AuEng99
Civil/Environmental
- Sep 22, 2012
- 32
I am fairly young engineer, but I still insist on checking a lot of the program's results with hand calcs. This is always beneficial, because I checked the RISA results for a plate loading on a metal building. I noticed that the plate load was not being distributed correctly. From the old days of hand calculations, I new that 0.02 ksf distributed over beams 20 feet long @ 3 feet O.C. should give a Shear (V) result that is (0.02 ksf * 3 feet * 20 ft)/2 however the results are not even close to this.
When I apply a distributed load of 0.02 ksf over the beam, I get a Shear (V) equal to approximately (0.02 ksf * 3 feet * 20 ft)/2.
Can anyone help me find out what I need to do to get the plate loading to show up correctly?
When I apply a distributed load of 0.02 ksf over the beam, I get a Shear (V) equal to approximately (0.02 ksf * 3 feet * 20 ft)/2.
Can anyone help me find out what I need to do to get the plate loading to show up correctly?