Chapter 5, Problem 6P

To determine

Provide the equation for moment along the length of the beam for origin at A and then for origin at D.

Verify the moment at point C is same for both cases.

Answer to Problem 6P

The equation for moment along AB for origin at A is $\underset{\_}{M_1=39.1x_1-1.2x_1^2}$.

The equation for moment along BC for origin at A is $\underset{\_}{M_2=21.1x_2+108-1.2x_2^2}$.

The equation for moment along BC for origin at A is $\underset{\_}{M_3=-17.3x_3+415.72}$.

The equation for moment along DC for origin at D is $\underset{\_}{M_4=17.3x_4}$.

The equation for moment along CB for origin at D is $\underset{\_}{M_5=17.3x_5-1.2{\left(x_5-8\right)}^2}$.

The equation for moment along BA for origin at D is $\underset{\_}{M_6=-0.7x_6-1.2{\left(x_6-8\right)}^2+324}$.

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.

Calculation:

Let $R_A$ be the vertical reaction at the roller at A and $R_D$ be the vertical reaction at the hinged support B.

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

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ A_y\times 24-18\times 18-2.4\times 16\times \left(\frac{16}{2}+8\right)=0\\ A_y=39.1\text{kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y+39.1-18-2.4\times 16=0\\ D_y=17.3\text{kips}(\uparrow )\end{array}$

Consider a section along the length of the beam for the origin at A.

Sketch the section along the length of the beam for the origin at A as shown in Figure 2.

Refer to Figure 2.

Along AB $0\le x_1<6$.

Summation of moments at section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ 39.1\times x_1-2.4\times x_1\times \frac{x_1}{2}-M_1=0\\ M_1=39.1x_1-1.2x_1^2\end{array}$

Hence, the equation for moment along AB for origin at A is $\underset{\_}{M_1=39.1x_1-1.2x_1^2}$.

Along BC $6<x_2<16$.

Summation of moments at section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ 39.1\times x_2-18\times \left(x_2-6\right)-2.4\times x_2\times \frac{x_2}{2}-M_2=0\\ M_2=21.1x_2+108-1.2x_2^2\end{array}$        (1)

Hence, the equation for moment along BC for origin at A is $\underset{\_}{M_2=21.1x_2+108-1.2x_2^2}$.

Along CD $16<x_3\le 24$.

Summation of moments at section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ 39.1\times x_3-18\times \left(x_3-6\right)-2.4\times 16\times \left(x_3-\frac{16}{2}\right)-M_3=0\\ M_3=-17.3x_3+415.72\end{array}$

Hence, the equation for moment along BC for origin at A is $\underset{\_}{M_3=-17.3x_3+415.72}$.

Consider a section along the length of the beam for the origin at D.

Sketch the section along the length of the beam for the origin at D as shown in Figure 3.

Refer to Figure 3.

Along DC $0\le x_4<8$.

Summation of moments at section is equal to 0.

$\begin{array}{l}\sum M_x=0\\ -17.3\times x_4+M_4=0\\ M_4=17.3x_4\end{array}$

Hence, the equation for moment along DC for origin at D is $\underset{\_}{M_4=17.3x_4}$.

Along CB $8<x_5<18$.

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

$\begin{array}{l}\sum M_x=0\\ -17.3\times x_5+2.4\times \left(x_5-8\right)\times \frac{\left(x_5-8\right)}{2}+M_5=0\\ M_5=17.3x_5-1.2{\left(x_5-8\right)}^2\end{array}$        (2)

Hence, the equation for moment along CB for origin at D is $\underset{\_}{M_5=17.3x_5-1.2{\left(x_5-8\right)}^2}$.

Along BA $18<x_6\le 24$.

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

$\begin{array}{l}\sum M_x=0\\ -17.3\times x_6+2.4\times \left(x_6-8\right)\times \frac{\left(x_6-8\right)}{2}+18\left(x_6-18\right)+M_6=0\\ M_6=-0.7x_6-1.2{\left(x_6-8\right)}^2+324\end{array}$

Hence, the equation for moment along BA for origin at D is $\underset{\_}{M_6=-0.7x_6-1.2{\left(x_6-8\right)}^2+324}$.

Calculate the bending moment at C$\left(x=16\text{ft}\right)$ from A using Equation (1):

$\begin{array}{c}{M}_{C\text{from }A}=21.1\times 16+108-1.2\times {16}^2\\ =138.4\text{kips}\cdot \text{ft}\end{array}$

Calculate the bending moment at C$\left(x=8\text{ft}\right)$ using Equation (2):

$\begin{array}{c}{M}_{C\text{ from }D}=17.3\times 8-1.2{\left(8-8\right)}^2\\ =138.4\text{kips}\cdot \text{ft}\end{array}$

Therefore, the moment at point C is same for both cases.