Chapter 5, Problem 20P

To determine

Sketch the shear and moment curves and label the maximum values of shear and moment, locate points of inflection, and draw the deflected shape.

Explanation of Solution

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.

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

Refer to Figure 1.

Calculate the length of the beam AB as shown in Figure 1.

$\begin{array}{c}{L}_{AB}=\sqrt{3^2+4^2}\\ =5\text{ft}\end{array}$

Use equilibrium equations:

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

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

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y+B_y-12-2\times 40-2\times 5=0\\ D_y+B_y=102\end{array}$        (1)

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -D_y\times 40-M_D+2\times 40\times \frac{40}{2}-2\times 5\times \frac{4}{2}-12\times 4=0\\ 40D_y+M_D=1532\end{array}$        (2)

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

Refer to Figure 2.

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ -D_y\times 20-M_D+2\times 20\times \frac{20}{2}=0\\ 20D_y+M_D=400\end{array}$        (3)

Solving Equations (2) and (3) to get the value of $D_y$ as shown below.

$D_y=56.6\text{kips}$

Substitute 56.6 kips for $D_y$ in Equations (3).

$\begin{array}{l}20\times 56.6+M_D=400\\ M_D=-732\text{k}\cdot \text{ft}\end{array}$

Substitute 56.6 kips for $D_y$ in Equations (1).

$\begin{array}{l}56.6+B_y=102\\ B_y=45.4\text{k}\end{array}$

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

Refer to Figure 3.

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

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

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ B_y-12-2\times 5=0\\ B_y=22\text{k}\end{array}$

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -M_B-2\times 5\times \frac{4}{2}-12\times 4=0\\ M_B=-68\text{k}\cdot \text{ft}\end{array}$

Calculate the axial forces as shown below.

$\begin{array}{l}\frac{12}{5}=\frac{A_A}{3}\\ A_A=7.2\text{ k}\\ \frac{2\times 5}{5}=\frac{A_B}{3}\\ A_{@B\text{Left}}=6\text{ k}\\ A_B=6+7.2\\ =13.2\text{ k}\end{array}$

Calculate the shearing forces as shown below.

$\begin{array}{l}\frac{12}{5}=\frac{V_A}{4}\\ V_A=9.6\text{ k}\\ \frac{2\times 5}{5}=\frac{V_B}{4}\\ V_B=8\text{ k}\end{array}$

Calculate the shear at each point as shown below.

For segment AB.

$\begin{array}{l}{V}_A=-9.6\text{k}\\ V_{@B\text{Left}}=-9.6-8\\ =-17.6\text{k}\end{array}$

For segment BCD.

$\begin{array}{c}{V}_B=45.4-22\\ =23.4\text{ k}\\ V_D=45.4-22-2\times 40\\ =56.6\text{k}\end{array}$

Calculate the moment at each point as shown below.

For segment AB.

$\begin{array}{c}{M}_A=0\\ M_B=-12\times 4-2\times 5\times \frac{4}{2}\\ =-68\text{k}\cdot \text{ft}\end{array}$

For segment BCD.

$\begin{array}{l}{M}_B=-68\text{k}\cdot \text{ft}\\ M_{@\text{ Left of }D}=-68-22\times 40+45.4\times 40-2\times 40\times \frac{40}{2}\\ =-732\text{k}\cdot \text{ft}\end{array}$

Calculate the point of zero shear force as shown below.

Consider a section at a distance x from B.

Sketch the Free Body Diagram of BCD as shown in Figure 4.

Refer to Figure 4.

Calculate the location of zero shear force as shown below.

Summation of shear at section x is equal to 0.

$\begin{array}{l}45.4-22-2\times x=0\\ 2x=23.4\\ x=11.7\text{ft from }B\end{array}$

Maximum bending moment occurs at the point of zero shear.

$\begin{array}{c}{M}_{\max }=-68-22\times 11.7+45.4\times 11.7-2\times 11.7\times \frac{11.7}{2}\\ =68.9\text{k}\cdot \text{ft}\end{array}$

Point of inflection is the location at which the bending moment changes its sign.

Here the BM is zero sign between BC and CD.

Refer to Figure 4.

Calculate the location of point of inflection where BM is zero as shown below.

Summation of moment at section x-x is equal to 0.

$\begin{array}{l}-68-22\times x+45.4\times x-2\times x\times \frac{x}{2}=0\\ x^2-23.4x+68=0\\ x=3.4\text{ft and 2}0\text{ft from }B\end{array}$

Hence, the point of inflection occurs at C and a distance of 3.4 ft from B.

Sketch the shear and moment curves for segment AB as shown in Figure 5.

Sketch the shear and moment curves for segment BCD as shown in Figure 6.

Sketch the deflected shape and the location of points of inflection of the beam as shown in Figure 7.