Chapter 9.6, Problem 144P

To determine

The magnitude $\left(y_K\right)$ and location $\left(x_K\right)$ of the largest downward deflection.

Answer to Problem 144P

The magnitude $\left(y_K\right)$ and location $\left(x_K\right)$ of the largest downward deflection are $\underset{\_}{0.212\text{in}\text{.}\left(\text{downward}\right)}$ and $\underset{\_}{5.15\text{ft}}$ respectively.

Explanation of Solution

Given information:

The elastic modulus (E) is $29\times {10}^6\text{psi}$.

The section of the beam is $\text{W}12\times 26$.

Calculation:

Refer Appendix C, “Properties of Rolled steel shapes”.

The moment of inertia (I) for the given section is $204\text{in}{\text{.}}^4$.

Use moment area method:

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

Determine the reaction of the support by taking moment at point B.

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ -\left(R_A\times 10\right)+\left[\left(5\times 4\right)\left(4+\frac{4}{2}\right)\right]+\left(8\times 4\times \frac{4}{2}\right)=0\\ R_A=18.4\text{kips}\end{array}$

Determine the reaction of the support by considering the vertical equilibrium condition:

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_A+R_B=\left(5\times 4\right)+\left(8\times 4\right)\\ R_A+R_B=52\end{array}$

Substitute $18.4\text{kips}$ for $R_A$.

$\begin{array}{l}{R}_B+18.4=52\\ R_B=33.6\text{kips}\end{array}$

Show the moment $\left(\frac{M}{EI}\right)$ diagram for the given loading of the beam as in Figure 2.

Calculate the moment $\left(M_1\right)$ using the formula:

$M_1=R_A\times 10$

Substitute $\text{18}\text{.4kips}$ for $R_A$.

$\begin{array}{c}{M}_1=18.4\times 10\\ =184\text{kips}\cdot \text{ft}\end{array}$

Calculate the ratio of $\frac{M_1}{EI}$:

$\text{Ratio}=\frac{M_1}{EI}$

Substitute $184\text{kips}\cdot \text{ft}$ for $M_1$.

$\frac{M_1}{EI}=\frac{184\text{kips}\cdot \text{ft}}{EI}$

Calculate the area $\left(A_1+A_2\right)$ due to the moment $\left(M_1\right)$ using the formula:

$A_1+A_2=\frac{1}{2}bh$

Here, b is the width of the triangle in area $\left(A_1+A_2\right)$ and h is the height of the triangle in area $\left(A_1+A_2\right)$.

Substitute $\text{10ft}$ for b and $\frac{184\text{kips}\cdot \text{ft}}{EI}$ for h.

$\begin{array}{c}{A}_1+A_2=\frac{1}{2}\times 10\times \left(\frac{184\text{kips}\cdot \text{ft}}{EI}\right)\\ =\frac{920}{EI}\text{kips}\cdot {\text{ft}}^2\end{array}$

Calculate the moment $\left(M_2\right)$ using the formula:

$M_2=R_A\times 6\text{ft}$

Substitute $\text{18}\text{.4kips}$ for $R_A$.

$\begin{array}{c}{M}_2=18.4\times 6\text{ft}\\ \text{=110}\text{.4kips}\cdot \text{ft}\end{array}$

Calculate the ratio of $\frac{M_2}{EI}$:

$\text{Ratio}=\frac{M_2}{EI}$

Substitute $\text{110}\text{.4kips}\cdot \text{ft}$ for $M_2$.

$\frac{M_2}{EI}=\frac{\text{110}\text{.4kips}\cdot \text{ft}}{EI}$

Calculate the area $\left(A_1\right)$ due to the moment $\left(M_2\right)$ using the formula:

$A_1=\frac{1}{2}{b}_1{h}_1$

Here, $b_1$ is the width of the rectangle in area $\left(A_1\right)$ and $h_1$ is the height of the rectangle in area $\left(A_1\right)$.

Substitute $\text{6ft}$ for $b_1$ and $\frac{\text{110}\text{.4kips}\cdot \text{ft}}{EI}$ for $h_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\times 6\times \left(\frac{\text{110}\text{.4kips}\cdot \text{ft}}{EI}\right)\\ =\frac{331.2}{EI}\text{kips}\cdot {\text{ft}}^2\end{array}$

Calculate the moment $\left(M_3\right)$ using the formula:

$\begin{array}{c}{M}_3=-5\times 8\times \frac{8}{2}\\ =-160\text{kips}\cdot \text{ft}\end{array}$

Calculate the ratio of $\frac{M_3}{EI}$ using the relation:

$\text{Ratio}=\frac{M_3}{EI}$

Substitute $-160\text{kips}\cdot \text{ft}$ for $M_3$.

$\frac{M_3}{EI}=\frac{-160\text{kips}\cdot \text{ft}}{EI}$

Calculate the area $\left(A_3+A_4\right)$ due to the moment $\left(M_3\right)$ using the formula:

$A_3+A_4=\frac{1}{3}{b}_3{h}_3$

Here, $b_3$ is the width of the parabola in area $\left(A_3+A_4\right)$ and $h_3$ is the height of the parabola in area $\left(A_3+A_4\right)$.

Substitute $\text{8ft}$ for $b_3$ and $\frac{-160\text{kips}\cdot \text{ft}}{EI}$ for $h_3$.

$\begin{array}{c}{A}_3+A_4=\frac{1}{3}\times 8\times \left(\frac{-160\text{kips}\cdot \text{ft}}{EI}\right)\\ =\frac{-426.667}{EI}\text{kips}\cdot {\text{ft}}^2\end{array}$

Calculate the moment $\left(M_4\right)$:

$\begin{array}{c}{M}_4=-\left(5\times 4\right)\frac{4}{2}\\ =-40\text{kips}\cdot \text{ft}\end{array}$

Calculate the ratio of $\frac{M_4}{EI}$ using the relation:

$\text{Ratio}=\frac{M_4}{EI}$

Substitute $-40\text{kips}\cdot \text{ft}$ for $M_4$.

$\frac{M_4}{EI}=\frac{-40\text{kips}\cdot \text{ft}}{EI}$

Calculate the area $\left(A_3\right)$ due to the moment $\left(M_4\right)$ using the formula:

$A_3=\frac{1}{3}{b}_4{h}_4$

Here, $b_4$ is the width of the parabola in area $\left(A_3\right)$ and $h_4$ is the height of the parabola in area $\left(A_3\right)$.

Substitute $4\text{ft}$ for $b_4$ and $\frac{-40\text{kips}\cdot \text{ft}}{EI}$ for $h_4$.

$\begin{array}{c}{A}_3=\frac{1}{3}\times 4\times \left(\frac{-40\text{kips}\cdot \text{ft}}{EI}\right)\\ =\frac{-53.333}{EI}\text{kips}\cdot {\text{ft}}^2\end{array}$

Show the moment $\left(\frac{M_5}{EI}\right)$ diagram for the given loading of the beam as in Figure 3.

Calculate the moment $\left(M_5\right)$ using the formula:

$\begin{array}{c}{M}_5=-\left(8-5\right)\text{kips}/\text{ft}\times 4\times \frac{4}{2}\\ =-24\text{kips}\cdot \text{ft}\end{array}$

Calculate the ratio of $\frac{M_5}{EI}$:

$\text{Ratio}=\frac{M_5}{EI}$

Substitute $-\text{24kips}\cdot \text{ft}$ for $M_5$.

$\frac{M_5}{EI}=\frac{-\text{24}\text{kips}\cdot \text{ft}}{EI}$

Calculate the area $\left(A_5\right)$ due to the moment $\left(M_5\right)$ using the formula:

$A_5=\frac{1}{3}{b}_5{h}_5$

Here, $b_5$ is the width of the parabola in area $\left(A_5\right)$ and $h_5$ is the height of the parabola in area $\left(A_5\right)$.

Substitute $\text{4ft}$ for $b_5$ and $\frac{-\text{24}\text{kips}\cdot \text{ft}}{EI}$ for $h_5$.

$\begin{array}{c}{A}_5=\frac{1}{3}\times 4\times \left(\frac{\text{-24kips}\cdot \text{ft}}{EI}\right)\\ =\frac{-32}{EI}\text{kips}\cdot {\text{ft}}^2\end{array}$

Calculate the slope at the end A related to the point B $\left(t_{B/A}\right)$ using the formula:

$t_{B/A}=\left[\left(A_1+A_2\right)\times \left(\frac{1}{3}\times 10\right)\right]+\left[\left(A_3+A_4\right)\times \left(\frac{1}{4}\times 8\right)\right]+\left[A_5\times \left(\frac{1}{4}\times 4\right)\right]$

Substitute $\frac{920}{EI}\text{kips}\cdot {\text{ft}}^2$ for $A_1+A_2$, $\frac{-426.667}{EI}\text{kips}\cdot {\text{ft}}^2$ for $A_3+A_4$, $\frac{-32}{EI}\text{kips}\cdot {\text{ft}}^2$ for $A_5$, $29\times {10}^3\text{ksi}$ for E, and $204\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}{t}_{B/A}=\left[\frac{920}{EI}\left(\frac{1}{3}\times 10\right)\right]+\left[\frac{-426.667}{EI}\left(\frac{1}{4}\times 8\right)\right]+\left[\frac{-32}{EI}\left(\frac{1}{4}\times 4\right)\right]\\ =\frac{3,066.667}{EI}-\frac{853.334}{EI}-\frac{32}{EI}\\ =\frac{3,066.667-853.334-32}{EI}\end{array}$

$\begin{array}{c}=\frac{\text{2,181}\text{.333kips}\cdot {\text{ft}}^2}{\frac{29\times {10}^3\times 204}{144\text{in}{\text{.}}^2/{\text{ft}}^2}}\\ =53.096\times {10}^{-3}\text{ft}\end{array}$

Calculate the slope $\left({\theta }_A\right)$ at point A using the formula:

${\theta }_A=\frac{-t_{B/A}}{L}$

Substitute $53.096\times {10}^{-3}\text{ft}$ for $t_{B/A}$ and $\text{10ft}$ for L.

$\begin{array}{c}{\theta }_A=-\frac{53.096\times {10}^{-3}\text{in}\text{.}}{\left(10\right)\text{in}\text{.}}\\ =-5.31\times {10}^{-3}\text{rad}\end{array}$

Calculate the slope $\left({\theta }_{E/A}\right)$ using the formula:

${\theta }_{E/A}=\left(A_1+A_3\right)$

Substitute $\frac{331.2}{EI}$ for $A_1$, $\frac{-53.333}{EI}$ for $A_3$, $29\times {10}^3\text{ksi}$ for E, and $204\text{in}{\text{.}}^4$ for I

$\begin{array}{c}{\theta }_{E/A}=\left(\frac{331.2}{EI}+\frac{-53.333}{EI}\right)\\ =\frac{1}{\frac{29\times {10}^3\times 204}{144\text{in}{\text{.}}^2/{\text{ft}}^2}}\left(331.2-53.33\right)\\ =6.7636\times {10}^{-3}\text{rad}\end{array}$

Calculate the slope $\left({\theta }_E\right)$ at point E using the formula:

${\theta }_E={\theta }_A+{\theta }_{E/A}$

Substitute $-5.31\times {10}^{-3}\text{rad}$ for ${\theta }_A$ and $6.7636\times {10}^{-3}\text{rad}$ for ${\theta }_{E/A}$

$\begin{array}{c}{\theta }_E=-5.31\times {10}^{-3}+6.7636\times {10}^{-3}\\ =1.4536\times {10}^{-3}\text{rad}\end{array}$

Let point K be location of the largest downward deflection.

Hence, the slope at point E is greater than zero, the point K of zeros slope lies to the left of the point E.

Calculate the moment $\left(M_6\right)$ using the relation:

$\begin{array}{c}{M}_6=18.4\times x_K\\ =18.4x_K\end{array}$

Calculate the ratio of $\frac{M_6}{EI}$ using the relation:

$\text{Ratio}=\frac{M_6}{EI}$

Substitute $18.4x_K$ for $M_6$.

$\frac{M_6}{EI}=\frac{18.4x_K}{EI}$

Calculate the moment $\left(M_7\right)$ using the relation:

$\begin{array}{c}{M}_7=-5\times \left(x_K-2\right)\times \frac{\left(x_K-2\right)}{2}\\ =-5\frac{{\left(x_K-2\right)}^2}{2}\end{array}$

Calculate the ratio of $\frac{M_7}{EI}$:

$\text{Ratio}=\frac{M_7}{EI}$

Substitute $-5\frac{{\left(x_K-2\right)}^2}{2}$ for $M_7$.

$\begin{array}{c}\frac{M_7}{EI}=\frac{-5\frac{{\left(x_K-2\right)}^2}{2}}{EI}\\ =-\frac{5{\left(x_K-2\right)}^2}{2EI}\end{array}$

Show the moment $\left(\frac{M}{EI}\right)$ diagram for section of beam AK as in Figure 4.

Calculate the area $\left(A_6\right)$ using the formula:

$A_6=\frac{1}{2}\left(\frac{M_6}{EI}\right)x_K$

Substitute $\frac{18.4x_K}{EI}$ for $\left(\frac{M_6}{EI}\right)$.

$\begin{array}{c}{A}_6=\frac{1}{2}\times \frac{18.4x_K}{EI}\times x_K^{}\\ =\frac{9.2x_K^2}{EI}\end{array}$

Calculate the area $\left(A_7\right)$ using the formula:

$A_7=\frac{1}{3}\left(\frac{M_7}{EI}\right)\left(x_K-2\right)$

Substitute $-\frac{5{\left(x_K-2\right)}^2}{2EI}$ for $\frac{M_7}{EI}$.

$\begin{array}{c}{A}_7=\frac{1}{3}\frac{5{\left(x_K-2\right)}^2}{2EI}\left(x_K-2\right)\\ =-\frac{5{\left(x_K-2\right)}^3}{6EI}\end{array}$

Calculate the slope $\left({\theta }_K\right)$ using the formula:

$\begin{array}{c}{\theta }_K={\theta }_A+{\theta }_{K/A}\\ ={\theta }_A+A_6+A_7\end{array}$

Substitute $\frac{9.2x_K^2}{EI}$ for $A_6$, $-\frac{5{\left(x_K-2\right)}^3}{6EI}$ for $A_7$ and  $-\frac{218.134}{EI}$ for ${\theta }_A$.

$\begin{array}{l}{\theta }_K=-\frac{218.134}{EI}+\frac{9.2x_K^2}{EI}-\frac{5{\left(x_K-2\right)}^3}{6EI}\\ 0=\frac{1}{EI}\left(-218.134+9.2x_K^2-\frac{5}{6}{\left(x_K-2\right)}^3\right)\\ 0=-218.134+9.2x_K^2-\frac{5}{6}{\left(x_K-2\right)}^3\end{array}$ (1)

Differentiate the Equation (1).

$f\left(x_K\right)=-218.134+9.2x_K^2-\frac{5}{6}{\left(x_K-2\right)}^3$ (2)

$\frac{df}{d{x}_K}=0+18.4x_K-2.5{\left(x_K-2\right)}^2$ (3).

Solve the value $x_K$ by iteration.

Iteration 1:

Substitute 5 for $x_K$ in Equation (2).

$\begin{array}{c}f\left(5\right)=-218.134+9.2{\left(5\right)}^2-\frac{5}{6}{\left(5-2\right)}^3\\ =-10.634\end{array}$

Substitute 5 for $x_K$ in Equation (3).

$\begin{array}{c}\frac{df}{d{x}_K}=0+18.4\left(5\right)-2.5{\left(5-2\right)}^2\\ =69.5\end{array}$

Iteration 2:

Calculate the value $x_K$ using the relation:

$x_K={\left(x_K\right)}_0-\frac{f}{\left(\frac{df}{d{x}_K}\right)}$

Substitute 5 for ${\left(x_K\right)}_0$, $-10.634$ for f and $69.5$ for $\left(\frac{df}{d{x}_K}\right)$.

$\begin{array}{c}{x}_K=5-\frac{-10.634}{69.5}\\ =5+0.153\\ =5.153\end{array}$

Similarly calculate the value $x_K$ and tabulated in the below Table 1.

$x_K$	f	$\left(\frac{df}{d{x}_K}\right)$
5	-10.634	69.5
5.153	-0.037	69.962
5.0005	-10.597	69.5.02
5.1525	0.001

The value of $x_K$ is $5.1525\text{ft}$.

Calculate the slope at the end A related to the point K $\left(t_{A/K}\right)$ using the formula:

$\left(t_{A/K}\right)=A_6\left(\frac{2}{3}{x}_K\right)+A_7\left(2+\frac{3}{4}\left(x_K-2\right)\right)$

Substitute $\frac{9.2x_K^2}{EI}$ for $A_6$, $-\frac{5{\left(x_K-2\right)}^3}{6EI}$ for $A_7$, $5.1525\text{ft}$ for $x_K$, $29\times {10}^3\text{ksi}$ for E and $204\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}\left(t_{A/K}\right)=\frac{9.2\times {5.1525}_{}^2}{EI}\left(\frac{2}{3}\times 5.1525\right)-\frac{5{\left(5.1525-2\right)}^3}{6EI}\times \left(2+\frac{3}{4}\left(5.1525-2\right)\right)\\ =\frac{838.978}{EI}-\frac{113.949}{EI}\\ =\frac{725.029}{\frac{29\times {10}^3\times 204}{144\text{in}{\text{.}}^2/{\text{ft}}^2}}\end{array}$

$\begin{array}{c}=0.017648\text{ft}\\ \text{=}0.017648\text{ft}\times \frac{12\text{in}\text{.}}{\text{1}\text{ft}}\\ =0.212\text{in}\text{.}\end{array}$

Calculate the magnitude $\left(y_K\right)$ of the largest downward deflection using the relation:

$y_K=-t_{A/K}$

Substitute $0.212\text{in}\text{.}$ for $t_{A/K}$.

$y_K=-0.212\text{in}\text{.}$

Thus, the magnitude $\left(y_K\right)$ and location $\left(x_K\right)$ of the largest downward deflection are $\underset{\_}{0.212\text{in}\text{.}\left(\text{downward}\right)}$ and $\underset{\_}{5.15\text{ft}}$ respectively.