Section Properties

Taken from Table 1-1 of Ref. 1:

• S_x = 54.6 in³
• r_ts = 2.84 in
• h_o = 9.42 in
• torsional constant, J = 1.39 in⁴
• warping constant, C_w = 2,070 in⁶

Taken from Appendix A of Ref. 2 (p.38):

• torsional constant, a = 62.1 in
• normalized warping function, W_no = 23.6 in²
• warping statical moment, S_w1 = 33.0 in⁴
• statical moment in flange (above web edge), Q_f = 13.0 in³
• statical moment in web, Q_w = 30.2 in³

Direct Bending Stress

Direct Shear Stress

In the web:

In the flanges:

Torsional Stresses

Evaluate the torsional function constants for ɑ = 0.5 for Case 3 using the charts in Appendix B of Ref. 2:

• at midspan (z/l = 0.5):
  • $\theta \times \frac{GJ}{T_u}\times \frac{1}{l}=\mathrm{+0.009}$, $\theta =\mathrm{+0.009}\frac{T_{u}l}{GJ}$
  • $\theta \text{'}\times \frac{GJ}{T_u}=0$, $\theta \text{'}=0$
  • $\theta \text{'}\text{'}\times \frac{GJ}{T_u}\times a=-0.44$, $\theta \text{'}\text{'}=-0.44\frac{T_u}{GJa}$
  • $\theta \text{'}\text{'}\text{'}\times \frac{GJ}{T_u}\times a^2=-0.50$, $\theta \text{'}\text{'}\text{'}=-0.50\frac{T_u}{GJ{a}^2}$
• at the support (z/l = 0):
  • $\theta \times \frac{GJ}{T_u}\times \frac{1}{l}=0$, $\theta =0$
  • $\theta \text{'}\times \frac{GJ}{T_u}=\mathrm{+0.28}$, $\theta \text{'}=\mathrm{+0.28}\frac{T_u}{GJ}$
  • $\theta \text{'}\text{'}\times \frac{GJ}{T_u}\times a=0$, $\theta \text{'}\text{'}=0$
  • $\theta \text{'}\text{'}\text{'}\times \frac{GJ}{T_u}\times a^2=-0.22$, $\theta \text{'}\text{'}\text{'}=-0.22\frac{T_u}{GJ{a}^2}$

The shear stress due to pure torsion (St. Venant's), ${\tau }_t=Gt\theta \text{'}$:

• At midspan, ${\tau }_t=0$
• At the support,
  • in the flange: ${\tau }_t=\mathrm{11,200}\left(0.340\right)\left(0.28\times -5.78{\left(10\right)}^{-3}\right)=-6.16\text{ ksi}$
  • in the web: ${\tau }_t=\mathrm{11,200}\left(0.560\right)\left(0.28\times -5.78{\left(10\right)}^{-3}\right)=-10.2\text{ ksi}$

The shear stress due to warping, ${\tau }_w=\frac{-E{S}_{\mathrm{w1}}\theta \text{'}\text{'}\text{'}}{t_f}$

• At midspan, ${\tau }_w=\frac{\mathrm{-29,000}\left(33.0\right)}{0.560}\left[\frac{-0.50\times -5.78{\left(10\right)}^{-3}}{{\left(62.1\right)}^2}\right]=-1.28\text{ ksi}$
• At the support, ${\tau }_w=\frac{\mathrm{-29,000}\left(33.0\right)}{0.560}\left[\frac{-0.22\times -5.78{\left(10\right)}^{-3}}{{\left(62.1\right)}^2}\right]=-0.56\text{ ksi}$

The normal stress due to warping, ${\sigma }_w=E{W}_{\mathrm{no}}\theta \text{'}\text{'}$

• At midspan, ${\sigma }_w=\mathrm{29,000}\left(23.6\right)\left[\frac{-0.44\times -5.78{\left(10\right)}^{-3}}{62.1}\right]=28.0\text{ ksi}$
• At the support, ${\sigma }_w=0$

Combined Stresses

The maximum magnitude normal stress, f_un, occurs at the midspan (where the normal stress is non-zero) within the flanges in which the bending and warping normal stresses are additive in magnitude:

The modification factor for σ_by and σ_w per Cl. 4.7.3 of Ref. 2 is given as:

where

$F_{\mathrm{cr}}^e$	=	$\frac{C_b{\pi }^{2}E}{{\left(\frac{L_b}{r_{\mathrm{ts}}}\right)}^2}\sqrt{1+0.078\frac{Jc}{S_x{h}_o}{\left(\frac{L_b}{r_{\mathrm{ts}}}\right)}^2}$
$C_b$	=	1.32 (for a simple span beam with a point load at midspan, simplified from eqn. F1-1 from Ref. 1)
$L_b$	=	= 180 in
c	=	1 for a doubly symmetric I section

Thus, the modification factor is: $\frac{0.9\left(127.4\right)}{0.9\left(127.4\right)-12.4}=1.121$

So the modified normal stress is taken as: $f_{\mathrm{un}}={\sigma }_b+{\sigma }_w\left(\frac{\varphi F_{\mathrm{cr}}^e}{\varphi F_{\mathrm{cr}}^e-{\sigma }_{\mathrm{bx}}}\right)=12.4+28.0\left(1.121\right)=43.8\text{ ksi}$

Check the limit state of yielding under normal stress per Cl. 4.7.1 of Ref. 2:

The stress ratio is then: $\frac{f_{\mathrm{un}}}{\varphi F_y}=\frac{43.8}{45}=0.973$

The maximum magnitude shear stress, f_uv, occurs at the support (where the torsion stress is non-zero):

• flanges:
  $f_{\mathrm{uv}}={\tau }_t+{\tau }_w+{\tau }_b=-10.2+-0.56+-0.64=-11.4\text{ ksi}$
• web:
  $f_{\mathrm{uv}}={\tau }_t+{\tau }_b=-6.16+-2.45=-8.61\text{ ksi}$

Check the limit state of yielding under shear stress per Cl. 4.7.1 of Ref. 2:

The stress ratio is then: $\frac{f_{\mathrm{uv}}}{\varphi 0.6F_y}=\frac{11.4}{27}=0.422$

Calculate Twist

Occurs at midspan: