Chapter 5, Problem 42P

To determine

Sketch the shear and moment curves for each member of the frame and draw the deflected shape.

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 clockwise moment as positive and the counter clockwise moment as negative.

Calculation:

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

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about F is equal to 0.

$\begin{array}{l}\sum M_F=0\\ D_y\times 8+D_x\times 6=0\\ D_x=-\frac{4}{3}{D}_y\end{array}$        (1)

Sketch the free body diagram of the segment BCD as shown in Figure 2.

Refer to Figure 2.

Use equilibrium equations:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ D_y\times 10-D_x\times 6+4\times 10\times \frac{10}{2}+35\times 3=0\\ D_y\times 10-\left(\frac{4}{3}{D}_y\right)\times 6+4\times 10\times \frac{10}{2}+35\times 3=0\\ D_y=-16.94\text{kips}(\uparrow )\end{array}$

From Equation (1).

$\begin{array}{c}{D}_x=-\frac{4}{3}\times 16.94\\ =22.59\text{kips}(\leftarrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ B_y-4\times 10+16.94=0\\ B_y=23.06\text{kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ B_x+35-22.59=0\\ B_x=12.41\text{kips}(\to )\end{array}$

Refer to Figure 1.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_y-16.94=0\\ F_y=16.94\text{kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ F_x+22.59=0\\ F_x=-22.59\text{kips}(\leftarrow )\end{array}$

Sketch the free body diagram of the segment AB as shown in Figure 3.

Refer to Figure 3.

Use equilibrium equations:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-23.06=0\\ A_y=23.06\text{kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x+12.41=0\\ A_x=-12.41\text{kips}(\leftarrow )\end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ 12.41\times 6+M_A=0\\ M_A=-74.46\text{kips}\cdot \text{ft}\end{array}$

Calculate the shear at each point of the frame as shown below.

$\begin{array}{l}{V}_A=12.41\text{kips}\\ V_{@\text{Left of }B}=12.41\text{kips}\\ V_{@\text{Left of }C}=12.41-35\\ =-22.59\text{kips}\\ V_C=23.06\text{kips}\end{array}$

$\begin{array}{l}{V}_D=23.06-4\times 10\\ =-16.94\text{kips}\\ V_E=23.06-4\times 10\\ =-16.94\text{kips}\end{array}$

$\begin{array}{l}{V}_{@\text{Right of }E}=22.59\text{kips}\\ V_F=22.59-22.59\\ =0\end{array}$

Maximum bending moment occurs between the member CD.

Consider the shear between CD at a distance x from C is equal to zero.

$\begin{array}{l}{V}_x=0\\ 23.06-4\times x=0\\ x=5.765\text{ft}\end{array}$

Calculate the moment at each point of the frame as shown below.

$\begin{array}{c}{M}_A=-74.46\text{kips}\cdot \text{ft}\\ M_B=-74.46+12.41\times 6\\ =0\end{array}$

$\begin{array}{c}{M}_{\text{@ }35\text{ kips}}=-74.46+12.41\times 9\\ =37.23\text{kips}\cdot \text{ft}\\ M_C=-74.46+12.41\times 12-35\times 3\\ =30.54\text{kips}\cdot \text{ft}\end{array}$

$\begin{array}{l}{M}_{\text{@ }5.765\text{ft from }C}=23.06\times 5.765-30.54-4\times 5.765\times \frac{5.765}{2}\\ =35.93\text{kips}\cdot \text{ft}\\ M_D=23.06\times 10-30.54-4\times 10\times \frac{10}{2}\\ =0\end{array}$

$\begin{array}{c}{M}_E=23.06\times 18-30.54-4\times 18\times \frac{18}{2}\\ =-135.5\text{kips}\cdot \text{ft}\\ M_E=-135.5+22.59\times 6\\ =0\end{array}$

Sketch the shear and moment curves of the frame as shown in Figure 4.

Sketch the deflected shape of the frame as shown in Figure 5.