bi=B−2tw (text_label - position: above the cross-section diagram)
Neutral axis location for force equilibrium: ap=4twFy+0.85fc′bi2FyHtw+0.85fc′bitf
| Steel Stress | Stress at y=ay | 0 |
| Steel Stress | Stress at y=2ay | Fy |
Neutral axis location for force equilibrium: ay=4twFy+0.35fc′bi2FyHtw+0.35fc′bitf
Neutral axis location for force equilibrium: acr=tw(Fn+Fy)+0.35fc′biFyHtw+(0.35fc′+Fy−Fn)bitf
Found in