Chapter 9.6, Problem 146P

For the beam and loading shown, determine (a) the slope at point B, (b) the deflection at point D. Use E = 200 GPa.

Fig. P9.128

To determine

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

Answer to Problem 146P

The magnitude $\left(y_K\right)$ of the largest downward deflection is $\underset{\_}{-1.841\times {10}^{-3}\text{m}}$.

Explanation of Solution

Given information:

The section of the beam is $\text{W410}\times 114$.

The young’s modulus of steel is $200\text{GPa}$.

Calculation:

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

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

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -4.8R_A+\left(40\times 4.8\times \frac{4.8}{2}\right)-\left(160\times 1.8\right)=0\\ R_A=\frac{-172.8}{-4.8}\\ R_A=36\text{kN}\end{array}$

Refer Appendix C “Properties of rolled steel shape (SI units)” for moment of inertia of section $\text{W}410\times 114$ and the value is $462\times {10}^6{\text{mm}}^{\text{4}}$.

Calculate the value (EI):

Substitute $200\text{GPa}$ for E and $462\times {10}^6{\text{mm}}^{\text{4}}$ for I.

$\begin{array}{c}EI=\left(200\text{GPa}\times \frac{1,000,000{\text{kN/m}}^{\text{2}}}{1\text{GPa}}\right)\times \left(462\times {10}^6{\text{mm}}^4\times \frac{1{\text{m}}^{\text{4}}}{1,{000}^4{\text{mm}}^{\text{4}}}\right)\\ =92,400\text{kN}\cdot {\text{m}}^{\text{2}}\end{array}$

Calculate the moment due to the reaction at A:

$M_1=R_A\left(L_{AB}\right)$

Substitute $36\text{kN}$ for $R_A$ and $4.8\text{m}$ for $L_{AB}$.

$\begin{array}{c}{M}_1=36\times 4.8\\ =172.8\text{kN}\cdot \text{m}\end{array}$

Calculate the $\frac{M_1}{EI}$ value;

Substitute $172.8\text{kN}\cdot \text{m}$ for $M_1$ and $92,400\text{kN}\cdot {\text{m}}^{\text{2}}$ for EI.

$\begin{array}{c}\frac{M_1}{EI}=\frac{172.8}{92,400}\\ =1.870\times {10}^{-3}{\text{m}}^{-\text{1}}\end{array}$

Show the $\left(\frac{M_1}{EI}\right)$ diagram in below Figure 2.

Calculate the area $\left(A_1\right)$ using the relation:

$A_1=\frac{1}{2}\left(L_{AB}\right)\left(\frac{M_1}{EI}\right)$

Substitute $4.8\text{m}$ for $L_{AB}$ and $1.870\times {10}^{-3}{\text{m}}^{-\text{1}}$ for $\left(\frac{M_1}{EI}\right)$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(4.8\right)\left(1.870\times {10}^{-3}\right)\\ =4.4883\times {10}^{-3}\end{array}$

Calculate the moment due UDL:

$M_2=-40\times L_{AB}\times \frac{L_{AB}}{2}$

Substitute $4.8\text{m}$ for $L_{AB}$.

$\begin{array}{c}{M}_2=-40\times 4.8\times \frac{4.8}{2}\\ =-460.8\text{kN}\cdot \text{m}\end{array}$

Calculate the $\frac{M_2}{EI}$ value:

Substitute $-460.8\text{kN}\cdot \text{m}$ for $M_2$ and $92,400\text{kN}\cdot {\text{m}}^{\text{2}}$ for EI.

$\begin{array}{c}\frac{M_2}{EI}=\frac{-460.8}{92400}\\ =-4.987\times {10}^{-3}{\text{m}}^{\text{-1}}\end{array}$

Show the $\left(\frac{M_2}{EI}\right)$ diagram in below Figure 3.

Calculate the area $\left(A_2\right)$ using the relation:

$A_2=\frac{1}{3}\left(L_{AB}\right)\left(\frac{M_2}{EI}\right)$

Substitute $4.8\text{m}$ for $L_{AB}$ and $-4.987\times {10}^{-3}{\text{m}}^{\text{-1}}$ for $\left(\frac{M_2}{EI}\right)$.

$\begin{array}{c}{A}_2=\frac{1}{3}\left(4.8\right)\left(-4.987\times {10}^{-3}\right)\\ =-7.9792\times {10}^{-3}\end{array}$

Calculate the moment due to the point load at D as below:

$M_3=-160\times L_{BD}$

Substitute $1.8$ for $L_{BD}$.

$\begin{array}{c}{M}_3=-160\times 1.8\\ =-288\text{kN}\cdot \text{m}\end{array}$

Calculate the $\frac{M_3}{EI}$ value:

Substitute $-288\text{kN}\cdot \text{m}$ for $M_3$ and $92,400\text{kN}\cdot {\text{m}}^{\text{2}}$ for EI.

$\begin{array}{c}\frac{M_3}{EI}=\frac{-288}{92400}\\ =-3.117\times {10}^{-3}{\text{m}}^{\text{-1}}\end{array}$

Show the $M_3$ diagram in below Figure 4.

Calculate the area $\left(A_3\right)$ using the relation:

$A_3=\frac{1}{2}\left(L_{BD}\right)\left(\frac{M_3}{EI}\right)$

Substitute $1.8\text{m}$ for $L_{BD}$ and $-3.117\times {10}^{-3}{\text{m}}^{\text{-1}}$$-3.117\times {10}^{-3}{\text{m}}^{\text{-1}}$ for $\left(\frac{M_3}{EI}\right)$.

$\begin{array}{c}{A}_3=\frac{1}{2}\left(1.8\right)\left(-3.117\times {10}^{-3}\right)\\ =-2.8052\times {10}^{-3}\end{array}$

Calculate the tangential deviation of B with respect to A using the relation:

$t_{B/A}=A_1\left(\frac{1}{3}\left(L_{AB}\right)\right)+A_2\left(\frac{1}{4}\times L_{AB}\right)$

Substitute $4.4883\times {10}^{-3}$ for $A_1$, $-7.9792\times {10}^{-3}$ for $A_2$, $4.8\text{m}$ for $L_{AB}$.

$\begin{array}{c}{t}_{B/A}=4.4883\times {10}^{-3}\left(\frac{1}{3}\left(4.8\right)\right)-7.9792\times {10}^{-3}\left(\frac{1}{4}\times 4.8\right)\\ =7.1813\times {10}^{-3}-9.5750\times {10}^{-3}\\ =-2.3937\times {10}^{-3}\end{array}$

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

${\theta }_A=\frac{t_{B/A}}{\left(L_{AB}\right)}$

Substitute $-2.3937\times {10}^{-3}$ for $t_{A/B}$ and $\text{4}\text{.8}\text{m}$ for $L_{AB}$.

$\begin{array}{c}{\theta }_A=\frac{-2.3937\times {10}^{-3}}{\left(4.8\right)}\\ =-4.987\times {10}^{-4}\text{rad}\end{array}$

Let point K is the maximum deflection.

Calculate the moment due to the reaction at A as below:

$M_1=R_A{x}_K$

Substitute $36\text{kN}$ for $R_A$.

$M_1=36x_K$

Calculate the ${\left(\frac{M_1}{EI}\right)}_{x_K}$ value as below;

Substitute $36x$ for $M_1$.

${\left(\frac{M_1}{EI}\right)}_{x_K}=\frac{36x_K}{EI}$

Calculate the moment due UDL:

$\begin{array}{c}{M}_2=-40\times x_K\times \frac{x_K}{2}\\ =-20x_K^2\end{array}$

Calculate the ${\left(\frac{M_2}{EI}\right)}_{x_K}$ value:

Substitute $-20x_K^2$ for $M_2$.

${\left(\frac{M_2}{EI}\right)}_{x_K}=\frac{-20x_K^2}{EI}$

Show the ${\left(\frac{M}{EI}\right)}_{x_K}$ diagram in Figure 5.

Calculate the area ${\left(A_1\right)}_{x_K}$ using the relation:

${\left(A_1\right)}_{x_K}=\frac{1}{2}\left(x\right)\left(\frac{M_1}{EI}\right)$

Substitute $\frac{36x_K}{EI}$ for $\left(\frac{M_1}{EI}\right)$.

$\begin{array}{c}{\left(A_1\right)}_{x_K}=\frac{1}{2}\left(x_K\right)\left(\frac{36x_K}{EI}\right)\\ =\frac{18x_K^2}{EI}\end{array}$

Calculate the area ${\left(A_2\right)}_{x_K}$ using the relation:

${\left(A_2\right)}_{x_K}=\frac{1}{3}\left(x_K\right)\left(\frac{M_2}{EI}\right)$

Substitute $\frac{-20x_K^2}{EI}$ for $\left(\frac{M_2}{EI}\right)$.

$\begin{array}{c}{\left(A_2\right)}_{x_K}=\frac{1}{3}\left(x_K\right)\frac{-20x_K^2}{EI}\\ =\frac{-20x_K^3}{3EI}\end{array}$

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

$\begin{array}{c}{\theta }_K={\theta }_A+{\theta }_{K/A}\\ ={\theta }_A+{\left(A_1\right)}_{x_K}+{\left(A_2\right)}_{x_K}\end{array}$

Substitute $\frac{18x_K^2}{EI}$ for ${\left(A_1\right)}_{x_K}$, $\frac{-20x_K^3}{3EI}$ for ${\left(A_2\right)}_{x_K}$, $-4.987\times {10}^{-4}\text{rad}$ for ${\theta }_A$, and $92,400\text{kN}\cdot {\text{m}}^{\text{2}}$ for EI.

$\begin{array}{l}{\theta }_K=-4.987\times {10}^{-3}+\frac{18x_K^2}{EI}-\frac{-20x_K^3}{3EI}\\ 0=\frac{1}{EI}\left(-4.987\times {10}^{-3}EI+18x_K^2-\frac{20}{3}{x}_K^3\right)\\ 0=-4.987\times {10}^{-4}\left(92,400\right)+18x_K^2-\frac{20}{3}{x}_K^3\\ 0=46.08+18x_K^2-\frac{20}{3}{x}_K^3\end{array}$ (1)

Differentiate the Equation (1).

$f\left(x_K\right)=46.08+18x_K^2-\frac{20}{3}{x}_K^3$ (2)

$\frac{df}{d{x}_K}=36x_K^{}-20x_K^2$ (3)

Solve the value $x_K$ by iteration.

Iteration 1:

Substitute 3 for $x_K$ in Equation (2).

$\begin{array}{c}f\left(3\right)=46.08+18{\left(3\right)}^2-\frac{20}{3}{\left(3\right)}^3\\ =28.08\end{array}$

Substitute 3 for $x_K$ in equation (3).

$\begin{array}{c}\frac{df}{d{x}_K}=36\left(3\right)-20{\left(3\right)}^2\\ =-72\end{array}$

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 3 for ${\left(x_K\right)}_0$, $28.08$ for f and $-72$ for $\left(\frac{df}{d{x}_K}\right)$.

$\begin{array}{c}{x}_K=3-\frac{28.08}{-72}\\ =3+0.39\\ =3.39\end{array}$

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

$x_K$	f	$\left(\frac{df}{d{x}_K}\right)$
3	28.08	-72
3.39	-6.78	-107.8
3.327	-0.188	-101.6
3.3251	0.005	-101.42
3.32514	0.0001

The value of $x_K$ is $3.32514\text{m}$.

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

$\left(t_{A/K}\right)={\left(A_1\right)}_{x_K}\left(\frac{2}{3}{x}_K\right)+{\left(A_2\right)}_{x_K}\left(2+\frac{3}{4}{x}_K\right)$

Substitute $\frac{18x_K^2}{EI}$ for ${\left(A_1\right)}_{x_K}$, $\frac{-20x_K^3}{3EI}$ for ${\left(A_2\right)}_{x_K}$, $3.32514\text{m}$ for $x_K$, and $92,400\text{kN}\cdot {\text{m}}^{\text{2}}$ for EI.

$\begin{array}{c}\left(t_{A/K}\right)=\frac{18{\left(3.32514\right)}^2}{EI}\left(\frac{2}{3}\left(3.32514\right)\right)+\frac{-20{\left(3.32514\right)}^3}{3EI}\left(\frac{3}{4}\left(3.32514\right)\right)\\ =\frac{441.175}{EI}-\frac{611.237}{EI}\\ =-\frac{170.62}{92,400}\\ =1.841\times {10}^{-3}\text{m}\end{array}$

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

$y_K=-t_{A/K}$

Substitute $1.841\times {10}^{-3}\text{m}$ for $t_{A/K}$.

$y_K=-1.841\times {10}^{-3}\text{m}$

Thus, the magnitude $\left(y_K\right)$ of the largest downward deflection is $\underset{\_}{-1.841\times {10}^{-3}\text{m}}$.