Chapter 5, Problem 55PA

To determine

Sketch the shear and moment curves produced by the total load for beams $B_1$ and $B_2$.

Explanation of Solution

Given information:

The thickness of the concrete slab is $10\text{ in}\text{.}$.

The weight of slab is $125\text{lb}/{\text{ft}}^{\text{2}}$.

The weight of light fixtures and utilities suspended from the bottom of the slab is $5\text{lb}/{\text{ft}}^{\text{2}}$.

The height of the masonry wall is 15 ft.

The weight of the hollow concrete block is $38\text{lb}/{\text{ft}}^{\text{2}}$.

The weight of the beams and their fireproofing is $80\text{lb}/\text{ft}$.

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:

Calculate the intensity of load acting on beam $B_1$ and $B_2$ as shown below.

$\begin{array}{c}\left(\begin{array}{l}\text{Trapezoidal loading}\\ \text{or Triangular loading}\end{array}\right)=\left(\begin{array}{l}\text{Total weight on slab}\times \text{center to center}\\ \text{ height of masonry wall}\end{array}\right)\\ =\left(125+5\right)\times \frac{14}{2}\\ =910\text{lb}/\text{ft}\end{array}$

Find the total load over the beam:

$\begin{array}{c}\text{Total load over the beam}=\text{Weight of masonry wall}+\text{Weight of the beam}\\ =38\times 14+80\\ =612\text{lb}/\text{ft}\end{array}$

Consider beam $B_1$ as AB and $B_2$ as CD.

Sketch the free body diagram for the loading on the beam $B_1$ as shown in Figure 1.

Refer to Figure 1.

Use equilibrium equations:

Due to the symmetry of the beam.

$\begin{array}{c}{A}_y=B_y=\frac{1}{2}\left(612\times 18+910\times 4+2\times \frac{1}{2}\times 7\times 910\right)\\ =10,513\text{lb}\times \frac{1\text{kips}}{1,000\text{lb}}\\ =10.5\text{kips}\end{array}$

Calculate the shear at each point as shown below.

$\begin{array}{l}{V}_A=10.5\text{kips}\\ V_{\text{@ 7 ft from}A}=10.5-\frac{1}{1,000}\left(612\times 7+\frac{1}{2}\times 7\times 910\right)\\ =3\text{kips}\\ V_{\text{@ Mid}\text{of}AB}=10.5-\frac{1}{1,000}\left(612\times 7+910\times 2+\frac{1}{2}\times 7\times 910\right)\\ =0\end{array}$

$\begin{array}{l}{V}_{@\text{ 11 ft from }A}=10.5-\frac{1}{1,000}\left(612\times 11+910\times 4+\frac{1}{2}\times 7\times 910\right)\\ =-3\text{kips}\\ V_B=10.5-\frac{1}{1,000}\left(612\times 18+910\times 4+2\times \frac{1}{2}\times 7\times 910\right)\\ =-10.5\text{kips}\end{array}$

Calculate the moment at each point as shown below.

$\begin{array}{l}{M}_A=0\\ M_{\text{@ 7 ft from}A}=10.5\times 7-\frac{1}{1,000}\left(612\times 7\times \frac{7}{2}+\frac{1}{2}\times 7\times 910\times \left(\frac{1}{3}\times 7\right)\right)\\ =51.07\text{kips}\cdot \text{ft}\\ M_{\text{@ Mid}\text{of}AB}=10.5\times 9-\frac{1}{1,000}\left(612\times 9\times \frac{9}{2}+910\times 2\times \frac{2}{2}+\frac{1}{2}\times 7\times 910\left(\frac{1}{3}\times 7+2\right)\right)\\ =54.09\text{kips}\cdot \text{ft}\end{array}$

Due to the symmetry of the beam,

$\begin{array}{l}{M}_{\text{@ 11 ft from}A}=51.07\text{kips}\cdot \text{ft}\\ M_B=0\end{array}$

Sketch the shear and moment curves of the beam $B_1$ as shown in Figure 2.

Sketch the free body diagram for the loading on the beam $B_2$ as shown in Figure 3.

Refer to Figure 3.

Use equilibrium equations:

Due to the symmetry of the beam.

$\begin{array}{c}{C}_y=D_y=\frac{1}{2}\left(612\times 14+2\times \frac{1}{2}\times 7\times 910\right)\\ =7,469\text{lb}\times \frac{1\text{kips}}{1,000\text{lb}}\\ =7.5\text{kips}\end{array}$

Calculate the shear at each point as shown below.

$\begin{array}{l}{V}_C=7.5\text{kips}\\ V_{\text{@ mid of}CD}=7.5-\frac{1}{1,000}\left(612\times 7+\frac{1}{2}\times 7\times 910\right)\\ =0\\ V_D=7.5-\frac{1}{1,000}\left(612\times 14+2\times \frac{1}{2}\times 7\times 910\right)\\ =-7.5\text{kips}\end{array}$

Calculate the moment at each point as shown below.

$\begin{array}{l}{M}_D=0\\ M_{\text{@ Mid}\text{of}CD}=7.5\times 7-\frac{1}{1,000}\left(612\times 7\times \frac{7}{2}+\frac{1}{2}\times 7\times 910\left(\frac{1}{3}\times 7\right)\right)\\ =30\text{kips}\cdot \text{ft}\\ M_D=7.5\times 14-\frac{1}{1,000}\left(612\times 14\times \frac{14}{2}+\frac{1}{2}\times 7\times 910\left(\frac{1}{3}\times 7+7\right)+\frac{1}{2}\times 7\times 910\left(\frac{2}{3}\times 7\right)\right)\\ =0\end{array}$

Sketch the shear and moment curves of the beam $B_2$ as shown in Figure 4.