Chapter 5, Problem 1P

(a)

To determine

Provide the equations for shear and moment between points B and C as a function of x for origin of x at point A.

Answer to Problem 1P

The shear between points B and C as a function of x for origin of x at point A is $\underset{\_}{V=-14\text{kips}}$.

The equation for moment between points B and C as a function of x for origin of x at point A is $\underset{\_}{M=80-14x\text{kip}\cdot \text{ft}}$.

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.

Let $A_y$ be the vertical reactions at the hinged support A and $C_y$ be the vertical reaction at the roller 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 A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ 25+15\times 30+0.5\times 10\times \left(20+\frac{10}{2}\right)-C_y\times 20+8\times 10=0\\ C_y=34\text{kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+34-8-0.5\times 10-15=0\\ A_y=-6\\ A_y=6\text{kips}(\downarrow )\end{array}$

Consider a section at a distance x from A.

Sketch the section at a distance x from A between B and C as shown in Figure 2.

Refer to Figure 2.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -6-8-V=0\\ V=-14\text{kips}\end{array}$

Hence, the shear between points B and C as a function of x for origin of x at point A is $\underset{\_}{V=-14\text{kips}}$.

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -M+\left(-14\right)\times x+8\times 10=0\\ M=80-14x\text{kip}\cdot \text{ft}\end{array}$

Hence, the equation for moment between points B and C as a function of x for origin of x at point A is $\underset{\_}{M=80-14x\text{kip}\cdot \text{ft}}$.

(b)

To determine

Provide the equations for shear and moment between points B and C as a function of x for origin of x at D.

Answer to Problem 1P

The shear between points B and C as a function of x for origin of x at D is $\underset{\_}{V=-14\text{kips}}$.

The equation for moment between points B and C as a function of x for origin of x at D is $\underset{\_}{M=14x-340\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Calculation:

Refer to part (a),

The reactions at supports.

$\begin{array}{l}{A}_y=6\text{kips}(\downarrow )\\ C_y=34\text{kips}(\uparrow )\end{array}$

Consider a section at a distance x from D.

Sketch the section at a distance x from D between B and C as shown in Figure 3.

Refer to Figure 3.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -15-0.5\times 10+34+V=0\\ V=-14\text{kips}\end{array}$

Hence, the shear between points B and C as a function of x for origin of x at D is $\underset{\_}{V=-14\text{kips}}$.

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ M+\left(-14\right)\times x+34\times 10-0.5\times 10\times \frac{10}{2}+25=0\\ M=14x-340\text{kip}\cdot \text{ft}\end{array}$

Therefore, the equation for moment between points B and C as a function of x for origin of x at D is $\underset{\_}{M=14x-340\text{kip}\cdot \text{ft}}$.