Chapter 5, Problem 12P

(a)

To determine

Provide the equations for shear and moment from the origin at end A.

Answer to Problem 12P

The equation for shear along for origin at A is $\underset{\_}{V=-8\text{kips}}$ for $0\le x<4$.

The equation for moment for origin at A is $\underset{\_}{M=-8x\text{kip}\cdot \text{ft}}$ for $0\le x<4$.

The equation for shear for origin at A is $\underset{\_}{V=38-3x\text{ kips}}$ for $4<x\le 20$.

The equation for moment for origin at A is $\underset{\_}{M=-1.5x^2+38x-160\text{kip}\cdot \text{ft}}$ for $4<x\le 20$.

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:

Let $R_B$ be the vertical reaction at the roller support B and $R_C$ be the vertical reaction at the hinged support C.

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

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -R_C\times 16+3\times 16\times \frac{16}{2}-8\times 4=0\\ R_C=22\text{kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -8-3\times 16+R_B+22=0\\ R_B=34\text{kips}(\uparrow )\end{array}$

Consider a section at a distance x from A.

For $0\le x<4$.

Sketch the section at a distance x from A as shown in Figure 2.

Refer to Figure 2.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -8-V=0\\ V=-8\text{ kips}\end{array}$

Hence, the equation for shear along for origin at A is $\underset{\_}{V=-8\text{kips}}$.

Summation of moments about the section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ -8\times x-M=0\\ M=-8x\text{kip}\cdot \text{ft}\end{array}$

Hence, the equation for moment for origin at A is $\underset{\_}{M=-8x\text{kip}\cdot \text{ft}}$.

For $4<x\le 20$.

Sketch the section at a distance x from A as shown in Figure 3.

Refer to Figure 3.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -8+34-3\left(x-4\right)-V=0\\ V=38-3x\text{ kips}\end{array}$        (1)

Hence, the equation for shear for origin at A is $\underset{\_}{V=38-3x\text{ kips}}$.

Summation of moments about the section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ -8\times x+34\left(x-4\right)-3\left(x-4\right)\frac{\left(x-4\right)}{2}-M=0\\ M=-1.5x^2+38x-160\text{kip}\cdot \text{ft}\end{array}$        (2)

Hence, the equation for moment for origin at A is $\underset{\_}{M=-1.5x^2+38x-160\text{kip}\cdot \text{ft}}$.

(b)

To determine

Calculate the moment at section 1 for the origin at end A.

Answer to Problem 12P

The moment at section 1 for origin at A is $\underset{\_}{M=60.5\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (a).

The equation for moment for origin at A is $M=-1.5x^2+38x-160\text{kip}\cdot \text{ft}$.

At section 1, the value of $x=9\text{ft}$.

Calculate the moment at section 1 for the origin at end A as below.

Substitute 9 ft for x in Equation (2).

$\begin{array}{c}{M}_1=-1.5\times {\left(9\right)}^2+38\times 9-160\\ =60.5\text{kip}\cdot \text{ft}\end{array}$

Hence, the moment at section 1 for the origin at end A is $\underset{\_}{M_1=60.5\text{kip}\cdot \text{ft}}$.

(c)

To determine

Calculate the location of the point of zero shear between B and C.

Answer to Problem 12P

The location of the point of zero shear between B and C is $\underset{\_}{x=12.67\text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (a).

The equation for shear for origin at A is $V=38-3x\text{ kips}$.

Calculate the location of the point of zero shear between B and C as shown below.

Substitute 0 for V in Equation (1).

$\begin{array}{l}38-3x=0\\ x=12.67\text{ft}\end{array}$

Hence, the location of the point of zero shear between B and C is $\underset{\_}{x=12.67\text{ft}}$.

(d)

To determine

Calculate the maximum moment between B and C.

Answer to Problem 12P

The maximum moment between B and C is $\underset{\_}{M_{\max }=80.67\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (a).

The equation for moment for origin at A is $M=-1.5x^2+38x-160\text{kip}\cdot \text{ft}$.

Refer to part (c).

The location of the point of zero shear between B and C is $x=12.67\text{ft}$.

The maximum moment occurs at the point of zero shear.

Calculate the maximum moment as shown below.

Substitute 12.67 ft for x in Equation (2).

$\begin{array}{c}{M}_{\max }=-1.5\times {12.67}^2+38\times 12.67-160\\ =80.67\text{kip}\cdot \text{ft}\end{array}$

Hence, the maximum moment between B and C is $\underset{\_}{M_{\max }=80.67\text{kip}\cdot \text{ft}}$.

(e)

To determine

Provide the equations for shear and moment for the origin at C.

Answer to Problem 12P

The equation for shear along for origin at C is $\underset{\_}{V=3x-22\text{kips}}$ for $0\le x<16$.

The equation for moment for origin at C is $\underset{\_}{M=22x-\frac{3}{2}{x}^2\text{kip}\cdot \text{ft}}$ for $0\le x<16$.

The equation for shear for origin at C is $\underset{\_}{V=-8\text{ kips}}$ for $16<x\le 20$.

The equation for moment for origin at C is $\underset{\_}{M=8x-160\text{kip}\cdot \text{ft}}$ for $16<x\le 20$.

Explanation of Solution

Calculation:

Refer to part (a).

The reaction at support B is  $R_B=34\text{kips}(\uparrow )$.

The reaction at support C is $R_C=22\text{kips}(\uparrow )$.

Consider a section at a distance x from C.

For $0\le x\le 16$.

Sketch the section at a distance x from C as shown in Figure 4.

Refer to Figure 4.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 22-3\times x+V=0\\ V=3x-22\text{ kips}\end{array}$        (3)

Hence, the equation for shear along for origin at C is $\underset{\_}{V=3x-22\text{kips}}$.

Summation of moments about the section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ -22\times x+3\times x\times \frac{x}{2}+M=0\\ M=22x-\frac{3}{2}{x}^2\text{kip}\cdot \text{ft}\end{array}$        (4)

Hence, the equation for moment for origin at C is $\underset{\_}{M=22x-\frac{3}{2}{x}^2\text{kip}\cdot \text{ft}}$.

For $16<x\le 20$.

Sketch the section at a distance x from C that is $20-x$ from A as shown in Figure 5.

Refer to Figure 3.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -8-V=0\\ V=-8\text{ kips}\end{array}$

Hence, the equation for shear for origin at C is $\underset{\_}{V=-8\text{ kips}}$.

Summation of moments about the section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ -8\times \left(20-x\right)-M=0\\ M=8x-160\text{kip}\cdot \text{ft}\end{array}$

Hence, the equation for moment for origin at C is $\underset{\_}{M=8x-160\text{kip}\cdot \text{ft}}$.

(f)

To determine

Calculate the moment at section 1 for the origin at end C.

Answer to Problem 12P

The moment at section 1 for origin at C is $\underset{\_}{M_1=60.5\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (e).

The equation for moment for origin at C is $M=22x-\frac{3}{2}{x}^2\text{kip}\cdot \text{ft}$.

At section 1, the value of $x=11\text{ft}$.

Calculate the moment at section 1 for the origin at end C as below.

Substitute 11 ft for x in Equation (2).

$\begin{array}{c}{M}_1=22\times 11-\frac{3}{2}\times {11}^2\\ =60.5\text{kip}\cdot \text{ft}\end{array}$

Hence, the moment at section 1 for the origin at end C is $\underset{\_}{M_1=60.5\text{kip}\cdot \text{ft}}$.

(g)

To determine

Calculate the location of the maximum moment and the value of maximum moment.

Answer to Problem 12P

The location of the maximum moment is $\underset{\_}{x=12.67\text{ft}}$.

The maximum moment is $\underset{\_}{M_{\max }=80.67\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (e).

The equation for moment for origin at C is $M=22x-\frac{3}{2}{x}^2\text{kip}\cdot \text{ft}$.

The equation for shear for origin at C is $V=3x-22\text{ kips}$.

The maximum moment occurs at the point of zero shear.

Calculate the location of the maximum moment as shown below.

Substitute 0 for V in Equation (3).

$\begin{array}{l}3x-22=0\\ x=7.33\text{ft}\end{array}$

Hence, the location of maximum moment is $\underset{\_}{x=7.33\text{ft}}$.

Calculate the maximum moment as below.

Substitute 7.33 ft for x in Equation (4).

$\begin{array}{c}{M}_{\max }=22\times 7.33-\frac{3}{2}\times {7.33}^2\\ =80.67\text{kip}\cdot \text{ft}\end{array}$

Hence, the maximum moment is $\underset{\_}{M_{\max }=80.67\text{kip}\cdot \text{ft}}$.

(h)

To determine

Provide the equations for shear and moment between B and C for the origin at B.

Answer to Problem 12P

The equation for shear between B and C for the origin at B is $\underset{\_}{V=26-3x\text{ kips}}$.

The equation for moment between B and C for the origin at B is $\underset{\_}{M=-\frac{3}{2}{x}^2+26x-32\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (a).

The reaction at support B is  $R_B=34\text{kips}(\uparrow )$.

The reaction at support C is $R_C=22\text{kips}(\uparrow )$.

Consider a section at a distance x from B.

Sketch the section at a distance x from B between B and C as shown in Figure 6.

Refer to Figure 4.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -8+34-3\times x-V=0\\ V=26-3x\text{ kips}\end{array}$

Hence, the equation for shear between B and C for the origin at B is $\underset{\_}{V=26-3x\text{ kips}}$.

Summation of moments about the section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ -8\times \left(4+x\right)+34\times x-3\times x\times \frac{x}{2}-M=0\\ M=-\frac{3}{2}{x}^2+26x-32\text{kip}\cdot \text{ft}\end{array}$        (5)

Hence, the equation for moment between B and C for the origin at B is $\underset{\_}{M=-\frac{3}{2}{x}^2+26x-32\text{kip}\cdot \text{ft}}$.

(i)

To determine

Calculate the moment at section 1 for the origin at end B.

Answer to Problem 12P

The moment at section 1 for origin at B is $\underset{\_}{M_1=60.5\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (h).

The equation for moment between B and C for the origin at B is $M=-\frac{3}{2}{x}^2+26x-32\text{kip}\cdot \text{ft}$.

At section 1, the value of $x=5\text{ft}$.

Calculate the moment at section 1 for the origin at end B as below.

Substitute 5 ft for x in Equation (2).

$\begin{array}{c}{M}_1=-\frac{3}{2}\times 5^2+26\times 5-32\\ =60.5\text{kip}\cdot \text{ft}\end{array}$

Therefore, the moment at section 1 for the origin at B is $\underset{\_}{M_1=60.5\text{kip}\cdot \text{ft}}$.