL4768wL4384+wL4192wL47680=MAL22+AyL36+27wL47682wL4768+4wL4768wL47680=MAL22+AyL36+28wL4768 (6)

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

Mechanics of Materials, 7th Edition, Chapter 9.2, Problem 30P , additional homework tip  3

Resolve the vertical component of forces as follows;

Fy=0Ay+By+wL2wL2=0Ay=By (7)

Take moment about the point A as follows;

MA=0MA+wL2(L4)wL2(3L4)+By(L)=0MA+wL283wL28+ByL=0MA=wL24ByL (8)

Substitute (wL24ByL) for MA and (By) for Ay in Equation (6).

(wL24ByL)L22+ByL36+28wL4768=0wL48+ByL32ByL36+7wL4192=024wL4192+3ByL36ByL36+7wL4192=0

2ByL36=17wL4192By=17wL64()

Therefore, the reaction at the roller support B is By=17wL64()_.

Substitute 17wL64 for By in Equation (7).

Ay=17wL64=17wL64()

Substitute 17wL64 for By in Equation (8).

MA=wL2417wL64(L)=16wL26417wL264=wL264(Clockwise)

At point C; x=L2;y=yC.

Substitute L2 for x, yC for y, wL3192 for C3, 17wL64 for Ay, wL264 for MA, and wL4768 for C4 in Equation (5).

EIyC=[(wL264)(L2)22+(17wL64)(L2)36+wL12(L2L4)3w24(L2L2)4+wL3192(L2)wL4768]=wL451217wL43,072+wL47680+wL4384wL4768=6wL43,07217wL43,072+8wL43,072=3wL43,072

yC=wL41,024EI()

Therefore, the deflection at point C of the beam is yC=wL41,024EI()_.

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