Chapter 3.10, Problem 137P

A 4-m-long steel member has a W310 × 60 cross section. Knowing that G = 77.2 GPa and that the allowable shearing stress is 40 MPa, determine (a) the largest torque T that can be applied, (b) the corresponding angle of twist. Refer to Appendix C for the dimensions of the cross section and neglect the effect of stress concentrations. (Hint: consider the web and flanges separately and obtain a relation between the torques exerted on the web and a flange, respectively, by expressing that the resulting angles of twist are equal.)

Fig. P3.137

To determine

Find the largest torque (T) that can be applied.

Answer to Problem 137P

The largest torque (T) is $\underset{\_}{1761.53\text{N}\cdot \text{m}}$.

Explanation of Solution

Given information:

The length of the steel member (L) is $4\text{m}$.

The provided section of the member is $\text{W}310\times 60$.

The allowable shearing stress $\left({\tau }_{all}\right)$ is $40\text{MPa}$.

The modulus rigidity of the steel (G) is $77.2\text{GPa}$.

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 $203\text{mm}$ or $\text{0}\text{.203m}$.

The thickness of the flange (b) is $13.1\text{mm}$ or $\text{0}\text{.0131m}$.

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

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

Substitute $203\text{mm}$ for a and $13.1\text{mm}$ for b.

$\begin{array}{c}\frac{a}{b}=\frac{203}{13.1}\\ =15.50\end{array}$

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 $13.1\text{mm}$ for b and $203\text{mm}$ for a.

$\begin{array}{c}\frac{b}{a}=\frac{13.1}{203}\\ =0.0645\end{array}$

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 0.0645 for $\frac{b}{a}$.

$\begin{array}{c}{c}_1=c_2=\frac{1}{3}\left(1-0.630\times 0.0645\right)\\ c_1=c_2=0.31979\end{array}$

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 0.31979 for $c_2$, $\text{0}\text{.203m}$ for a, $\text{0}\text{.0131m}$ for b, $\text{4m}$ for L and $77.2\text{GPa}$ for G.

$\begin{array}{c}{\varphi }_f=\frac{T_f\left(4\right)}{0.31979\times \left(0.203\right)\times {\left(0.0131\right)}^3\times 77.2\times {10}^9}\\ =3.5503\times {10}^{-4}{T}_f\end{array}$

Consider web:

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

The thickness of the web (b) is $7.49\text{mm}$ or $0.00749\text{m}$.

The depth of the member (D) is $302\text{mm}$ or $0.302\text{m}$.

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

$a=D-2b_f$

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

Substitute $302\text{mm}$ for D and $13.1\text{mm}$ for $b_f$.

$\begin{array}{c}a=302-2\left(13.1\right)\\ =275.8\text{mm}\\ =\frac{275.8\text{mm}}{1000}\\ =0.2758\text{m}\end{array}$

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

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

Substitute $275.8\text{mm}$ for a and $7.49\text{mm}$ for b.

$\begin{array}{c}\frac{a}{b}=\frac{275.8}{7.49}\\ =36.822\end{array}$

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 $7.49\text{mm}$ for b and $275.8\text{mm}$ for a.

$\begin{array}{c}\frac{b}{a}=\frac{7.49}{275.8}\\ =0.02716\end{array}$

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 0.02716 for $\frac{b}{a}$.

$\begin{array}{c}{c}_1=c_2=\frac{1}{3}\left(1-0.630\times 0.02716\right)\\ c_1=c_2=0.32763\end{array}$

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 0.32763 for $c_2$, $0.2758\text{m}$ for a, $0.00749\text{m}$ for b, $4\text{m}$ for L, and $77.2\text{GPa}$ for G.

$\begin{array}{c}{\varphi }_w=\frac{T_w\left(4\right)}{0.32763\times \left(0.2758\right)\times {\left(0.00749\right)}^3\times 77.2\times {10}^9}\\ =1.36464\times {10}^{-3}{T}_w\end{array}$

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

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

Substitute $3.5503\times {10}^{-4}{T}_f$ for ${\varphi }_f$ and $1.36464\times {10}^{-3}{T}_w$ for ${\varphi }_w$.

$\begin{array}{l}3.5503\times {10}^{-4}{T}_f=1.36464\times {10}^{-3}{T}_w\\ T_f=\frac{1.36464\times {10}^{-3}{T}_w}{3.5503\times {10}^{-4}}\end{array}$

$T_f=3.8437T_w$ (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 $3.8437T_w$ for $T_f$.

$\begin{array}{l}2\left(3.8437T_w\right)+T_w=T\\ 7.6874T_w+T_w=T\\ 8.6874T_w=T\\ T_w=\frac{T}{8.6874}\end{array}$

$T_w=0.11511T$ (2)

Substitute $0.11511T$ for $T_w$ in Equation (1).

$T_f=3.8437\left(0.11511T\right)$

$T_f=0.44244T$ (3)

Calculate the torque in the flange $\left(T_f\right)$ using the formula:

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

Substitute $40\text{MPa}$ for ${\tau }_{\max }$, 0.31979 for $c_1$, $0.203\text{m}$ for a, and $\text{0}\text{.0131m}$ for b.

$\begin{array}{l}40\times {10}^6=\frac{T_f}{0.31979\times 0.203\times {0.0131}^2}\\ T_f=\left(40\times {10}^6\right)\times 0.32763\times 0.203\times {0.0131}^2\\ T_f=445.62\text{N}\cdot \text{m}\end{array}$

Substitute $445.62\text{N}\cdot \text{m}$ for $T_f$ in the equation (3).

$\begin{array}{l}445.62=0.44244T\\ T=\frac{445.62}{0.44244}\\ T=1,007.2\text{N}\cdot \text{m}\end{array}$

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

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

Substitute $40\text{MPa}$ for ${\tau }_{\max }$, 0.32763 for $c_1$, $0.2758\text{m}$ for a, and $\text{0}\text{.00749m}$ for b.

$\begin{array}{l}40\times {10}^6=\frac{T_w}{0.32763\times 0.2758\times {0.00749}^2}\\ T_w=\left(40\times {10}^6\right)\times 0.32763\times 0.2758\times {0.00749}^2\\ T_w=202.77\text{N}\cdot \text{m}\end{array}$

Substitute $202.77\text{N}\cdot \text{m}$ for $T_w$ in the equation (2).

$\begin{array}{l}202.77=0.11511T\\ T=\frac{202.77}{0.11511}\\ T=1761.53\text{N}\cdot \text{m}\end{array}$

Hence, take the lesser value from the torque produced in flange and web respectively.

Therefore, the largest torque (T) is $\underset{\_}{1761.53\text{N}\cdot \text{m}}$.

To determine

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

Answer to Problem 137P

The angle $\left(\varphi \right)$ of twist is $\underset{\_}{9.06°}$.

Explanation of Solution

Given information:

The length of the steel member (L) is $4\text{m}$.

The provided section of the member is $\text{W}310\times 60$.

The allowable shearing stress $\left({\tau }_{all}\right)$ is $40\text{MPa}$.

The modulus rigidity of the steel (G) is $77.2\text{GPa}$.

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:

${\varphi }_f=3.5503\times {10}^{-4}{T}_f$

Substitute $445.62\text{N}\cdot \text{m}$ for $T_f$.

$\begin{array}{c}{\varphi }_f=3.5503\times {10}^{-4}\left(445.62\right)\\ =0.15821\text{radians}\\ =0.15821\times \frac{180}{\pi }\\ =9.06°\end{array}$

Therefore, the angle of twist of the section is $\underset{\_}{9.06°}$.