Chapter 9.5, Problem 107P

Two cover plates are welded to the rolled-steel beam as shown. Using E = 200 GPa, determine (a) the slope at end A, (b) the deflection at end A.

Fig. P9.107

To determine

Find the slope $\left({\theta }_A\right)$ at end A.

Answer to Problem 107P

The slope $\left({\theta }_A\right)$ at end A is $\underset{\_}{3.43\times {10}^{-3}\text{rad}}$.

Explanation of Solution

Given information:

The elastic modulus (E) is $200\text{GPa}$.

The section of the beam is $\text{W410}\times 60$.

The dimension of the top plate and bottom plate is $12\times 200\text{mm}$ respectively.

Calculation:

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

The moment of inertia (I) for the given section is $216\times {10}^6{\text{mm}}^{\text{4}}$ or $216\times {10}^{-6}{\text{m}}^{\text{4}}$.

The depth of the section (D) is $406\text{mm}$.

The width of the section (b) is $178\text{mm}$.

Use moment area method:

Consider from bottom.

Calculate the neutral axis $\left(\overline{x}\right)$ using the formula:

$\overline{x}=\frac{A_1{x}_1+A_2{x}_2+A_3{x}_3}{A_1+A_2+A_3}$

Substitute $12\times 200\text{mm}$ for $A_1$, $406\times 178\text{mm}$ for $A_2$, $12\times 200\text{mm}$ for $A_3$, $6\text{mm}$ for $x_1$, $\left(12+\frac{406}{2}\right)\text{mm}$ for $x_2$, and $\left(12+406+\frac{12}{2}\right)\text{mm}$ for $x_3$.

$\begin{array}{c}\overline{x}=\frac{\left[\left(12\times 200\right)6\right]+\left[\left(406\times 178\right)\left(12+\frac{406}{2}\right)\right]+\left[\left(12\times 200\right)\left(12+406+\frac{12}{2}\right)\right]}{\left(12\times 200\right)+\left(406\times 178\right)+\left(12\times 200\right)}\\ =215\text{mm}\end{array}$

Top plate:

Calculate the area of the top plate $\left(A_{top}\right)$ using the formula:

Since the dimension of the top plate is $12\times 20\text{mm}$.

$\begin{array}{c}{A}_{top}=12\times 200\\ =2,400{\text{mm}}^{\text{2}}\end{array}$

Calculate the depth of neutral axis (d) using the formula:

$d_{top}=\overline{x}-\frac{12}{2}$

Substitute $215\text{mm}$ for $\overline{x}$.

$\begin{array}{c}{d}_{top}=215-\frac{12}{2}\\ =209\text{mm}\end{array}$

Calculate the product of $A{d}^2$ using the relation:

$\text{Product}=A{d}^2$

Substitute $2,400{\text{mm}}^{\text{2}}$ for A and $209\text{mm}$ for d.

$\begin{array}{c}A{d}^2=2,400\times {209}^2\\ =104.834\times {10}^6{\text{mm}}^4\end{array}$

Calculate the moment of inertia (I) using the formula:

$\overline{I_{top}}=\frac{b{h}^3}{12}$

Here, b is the width the top plate and h is the height of the top plate.

Substitute 200 mm for b and 12 mm for h.

$\begin{array}{c}\overline{I_{top}}=\frac{200\times {12}^3}{12}\\ =28,800{\text{mm}}^4\end{array}$

Bottom plate:

Top plate:

Calculate the area of the bottom plate $\left(A_{bottom}\right)$ using the formula:

Since the dimension of the bottom plate is $12\times 20\text{mm}$.

$\begin{array}{c}{A}_{bottom}=12\times 200\\ =2,400{\text{mm}}^{\text{2}}\end{array}$

Calculate the depth of neutral axis (d) using the formula:

$d_{top}=\overline{x}-\frac{12}{2}$

Substitute $215\text{mm}$ for $\overline{x}$.

$\begin{array}{c}{d}_{top}=215-\frac{12}{2}\\ =209\text{mm}\end{array}$

Calculate the product of $A{d}^2$ using the relation:

$\text{Product}=A{d}^2$

Substitute $2,400{\text{mm}}^{\text{2}}$ for A and $209\text{mm}$ for d.

$\begin{array}{c}A{d}^2=2,400\times {209}^2\\ =104.834\times {10}^6{\text{mm}}^4\end{array}$

Calculate the moment of inertia (I) using the formula:

$\overline{I_{bottom}}=\frac{b{h}^3}{12}$

Here, b is the width the top plate and h is the height of the top plate.

Substitute $200\text{mm}$ for b and $12\text{mm}$ for h.

$\begin{array}{c}\overline{I_{bottom}}=\frac{200\times {12}^3}{12}\\ =28,800{\text{mm}}^4\end{array}$

Tabulate the calculated values and compute the moment of inertia (I) as in Table 1.

Segments	Area, A $\left({\text{mm}}^2\right)$	Depth, d (mm)	$A{d}^2$$\left({\text{mm}}^4\right)$	$\overline{I}$$\left({\text{mm}}^4\right)$
Top plate	2400	209	$104.834\times {10}^6$	28,800
$\text{W410}\times 60$				$216\times {10}^6$
Bottom plate	2400	209	$104.834\times {10}^6$	28,000
Summation			$209.668\times {10}^6$	$216\times {10}^6$

Take the greater value of moment of inertia from the three segments is $216\times {10}^6{\text{mm}}^4$.

Calculate the moment of inertia (I) using the relation:

$I=\overline{I}+A{d}^2$

Substitute $216\times {10}^6{\text{mm}}^4$ for $\overline{I}$ and $209.668\times {10}^6{\text{mm}}^2$ for $A{d}^2$.

$\begin{array}{c}I=\left(216\times {10}^6\right)+\left(209.668\times {10}^6\right)\\ =425.7\times {10}^6{\text{mm}}^4\end{array}$

Show the free body diagram of beam by considering the point load as in Figure 1.

Draw the moment diagram of the above beam as in Figure 2.

Calculate the moment $\left(M_1\right)$ by taking moment about point B:

$\begin{array}{c}{M}_1=40\times 0.6\\ =24\text{kN}\cdot \text{m}\end{array}$

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

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

Substitute $24\text{kN}\cdot \text{m}$ for $M_1$, $200\text{GPa}$ for E, and $216\times {10}^{-6}{\text{m}}^{\text{4}}$ for I.

$\begin{array}{c}\frac{M_1}{EI}=-\frac{\text{24}\text{kN}\cdot \text{m}}{200\times {10}^6\times 216\times {10}^{-6}{\text{m}}^{\text{4}}}\\ =-0.55556\times {10}^{-3}{\text{m}}^{-\text{1}}\end{array}$

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

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

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

Substitute 0.6 mm for $b_1$ and $-0.55556\times {10}^{-3}{\text{m}}^{-1}$ for $h_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\times 0.6\times \left(-0.55556\times {10}^{-3}{\text{m}}^{\text{-1}}\right)\\ =-0.166668\times {10}^{-3}\end{array}$

Calculate the moment $\left(M_2\right)$ by taking moment about point C:

$\begin{array}{c}{M}_2=40\times 2.7\\ =108\text{kN}\cdot \text{m}\end{array}$

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

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

Substitute $108\text{kN}\cdot \text{m}$ for $M_2$, $200\text{GPa}$ for E, and $425.7\times {10}^6{\text{mm}}^4$ for I.

$\begin{array}{c}\frac{M_2}{EI}=-\frac{\text{108}\text{kN}\cdot \text{m}}{200\times {10}^6\times 425.7\times {10}^6{\text{m}}^4}\\ =-1.2685\times {10}^{-3}{\text{m}}^{-1}\end{array}$

Calculate the area $\left(A_2\right)$ due to the moment $\left(M_2\right)$ using the formula:

$A_2=\frac{1}{2}{b}_2{h}_2$

Here, $b_2$ is the width of the triangle in area $\left(A_2\right)$ and $h_2$ is the height of the triangle in area $\left(A_2\right)$.

Substitute 2.1 m for $b_2$ and $-1.2685\times {10}^{-3}{\text{m}}^{-1}$ for $h_2$.

$\begin{array}{c}{A}_2=\frac{1}{2}\times 2.1\times \left(-1.2685\times {10}^{-3}{\text{m}}^{-\text{1}}\right)\\ =-1.3319\times {10}^{-3}\end{array}$

Calculate the area $\left(A_3\right)$ by taking ordinate using the formula:

$A_3=\frac{0.6}{2.7}{A}_2$

Substitute $-1.3319\times {10}^{-3}$ for $A_2$.

$\begin{array}{c}{A}_3=\frac{0.6}{2.7}\left(-1.3319\times {10}^{-3}\right)\\ =-0.29597\times {10}^{-3}\end{array}$

Show the free body diagram of beam by considering the uniformly distributed load as in Figure 3.

Draw the moment diagram of the above beam as in Figure 4.

Calculate the moment $\left(M_4\right)$ by taking moment about point C:

$\begin{array}{c}{M}_4=-90\times 2.1\times \frac{2.1}{2}\\ =-198.45\text{kN}\cdot \text{m}\end{array}$

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

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

Substitute $-198.45\text{kN}\cdot \text{m}$ for $M_4$, $200\text{GPa}$ for E, and $425.7\times {10}^6{\text{mm}}^4$ for I.

$\begin{array}{c}\frac{M_4}{EI}=-\frac{\text{198}\text{.45kN}\cdot \text{m}}{200\times {10}^6\times 425.7\times {10}^6{\text{m}}^4}\\ =-2.3308\times {10}^{-3}{\text{m}}^{-\text{1}}\end{array}$

Calculate the area $\left(A_4\right)$ due to the moment $\left(M_4\right)$ using the formula:

$A_4=\frac{1}{2}{b}_4{h}_4$

Here, $b_4$ is the width of the triangle in area $\left(A_4\right)$ and $h_4$ is the height of the triangle in area $\left(A_4\right)$.

Substitute $2.1\text{m}$ for $b_4$ and $-2.3308\times {10}^{-3}{\text{m}}^{-\text{1}}$ for $h_4$.

$\begin{array}{c}{A}_4=\frac{1}{3}\times 2.1\times \left(-2.3308\times {10}^{-3}{\text{m}}^{\text{-1}}\right)\\ =-1.6315\times {10}^{-3}\end{array}$

Show the tangent slope and deflection at point A related to reference tangent as in Figure 5.

Since the support C has fixed support, the slope $\left({\theta }_C\right)$ and deflection $\left(y_C\right)$ at the point A is zero respectively.

Calculate the slope at the end A related to the fixed end C $\left({\theta }_{A/C}\right)$ using the formula:

${\theta }_{A/C}=A_1+A_2+A_3+A_4$

Substitute $-0.166668\times {10}^{-3}$ for $A_1$, $-1.3319\times {10}^{-3}$ for $A_2$, $-0.29597\times {10}^{-3}$ for $A_3$, and $-1.6315\times {10}^{-3}$ for $A_4$.

$\begin{array}{c}{\theta }_{A/C}=\left(-0.166668\times {10}^{-3}\right)+\left(-1.3319\times {10}^{-3}\right)+\left(-0.29597\times {10}^{-3}\right)+\left(-1.6315\times {10}^{-3}\right)\\ =-3.43\times {10}^{-3}\text{rad}\end{array}$

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

${\theta }_A={\theta }_C-{\theta }_{C/A}$

Substitute 0 for ${\theta }_C$ and $-3.43\times {10}^{-3}\text{rad}$ for ${\theta }_{C/A}$.

$\begin{array}{c}{\theta }_A=0-\left(-3.43\times {10}^{-3}\text{rad}\right)\\ =3.43\times {10}^{-3}\text{rad}\end{array}$

Thus, the slope $\left({\theta }_A\right)$ at point A is $\underset{\_}{3.43\times {10}^{-3}\text{rad}}$.

To determine

Find the deflection $\left(y_A\right)$ at point A.

Answer to Problem 107P

The deflection $\left(y_A\right)$ at point A is $\underset{\_}{6.66\text{mm}(\downarrow )}$.

Explanation of Solution

Given information:

The elastic modulus (E) is $200\text{GPa}$.

The section of the beam is $\text{W410}\times 60$.

The dimension of the top plate and bottom plate is $12\times 200\text{mm}$ respectively.

Calculation:

Calculate the deflection at end A related to the fixed end C $\left(t_{A/C}\right)$ using the formula:

$t_{A/C}=\left(A_1\times 0.4\right)+\left(A_2\times 2\right)+\left(A_3\times 1.3\right)+\left(A_4\times 2.175\right)$

Substitute $-0.166668\times {10}^{-3}$ for $A_1$, $-1.3319\times {10}^{-3}$ for $A_2$, $-0.29597\times {10}^{-3}$ for $A_3$ and $-1.6315\times {10}^{-3}$ for $A_4$.

$\begin{array}{c}{t}_{A/C}=\left\{\begin{array}{c}\left(-0.166668\times {10}^{-3}\times 0.4\right)+\left(-1.3319\times {10}^{-3}\times 2\right)\\ +\left(-0.29597\times {10}^{-3}\times 1.3\right)+\left(-1.6315\times {10}^{-3}\times 2.175\right)\end{array}\right\}\\ =-0.06667\times {10}^{-3}-2.664\times {10}^{-3}-0.385\times {10}^{-3}-3.5485\times {10}^{-3}\\ =-6.66\times {10}^{-3}\text{m}\end{array}$

Calculate the deflection at the point A $\left(y_A\right)$ using the formula:

$y_A=y_C+\left({\theta }_{C}L\right)+t_{A/C}$

Substitute 0 for $y_{AC}$, 0 for ${\theta }_C$ and $-6.66\times {10}^{-3}\text{m}$ for $t_{C/A}$.

$\begin{array}{c}{y}_C=0+\left(0\right)+\left(-6.66\times {10}^{-3}\text{m}\right)\\ =-6.66\times {10}^{-3}\text{m}\times 1000\text{mm}\\ =6.66\text{mm}(\downarrow )\end{array}$

Thus, the deflection $\left(y_A\right)$ at point A is $\underset{\_}{6.66\text{mm}(\downarrow )}$.