Chapter 4, Problem 21P

Consider the uniformly loaded simply supported steel beam with an overhang as shown. The second-area moment of the beam is I = 0.05 in⁴. Use superposition (with Table A–9 and the results of Prob. 4–20) to determine the reactions and the deflection equations of the beam. Plot the deflections.

Problem 4–21

To determine

The net reaction at $A$.

The net reaction at $B$.

The expression for the deflection in the beam of portion $AB$ using superposition.

The expression for the deflection in the beam of portion $BC$ using superposition.

The plot of deflection verses length of the beam.

Answer to Problem 21P

The net reaction at $A$ is $\text{420}\text{lbf}$.

The net reaction at $B$ is $980\text{lbf}$.

The expression for the deflection in the beam of portion $AB$ using superposition is

$y_{AB}=\left[2.777\times {10}^{-6}x\left(\left(20\right)x^2-x^3-1000\right)+\left(\left(8.888\times {10}^{-6}\right)x\left(100-x^2\right)\right)\right]$.

The expression for the deflection in the beam of portion $BC$ using superposition is

$y_{BC}=\left\{\left(2.777\times {10}^{-3}\right)\left(x-10\right)-\left(2.777\times {10}^{-6}\right)\left[{\left(14+x\right)}^4+896x-9216\right]\right\}$.

The plot of deflection verses length of the beam is

Explanation of Solution

Write the expression for the reaction at $A$ (one end) due to uniform load between the supports.

    $R_1=\frac{wl}{2}$ (I)

Here, the uniform load on the beam is $w$, the length of the beam is $l$ and the reaction at the $A$ is $R_1$.

Write the expression for the reaction at $A$ (other end) due to uniform load between the supports.

    $R_2=\frac{wl}{2}$                                                                                          (II)

Here, the reaction at the $A$ is $R_2$.

Write the expression for the deflection of the beam between $A$ and $B$.

    $y_{AB}=\frac{wx}{24EI}\left(2l{x}^2-x^3-l^3\right)$                                                           (III)

Here, the moment of inertia of the beam is $I$ and the modulus of the elasticity is $E$

Write the expression for the slope of the deflection in the beam.

    ${\theta }_{AB}=\frac{d{y}_{AB}}{dx}$                                                                                  (IV)

Substitute $\frac{wx}{24EI}\left(2l{x}^2-x^3-l^3\right)$ for $y_{AB}$ in Equation (IV).

    $\begin{array}{c}{\theta }_{AB}=\frac{d{y}_{AB}}{dx}\\ =\frac{d\left(\frac{wx}{24EI}\left(2l{x}^2-x^3-l^3\right)\right)}{dx}\\ =\frac{w}{24EI}\left(6l{x}^2-4x^3-l^3\right)\end{array}$                                                     (V)

Substitute $l$ for $x$ in Equation (V).

    $\begin{array}{c}{\theta }_{AB}=\frac{w}{24EI}\left(6l\left(l^2\right)-4l^3-l^3\right)\\ =\frac{w{l}^3}{24EI}\end{array}$                                                       (VI)

Write the expression for the deflection of the beam for $BC$ portion.

    $y_{BC}={{\theta }_{AB}|}_{x=l}\left(x-l\right)$                                                                     (VII)

Substitute $\frac{w{l}^3}{24EI}$ for ${\theta }_{AB}$ in Equation (VII).

    $\begin{array}{c}{y}_{BC}={{\theta }_{AB}|}_{x=l}\left(x-l\right)\\ =\frac{w{l}^3}{24EI}\left(x-l\right)\end{array}$                                                                  (VIII)

Write the expression for the reaction at one end of the overhang section.

    $R_A=\frac{w{a}^2}{2l}$                                                                                    (IX)

Here, the distance between $B$ and $C$ is $a$.

Write the expression for the reaction at other end of the overhang section.

    $R_B=\frac{wa}{2l}\left(2l-a\right)$                                                                            (X)

Write the expression for the deflection in the beam section $AB$ due to overhung load.

    $y_{AB}=\frac{w{a}^{2}x}{12EIl}\left(l^2-x^2\right)$                                                                   (XI)

Write the expression for the deflection in the beam section $BC$ due to overhung load.

    $y_{BC}=-\frac{w}{24EI}\left[{\left(l+a-x\right)}^4-4a^2\left(l-x\right)\left(l+a\right)-a^4\right]$               (XII)

Write the expression for the net reaction at $A$.

    $R_{nA}=R_1-R_A$                                                                          (XIII)

Write the expression for the net reaction at $B$.

    $R_{nB}=R_2+R_B$                                                                          (XIV)

Add equation (III) and (XI).

    $\begin{array}{c}{y}_{AB}=\frac{wx}{24EI}\left(2l{x}^2-x^3-l^3\right)+\frac{w{a}^{2}x}{12EIl}\left(l^2-x^2\right)\\ =\frac{wx}{12EI}\left[\frac{1}{2}\left(2l{x}^2-x^3-l^3\right)+\frac{a^2}{l}\left(l^2-x^2\right)\right]\end{array}$                              (XV)

Add equation (VII) and (XII).

    $\begin{array}{c}{y}_{BC}=\frac{w{l}^3}{24EI}\left(x-l\right)-\frac{w}{24EI}\left[{\left(l+a-x\right)}^4-4a^2\left(l-x\right)\left(l+a\right)-a^4\right]\\ =\frac{w}{24EI}\left[l^3\left(x-l\right)-\left[{\left(l+a-x\right)}^4-4a^2\left(l-x\right)\left(l+a\right)-a^4\right]\right]\end{array}$ (XVI)

Conclusion:

Substitute $100\text{lbf}/\text{in}$ for $w$ and $10\text{in}$ for $l$ in Equation (I).

    $\begin{array}{c}{R}_1=\frac{\left(100\text{lbf}/\text{in}\right)\left(10\text{in}\right)}{2}\\ =500\text{lbf}\end{array}$

Substitute $100\text{lbf}/\text{in}$ for $w$ and $10\text{in}$ for $l$ in Equation (II).

    $\begin{array}{c}{R}_2=\frac{\left(100\text{lbf}/\text{in}\right)\left(10\text{in}\right)}{2}\\ =500\text{lbf}\end{array}$

Substitute $100\text{lbf}/\text{in}$ for $w$,  $4\text{in}$ for $a$ and $10\text{in}$ for $l$ in Equation (IX).

    $\begin{array}{c}{R}_A=\frac{\left(100\text{lbf}/\text{in}\right){\left(4\text{in}\right)}^2}{2\left(10\text{in}\right)}\\ =\frac{1600\text{lbf}/\text{in}\left({\text{in}}^2\right)}{20\text{in}}\\ =80\text{lbf}\end{array}$

Substitute $100\text{lbf}/\text{in}$ for $w$, $4\text{in}$ for $a$ and $10\text{in}$ for $l$ in Equation (X).

    $\begin{array}{c}{R}_B=\frac{\left(100\text{lbf}/\text{in}\right)\left(4\text{in}\right)}{2\left(10\text{in}\right)}\left(2\left(10\text{in}\right)+4\text{in}\right)\\ =\frac{9600\text{lbf}/\text{in}\left({\text{in}}^2\right)}{20\text{in}}\\ =480\text{lbf}\end{array}$

Substitute $80\text{lbf}$ for $R_A$ and $500\text{lbf}$ for $R_1$ in Equation (XIII).

    $\begin{array}{c}{R}_{nA}=500\text{lbf}-80\text{lbf}\\ =\text{420}\text{lbf}\end{array}$

Thus, the net reaction at $A$ is $\text{420}\text{lbf}$.

Substitute $480\text{lbf}$ for $R_B$ and $500\text{lbf}$ for $R_2$ in Equation (XIV).

    $\begin{array}{c}{R}_{nB}=500\text{lbf+}480\text{lbf}\\ =98\text{0}\text{lbf}\end{array}$

Thus, the net reaction at $B$ is $98\text{0}\text{lbf}$.

Substitute $100\text{lbf}/\text{in}$ for $w$, $30\times {10}^6\text{psi}$ for $E$, $0.05{\text{in}}^4$ for $I$, $4\text{in}$ for $a$ and $10\text{in}$ for $l$ in Equation (XV).

    $\begin{array}{c}{y}_{AB}=\frac{\left(100\text{lbf}/\text{in}\right)x}{12\left(30\times {10}^6\text{psi}\right)\left(0.05{\text{in}}^4\right)}\left[\begin{array}{l}\frac{1}{2}\left(2\left(10\text{in}\right)x^2-x^3-{\left(10\text{in}\right)}^3\right)\\ -\frac{{\left(4\text{in}\right)}^2}{\left(10\text{in}\right)}\left({\left(10\text{in}\right)}^2-x^2\right)\end{array}\right]\\ =5.555\times {10}^{-6}x\left[\begin{array}{l}\left(0.5\left(20\text{in}\right)x^2-x^3-1000{\text{in}}^3\right)\\ -\left(1.6\text{in}\left(100{\text{in}}^2-x^2\right)\right)\end{array}\right]\\ =\left[2.777\times {10}^{-6}x\left(\left(20\right)x^2-x^3-1000\right)+\left(\left(8.888\times {10}^{-6}\right)x\left(100-x^2\right)\right)\right]\end{array}$ (XVII)

Thus, the expression for the deflection in the beam of portion $AB$ using superposition is $y_{AB}=\left[2.777\times {10}^{-6}x\left(\left(20\right)x^2-x^3-1000\right)+\left(\left(8.888\times {10}^{-6}\right)x\left(100-x^2\right)\right)\right]$.

Substitute $100\text{lbf}/\text{in}$ for $w$, $30\times {10}^6\text{psi}$ for $E$, $0.05{\text{in}}^4$ for $I$, $4\text{in}$ for $a$ and $10\text{in}$ for $l$ in Equation (XVI).

    $\begin{array}{c}{y}_{BC}=\frac{\left(100\text{lbf}/\text{in}\right)}{24\left(30\times {10}^6\text{psi}\right)\left(0.05{\text{in}}^4\right)}\left[\begin{array}{l}\left({\left(10\text{in}\right)}^3\left(x-10\text{in}\right)\right)\\ -\left[\begin{array}{l}{\left(10\text{in}+4\text{in}-x\right)}^4\\ -4{\left(4\text{in}\right)}^2\left(10\text{in}-x\right)\left(10\text{in}+4\text{in}\right)\\ -{\left(4in\right)}^4\end{array}\right]\end{array}\right]\\ =2.777\times {10}^{-6}\left[\left(\left(1000\right)\left(x-10\right)\right)-\left[{\left(14+x\right)}^4-896\left(10-x\right)-256\right]\right]\\ =\left\{\left(2.777\times {10}^{-3}\right)\left(x-10\right)-\left(2.777\times {10}^{-6}\right)\left[{\left(14+x\right)}^4+896x-9216\right]\right\}\end{array}$. (XVIII)

Thus, the expression for the deflection in the beam of portion $BC$ using superposition is $y_{BC}=\left\{\left(2.777\times {10}^{-3}\right)\left(x-10\right)-\left(2.777\times {10}^{-6}\right)\left[{\left(14+x\right)}^4+896x-9216\right]\right\}$.

Substitute different values of $x$ to calculate the deflection at different points of the beam.

Substitute $0$ for $x$ in Equation (XVII).

    $\begin{array}{c}{y}_{AB}=\left[2.777\times {10}^{-6}\left(0\right)\left(\left(20\right){\left(0\right)}^2-{\left(0\right)}^3-1000\right)+\left(\left(8.888\times {10}^{-6}\right)\left(0\right)\left(100-{\left(0\right)}^2\right)\right)\right]\\ =0\end{array}$

Substitute $1$ for $x$ in Equation (XVII).

    $\begin{array}{c}{y}_{AB}=\left[2.777\times {10}^{-6}\left(1\right)\left(\left(20\right){\left(1\right)}^2-{\left(1\right)}^3-1000\right)+\left(\left(8.888\times {10}^{-6}\right)\left(1\right)\left(100-{\left(1\right)}^2\right)\right)\right]\\ =\left(-2.7242\times {10}^{-3}\right)+\left(8.7991\times {10}^{-4}\right)\\ =-0.001844\end{array}$

Similarly,

Use excel spread sheet to calculate the $y$ values for different values of $x$.

The Table-(1) shows the different values of the deflection at different point of the beam.

Table-(1)

S. No.	length $\left(x\right)$ $\text{in}$	deflection $\left(y\right)$ $\text{in}$
1	$0$	$0.0000$
2	$0.5$	$-0.00093$
3	$1.0$	$-0.00184$
4	$1.5$	$-0.00269$
5	$2$	$-0.00344$
6	$2.5$	$-0.00410$
7	$3$	$-0.00463$
8	$3.5$	$-0.00502$
9	$4$	$-0.00528$
10	$4.5$	$-0.00538$
11	$5$	$-0.00534$
12	$5.5$	$-0.00516$
13	$6$	$-0.00485$
14	$6.5$	$-0.00442$
15	$7$	$-0.00388$
16	$7.5$	$-0.00326$
17	$8$	$-0.00259$
18	$8.5$	$-0.00189$
19	$9$	$-0.00120$
20	$9.5$	$-0.00055$
21	$10$	$0.00000$
22	$10.5$	$0.00043$
23	$11$	$0.00077$
24	$11.5$	$0.00103$
25	$12$	$0.00124$
26	$12.5$	$0.00141$
27	$13$	$0.00157$
28	$13.5$	$0.00172$
29	$14$	$0.00186$

Draw the plot of length of the beam verses deflection of the beam.

Figure-(1)