Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Answer to Problem 29P

The horizontal reaction at A is $\underset{\_}{A_x=0}$.

The vertical reaction at A is $\underset{\_}{A_y=7.5\text{k}\uparrow }$.

The vertical reaction at B is $\underset{\_}{B_y=90\text{k}\uparrow }$.

The vertical reaction at C is $\underset{\_}{C_y=90\text{k}\uparrow }$.

The vertical reaction at D is $\underset{\_}{D_y=7.5\text{k}\uparrow }$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reaction forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Let $A_x$ and $A_y$ be the horizontal and vertical reactions at the hinged support A.

Let $B_y$, $C_y$, and $D_y$ be the vertical reaction at the roller support B, C, and D.

Sketch the free body diagram of the given structure as shown in Figure 1.

Use equilibrium equations:

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ A_x=0\end{array}$

Therefore, the horizontal reaction at A is $\underset{\_}{A_x=0}$.

For the member AE, the summation of moments about E is equal to 0.

$\begin{array}{l}\sum M_E^{AE}=0\\ -A_y\left(15\right)+\frac{1}{2}\left(3\right)\left(15\right)\left(\frac{1}{3}\times 15\right)=0\\ -A_y\left(15\right)+112.5=0\\ A_y=7.5\text{k}\uparrow \end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=7.5\text{k}\uparrow }$.

For the member DF, the summation of moments about F is equal to 0.

$\begin{array}{l}\sum M_F^{DF}=0\\ D_y\left(15\right)-\frac{1}{2}\left(3\right)\left(15\right)\left(\frac{1}{3}\times 15\right)=0\\ D_y\left(15\right)-112.5=0\\ D_y=7.5\text{k}\uparrow \end{array}$

Therefore, the vertical reaction at D is $\underset{\_}{D_y=7.5\text{k}\uparrow }$.

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ -A_y\left(50\right)-B_y\left(20\right)+\frac{1}{2}\left(3\right)\left(15\right)\left(\frac{15}{3}+35\right)+3\left(50\right)\left(10\right)-\frac{1}{2}\left(3\right)\left(15\right)\left(20\right)+D_y\left(30\right)=0\end{array}$

Substitute 7.5 k for $A_y$ and 7.5 k for $D_y$.

$\begin{array}{l}-7.5\left(50\right)-B_y\left(20\right)+\frac{1}{2}\left(3\right)\left(15\right)\left(\frac{15}{3}+35\right)+3\left(50\right)\left(10\right)-\frac{1}{2}\left(3\right)\left(15\right)\left(\frac{15}{3}+15\right)+7.5\left(30\right)=0\\ -20B_y+1,800=0\\ B_y=90\text{k}\uparrow \end{array}$

Therefore, the vertical reaction at B is $\underset{\_}{B_y=90\text{k}\uparrow }$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+B_y+C_y+D_y-\frac{1}{2}\left(3\right)\left(15\right)-3\left(50\right)-\frac{1}{2}\left(3\right)\left(15\right)=0\end{array}$

Substitute 7.5 k for $A_y$, 90 k for $B_y$, and 7.5 k for $D_y$.

$\begin{array}{l}7.5+90+C_y+7.5-\frac{1}{2}\left(3\right)\left(15\right)-3\left(50\right)-\frac{1}{2}\left(3\right)\left(15\right)=0\\ C_y-90=0\\ C_y=90\text{k}\uparrow \end{array}$

Therefore, the vertical reaction at C is $\underset{\_}{C_y=90\text{k}\uparrow }$.