I work for the same problem recently, and I choose the Chapter 11 of Taiwanese LSD Code to solve the yield problem. I can calculate the limitation of the flange or web for the H beam, and the formulas are just to control its allowable reaction force. Of course, the reaction force may be too large for the H beam, but I can choose the larger H beam for the engineering construction design. The ASTM A572 Gr.50 H beam, RH900x300x16x28, is selected, and the formulas 11.2-1 & 11.2-2 are also selected for the present situations. If you wanna know how to define N, you can see the Figure C11.2-1 where is located on the Taiwanese LSD Code. Besides, the parameter k is NOT tf, but you have to include the circular region which is k = tf + R. In my engineering case, I don't have to add the stiffening plates, so that I just provide the fundamental formulas. The steel material is not JIS or CNS standard, so that I don't have to reduce the yield stress Fyf for the flange whose thickness is less than 1.5in. Hence, I can check whether the size of H beam is enough or not, so that the yield problem has been solved.
F_{yw}t_{w} "\phi R_{n}=1.0\times \left ( 5k+N \right )F_{yw}t_{w}")
+40 \right ]\times 3.52\times 1.6=354.816tf>R_{u}=82.392tf "=1.0\times \left [ 5\times \left ( 2.8+1.8 \right )+40 \right ]\times 3.52\times 1.6=354.816tf>R_{u}=82.392tf")
\times \left ( \frac{1.6}{2.8} \right )^{1.5} \right ]\times \sqrt{\frac{3.52\times 2.8}{1.6}} "=0.75\times 36\times 1.6^{2}\times \left [ 1+3\times \left ( \frac{40}{90} \right )\times \left ( \frac{1.6}{2.8} \right )^{1.5} \right ]\times \sqrt{\frac{3.52\times 2.8}{1.6}}")
Reference
Reference
- 內政部營建署（2010）鋼結構極限設計法規範及解說，內政部營建署。
