Chapter 3, Problem 12P

Repeat Prob. 3–8 using singularity functions exclusively (including reactions).

To determine

The reaction at the supports.

The shear force diagram.

The bending moment diagram.

Answer to Problem 12P

The reaction force $R_1$ is $200\text{lbf}$.

The reaction force $R_1$ is $440\text{lbf}$.

The reaction force $R_1$ is $160\text{lbf}$.

The shear force diagram is shown is below.

The bending moment diagram is shown below.

Explanation of Solution

The following figure shows the load applied to a beam.

Figure-(1)

Refer to Table 3-1 “Singularity (Macaulay) functions”.

Write the expression for load intensity acting on the beam.

    $q=\left[\begin{array}{l}{R}_1{\langle x\rangle }^{-1}-F_A{\langle x-a\rangle }^{-1}+R_2{\langle x-b\rangle }^{-1}-w{\langle x-b\rangle }^0+w{\langle x-l\rangle }^0\\ +R_3{\langle x-l\rangle }^{-1}\end{array}\right]$             (I)

Here, the reaction force at point O is $R_1$, the load at point A is $F_A$, the reaction force at point C is $R_2$, the uniform load is $w$, the reaction at point D is $R_3$, the length of the beam is $l$, the distance between the point O and A is $a$, the distance between the point O and C is $b$, the load intensity on the beam is $q$ and the distance from the left end is $x$.

Write the expression for the shear force.

    $V={\displaystyle \int qdx}$                                                                                                   (II)

Here, the net shear force is $V$.

Substitute the value from Equation (I) to Equation (II).

    $\begin{array}{c}V={\displaystyle \int \left[\begin{array}{l}{R}_1{\langle x\rangle }^{-1}-F_A{\langle x-a\rangle }^{-1}+R_2{\langle x-b\rangle }^{-1}-w{\langle x-b\rangle }^0+w{\langle x-l\rangle }^0\\ +R_3{\langle x-l\rangle }^{-1}\end{array}\right]dx}\\ =\left[\begin{array}{c}{R}_1{\langle x\rangle }^0-F_A{\langle x-a\rangle }^0+R_2{\langle x-b\rangle }^0-w{\langle x-b\rangle }^1+w{\langle x-l\rangle }^1\\ +R_3{\langle x-l\rangle }^0\end{array}\right]\end{array}$     (III)

Write the expression for the moment.

    $M={\displaystyle \int Vdx}$                                                                                                  (IV)

Here, the net bending moment is $M$.

Substitute the value from Equation (III) to Equation (IV).

    $\begin{array}{c}M={\displaystyle \int \left[\begin{array}{c}{R}_1{\langle x\rangle }^0-F_A{\langle x-a\rangle }^0+R_2{\langle x-b\rangle }^0-w{\langle x-b\rangle }^1+w{\langle x-l\rangle }^1\\ +R_3{\langle x-l\rangle }^0\end{array}\right]}dx\\ =\left[\begin{array}{c}{R}_1{\langle x\rangle }^1-F_A{\langle x-a\rangle }^1+R_2{\langle x-b\rangle }^1-\frac{w}{2}{\langle x-b\rangle }^2+\frac{w}{2}{\langle x-l\rangle }^2\\ +R_3{\langle x-l\rangle }^1\end{array}\right]\end{array}$          (V)

Conclusion:

Substitute $0\text{lbf}\cdot \text{in}$ for $M$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$ and $8\text{in}$ for $x$ in Equation (V).

    $\begin{array}{c}\left(0\text{lbf}\cdot \text{in}\right)=R_1{\langle \left(8\text{in}\right)\rangle }^1-\left(400\text{lbf}\right){\langle \left(8\text{in}\right)-\left(4\text{in}\right)\rangle }^1+0-0+0\\ \left(0\text{lbf}\cdot \text{in}\right)=\left(8\text{in}\right)R_1-\left(400\text{lbf}\right)\left(4\text{in}\right)\\ R_1=\frac{\left(1600\text{lbf}\cdot \text{in}\right)}{\left(8\text{in}\right)}\\ =200\text{lbf}\end{array}$

Thus, the reaction force $R_1$ is $200\text{lbf}$.

Substitute $0\text{lbf}$ for $V$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$ and $20\text{in}$ for $x$ in Equation (III).

    $\begin{array}{l}\left(0\text{lbf}\right)=\left[\begin{array}{l}\left(200\text{lbf}\right){\langle \left(20\text{in}\right)\rangle }^0-\left(400\text{lbf}\right){\langle 20-4\rangle }^0+R_2{\langle 20-10\rangle }^0\\ -40{\langle 20-10\rangle }^1+\left(40\text{lbf}/\text{in}\right){\langle 20\text{in}-20\rangle }^1+R_3{\langle 20-20\rangle }^0\end{array}\right]\\ \left(0\text{lbf}\right)=200\text{lbf}-400\text{lbf}+R_2-40\text{lbf}/\text{in}\left(10\text{in}\right)+R_3\\ R_2+R_3=600\text{lbf}\end{array}$        (VI)

Substitute $0$ for $M$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$ and $20\text{in}$ for $x$ in Equation (V).

    $\begin{array}{l}0=\left[\begin{array}{l}200\text{lbf}\times {\langle 20\text{in}\rangle }^1-400\text{lbf}{\langle 20\text{in}-4\text{in}\rangle }^1+R_2{\langle 20\text{in}-10\text{in}\rangle }^1\\ -\left(\frac{40}{2}\text{lbf}/\text{in}\right){\langle 20-10\rangle }^2+\left(\frac{40\text{lbf}/\text{in}}{2}\right){\langle 20-20\rangle }^2+R_3{\langle 20-20\rangle }^1\end{array}\right]\\ 0=200\text{lbf}\left(20\text{in}\right)-400\text{lbf}\left(16\text{in}\right)+R_2\left(10\text{in}\right)-20\text{lbf}\left(100\text{in}\right)\\ R_2=440\text{lbf}\end{array}$

Thus, the reaction force $R_2$ is $440\text{lbf}$.

Substitute $440\text{lbf}$ for $R_2$ in Equation (VI).

    $\begin{array}{c}{R}_3=600\text{lbf}-440\text{lbf}\\ =160\text{lbf}\end{array}$

Thus, the reaction force $R_3$ is $160\text{lbf}$.

Substitute the value of $x$ in the range of $0\le x\le 4$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$, $440\text{lbf}$ for $R_2$ and $160\text{lbf}$ for $R_3$ in Equation (III).

    $\begin{array}{c}V=200\text{lbf}{\langle x\rangle }^0\\ =200\text{lbf}\end{array}$

The magnitude of shear force remains constant.

Substitute the value of $x$ in the range of $0\le x\le 4$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$, $440\text{lbf}$ for $R_2$ and $160\text{lbf}$ for $R_3$ in Equation (V).

    $\begin{array}{c}M=200\text{lbf}{\langle x\rangle }^1-0+0\\ =200x\end{array}$                                                                       (VII)

At $x=2\text{in}$, calculate the bending  moment.

Substitute $2\text{in}$ for $x$ in Equation (VII).

    $\begin{array}{c}M=200\text{lbf}\times 2\text{in}\\ =400\text{lbf}\cdot \text{in}\end{array}$

At $x=3\text{in}$, calculate the bending  moment.

Substitute $3\text{in}$ for $x$ in Equation (VII).

    $\begin{array}{c}M=200\text{lbf}\times 3\text{in}\\ =600\text{lbf}\cdot \text{in}\end{array}$

Similarly for the other values of $x$ bending moment can be calculated.

The Table-1 shows the variation of moment.

Table-1

$x\left(\text{in}\right)$	$M\left(\text{lbf}\cdot \text{in}\right)$
$0$	$0$
$2$	$400$
$3$	$600$

Substitute the value of $x$ in the range of $4\le x\le 10$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$, $440\text{lbf}$ for $R_2$ and $160\text{lbf}$ for $R_3$ in Equation (III).

    $\begin{array}{c}V=200{\langle x\rangle }^0-400{\langle x-4\rangle }^0\\ =200-400\\ =-200\text{lbf}\end{array}$

The magnitude of shear force remains constant.

Substitute the value of $x$ in the range of $4\le x\le 10$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$, $440\text{lbf}$ for $R_2$ and $160\text{lbf}$ for $R_3$ in Equation (V).

    $\begin{array}{c}M=200{\langle x\rangle }^1-400{\langle x-4\rangle }^1\\ =200x-400x+1600\\ =-200x+1600\end{array}$                                                      (VIII)

At $x=4\text{in}$, calculate the bending moment.

Substitute $4\text{in}$ for $x$ in Equation (VIII).

    $\begin{array}{c}M=-200\text{lbf}\times 4\text{in}+1600\text{lbf}\cdot \text{in}\\ =-800\text{lbf}\cdot \text{in}+1600\text{lbf}\cdot \text{in}\\ =800\text{lbf}\cdot \text{in}\end{array}$

At $x=6\text{in}$, calculate the bending moment.

Substitute $6\text{in}$ for $x$ in Equation (VIII).

    $\begin{array}{c}M=-200\text{lbf}\times 6\text{in}+1600\text{lbf}\cdot \text{in}\\ =-1200\text{lbf}\cdot \text{in}+1600\text{lbf}\cdot \text{in}\\ =400\text{lbf}\cdot \text{in}\end{array}$

Similarly for the other values of $x$ bending moment can be calculated.

Table-2 shows the variation of moment.

Table-2

$x\left(\text{in}\right)$	$M\left(\text{lbf}\cdot \text{in}\right)$
$4$	$800$
$6$	$400$
$10$	$-400$

Substitute the value of $x$ in the range of $10\le x\le 20$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$, $440\text{lbf}$ for $R_2$ and $160\text{lbf}$ for $R_3$ in Equation (III).

    $\begin{array}{c}V=200\text{lbf}{\langle x\rangle }^0-400\text{lbf}{\langle x-4\rangle }^0+440\text{lbf}{\langle x-10\rangle }^0-40\text{lbf}/\text{in}{\langle x-10\rangle }^1\\ V=200\text{lbf}-400\text{lbf}+440\text{lbf}-40\text{lbf}/\text{in}\left(x-10\right)\\ =640-40x\end{array}$ (IX)

At $x=12\text{in}$, calculate the shear force.

Substitute $12\text{in}$ for $x$ in Equation (IX).

    $\begin{array}{c}V=640-40\left(12\right)\\ =160\text{lbf}\end{array}$

At $x=16\text{in}$, calculate the shear force.

Substitute $16\text{in}$ for $x$ in Equation (IX).

    $\begin{array}{c}V=640\text{lbf}-\left(40\text{lbf}/\text{in}\right)\left(16\text{in}\right)\\ =640\text{lbf}-640\text{lbf}\\ =0\text{lbf}\end{array}$

Similarly for the other values of $x$ shear force can be calculated.

The Table-3 shows the variation of shear force.

Table-3

$x\left(\text{in}\right)$	$V\left(\text{lbf}\right)$
$10$	$240$
$12$	$160$
$16$	$0$

Substitute the value of $x$ in the range of $10\le x\le 20$, $400\text{lbf}$ for $F_A$, $40\text{lbf}/\text{in}$ for $w$, $20\text{in}$ for $l$, $4\text{in}$ for $a$, $10\text{in}$ for $b$, $200\text{lbf}$ for $R_1$, $440\text{lbf}$ for $R_2$ and $160\text{lbf}$ for $R_3$ in Equation (V).

    $\begin{array}{c}M=200{\langle x\rangle }^1-400{\langle x-4\rangle }^1+440{\langle x-10\rangle }^1-\frac{40}{2}{\langle x-10\rangle }^2\\ =200x-400\left(x-4\right)+440\left(x-10\right)-20{\left(x-10\right)}^2\\ =-20x^2+640x-4800\end{array}$                   (X)

At $x=12\text{in}$, calculate the bending moment.

Substitute $12\text{in}$ for $x$ in Equation (X).

    $\begin{array}{c}M=\left(-20\text{lbf}/\text{in}\right){\left(12\right)}^2+\left(640\text{lbf}\right)\left(12\text{in}\right)-4800\text{lbf}\cdot \text{in}\\ \text{=}\left(-20\text{lbf}/\text{in}\right){\left(12\right)}^2+7680\text{lbf}\cdot \text{in}-4800\text{lbf}\cdot \text{in}\\ =0\text{lbf}\cdot \text{in}\end{array}$

At $x=16\text{in}$, calculate the bending moment.

Substitute $16\text{in}$ for $x$ in Equation (X).

    $\begin{array}{c}M=\left(-20\text{lbf}/\text{in}\right){\left(16\text{in}\right)}^2+\left(640\text{lbf}\right)\left(16\text{in}\right)-4800\text{lbf}\cdot \text{in}\\ =\left(-20\text{lbf}/\text{in}\right){\left(16\text{in}\right)}^2+10240\text{lbf}\cdot \text{in}-4800\text{lbf}\cdot \text{in}\\ =320\text{lbf}\cdot \text{in}\end{array}$

Similarly for the other values of $x$ bending moment can be calculated.

Differentiate Equation (X).

    $\begin{array}{c}\frac{dM}{dx}=\frac{d}{dx}\left(-20x^2+640x-4800\right)\\ =-40x+640\end{array}$

Equate the above expression to zero to obtain the location of maxima.

    $\begin{array}{l}-40x+640=0\\ x=\frac{640\text{in}}{40}\\ x=16\text{in}\end{array}$

The Table-4 shows the variation of moment.

Table-4

$x\left(\text{in}\right)$	$M\left(\text{lbf}\cdot \text{in}\right)$
$10$	$-400$
$12$	$0$
$16$	$320$

The shear force diagram is shown in Figure-(2).

Figure-(2)

The bending moment diagram is shown in Figure-(3).

Figure-(3)