Chapter 6, Problem 6.15P

The structure is supported by a pin at C and a cable attached to A. The cable runs over the small pulley D. Find the internal force systems acting on sections 1 and 2.

To determine

Internal force system acting on section 1 and 2

Answer to Problem 6.15P

The horizontal force $P_1$ at section 1 is $3411.6lb$ to the left.

The vertical force $V_1$ at section 1 is $384.15lb$ downwards.

The bending moment $M_1$ at section 1 is $13615.62lb.ft$ clockwise.

The horizontal force $P_2$ at section 2 is equal to zero.

The vertical force $V_2$ at section 2 is $1866.67lb$ downwards.

The bending moment $M_2$ at section 2 is $16800lb.ft$ clockwise.

Explanation of Solution

Given information:

If all external forces are known, we can use equilibrium analysis to find internal forces.

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:

FBD of entire body

Assume $C_x,C_y$ as the reaction force on point C and 'T' as the tension on the cable.

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

  ${\displaystyle \sum M_C=0}$

  $\begin{array}{l}\left(2800lb\right)\left(15ft\right) - \left(9ft\right)T=0\\ T=4666.67lb\end{array}$

FBD of section 1

Assume $P_1,V_1,M_1$ as the horizontal force, vertical force and bending moment about point 1.

For the equilibrium of section AB, the bending moment about point B is equal to zero.

  ${\displaystyle \sum M_B=0}$

  $\begin{array}{l}\left(2800lb\right)\left(6ft\right) -\left(\frac{5.6}{\sqrt{{5.6}^2+6^2}}ft\right)\left(4666.67lb\right)+M_1=0\\ M_1=-13615.62lb.ft\end{array}$

Write equilibrium equation in horizontal direction.

  $\to {\displaystyle \sum F_x}=0$

  $\begin{array}{l}\left(\frac{6}{\sqrt{{5.6}^2+6^2}}ft\right)\left(4666.67lb\right)+P_1=0\\ P_1=-3411.6lb\end{array}$

Write equilibrium equation in vertical direction.

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

  $\begin{array}{l}\left(\frac{5.6}{\sqrt{{5.6}^2+6^2}}ft\right)\left(4666.67lb\right)-V_1-2800lb=0\\ V_1=384.15lb\end{array}$

FBD of section 2

Assume $P_2,V_2,M_2$ as the horizontal force, vertical force and bending moment about point 2.

For the equilibrium of section AB, the bending moment about point B is equal to zero.

  ${\displaystyle \sum M_B=0}$

  $\begin{array}{l}\left(2800lb\right)\left(6ft\right) +M_2=0\\ M_2=-16800lb.ft\end{array}$

Write equilibrium equation in horizontal direction.

  $\to {\displaystyle \sum F_x}=0$

  $P_2=0$

Write equilibrium equation in vertical direction.

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

  $\begin{array}{l}4666.67lb-V_2-2800lb=0\\ V_2=1866.67lb\end{array}$

Conclusion:

The horizontal force $P_1$ at section 1 is $3411.6lb$ to the left.

The vertical force $V_1$ at section 1 is $384.15lb$ downwards.

The bending moment $M_1$ at section 1 is $13615.62lb.ft$ clockwise.

The horizontal force $P_2$ at section 2 is equal to zero.

The vertical force $V_2$ at section 2 is $1866.67lb$ downwards.

The bending moment $M_2$ at section 2 is $16800lb.ft$ clockwise.