Chapter 4, Problem 4.155P

Assuming that $P=48000\text{lb}$ and that it may be applied at any joint on the line FJ, determine the location of P that would cause (a) maximum tension in member HI; (b) maximum compression in member CI; and (c) maximum tension in member CI. Also determine the magnitude of the indicated force in each case.

To determine

(a)

Location of force 'P' that would cause maximum tension in member HI.

Answer to Problem 4.155P

The maximum tension occurs at HI, when force 'P' acts at H.

The magnitude of maximum tension $P_{HI}$ is $48000lb$.

Explanation of Solution

Given information:

Assume $P=48000lb$.

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

Assume $E_y$ as the vertical reaction at point E.

Consider entire body

Force 'P' at point J

  $\uparrow E_y=P$

Force 'P' at point I

  $\uparrow E_y=0.75P$

Force 'P' at point H

  $\uparrow E_y=0.5P$

Force 'P' at point G

  $\uparrow E_y=0.25P$

Force 'P' at point F

  $\uparrow E_y=0$

FBD of below section

Assume $P_{CD},P_{CI},P_{HI}$ as the forces acting on member CD, CI and HI respectively.

If force 'P' acts at point J

Write equilibrium equation in vertical direction.

  $\uparrow {\displaystyle \sum F_y}=0$

  $P_{CI}=0$

For the equilibrium of above section, the bending moment about point C is equal to zero.

  ${\displaystyle \sum M_C=0}$

  $P_{HI}=0$

If force 'P' acts at point I

  $E_y-P+\left(\frac{1}{\sqrt{2}}\right)P_{CI}=0$

Solve

  $\begin{array}{l}{P}_{CI}=\sqrt{2}\left(P-E_y\right)\\ =\sqrt{2}\left(P-0.75P\right)\\ =0.353P\end{array}$

For the equilibrium of above section, the bending moment about point C is equal to zero.

  ${\displaystyle \sum M_C=0}$

  $E_y\left(2a\right)-P\left(a\right)-P_{HI}\left(a\right)=0$

Solve

  $\begin{array}{l}{P}_{HI}=2\left(0.75P\right)-P\\ =0.5P\end{array}$

If force 'P' acts at points G, H and F,

Write equilibrium equation in vertical direction.

  $\uparrow {\displaystyle \sum F_y}=0$

  $\begin{array}{l}{E}_y+\left(\frac{1}{\sqrt{2}}\right)P_{CI}=0\\ P_{CI}=-\sqrt{2}{E}_y\end{array}$

For the equilibrium of above section, the bending moment about point C is equal to zero.

  ${\displaystyle \sum M_C=0}$

  $\begin{array}{l}{E}_y\left(2a\right)-P_{HI}\left(a\right)=0\\ P_{HI}=2E_y\end{array}$

The maximum tension occurs at HI, when force 'P' acts at H.

  $\begin{array}{l}{P}_{HI}=2E_y\\ =2\left(0.5P\right)\\ =P\\ =48000lb\end{array}$

Conclusion:

The maximum tension occurs at HI, when force 'P' acts at H.

The magnitude of maximum tension $P_{HI}$ is $48000lb$.

To determine

(b)

Location of force 'P' that would cause maximum compression in member CI.

Answer to Problem 4.155P

The maximum compression occurs at CI, when force 'P' acts at H.

The magnitude of maximum compression $P_{CI}$ is $33941.12lb$.

Explanation of Solution

Given information:

Assume $P=48000lb$.

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

According to sub part a

Force 'P' at point H

  $\uparrow E_y=0.5P$

The force $P_{CI}$ in member CI

  $P_{CI}=-\sqrt{2}{E}_y$

The maximum compression occurs at CI, when force 'P' acts at H.

  $\begin{array}{l}{P}_{CI}=\sqrt{2}{E}_y\\ =\sqrt{2}\left(0.5P\right)\\ =0.7071P\\ =33941.12lb\end{array}$

Conclusion:

The maximum compression occurs at CI, when force 'P' acts at H.

The magnitude of maximum compression $P_{CI}$ is $33941.12lb$.

To determine

(c)

Location of force 'P' that would cause maximum tension in member CI

Answer to Problem 4.155P

The maximum compression occurs at CI, when force 'P' acts at I.

The magnitude of maximum tension $P_{CI}$ is $16944lb$.

Explanation of Solution

Given information:

Assume $P=48000lb$.

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

According to sub part a

If force 'P' acts at point I

  $E_y-P+\left(\frac{1}{\sqrt{2}}\right)P_{CI}=0$

Solve

  $\begin{array}{l}{P}_{CI}=\sqrt{2}\left(P-E_y\right)\\ =\sqrt{2}\left(P-0.75P\right)\\ =0.353P\\ =16944lb\end{array}$

The maximum tension occurs at member CI when force 'P' acts at point I.

Conclusion:

The maximum compression occurs at CI, when force 'P' acts at I.

The magnitude of maximum tension $P_{CI}$ is $16944lb$.