Chapter 3, Problem 28P

To determine

Sketch the free body diagram of each segment above support A and calculate the internal forces on each free body.

Answer to Problem 28P

The horizontal reaction at A is $\underset{\_}{A_x=6\text{kips}(\leftarrow )}$.

The vertical reaction at A is $\underset{\_}{A_y=21\text{kips}(\uparrow )}$.

The moment at A is $\underset{\_}{M_A=113.75\text{kip}\cdot \text{ft}}$.

The vertical force at B for AB is $\underset{\_}{B_y=19\text{kips}(\uparrow )}$.

The internal moment at B for AB is $\underset{\_}{M_B=74.25\text{kip}\cdot \text{ft}}$.

The vertical force at joint B is $\underset{\_}{B_y{}^{\prime }=21\text{kips}(\uparrow )}$.

The internal moment at joint B is $\underset{\_}{M_B{}^{\prime }=83.75\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions 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 counter clockwise moment as negative and the clockwise moment as positive.

Calculation:

Let $A_x$ and $A_y$ be the horizontal and vertical reaction at the support A and $M_A$ is the moment at support A.

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

Refer to Figure 1.

Use equilibrium equations.

Find the horizontal reaction at A:

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x+6=0\\ A_x=6\text{kips}(\leftarrow )\end{array}$

Hence, the horizontal reaction at A is $\underset{\_}{A_x=6\text{kips}(\leftarrow )}$.

Find the vertical reaction at A:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-2\times 6.5-8=0\\ A_y=21\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at A is $\underset{\_}{A_y=21\text{kips}(\uparrow )}$.

Find the moment at A:

Summation of moment about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ 8\times 6+2\times 6.5\times \left(6-\frac{6.5}{2}\right)+6\times 5-M_A=0\\ M_A=113.75\text{kip}\cdot \text{ft}\end{array}$

Hence, the moment at A is $\underset{\_}{M_A=113.75\text{kip}\cdot \text{ft}}$.

Sketch the free body diagram of the beam BC as shown in Figure 2.

Refer to Figure 2.

Use equilibrium equations.

Find the vertical force at B:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ B_y-2\times 5.5-8=0\\ B_y=19\text{kips}(\uparrow )\end{array}$

Hence, the vertical force at B is $\underset{\_}{B_y=19\text{kips}(\uparrow )}$.

Find the internal moment at B:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ 8\times 5.5+2\times 5.5\times \frac{5.5}{2}-M_B=0\\ M_B=74.25\text{kip}\cdot \text{ft}\end{array}$

Hence, the internal moment at B is $\underset{\_}{M_B=74.25\text{kip}\cdot \text{ft}}$.

Sketch the free body diagram of the joint B as shown in Figure 3.

Refer to Figure 3.

Use equilibrium equations.

Find the vertical force at joint B:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ B_y{}^{\prime }-19-2\times 1=0\\ B_y{}^{\prime }=21\text{kips}(\uparrow )\end{array}$

Hence, the vertical force at joint B is $\underset{\_}{B_y{}^{\prime }=21\text{kips}(\uparrow )}$.

Find the internal moment at joint B:

Summation of moments about B’ is equal to 0.

$\begin{array}{l}\sum M_B{}^{\prime }=0\\ 74.25+19\times \frac{1}{2}-M_B{}^{\prime }=0\\ M_B{}^{\prime }=83.75\text{kip}\cdot \text{ft}\end{array}$

Hence, the internal moment at joint B is $\underset{\_}{M_B{}^{\prime }=83.75\text{kip}\cdot \text{ft}}$.

Sketch the free body diagram of the column AB as shown in Figure 4.

Refer to Figure 1.

Use equilibrium equations.

Find the vertical force at A:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y=21\text{kips}(\uparrow )\end{array}$

Hence, the vertical force at A is $\underset{\_}{A_y=21\text{kips}(\uparrow )}$.

Find the moment at A:

Summation of moment about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ 83.75+6\times 5-M_A=0\\ M_A=113.75\text{kip}\cdot \text{ft}\end{array}$

Therefore, the moment at A is $\underset{\_}{M_A=113.75\text{kip}\cdot \text{ft}}$.