ElyasCivil
Civil/Environmental
- Jun 6, 2020
- 17
Hello guys how are you?
I am having a bit of a hard time visualizing and understanding a point regarding the design of steel moment frames for seismic loads.
I was told by my prof that beams are designed for
1.4 D
1.2 D+ 1.6L
1.2D + L +E
So here is my problem:
Lets say I have a compact I section and I am bringing in 1.2D + L +E (lateral)and other combos, what do I assume is the capacity of my beam? (phi Mp)????
So now lets say my beam with a capacity of (phi Mp) has resisted 1.2D + L +E (lateral) and other combos. Now I go towards the connection and you say that the beam must resist
Mu=2Mpr/Lh +Vgravity. I quote AISC "This calculation shall assume the moment at the center of each reduced
beam section is Mpr and shall include gravity loads acting on the beam based on the load combination 1.2D + f1L + 0.2S,"
I asked the prof how come in the connection load combo 1.2D + f1L + 0.2S you EXCLUDED the earthquake load E he said because the earthquake load is already included in the plastic hinges and did not provide me with any further explanation. I am confused. Gravity loads can also cause plastic hinges. So what does he mean by this?
Is AISC assuming that the earthquake is causing the plastic hinges? Is that what the prof means. I tried talking to him but he left me in the dark.
Any help based on your expertise would be greatly appreciated and if you could refer me to a source that would be awesome.
Have a great day
I am having a bit of a hard time visualizing and understanding a point regarding the design of steel moment frames for seismic loads.
I was told by my prof that beams are designed for
1.4 D
1.2 D+ 1.6L
1.2D + L +E
So here is my problem:
Lets say I have a compact I section and I am bringing in 1.2D + L +E (lateral)and other combos, what do I assume is the capacity of my beam? (phi Mp)????
So now lets say my beam with a capacity of (phi Mp) has resisted 1.2D + L +E (lateral) and other combos. Now I go towards the connection and you say that the beam must resist
Mu=2Mpr/Lh +Vgravity. I quote AISC "This calculation shall assume the moment at the center of each reduced
beam section is Mpr and shall include gravity loads acting on the beam based on the load combination 1.2D + f1L + 0.2S,"
I asked the prof how come in the connection load combo 1.2D + f1L + 0.2S you EXCLUDED the earthquake load E he said because the earthquake load is already included in the plastic hinges and did not provide me with any further explanation. I am confused. Gravity loads can also cause plastic hinges. So what does he mean by this?
Is AISC assuming that the earthquake is causing the plastic hinges? Is that what the prof means. I tried talking to him but he left me in the dark.
Any help based on your expertise would be greatly appreciated and if you could refer me to a source that would be awesome.
Have a great day