Chapter 3, Problem 42P

To determine

Calculate the support reactions for the given structure.

Answer to Problem 42P

The horizontal reaction at A is $\underset{\_}{A_x=309.25\text{kN}\leftarrow }$.

The vertical reaction at A is $\underset{\_}{A_y=170.25\text{kN}\downarrow }$.

The moment at A is $\underset{\_}{M_A=1,138.5\text{kN}\cdot \text{m}}$ acting in the counterclockwise direction.

The horizontal reaction at B is $\underset{\_}{B_x=50.75\text{kN}\leftarrow }$.

The vertical reaction at B is $\underset{\_}{B_y=129.75\text{kN}\downarrow }$.

The moment at B is $\underset{\_}{M_B=76.5\text{kN}\cdot \text{m}}$ acting in the counterclockwise direction.

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 $A_x$, $A_y$, and $M_A$ be the horizontal reaction, vertical reaction, and moment at the fixed support A.

Let $B_x$, $B_y$, and $M_B$ be the horizontal reaction, vertical reaction, and moment at the fixed support B.

Let $C_x$ and $C_y$ be the horizontal and vertical reaction at the internal hinge C.

Let $D_x$ and $D_y$ be the horizontal and vertical reaction at the internal hinge D.

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

Find the angle of the inclined member with respect to horizontal axis:

Refer Figure 1.

$\begin{array}{c}\tan \theta =\frac{3}{10}\\ \theta ={\tan }^{-1}\left(\frac{3}{10}\right)\\ =16.7°\end{array}$

Convert the $15\text{kN/m}$ uniformly distributed load into point load:

$\begin{array}{c}15\text{kN/m =15}\times \frac{3}{\sin 16.7°}\\ =156.6\text{kN}\end{array}$

Use Figure 1 to find the length of the inclined member:

$\begin{array}{c}L=\sqrt{3^2+{10}^2}\\ =10.44\text{m}\end{array}$

Use equilibrium equations:

Find the forces at the internal hinges C and D:

Consider the free body diagram of the portion CED.

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ C_y\left(20\right)-\left(25+15\right)\left(4.5\right)\left(\frac{4.5}{2}\right)-156.6\left(\cos 16.7°\right)\left(15+5\right)=0\\ 20C_y=3,404.9\\ C_y=170.25\text{kN}\end{array}$

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

$\begin{array}{l}\sum M_E^{CE}=0\\ C_y\left(10\right)-C_x\left(7.5\right)+25\left(4.5\right)\left(\frac{4.5}{2}+3\right)-156.6\left(\frac{10.44}{2}\right)=0\end{array}$

Substitute $170.25\text{kN}$ for $C_y$.

$\begin{array}{l}170.25\left(10\right)-C_x\left(7.5\right)+25\left(4.5\right)\left(\frac{4.5}{2}+3\right)-156.6\left(\frac{10.44}{2}\right)=0\\ -7.5C_x+1,475.673=0\\ C_x=196.75\text{kN}\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -C_x+\left(25+15\right)\left(4.5\right)+D_x=0\end{array}$

Substitute $196.75\text{kN}$ for $C_x$.

$\begin{array}{l}-196.75+\left(25+15\right)\left(4.5\right)+D_x=0\\ D_x=16.75\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -C_y+2\left(156.6\right)\left(\cos 16.7°\right)-D_y=0\end{array}$

Substitute $170.25\text{kN}$ for $C_y$.

$\begin{array}{l}-170.25+2\left(156.6\right)\left(\cos 16.7°\right)-D_y=0\\ -D_y=-129.75\\ D_y=129.75\text{kN}\end{array}$

Find the reactions at the support A:

Consider the equilibrium of the portion AC.

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x+C_x+25\left(4.5\right)=0\end{array}$

Substitute $196.75\text{kN}$ for $C_x$.

$\begin{array}{l}-A_x+196.75+25\left(4.5\right)=0\\ A_x=309.25\text{kN}\leftarrow \end{array}$

Therefore, the horizontal reaction at A is $\underset{\_}{A_x=309.25\text{kN}\leftarrow }$.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+C_y=0\end{array}$

Substitute $170.25\text{kN}$ for $C_y$.

$\begin{array}{l}-A_y+170.25=0\\ A_y=170.25\text{kN}\downarrow \end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=170.25\text{kN}\downarrow }$.

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ M_A-C_x\left(4.5\right)-25\left(4.5\right)\left(\frac{4.5}{2}\right)=0\end{array}$

Substitute $196.75\text{kN}$ for $C_x$.

$\begin{array}{l}{M}_A-196.75\left(4.5\right)-25\left(4.5\right)\left(\frac{4.5}{2}\right)=0\\ M_A=1,138.5\text{kN}\cdot \text{m}\end{array}$

Therefore, the moment at A is $\underset{\_}{M_A=1,138.5\text{kN}\cdot \text{m}}$ acting in the counterclockwise direction.

Find the reactions at the support B:

Consider the equilibrium of the portion BD.

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

$\begin{array}{l}+\to \sum F_x=0\\ -B_x-D_x+15\left(4.5\right)=0\end{array}$

Substitute $16.75\text{kN}$ for $D_x$.

$\begin{array}{l}-B_x-16.75+15\left(4.5\right)=0\\ -B_x=-50.75\\ B_x=50.75\text{kN}\leftarrow \end{array}$

Therefore, the horizontal reaction at B is $\underset{\_}{B_x=50.75\text{kN}\leftarrow }$.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -B_y+D_y=0\end{array}$

Substitute $129.75\text{kN}$ for $D_y$.

$\begin{array}{l}-B_y+129.75=0\\ B_y=129.75\text{kN}\downarrow \end{array}$

Therefore, the vertical reaction at B is $\underset{\_}{B_y=129.75\text{kN}\downarrow }$.

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ M_B+D_x\left(4.5\right)-15\left(4.5\right)\left(\frac{4.5}{2}\right)=0\end{array}$

Substitute $16.75\text{kN}$ for $D_x$.

$\begin{array}{l}{M}_B+16.75\left(4.5\right)-15\left(4.5\right)\left(\frac{4.5}{2}\right)=0\\ M_B=76.5\text{kN}\cdot \text{m}\end{array}$

Therefore, the moment at B is $\underset{\_}{M_B=76.5\text{kN}\cdot \text{m}}$ acting in the counterclockwise direction.