Chapter 3, Problem 33P

To determine

Calculate the support reactions for the given structure.

Answer to Problem 33P

The horizontal reaction at A is $\underset{\_}{A_x=55\text{kN}\leftarrow }$.

The vertical reaction at A is $\underset{\_}{A_y=216.11\text{kN}\uparrow }$.

The horizontal reaction at B is $\underset{\_}{B_x=45\text{kN}\leftarrow }$.

The vertical reaction at B is $\underset{\_}{B_y=183.89\text{kN}\uparrow }$.

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 $B_x$ and $B_y$ be the horizontal reaction and vertical reaction at the hinged support B.

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

Use equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ \left(\begin{array}{l}-\left(100\right)\left(5\right)+\frac{1}{2}\left(20\right)\left(4\right)\left(\frac{2}{3}\times 4\right)-20\left(12\right)\left(\frac{12}{2}\right)+20\left(4\right)\left(\frac{4}{2}\right)\\ -\frac{1}{2}\left(20\right)\left(4\right)\left(12+\frac{4}{3}\right)+B_y\left(12\right)\end{array}\right)=0\\ -500+106.667-1440+160-533.333+12B_y=0\\ B_y=183.89\text{kN}\uparrow \end{array}$

Therefore, the vertical reaction at B is $\underset{\_}{B_y=183.89\text{kN}\uparrow }$.

For the member BC, the summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C^{BC}=0\\ -\left(20\right)\left(6\right)\left(\frac{6}{2}\right)-\frac{1}{2}\left(20\right)\left(4\right)\left(\frac{4}{3}+6\right)+B_y\left(6\right)-B_x\left(10\right)=0\end{array}$

Substitute $183.89\text{kN}$ for $B_y$.

$\begin{array}{l}-\left(20\right)\left(6\right)\left(\frac{6}{2}\right)-\frac{1}{2}\left(20\right)\left(4\right)\left(\frac{4}{3}+6\right)+\left(183.89\right)\left(6\right)-B_x\left(10\right)=0\\ 10B_x+450.007=0\\ B_x=-45\text{kN}\\ B_x=45\text{kN}\leftarrow \end{array}$

Therefore, the horizontal reaction at B is $\underset{\_}{B_x=45\text{kN}\leftarrow }$.

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x+B_x+100=0\end{array}$

Substitute $-45\text{kN}$ for $B_x$.

$\begin{array}{l}{A}_x-45+100=0\\ 55+A_x=0\\ A_x=-55\text{kN}\\ A_x=55\text{kN}\leftarrow \end{array}$

Therefore, the horizontal reaction at A is $\underset{\_}{A_x=55\text{kN}\leftarrow }$.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+B_y-\frac{1}{2}\left(20\right)\left(4\right)-20\left(16\right)-\frac{1}{2}\left(20\right)\left(4\right)=0\end{array}$

Substitute $183.89\text{kN}$ for $B_y$.

$\begin{array}{l}{A}_y+183.89-\frac{1}{2}\left(20\right)\left(4\right)-20\left(16\right)-\frac{1}{2}\left(20\right)\left(4\right)=0\\ A_y-216.11=0\\ A_y=216.11\text{kN}\uparrow \end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=216.11\text{kN}\uparrow }$.