Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Textbook Question
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Chapter 9, Problem 9.5.23P

The overhanging beam A BCD supports two concentrated loads P and Q (see figure),

  1. For what ratio PIQ will the deflection at point B be zero?
  2. For what ratio will the deflection at point D be zero?
  3. If Q is replaced by a uniform load with intensity q (on the overhang), repeat parts (a) and (b), but find ratio Pl(qa).

  Chapter 9, Problem 9.5.23P, The overhanging beam A BCD supports two concentrated loads P and Q (see figure), For what ratio PIQ

(a)

Expert Solution
Check Mark
To determine

Ratio P/Q for which deflection at B is zero .

Answer to Problem 9.5.23P

Ratio P/Q for which deflection at B is zero is P/Q=9a/4L.

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.23P , additional homework tip  1

The deflection at B is zero.

Calculation:

Consider the following free body diagram,

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.23P , additional homework tip  2

Take equilibrium of forces in horizontal direction as,

  Fx=0Ax=0

Take equilibrium of forces in vertical direction as,

  Fy=0Cy=P+Q

Take equilibrium of moments about A as,

  MA=0MAPL/2+CyLQ(L+a)=0MA=PL/2CyL+Q(L+a)MA=PL/2(P+Q)L+Q(L+a)

The bending moment at distance x from point A is given by,

  M={MA                                                xL/2MAP(xL/2)                          xLMAP(xL/2)+Cy(xL)       xL+a

The deflection and bending moment is related by following differential equation

  d2vdx2=MEI          ....(1)

Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

  θ=dvdx=1EI{MAx+C1                                                            xL/2MAxP(xL/2)2/2+C2                                xLMAxP(xL/2)2/2+Cy(xL)2/2+C3       xL+a

Integrate angle of rotation with respect to x get deflections as,

  δ=v=1EI{MAx2/2+C1x+C4                                                            xL/2MAx2/2P(xL/2)3/6+C2x+C5                                xLMAx2/2P(xL/2)3/6+Cy(xL)3/6+C3x+C6       xL+a

The following conditions are used to evaluate integration constants,

  limx(L/2)θ=limx(L/2)+θ                  ....(1)limx(L)θ=limx(L)+θ                ....(2)limx(L/2)δ=limx(L/2)+δ                  ....(3)limx(L)δ=limx(L)+δ                ....(4)θ(0)=0                               ....(5)δ(L)=0                             ....(6)

Since deflection at B is zero, hence

  δB=δ(L/2)=0

Now substitute values of constants and solve the above equation to get

  P/Q=9a/4L .

Conclusion:

Therefore, the ratio P/Q for which deflection at B is zero is P/Q=9a/4L .

(b)

Expert Solution
Check Mark
To determine

Ratio P/Q for which deflection at D is zero .

Answer to Problem 9.5.23P

Ratio P/Q for which deflection at D is zero is P/Q=8a(3L+a)/9L2.

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.23P , additional homework tip  3

The deflection at D is zero.

Calculation:

Consider the following free body diagram,

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.23P , additional homework tip  4

Take equilibrium of forces in horizontal direction as,

  Fx=0Ax=0

Take equilibrium of forces in vertical direction as,

  Fy=0Cy=P+Q

Take equilibrium of moments about A as,

  MA=0MAPL/2+CyLQ(L+a)=0MA=PL/2CyL+Q(L+a)MA=PL/2(P+Q)L+Q(L+a)

The bending moment at distance x from point A is given by,

  M={MA                                                xL/2MAP(xL/2)                          xLMAP(xL/2)+Cy(xL)       xL+a

The deflection and bending moment is related by following differential equation

  d2vdx2=MEI          ....(1)

Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

  θ=dvdx=1EI{MAx+C1                                                            xL/2MAxP(xL/2)2/2+C2                                xLMAxP(xL/2)2/2+Cy(xL)2/2+C3       xL+a

Integrate angle of rotation with respect to x get deflections as,

  δ=v=1EI{MAx2/2+C1x+C4                                                            xL/2MAx2/2P(xL/2)3/6+C2x+C5                                xLMAx2/2P(xL/2)3/6+Cy(xL)3/6+C3x+C6       xL+a

The following conditions are used to evaluate integration constants,

  limx(L/2)θ=limx(L/2)+θ                  ....(1)limx(L)θ=limx(L)+θ                ....(2)limx(L/2)δ=limx(L/2)+δ                  ....(3)limx(L)δ=limx(L)+δ                ....(4)θ(0)=0                               ....(5)δ(L)=0                             ....(6)

Since deflection at D is zero, hence

  δD=δ(L+a)=0

Now substitute values of constants and solve the above equation to get

  P/Q=8a(3L+a)/9L2 .

Conclusion:

Therefore, the ratio P/Q for which deflection at D is zero is P/Q=8a(3L+a)/9L2 .

(c)

Expert Solution
Check Mark
To determine

Ratio P/Q for which deflection at B and D is zero .

Answer to Problem 9.5.23P

  P/qa=9a/8L for δB=0 and P/qa=a(4L+a)/3L2 for δD=0 .

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.23P , additional homework tip  5

The deflection at B and D is zero.

Calculation:

Consider the following free body diagram,

  Mechanics of Materials (MindTap Course List), Chapter 9, Problem 9.5.23P , additional homework tip  6

Take equilibrium of forces in horizontal direction as,

  Fx=0Ax=0

Take equilibrium of forces in vertical direction as,

  Fy=0Cy=P+qa

Take equilibrium of moments about A as,

  MA=0MAPL/2+CyLqa(L+a/2)=0MA=[PL/2+CyLqa(L+a/2)]

The bending moment at distance x from point A is given by,

  M={MA                                                                      xL/2MAP(xL/2)                                                xLMAP(xL/2)+Cy(xL)q(xL)2/2       xL+a

The deflection and bending moment is related by following differential equation

  d2vdx2=MEI          ....(1)

Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

  θ=dvdx=1EI{MAx+C1                                                                                 xL/2MAxP(xL/2)2/2+C2                                                      xLMAxP(xL/2)2/2+Cy(xL)2/2q(xL)3/6+C3       xL+a

Integrate angle of rotation with respect to x get deflections as,

  δ=v=1EI{MAx2/2+C1x+C4                                                                                     xL/2MAx2/2P(xL/2)3/6+C2x+C5                                                          xLMAx2/2P(xL/2)3/6+Cy(xL)3/6q(xL)4/24+C3x+C6       xL+a

The following conditions are used to evaluate integration constants,

  limx(L/2)θ=limx(L/2)+θ                  ....(1)limx(L)θ=limx(L)+θ                ....(2)limx(L/2)δ=limx(L/2)+δ                  ....(3)limx(L)δ=limx(L)+δ                ....(4)θ(0)=0                               ....(5)δ(L)=0                             ....(6)

Substitute values of constants to get the expression for deflection.

Since deflection at B is zero, hence

  δB=δ(L/2)=0

Solve the above equation to get

  P/qa=9a/8L .

For deflection at D is zero,

  δD=δ(L+a)=0

Solve the above equation to get

  P/qa=a(4L+a)/3L2 .

Conclusion:

Therefore, the ratio P/Q P/qa=9a/8L for δB=0 and P/qa=a(4L+a)/3L2 for δD=0 

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Chapter 9 Solutions

Mechanics of Materials (MindTap Course List)

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