Question 3.14: A steel angle, L178 × 102 × 19, and a steel wide-flange beam...

The angle and wide-flange steel shapes have the same cross-sectional area [A = 4960 \mathrm{mm}^{2}; see Tables F-1(b) and F-5(b)] but thicknesses of flange and web components of each section are quite different. First, we consider the angle section.

(a) Steel angle section. We can approximate the unequal leg angle as one long rectangle with length b_{L}= 280 mm and constant thickness t_{L} = 19 mm, so b_{L} / t_{L} = 14.7. From Table 3-1, we estimate coefficients k_{1}=k_{2} to be approximately 0.319. The maximum allowable torques can be obtained from Eqs. (3-72) \tau_{\max }=\frac{T}{k_{1} b t^{2}} and (3-73) \phi=\frac{T L}{\left(k_{2} b t^{3}\right) G}=\frac{T L}{G J_{r}} based on the given allowable shear stress and allow-able twist rotation, respectively, as

Alternatively, we can compute the torsion constant for the angle J_{L},

J_{L}=K_{1} b_{L} t_{L}^{3}=6.128 \times 10^{5} \mathrm{~mm}^{4}    (c)

then use Eqs. (3-74) J_{f}=k_{1} b_{f} t_{f}^{3} \quad J_{w}=k_{1}\left(b_{w}-2 t_{f}\right)\left(t_{w}^{3}\right) and (3-76) to find maximum allowable torque values.
From Eq. (3-76a) \tau_{\max }=\frac{2 T\left(\frac{t}{2}\right)}{J}, we find T_{\max 1} and from Eq. (3-76b) \phi=\frac{T L}{G J} , we obtain T_{\max 2}:

For the angle, the lesser value controls, so T_{\max } = 1222 N.m

(b) Steel W-shape. The two flanges and the web are separate rectangles which together resist the applied torsional moment. However, the dimensions (b, t) of each of these rectangles are different: for a W 360 × 39, each flange has a width of b_{f} = 128 mm and a thickness of t_{f} = 10.7 mm [see Table F-1(b)]. The web has thickness t_{w} = 6.48 mm [Table F-5(b)] and, conservatively, b_{w} = (d_{w} – 2t_{f}) = (353 mm – 2(10.7 mm)) = 331.6 mm. Based on the b/t ratios, we can find separate coefficients k1 for the flanges and web from Table 3-1

Table 3-1, Dimensionless Coefficients for Rectangular Bars

then compute the torsion constants J for each component using Eqs. (3-74) J_{f}=k_{1} b_{f} t_{f}^{3} \quad J_{w}=k_{1}\left(b_{w}-2 t_{f}\right)\left(t_{w}^{3}\right) as: For the flanges:

so an estimated value for k_{1 f} = 0.316. Then we have

J_{f}=k_{1 f} b_{f} t_{f}^{3}=0.316(128 \mathrm{~mm})\left[(10.7 \mathrm{~mm})^{3}\right]=4.955 \times 10^{4} \mathrm{~mm}^{4}    (d)

For the web:

The torsion constant for the entire W360 × 39 section is obtained by adding web and flange contributions [Eqs. (d) and (e)]:

J_{W}=2 J_{f}+J_{w}=[2(4.955)+2.968]\left(10^{4}\right) \mathrm{mm}^{4}=1.288 \times 10^{5} \mathrm{~mm}^{4}       (f)

Now, we use Eq. 3-76a and the allowable shear stress \tau_{a} to compute the maximum allowable torque based on both flange and web maximum

shear stresses:

T_{\max f}=\tau_{a} \frac{J_{W}}{t_{f}}=45 \mathrm{MPa}\left(\frac{1.288 \times 10^{5} \mathrm{~mm}^{4}}{10.7 \mathrm{~mm}}\right)=542 \mathrm{~N} \cdot \mathrm{m}        (g)

T_{\max w}=\tau_{a} \frac{J_{w}}{t_{w}}=45 \mathrm{MPa}\left(\frac{1.288 \times 10^{5} \mathrm{~mm}^{4}}{6.48 \mathrm{~mm}}\right)=894 \mathrm{~N} \cdot \mathrm{m}       (h)

Note that since the flanges have greater thickness than the web, the maximum shear stress will be in the flanges. So, a calculation of  T_{\max } based on the maximum web shear stress using Eq. (h) is not necessary.
Finally, we use Eq. (3-76b) to compute T_{\max } based on the allowable angle of twist:

For the W-shape, the most restrictive requirement is the allowable twist rotation so T_{\max } = 257 N . m governs [Eq. (i)].
It is interesting to note that, even though both angle and W-shapes have the same cross-sectional area, the W-shape is considerably weaker in torsion, because its component rectangles are much thinner (t_{w } = 6.48 mm, tf = 10.7 mm) than the angle section (t_{L } = 19 mm). However, we will see in Chapter 5 that, although weak in torsion, the W-shape has a considerable advantage in resisting bending and transverse shear stresses.