Chapter 5, Problem 27P

5.12 through 5.28 Determine the equations for shear and bending moment for the beam shown. Use the resulting equations to draw the shear and bending moment diagrams.

To determine

Find the equations for the shear and bending moment of the beam.

Plot the shear and bending moment diagram.

Answer to Problem 27P

The equation of shear is,

$\underset{\_}{S=0.75x^2-25x\text{for}0\le x<2\text{m}}$, $\underset{\_}{S=0.75x^2-25x+95\text{for}2\text{m}<x<7\text{m}}$, and $\underset{\_}{S=0.75x^2-25x+175\text{for}7\text{m}<x\le 10\text{m}}$

The equation of bending moment is, $\underset{\_}{M=-12.5x^2+0.25x^3\text{for}0\le x<2\text{m}}$, $\underset{\_}{M=\left(-12.5x^2+0.25x^3+95x-190\right)\text{for}2\text{m}<x<7\text{m}}$, and $\underset{\_}{M=\left(-12.5x^2+0.25x^3+175x-750\right)\text{for}7\text{m}<x\le 10\text{m}}$

Explanation of Solution

Sign conversion:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Apply the following sign convention for calculating the shear and bending moments.

• When the portion of the beam considered is left of the section, then the external force acting to the left are considered as positive.
• When the portion of the beam considered is right of the section, then the external force acting to the right are considered as positive.
• When the portion of the beam considered is left of the section, then the external force acting upward are considered as positive.
• When the portion of the beam considered is right of the section, then the external force acting downward are considered as positive.
• When the portion of the beam considered is left of the section, then the clockwise moments are considered as positive.
• When the portion of the beam considered is right of the section, then the counterclockwise moments are considered as positive.

Calculation:

Show the free-body diagram of the entire beam as in Figure 1.

Find the horizontal reaction at point b by resolving the horizontal equilibrium.

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

Find the vertical reaction at point C by taking moment about point B.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_B=0}\\ 22\left(2\right)\left(\frac{2}{2}\right)+\frac{1}{2}\left(25-22\right)\left(2\right)\left(\frac{2}{3}\times 2\right)+C_y\left(5\right)-10\left(8\right)\left(\frac{8}{2}\right)-\frac{1}{2}\left(22-10\right)\left(8\right)\left(\frac{8}{3}\right)=0\\ 44+4+5C_y-320-128=0\\ C_y=80\text{kN}\uparrow \end{array}$

Find the vertical reaction at point B by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ B_y-10\left(10\right)-\frac{1}{2}\times \left(25-10\right)\times 10+C_y=0\\ B_y-100-75+80=0\\ B_y=95\text{kN}\uparrow \end{array}$

Pass the sections aa, bb, and cc at a distance x from left end of the beam.

Show the section aa, bb, and cc in the entire beam as in Figure 2.

Consider section AB:

Consider the left side of the section aa for calculation of shear and bending moment.

Show the free-body diagram of the left side of the section aa as in Figure 3.

Find the shear at section aa by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -\left(25-1.5x\right)\left(x\right)-\frac{1}{2}\left(25-25+1.5x\right)\left(x\right)-S=0\\ -25x+1.5x^2-0.75x^2-S=0\\ S=0.75x^2-25x\text{for}0\le x<2\text{m}\end{array}$

Find the moment at section aa by taking moment about the section aa.

$\begin{array}{l}+\circlearrowright {\displaystyle \sum M=0}\\ -\left(25-1.5x\right)\left(x\right)\left(\frac{x}{2}\right)-\frac{1}{2}\left(25-25+1.5x\right)\left(x\right)\left(\frac{2x}{3}\right)-M=0\\ -12.5x^2+0.75x^3-0.5x^3-M=0\\ M=-12.5x^2+0.25x^3\text{for}0\le x<2\text{m}\end{array}$

Consider section BC:

Consider the left side of the section bb for calculation of shear and bending moment.

Show the free-body diagram of the left side of the section bb as in Figure 4.

Find the shear at section bb by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ \left(\begin{array}{l}-\left(25-1.5x\right)\left(x\right)-\frac{1}{2}\left(25-25+1.5x\right)\left(x\right)\\ +95-S\end{array}\right)=0\\ -25x+1.5x^2-0.75x^2+95-S=0\\ S=0.75x^2-25x+95\text{for}2\text{m}<x<7\text{m}\end{array}$ (1)

Find the moment at section bb by taking moment about the section bb.

$\begin{array}{l}+\circlearrowright {\displaystyle \sum M=0}\\ \left(\begin{array}{l}-\left(25-1.5x\right)\left(x\right)\left(\frac{x}{2}\right)\\ -\frac{1}{2}\left(25-25+1.5x\right)\left(x\right)\left(\frac{2x}{3}\right)\\ +95\left(x-2\right)-M\end{array}\right)=0\\ \left(\begin{array}{l}-12.5x^2+0.75x^3-0.5x^3+95x\\ -190-M\end{array}\right)=0\\ M=\left(\begin{array}{l}-12.5x^2+0.25x^3\\ +95x-190\end{array}\right)\text{for}2\text{m}<x<7\text{m}\end{array}$ (2)

Consider section CD:

Consider the left side of the section cc for calculation of shear and bending moment.

Show the free-body diagram of the left side of the section cc as in Figure 5.

Find the shear at section cc by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ \left(\begin{array}{l}-\left(25-1.5x\right)\left(x\right)-\frac{1}{2}\left(25-25+1.5x\right)\left(x\right)\\ +95+80-S\end{array}\right)=0\\ -25x+1.5x^2-0.75x^2+175-S=0\\ S=0.75x^2-25x+175\text{for}7\text{m}<x\le 10\text{m}\end{array}$

Find the moment at section cc by taking moment about the section cc.

$\begin{array}{l}+\circlearrowright {\displaystyle \sum M=0}\\ \left(\begin{array}{l}-\left(25-1.5x\right)\left(x\right)\left(\frac{x}{2}\right)\\ -\frac{1}{2}\left(25-25+1.5x\right)\left(x\right)\left(\frac{2x}{3}\right)\\ +95\left(x-2\right)+80\left(x-7\right)-M\end{array}\right)=0\\ \left(\begin{array}{l}-12.5x^2+0.75x^3-0.5x^3+95x\\ -190+80x-560-M\end{array}\right)=0\\ M=\left(\begin{array}{l}-12.5x^2+0.25x^3\\ +175x-750\end{array}\right)\text{for}7\text{m}<x\le 10\text{m}\end{array}$

Thus, the equation of shear is,

$\underset{\_}{S=0.75x^2-25x\text{for}0\le x<2\text{m}}$, $\underset{\_}{S=0.75x^2-25x+95\text{for}2\text{m}<x<7\text{m}}$

$\underset{\_}{S=0.75x^2-25x+175\text{for}7\text{m}<x\le 10\text{m}}$

Thus, the equation of bending moment is,

$\underset{\_}{M=-12.5x^2+0.25x^3\text{for}0\le x<2\text{m}}$, $\underset{\_}{M=\left(-12.5x^2+0.25x^3+95x-190\right)\text{for}2\text{m}<x<7\text{m}}$, and $\underset{\_}{M=\left(-12.5x^2+0.25x^3+175x-750\right)\text{for}7\text{m}<x\le 10\text{m}}$

The maximum bending moment occurs where the shear is 0.

Equate Equation (1) to 0.

$\begin{array}{l}0=0.75x^2-25x+95\\ x=4.37\text{m}\end{array}$

Substitute 4.37 m for x in Equation (2).

$\begin{array}{c}M=-12.5{\left(4.37\right)}^2+0.25{\left(4.37\right)}^3+95\left(4.37\right)-190\\ =7.3\text{kN-m}\end{array}$

Tabulate the shear and bending moment values at important locations of the beam as in Table 1.

Distance x, m (From left)	Shear, kN	Bending moment, kN-m
0	0	0
2–	–47	–48
2+	48	–48
4.37	0	7.3
7–	–43.25	–51.75
7+	36.75	–51.75
10	0	0

Plot the shear and bending moment diagram of the beam as in Figure 6.