Question 3.14: A steel angle and a steel wide-flange beam, each of length L......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
The angle and wide-flange steel shapes have the same cross-sectional area but the thicknesses of flange and web components of each section are quite different. First ,consider the angle section.
Part (a): Steel angle section.
1, 2. Conceptualize, Categorize: 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, estimate coefficients k_1 = k_2 to be approximately 0.319.
3. Analyze: The maximum allowable torques can be obtained from Eqs. (3-72) and (3-73) based on the given allowable shear stress and allowable twist rotation, respectively, as

\tau_{\mathrm{max}}={\frac{T}{k_{1}b t^{2}}}\quad\quad(3-72)
\phi={\frac{T L}{(k_{2}b t^{3})G}}={\frac{T L}{G J_r}}\quad\quad(3-73)
T_{max1} = τ_ak_1b_Lt^2_L = 45\, MPa(0.319)(280 \,mm)[(19 \,mm)^2] = 1451\, N·m \quad (a)
T_{max2} = ɸ_a(k_2b_Lt^3_L)\frac{G}{L} = {\Bigg\lgroup} \frac{5π}{180} rad{\Bigg\rgroup}(0.319)(280 \,mm)[(19\, mm)^3]\frac{80\, GPa}{3500 \,mm} \quad (b)
\quad\quad = 1222\, N·m
Alternatively, compute the torsion constant for the angle J_L as
J_L = k_2b_Lt^3_L  =  6.128 × 10^5\, mm^4 \quad\quad\quad (c)
then use Eqs. (3-74)   and   (3-76) to find the maximum allowable torque values.

J_{f}\,=k_{2}b_{f}t_{f}^{3}\quad J_{w}\,=k_{2}(b_{w}-2t_{f})\bigl(t_{w}^{3}\bigr)\quad\quad(3-74a,b)

\tau_{\mathrm{max}}={\frac{2T\left({\frac{t}{2}}\right)}{J}}\ \ \mathrm{and}\ \ \phi={\frac{T L}{G J}}\quad\quad(3-76a,b)
From Eq. (3-76a), find T_{max1}, and from Eq. (3-76b), obtain T_{max2}:
T_{max1} = \frac{τ_aJ_L}{t_L} = 1451\, N·m \, and\,  T_{max2} = \frac{GJ_L}{L}ɸ_a = 1222 \,N·m
4. Finalize: For the angle, the lesser value controls, so T_{max} = 1222 \,N·m.
Part (b): Steel Wide-flange shape.
1, 2. Conceptualize, Categorize: The two flanges and the web are separate rectangles that together resist the applied torsional moment. However, the dimensions (b, t) of each of these rectangles are different: each flange has a width of b_f = 128\, mm and a thickness of t_f = 10.7 \,mm. The web has thickness t_w = 6.48\, mm and, conservatively,\, b_w = (d_w – 2t_f) = (353\, mm  –  2(10.7 \,mm)) = 331.6\, mm.
Based on the b/t ratios, find separate coefficients k_2 for the flanges and web from Table 3-1, then compute the torsion constants J for each component using Eqs. (3-74) as
For the flanges:
\quad\quad \frac{b_f}{t_f} = 11.963
so an estimated value for k_{2f} = 0.316. Thus,
J_f = k_{2f}b_ft^3_f = 0.316(128\, mm)[(10.7 \,mm)^3] = 4.955 × 10^4\, mm^4 \quad\quad (d)
For the web:
\quad\quad \frac{d_w – 2t_f}{t_w} = 51.173
and k_{2w} is estimated as k_{2w} = 0.329, so
J_w = k_{2w}(d_w – 2t_f)(t^3_w) = 0.329[353 \,mm – 2(10.7 \,mm)][(6.48\, mm)^3]
\quad\quad\quad\quad\quad = 2.968 × 10^4 \,mm^4\quad\quad\quad (e)
The torsion constant for the entire wide flange section is obtained by adding web and flange contributions [Eqs. (d)   and   (e)]:
J_w = 2J_f + J_w = [2(4.955) + 2.968](10^4)\, mm^4 = 1.288 × 10^5 \,mm^4\quad (f)
3. Analyze: Now, use Eq. (3-76a) and the allowable shear stress τ_a to compute the maximum allowable torque based on both flange and web maximum shear stresses:
T_{max f} = τ_a\frac{J_W}{t_f} = 45\, MPa {\Bigg\lgroup}\frac{1.288 × 10^5 \,mm^4}{10.7\, mm} {\Bigg\rgroup}= 542\, N·m \quad (g)
T_{max w} = τ_a\frac{J_W}{t_w} = 45\, MPa {\Bigg\lgroup}\frac{1.288 × 10^5 \,mm^4}{6.48\, mm} {\Bigg\rgroup}= 984\, N·m \quad (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, use Eq. (3-76b) to compute T_{max} based on the allowable angle of twist:

4. Finalize: For the wide-flange 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 wide-flange shapes have the same cross-sectional area, the wide flange shape is considerably weaker in torsion, because its component rectangles are much thinner (t_w = 6.48 \,mm, t_f = 10.7 \,mm) than the angle section (t_L= 19\, mm).
However, Chapter 5 shows that, although weak in torsion, the wide flange shape has a considerable advantage in resisting bending and transverse shear stresses.