Chapter 9.5, Problem 108P

Two cover plates are welded to the rolled-steel beam as shown. Using E = 29 × 10⁶ psi, determine (a) the slope at end C, (b) the deflection at end C.

Fig. P9.108

To determine

Find the slope $\left({\theta }_C\right)$ at end C.

Answer to Problem 108P

The slope $\left({\theta }_C\right)$ at end C is $\underset{\_}{-2.85\times {10}^{-3}\text{rad}}$.

Explanation of Solution

Given information:

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

The section of the beam is $\text{W10}\times 45$.

The dimension of the top plate and bottom plate is $\frac{1}{2}\times 9\text{in}\text{.}$ respectively.

Calculation:

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

The moment of inertia (I) for the given section is $248\text{in}{\text{.}}^4$.

The depth of the section (D) is $10.1\text{in}\text{.}$

The width of the section (b) is $8.02\text{in}\text{.}$

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 $\frac{1}{2}\times 9\text{in}\text{.}$ for $A_1$, $\left(10.1\times 8.02\text{in}\text{.}\right)$ for $A_2$, $\frac{1}{2}\times 9\text{in}\text{.}$ for $A_3$, $\text{0}\text{.25in}\text{.}$ for $x_1$, $\left(0.5+\frac{10.1}{2}\right)\text{in}\text{.}$ for $x_2$, and $\left(0.5+10.1+\frac{0.5}{2}\right)\text{in}\text{.}$ for $x_3$.

$\begin{array}{c}\overline{x}=\frac{\left[\left(\frac{1}{2}\times 9\right)0.25\right]+\left[\left(10.1\times 8.02\right)\left(0.5+\frac{10.1}{2}\right)\right]+\left[\left(\frac{1}{2}\times 9\right)\left(0.5+10.1+\frac{0.5}{2}\right)\right]}{\left(\frac{1}{2}\times 9\right)+\left(10.1\times 8.02\right)+\left(\frac{1}{2}\times 9\right)}\\ =5.55\text{in}\text{.}\end{array}$

Top plate:

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

Since the dimension of the top plate is $\frac{1}{2}\times 9\text{in}\text{.}$

$\begin{array}{c}{A}_{top}=\frac{1}{2}\times 9\text{in}\text{.}\\ =4.5\text{in}{\text{.}}^{\text{2}}\end{array}$

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

$d_{top}=\overline{x}-\frac{0.5}{2}$

Substitute $5.5\text{in}\text{.}$ for $\overline{x}$.

$\begin{array}{c}{d}_{top}=5.5-\frac{0.5}{2}\\ =5.3\text{in}\text{.}\end{array}$

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

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

Substitute $4.5\text{in}{\text{.}}^{\text{2}}$ for A and $5.3\text{in}\text{.}$ for d.

$\begin{array}{c}A{d}^2=4.5\times {5.3}^2\\ =126.405\text{in}{\text{.}}^4\end{array}$

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 $\text{9in}\text{.}$ for b and $\text{0}\text{.5in}\text{.}$ for h.

$\begin{array}{c}\overline{I_{top}}=\frac{9\times {0.5}^3}{12}\\ =0.09375\text{in}{\text{.}}^4\end{array}$

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 $\frac{1}{2}\times 9\text{in}\text{.}$.

$\begin{array}{c}{A}_{bottom}=\frac{1}{2}\times 9\text{in}\text{.}\\ =4.5\text{in}{\text{.}}^{\text{2}}\end{array}$

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

$d_{bottom}=\overline{x}-\frac{0.5}{2}$

Substitute $5.5\text{in}\text{.}$ for $\overline{x}$.

$\begin{array}{c}{d}_{bottom}=5.5-\frac{0.5}{2}\\ =5.3\text{in}\text{.}\end{array}$

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

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

Substitute $4.5\text{in}{\text{.}}^{\text{2}}$ for A and $5.3\text{in}\text{.}$ for d.

$\begin{array}{c}A{d}^2=4.5\times {5.3}^2\\ =126.405\text{in}{\text{.}}^4\end{array}$

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 $\text{9in}\text{.}$ for b and $\text{0}\text{.5in}\text{.}$ for h.

$\begin{array}{c}\overline{I_{bottom}}=\frac{9\times {0.5}^3}{12}\\ =0.09375\text{in}{\text{.}}^4\end{array}$

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

Segments	Area, A $\left(\text{in}{\text{.}}^2\right)$	Depth, d (in.)	$A{d}^2$$\left(\text{in}{\text{.}}^4\right)$	$\overline{I}$$\left(\text{in}{\text{.}}^4\right)$
Top plate	4.5	5.3	126.405	0.09375
$\text{W410}\times 60$				248
Bottom plate	4.5	5.3	126.405	0.09375
Summation			252.81	248

Take the greater value of moment of inertia from the three segments is $248\text{in}{\text{.}}^4$.

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

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

Substitute $248\text{in}{\text{.}}^4$ for $\overline{I}$ and $252.81\text{in}{\text{.}}^2$ for $A{d}^2$.

$\begin{array}{c}I=\left(248\right)+\left(252.81\right)\\ =501\text{in}{\text{.}}^4\end{array}$

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 A:

$\begin{array}{c}{M}_1=15\times 6\\ =90\text{kip}\cdot \text{ft}\end{array}$

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

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

Substitute $90\text{kip}\cdot \text{ft}$ for $M_1$, $29\times {10}^6\text{ksi}$ for E, and $501\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}\frac{M_1}{EI}=-\frac{\text{90}\text{kip}\cdot \text{ft}}{\left(\frac{29\times {10}^3\text{ksi}\times 501}{144}\right)\text{psf}}\\ =-892.01\times {10}^{-6}{\text{ft}}^{-\text{1}}\end{array}$

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 4.5 ft for $b_1$ and $-892.01\times {10}^{-6}{\text{ft}}^{-\text{1}}$ for $h_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\times 4.5\times \left(-892.01\times {10}^{-6}{\text{ft}}^{\text{-1}}\right)\\ =-2.007\times {10}^{-3}\end{array}$

Calculate the area $\left(A_2\right)$ by taking ordinate using the formula:

$A_2=\frac{1.5}{6}{A}_1$

Substitute $-2.007\times {10}^{-3}$ for $A_1$.

$\begin{array}{c}{A}_2=\frac{1.5}{6}\left(-2.007\times {10}^{-3}\right)\\ =-0.50175\times {10}^{-3}\end{array}$

Calculate the moment $\left(M_3\right)$ by taking moment about point B:

$\begin{array}{c}{M}_3=15\times 1.5\\ =22.5\text{kip}\cdot \text{ft}\end{array}$

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

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

Substitute $22.5\text{kip}\cdot \text{ft}$ for $M_3$, $29\times {10}^6\text{ksi}$ for E, and $248\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}\frac{M_3}{EI}=-\frac{22.5\text{kip}\cdot \text{ft}}{\frac{29\times {10}^3\times 248}{144}}\\ =-450.50\times {10}^{-6}{\text{ft}}^{-\text{1}}\end{array}$

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

$A_3=\frac{1}{2}{b}_3{h}_3$

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

Substitute 1.5 m for $b_3$ and $-450.50\times {10}^{-6}{\text{ft}}^{-\text{1}}$ for $h_3$.

$\begin{array}{c}{A}_3=\frac{1}{2}\times 1.5\times \left(-450.50\times {10}^{-6}\right)\\ =-0.33788\times {10}^{-3}\end{array}$

Show the tangent slope and deflection at point C related to reference tangent as in Figure 3.

Since the support A has fixed support, the slope $\left({\theta }_A\right)$ and deflection $\left(y_A\right)$ is zero respectively.

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

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

Substitute $-2.007\times {10}^{-3}$ for $A_1$, $-0.50175\times {10}^{-3}$ for $A_2$, and $-0.33788\times {10}^{-3}$ for $A_3$.

$\begin{array}{c}{\theta }_{C/A}=\left(-2.007\times {10}^{-3}\right)+\left(-0.50175\times {10}^{-3}\right)+\left(-0.33788\times {10}^{-3}\right)\\ =-2.85\times {10}^{-3}\text{rad}\end{array}$

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

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

Substitute 0 for ${\theta }_C$ and $-2.85\times {10}^{-3}\text{rad}$ for ${\theta }_{C/A}$.

$\begin{array}{c}{\theta }_C=0-\left(-2.85\times {10}^{-3}\text{rad}\right)\\ =2.85\times {10}^{-3}\text{rad}\end{array}$

Thus, the slope $\left({\theta }_C\right)$ at point C is $\underset{\_}{-2.85\times {10}^{-3}\text{rad}}$.

To determine

Find the deflection $\left(y_C\right)$ at point C.

Answer to Problem 108P

The deflection $\left(y_C\right)$ at point C is $\underset{\_}{0.1305\text{in}\text{.}(\downarrow )}$.

Explanation of Solution

Given information:

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

The section of the beam is $\text{W10}\times 45$.

The dimension of the top plate and bottom plate is $\frac{1}{2}\times 9\text{in}\text{.}$ respectively.

Calculation:

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

$t_{C/A}=\left(A_1\times 4.5\right)+\left(A_2\times 3\right)+\left(A_3\times 1\right)$

Substitute $-2.007\times {10}^{-3}$ for $A_1$, $-0.50175\times {10}^{-3}$ for $A_2$ and $-0.33788\times {10}^{-3}$ for $A_3$.

$\begin{array}{c}{t}_{C/A}=\left(-2.007\times {10}^{-3}\times 4.5\right)+\left(-0.50175\times {10}^{-3}\times 3\right)+\left(-0.33788\times {10}^{-3}\times 1\right)\\ =-9.0315\times {10}^{-3}-1.5053\times {10}^{-3}-0.33788\times {10}^{-3}\\ =-10.8746\times {10}^{-3}\text{ft}\end{array}$

Calculate the deflection at the point C $\left(y_C\right)$ using the formula:

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

Substitute 0 for $y_A$, 0 for ${\theta }_A$, and $-10.8746\times {10}^{-3}\text{ft}$ for $t_{C/A}$.

$\begin{array}{c}{y}_C=0+\left(0\right)+\left(-10.8746\times {10}^{-3}\text{ft}\right)\\ =-10.8746\times {10}^{-3}\text{ft}\times 12\text{in}\text{.}\\ =0.1305\text{in}\text{.}(\downarrow )\end{array}$

Thus, the deflection $\left(y_C\right)$ at point C is $\underset{\_}{0.1305\text{in}\text{.}(\downarrow )}$.