Chapter 3.10, Problem 138P

(a)

To determine

Find the maximum shearing stress $\left({\tau }_{\max }\right)$ along the line a-a.

Answer to Problem 138P

The maximum shearing stress $\left({\tau }_{\max }\right)$ along the line a-a is $\underset{\_}{4.55\text{ksi}}$.

Explanation of Solution

Given information:

The length of the steel member (L) is $8\text{ft}$.

The provided section of the member is $\text{W8}\times 31$.

The torque in the member (T) is $\text{5kip}\cdot \text{in}\text{.}$.

The modulus rigidity of the steel (G) is $11.2\times {10}^6\text{psi}$.

Assume the angle of twist in flange and web is equal.

Calculation:

Consider flange:

Refer Appendix C, “Properties of Rolled-Steel shapes”.

The width of the flange (a) is $\text{8}\text{in}\text{.}$

The thickness of the flange (b) is $\text{0}\text{.435in}\text{.}$

Calculate the ratio of width to thickness of the steel $\frac{a}{b}$.

$\text{Ratio}=\frac{a}{b}$

Substitute $\text{8in}\text{.}$ for a and $\text{0}\text{.435in}\text{.}$ for b.

$\begin{array}{c}\frac{a}{b}=\frac{8}{0.435}\\ =18.391\end{array}$

Hence, the ratio of $\frac{a}{b}$ is greater than 5. Therefore, use the below formula for calculating the coefficient of rectangular bar.

Calculate the ratio of thickness to width of the steel $\frac{b}{a}$.

$\text{Ratio}=\frac{b}{a}$

Substitute $\text{0}\text{.435in}\text{.}$ for b and $\text{8in}\text{.}$ for a.

$\begin{array}{c}\frac{b}{a}=\frac{0.435}{8}\\ =0.0544\end{array}$

Calculate the coefficient for rectangular bar $\left(c_1\right)$ and $\left(c_2\right)$ using the formula:

$c_1=c_2=\frac{1}{3}\left(1-0.630\frac{b}{a}\right)$

Substitute 0.0544 for $\frac{b}{a}$.

$\begin{array}{c}{c}_1=c_2=\frac{1}{3}\left(1-0.630\times 0.0544\right)\\ c_1=c_2=0.32191\end{array}$

Calculate the angle of twist in flange $\left({\varphi }_f\right)$ using the formula:

${\varphi }_f=\frac{T_{f}L}{c_{2}a{b}^{3}G}$

Here, $T_f$ is the torque in flange, L is the length of the member and G is the modulus of rigidity.

Substitute 0.32191 for $c_2$, $\text{8in}\text{.}$ for a, $\text{0}\text{.435in}\text{.}$ for b, $\text{8ft}$ for L and $11.2\times {10}^6\text{psi}$ for G.

$\begin{array}{c}{\varphi }_f=\frac{T_f\left(8\times 12\right)}{0.32191\times \left(\text{8ft}\right)\times {\left(\text{0}\text{.435ft}\right)}^3\times 11.2\times {10}^6}\\ =4.044\times {10}^{-5}{T}_f\end{array}$

Consider web:

Refer Appendix C, “Properties of Rolled-Steel shapes”.

The thickness of the web (b) is $0.285\text{in}\text{.}$.

The depth of the member (D) is $\text{8}\text{in}\text{.}$

Calculate the width of the web (a) using the formula:

$a=D-2b_f$

Here, $b_f$ is the thickness of the flange.

Substitute $\text{8in}\text{.}$ for D and $\text{0}\text{.435in}\text{.}$ for $b_f$.

$\begin{array}{c}a=8-2\left(\text{0}\text{.435}\right)\\ =7.13\text{in}\text{.}\end{array}$

Calculate the ratio of width to thickness of the steel $\frac{a}{b}$.

$\text{Ratio}=\frac{a}{b}$

Substitute $\text{7}\text{.13in}\text{.}$ for a and $\text{0}\text{.285in}\text{.}$ for b.

$\begin{array}{c}\frac{a}{b}=\frac{7.13}{0.285}\\ =25.0175\end{array}$

Hence, the ratio of $\frac{a}{b}$ is greater than 5. Therefore, use the below formula for calculating the coefficient of rectangular bar.

Calculate the ratio of thickness to width of the steel $\frac{b}{a}$.

$\text{Ratio}=\frac{b}{a}$

Substitute $\text{0}\text{.285in}\text{.}$ for b and $\text{7}\text{.13in}\text{.}$ for a.

$\begin{array}{c}\frac{b}{a}=\frac{0.285}{7.13}\\ =0.039972\end{array}$

Calculate the coefficient for rectangular bar $\left(c_1\right)$ and $\left(c_2\right)$ using the formula:

$c_1=c_2=\frac{1}{3}\left(1-0.630\frac{b}{a}\right)$

Substitute 0.039972 for $\frac{b}{a}$.

$\begin{array}{c}{c}_1=c_2=\frac{1}{3}\left(1-0.630\times 0.039972\right)\\ c_1=c_2=0.32494\end{array}$

Calculate the angle of twist in web $\left({\varphi }_w\right)$ using the formula:

${\varphi }_w=\frac{T_{w}L}{c_{2}a{b}^{3}G}$

Substitute 0.32494 for $c_2$, $\text{7}\text{.13in}\text{.}$ for a, $\text{0}\text{.285in}\text{.}$ for b, $\text{8ft}$ for L and $11.2\times {10}^6\text{psi}$ for G.

$\begin{array}{c}{\varphi }_w=\frac{T_w\left(8\times 12\right)}{0.32494\times \left(7.13\right)\times {\left(0.285\right)}^3\times 11.2\times {10}^6}\\ =1.5982\times {10}^{-4}{T}_w\end{array}$

Since the angle of twist in flange and web is equal, therefore,

${\varphi }_f={\varphi }_w$

Substitute $4.044\times {10}^{-5}{T}_f$ for ${\varphi }_f$ and $1.5982\times {10}^{-4}{T}_w$ for ${\varphi }_w$.

$\begin{array}{l}4.044\times {10}^{-5}{T}_f=1.5982\times {10}^{-4}{T}_w\\ T_f=\frac{1.5982\times {10}^{-4}{T}_w}{4.044\times {10}^{-5}}\end{array}$

$T_f=3.952T_w$ (1)

By taking the sum of torque exerted on two flanges and web in the member is equal to the total torque T applied to member. Therefore,

$2T_f+T_w=T$

Substitute $3.952T_w$ for $T_f$ and $\text{5kip}\cdot \text{in}\text{.}$ for T.

$\begin{array}{l}2\left(3.952T_w\right)+T_w=5\times {10}^3\\ 7.904T_w+T_w=5,000\\ 8.904T_w=5,000\end{array}$

$\begin{array}{c}{T}_w=\frac{5,000}{8.904}\\ T_w=561.5\text{lb}\cdot \text{in}\text{.}\end{array}$

Substitute $561.5\text{lb}\cdot \text{in}\text{.}$ for $T_w$ in Equation (1).

$\begin{array}{c}{T}_f=3.952\left(561.5\right)\\ =2,219.05\text{lb}\cdot \text{in}\text{.}\end{array}$

Calculate the maximum shearing stress $\left({\tau }_{\max }\right)$ along line a-a using the formula:

${\tau }_{\max }=\frac{T_f}{c_{1}a{b}^2}$

Substitute $2,219.05\text{lb}\cdot \text{in}\text{.}$ for $T_f$, 0.32191 for $c_1$, $\text{8in}\text{.}$ for a, and $\text{0}\text{.435in}\text{.}$ for b.

$\begin{array}{c}{\tau }_{\max }=\frac{2,219.05}{0.32191\times 8\times {0.435}^2}\\ =4,553\text{psi}\\ =4.55\text{ksi}\end{array}$

Therefore, maximum shearing stress $\left({\tau }_{\max }\right)$ along the line a-a is $\underset{\_}{4.55\text{ksi}}$.

(b)

To determine

Find the maximum shearing stress $\left({\tau }_{\max }\right)$ along the line b-b.

Answer to Problem 138P

The maximum shearing stress $\left({\tau }_{\max }\right)$ along the line b-b is $\underset{\_}{\text{2}\text{.98}\text{ksi}}$.

Explanation of Solution

Given information:

The length of the steel member (L) is $8\text{ft}$.

The provided section of the member is $\text{W8}\times 31$.

The torque in the member (T) is $\text{5}\text{kip}\cdot \text{in}\text{.}$.

The modulus rigidity of the steel (G) is $11.2\times {10}^6\text{psi}$.

Assume the angle of twist in flange and web is equal.

Calculation:

Calculate the torque in the web $\left(T_w\right)$ using the formula:

${\tau }_{\max }=\frac{T_w}{c_{1}a{b}^2}$

Substitute $561.5\text{lb}\cdot \text{in}\text{.}$ for $T_w$, 0.32494 for $c_1$, $\text{7}\text{.13in}\text{.}$ for a, and $\text{0}\text{.285in}\text{.}$ for b.

$\begin{array}{c}{\tau }_{\max }=\frac{561.5}{0.32494\times 7.13\times {0.285}^2}\\ =2983.8\text{psi}\\ =2.98\text{ksi}\end{array}$

The maximum shearing stress $\left({\tau }_{\max }\right)$ along the line b-b is $\underset{\_}{\text{2}\text{.98}\text{ksi}}$.

(c)

To determine

Find the angle $\left(\varphi \right)$ of twist.

Answer to Problem 138P

The angle $\left(\varphi \right)$ of twist is $\underset{\_}{5.14°}$.

Explanation of Solution

Given information:

The length of the steel member (L) is $8\text{ft}$.

The provided section of the member is $\text{W8}\times 31$.

The torque in the member (T) is $\text{5kip}\cdot \text{in}\text{.}$.

The modulus rigidity of the steel (G) is $11.2\times {10}^6\text{psi}$.

Assume the angle of twist in flange and web is equal.

Calculation:

From the above calculation of angle of twist, take the critical angle to compute the angle of twist.

Calculate the angle of twist $\left({\varphi }_f\right)$ using the relation:

Consider the torque equation,

$2T_f+T_w=5\times {10}^3$

Substitute $\frac{\varphi c_{2}a{b}^{3}G}{L}$ for $T_f$ and $\frac{\varphi c_{2}a{b}^{3}G}{L}$ for $T_w$.

$\begin{array}{l}2\left(\frac{\varphi c_{2}a{b}^{3}G}{L}\right)+\left(\frac{\varphi c_{2}a{b}^{3}G}{L}\right)=5\times {10}^3\\ \frac{\varphi G}{L}\left[2{\left(c_{2}a{b}^3\right)}_f+{\left(c_{2}a{b}^3\right)}_w\right]=5\times {10}^3\\ \frac{\varphi G}{L}\left[2K_f+K_w\right]=5,000\end{array}$

$\varphi =\frac{5,000L}{G\left(2K_f+K_w\right)}$ (2)

Assume $K_f={\left(c_{2}a{b}^3\right)}_f$ and $K_w={\left(c_{2}a{b}^3\right)}_w$.

Calculate the value of $K_f$.

$K_f={\left(c_{2}a{b}^3\right)}_f$

Substitute 0.32191 for $c_2$, $\text{8in}\text{.}$ for a, and $\text{0}\text{.435in}\text{.}$ for b.

$\begin{array}{c}{K}_f=0.32191\times 8\times {0.435}^3\\ =0.21198\text{in}{\text{.}}^4\end{array}$

Calculate the value of $K_w$.

$K_w={\left(c_{2}a{b}^3\right)}_w$

Substitute 0.32494 for $c_1$, $\text{7}\text{.13in}\text{.}$ for a, and $\text{0}\text{.285in}\text{.}$ for b.

$\begin{array}{c}{K}_w=0.32494\times 7.13\times {0.285}^3\\ =0.053632\text{in}{\text{.}}^4\end{array}$

Find the angle of twist:

Substitute $11.2\times {10}^6\text{psi}$ for G, $8\text{ft}$ for L, $0.21198\text{in}{\text{.}}^4$ for $K_f$ and $0.053632\text{in}{\text{.}}^4$ for $K_w$.

$\begin{array}{c}\varphi =\frac{5000\left(8\times 12\right)\text{in}\text{.}}{11.2\times {10}^6\left[2\left(0.21198\right)+0.053632\right]}\\ =0.0897\text{rad}\times \frac{180}{\pi }\\ =5.14°\end{array}$

Therefore, the angle of twist of the section is $\underset{\_}{5.14°}$.