Problem

Draw the shear and moment diagrams for the beam ABCDE All pulleys have a radius of 1 ft....

Draw the shear and moment diagrams for the beam ABCDE All pulleys have a radius of 1 ft. Neglect the weight of the beam and pulley arrangement. The load weighs 500 lb.

Step-by-Step Solution

Solution 1

Step #1 of 9

Radius of pulley, $r=1 \mathrm{ft}$

Weight of load, $w=500 \mathrm{lb}$

Let tension in the cable be $T$,

Let reaction force at $A$ be $R_{A}$

Let reaction force at $B$ be $R_{E}$

Step #2 of 9

Free body diagram of the beam:

Step #3 of 9

Considering the moment about $A$, we get

$$ \begin{aligned} \sum M_{A} &=0 \\ \left(R_{E} \times 15\right) &=(500 \times 7)+(500 \times 3) \\ R_{E} &=333.33 \mathrm{lb} \end{aligned} $$

By applying the equation of equilibrium in $y$-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A}+R_{E} &=500 \\ R_{A} &=500-R_{E} \\ R_{A} &=500-333.33 \\ R_{A} &=166.67 \mathrm{lb} \end{aligned} $$

Step #4 of 9

Free body diagram of the beam without pulleys:

Step #5 of 9

For $A B$ :

By applying the equation of equilibrium in $y$-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A}-v &=0 \\ v &=166.67 \mathrm{lb} \end{aligned} $$

Considering the moment, we get

$$ \begin{aligned} \sum M_{A} &=0 \\ M &=R_{A} \times x \\ M &=166.67 \times x \end{aligned} $$

Step #6 of 9

For $B C$ :

By applying the equation of equilibrium in $y$-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A} &=1000+v \\ 166.67 &=1000+v \\ v &=-833.33 \mathrm{lb} \end{aligned} $$

Considering the moment, we get

$$ \begin{aligned} \sum M &=0 \\ M &=\left(R_{A} \times x\right)-1000(x-8) \\ M &=166.67 x-1000 x+8000 \\ M &=-833.33 x+8000 \end{aligned} $$

Step #7 of 9

For $C D:$

By applying the equation of equilibrium in $y$-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A} &=1000-1000+v \\ v &=R_{A} \\ v &=166.67 \mathrm{lb} \end{aligned} $$

Considering the moment, we get

$$ \begin{aligned} \sum M_{A} &=0 \\ M &=\left(R_{A} \times x\right)-1000(x-8)+1000(x-10) \\ M &=166.67 x-1000 x+8000+1000 x-10000 \\ M &=166.67 x-2000 \end{aligned} $$

Step #8 of 9

For $D E$ :

By applying the equation of equilibrium in $y$-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A} &=1000-1000+500+v \\ v &=166.67-500 \\ v &=-333.33 \mathrm{lb} \end{aligned} $$

Considering the moment, we get

$$ \begin{aligned} \sum M_{A} &=0 \\ M &=\left(R_{A} \times x\right)-1000(x-8)+1000(x-10)-500(x-12)+1000 \\ M &=166.67 x-1000 x+8000+1000 x-10000-500 x+6000+1000 \\ M &=-333.33 x+5000 \end{aligned} $$

Step #9 of 9

Shear and moment diagrams: