Chapter 5.4, Problem 109P

(a)

To determine

Write the equations for shear force and bending moments based on singularity function.

Answer to Problem 109P

The equation of shear force as a singularity function is;

$\underset{\_}{V=\text{13}-3x+3{\langle x-3\rangle }^1-8{\langle x-7\rangle }^0-3{\langle x-11\rangle }^1\text{kips}}$.

The equation of bending moment as a singularity function is;

$\underset{\_}{M=13x-1.5x^2+1.5{\langle x-3\rangle }^2-8{\langle x-7\rangle }^1-1.5{\langle x-11\rangle }^2\text{kips-ft}}$.

Explanation of Solution

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

Determine the vertical reaction at point B by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B_y\left(14\right)-\left(3\times 3\right)\times 1.5-8\left(7\right)-\left(3\times 3\right)\times 12.5=0\\ 14B_y-13.5-56-112.5=0\\ B_y=13\text{kips}\end{array}$

Determine the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\left(3\times 3\right)-8-\left(3\times 3\right)+B_y=0\\ A_y-9-8-9+13=0\\ A_y=13\text{kips}\end{array}$

Write the equation of the load function as follows;

$w\left(x\right)=3-3{\langle x-3\rangle }^0+3{\langle x-11\rangle }^0$ (1)

The equation for shear force as a function of load is,

$w\left(x\right)=-\frac{dV}{dx}$

Integrate the equation (1) to find V;

$\begin{array}{c}V=A_y-8{\langle x-7\rangle }^0+{\displaystyle \int w\left(x\right)dx}\\ =13-8{\langle x-7\rangle }^0-3x+3{\langle x-3\rangle }^1-3{\langle x-11\rangle }^1\\ =\text{13}-3x+3{\langle x-3\rangle }^1-8{\langle x-7\rangle }^0-3{\langle x-11\rangle }^1\text{kips}\end{array}$ (2)

The equation for bending moment as a function of shear force is,

$V\left(x\right)=\frac{dM}{dx}$

Integrate the equation (1) to find M;

$\begin{array}{c}M={\displaystyle \int Vdx}\\ =\text{13}x-\frac{3x^2}{2}+\frac{3{\langle x-3\rangle }^2}{2}-8{\langle x-7\rangle }^1-\frac{3{\langle x-11\rangle }^2}{2}\\ =13x-1.5x^2+1.5{\langle x-3\rangle }^2-8{\langle x-7\rangle }^1-1.5{\langle x-11\rangle }^2\text{kips-ft}\end{array}$ (3)

Therefore,

The equation of shear force as a singularity function is,

$\underset{\_}{V=\text{13}-3x+3{\langle x-3\rangle }^1-8{\langle x-7\rangle }^0-3{\langle x-11\rangle }^1\text{kips}}$.

The equation of bending moment as a singularity function is,

$\underset{\_}{M=13x-1.5x^2+1.5{\langle x-3\rangle }^2-8{\langle x-7\rangle }^1-1.5{\langle x-11\rangle }^2\text{kips-ft}}$.

(b)

To determine

The maximum bending moment using the singularity function.

Answer to Problem 109P

The maximum bending moment in the beam is $\underset{\_}{41.5\text{kips-ft}}$.

Explanation of Solution

Refer to Equation (2), find the location of maximum bending moment where the shear force changes sign. i.e., $V=0$.

Point A $\left(x=0\text{ft}\right)$:

Substitute 0 ft for x in Equation (2).

$\begin{array}{c}{V}_A=\text{13}-3\left(0\right)+3{\langle 0-3\rangle }^1-8{\langle 0-7\rangle }^0-3{\langle 0-11\rangle }^1\\ =13-0+0-0-0\\ =\text{13 kips}\end{array}$

Point C $\left(x=3\text{ft}\right)$:

Substitute 3 ft for x in Equation (2).

$\begin{array}{c}{V}_C=\text{13}-3\left(3\right)+3{\langle 3-3\rangle }^1-8{\langle 3-7\rangle }^0-3{\langle 3-11\rangle }^1\\ =13-9+0-0-0\\ =\text{4 kips}\end{array}$

Point D $\left(x=7\text{ft}\right)$

Substitute 7 ft for x in Equation (2).

$\begin{array}{c}{V}_D^{Left}=\text{13}-3\left(7\right)+3{\langle 7-3\rangle }^1-8{\langle 7-7\rangle }^0-3{\langle 7-11\rangle }^1\\ =13-21+12-0-0\\ =\text{4 kips}\end{array}$

$\begin{array}{c}{V}_D^{Right}=\text{13}-3\left(7\right)+3{\langle 7-3\rangle }^1-8{\langle 7-7\rangle }^0-3{\langle 7-11\rangle }^1\\ =13-21+12-8-0\\ =-\text{4 kips}\end{array}$

Point E $\left(x=11\text{ft}\right)$:

Substitute 11 ft for x in Equation (2).

$\begin{array}{c}{V}_E=\text{13}-3\left(11\right)+3{\langle 11-3\rangle }^1-8{\langle 11-7\rangle }^0-3{\langle 11-11\rangle }^1\\ =13-33+24-8-0\\ =-\text{4 kips}\end{array}$

Point B $\left(x=14\text{ft}\right)$:

Substitute 14 ft for x in Equation (2).

$\begin{array}{c}{V}_B=\text{13}-3\left(14\right)+3{\langle 14-3\rangle }^1-8{\langle 14-7\rangle }^0-3{\langle 14-11\rangle }^1\\ =13-42+33-8-9\\ =-\text{13 kips}\end{array}$

Refer to the calculated shear force values, the shear force changes at point D.

Refer to Equation (3).

Substitute 7 ft for x in Equation (3).

$\begin{array}{c}{M}_{\max }=13\left(7\right)-1.5{\left(7\right)}^2+1.5{\langle 7-3\rangle }^2-8{\langle 7-7\rangle }^1-1.5{\langle 7-11\rangle }^2\\ =91-73.5+24-0-0\\ =41.5\text{kips-ft}\end{array}$

Therefore, the maximum bending moment in the beam is $\underset{\_}{41.5\text{kips-ft}}$.