Chapter 3, Problem 41P

To determine

Calculate the support reactions for the given structure.

Answer to Problem 41P

The horizontal reaction at A is $\underset{\_}{A_x=62.5\text{kN}\leftarrow }$.

The vertical reaction at A is $\underset{\_}{A_y=62.5\text{kN}\downarrow }$.

The moment at A is $\underset{\_}{M_A=187.5\text{kN}\cdot \text{m}}$ acting in the counterclockwise direction.

The horizontal reaction at B is $\underset{\_}{B_x=62.5\text{kN}\leftarrow }$.

The vertical reaction at B is $\underset{\_}{B_y=62.5\text{kN}\uparrow }$.

The moment at B is $\underset{\_}{M_B=187.5\text{kN}\cdot \text{m}}$ 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.

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.

$\begin{array}{l}\sum M_D=0\\ C_y\left(6\right)-125\left(3\right)=0\\ 6C_y=375\\ C_y=62.5\text{kN}\end{array}$

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

$\begin{array}{l}\sum M_E^{CE}=0\\ C_y\left(3\right)-C_x\left(3\right)=0\end{array}$

Substitute $62.5\text{kN}$ for $C_y$.

$\begin{array}{l}62.5\left(3\right)-C_x\left(3\right)=0\\ C_x=62.5\text{kN}\end{array}$

For the member CED, the summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ -C_x+125-D_x=0\end{array}$

Substitute $62.5\text{kN}$ for $C_x$.

$\begin{array}{l}-62.5+125-D_x=0\\ D_x=62.5\text{kN}\end{array}$

For the member CED, the summation of forces along y-direction is equal to 0.

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

Substitute $62.5\text{kN}$ for $C_y$.

$\begin{array}{l}-62.5+D_y=0\\ D_y=62.5\text{kN}\end{array}$

Find the reactions at the support A:

Consider the equilibrium of the portion AC.

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

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

Substitute $62.5\text{kN}$ for $C_x$.

$\begin{array}{l}-A_x+62.5=0\\ A_x=62.5\text{kN}\leftarrow \end{array}$

Therefore, the horizontal reaction at A is $\underset{\_}{A_x=62.5\text{kN}\leftarrow }$.

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 $62.5\text{kN}$ for $C_y$.

$\begin{array}{l}-A_y+62.5=0\\ A_y=62.5\text{kN}\downarrow \end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=62.5\text{kN}\downarrow }$.

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ M_A-C_x\left(3\right)=0\end{array}$

Substitute $62.5\text{kN}$ for $C_x$.

$\begin{array}{l}{M}_A-62.5\left(3\right)=0\\ M_A=187.5\text{kN}\cdot \text{m}\end{array}$

Therefore, the moment at A is $\underset{\_}{M_A=187.5\text{kN}\cdot \text{m}}$ 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.

$\begin{array}{l}+\to \sum F_x=0\\ -B_x+D_x=0\end{array}$

Substitute $62.5\text{kN}$ for $D_x$.

$\begin{array}{l}-B_x+62.5=0\\ B_x=62.5\text{kN}\leftarrow \end{array}$

Therefore, the horizontal reaction at B is $\underset{\_}{B_x=62.5\text{kN}\leftarrow }$.

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 $62.5\text{kN}$ for $D_y$.

$\begin{array}{l}{B}_y-62.5=0\\ B_y=62.5\text{kN}\uparrow \end{array}$

Therefore, the vertical reaction at B is $\underset{\_}{B_y=62.5\text{kN}\uparrow }$.

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ M_B-D_x\left(3\right)=0\end{array}$

Substitute $62.5\text{kN}$ for $D_x$.

$\begin{array}{l}{M}_B-62.5\left(3\right)=0\\ M_B=187.5\text{kN}\cdot \text{m}\end{array}$

Therefore, the moment at B is $\underset{\_}{M_B=187.5\text{kN}\cdot \text{m}}$ acting in the counterclockwise direction.