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 29P

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

Chapter 9.2, Problem 29P, 9.29 and 9.30 Determine the reaction at the roller support and the deflection at point C. Fig. P9.29

Fig. P9.29

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 29P

The reaction at the roller support A is Ay=41wL128()_.

The deflection at the point C is yC=19wL46,144EI()_.

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 29P , additional homework tip  1

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

M=0MAy(x)+wx×x2=0M=Ayxwx22

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 (Ayxwx22) for M(x) in Equation (1).

d2ydx2=Ayxwx22EIEId2ydx2=Ayxwx22

Integrate the equation with respect to x;

EIdydx=Ayx22wx36+C1 (2)

Integrate the Equation (2) with respect to x.

EIy=Ayx36wx424+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 29P , additional homework tip  2

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

M=0MAy(x)+wL2×(xL4)=0MAyx+wL2(xL4)=0M=AyxwL2(xL4)

Substitute (AyxwL2(xL4)) for M(x) in Equation (1).

d2ydx2=AyxwL2(xL4)EIEId2ydx2=AyxwL2(xL4)

Integrate the equation with respect to x;

EIdydx=Ayx22wL4(xL4)2+C3 (4)

Integrate the Equation (4) with respect to x.

EIy=Ayx36wL12(xL4)3+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)36w(0)424+C1(0)+C20=00+0+C2C2=0

Boundary condition 2:

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

Substitute L for x and 0 for dydx in Equation (4).

EI(0)=Ay(L)22wL4(LL4)2+C30=AyL229wL364+C3C3=AyL22+9wL364

Boundary condition 3:

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

Equate Equation (2) and (4).

Ayx22wx36+C1=Ayx22wL4(xL4)2+C3wx36+C1=wL4(xL4)2+C3

Substitute L2  and x and (AyL22+9wL364) for C3.

w(L2)36+C1=wL4(L2L4)2AyL22+9wL364wL348+C1=wL364AyL22+9wL364C1=3wL3192AyL22+27wL3192+4wL3192=AyL22+28wL3192

Boundary condition 4:

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

Equate Equation (3) and (5).

Ayx36wx424+C1x+C2=Ayx36wLx312+wL2x216+C3x+C4wx424+C1x+C2=wLx312+wL2x216+C3x+C4

Substitute L2  for x, 0 for C2, (AyL22+28wL3192) for C1, and (AyL22+9wL364) for C3.

w(L2)424+(AyL22+28wL3192)L2+0=wL12(L2L4)3+(AyL22+9wL364)(L2)+C4wL4384AyL34+28wL4384=wL4768AyL34+9wL4128+C42wL4768+56wL4384=wL4768+54wL4768+C4C4=wL4768

Boundary condition 5:

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

Substitute L for x, 0 for y, (AyL22+9wL364) for C3, and wL4768 for C4 in Equation (5).

EI(0)=AyL36wL12(LL4)3+(AyL22+9wL364)(L)+wL47680=AyL3627wL4768AyL32+9wL464+wL47680=AyL3627wL47683AyL36+108wL4768+wL47682AyL36=82wL4768

Ay=41wL128()

Therefore, the reaction at the roller support A is Ay=41wL128()_.

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

Substitute L2 for x, yC for y, 41wL128 for Ay, (AyL22+9wL364) for C3, and wL4768 for C4 in Equation (5).

EIyC=41wL128(L2)36wL12(L2L4)3+((41wL128)L22+9wL364)(L2)+wL4768=41wL46,144wL476841wL4512+9wL4128+wL4768=41wL46,1448wL46,144492wL46,144+432wL46,144+8wL46,144yC=19wL46,144EI()

Therefore, the deflection at the point C is yC=19wL46,144EI()_.

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