Problem Two - (a) From Table 1-1 find the Plastic Section Modulus Z for W16x100

Question

Problem Two - (a) From Table 1-1 find the Plastic Section Modulus Z for W16x100 and W14x550 (b) From Table 3-2, which is sorted using Z values, find using values of part (a) the sections WI 6x100 and WI 4x550 and record the LRFD limits bMpx and the ASD limits Mrs/. (c) Use the Z values of the sections and a yield limit of Fu = 50 ksi, find Mpz. Afterward, use 0.9 and S2b 1.67 to see if you could duplicate the values from Table 3-2. Note: to get the exact values you need to do round off of calculations properly. For example, 7625 should be recorded as 7630, 3 digits only. Using 0.6 instead of 1/1.67 might make your calculation off a little as well. This is iust an educational process.

Explanation / Answer

a) From Table 1-1, Z of w16x100 is =200 in3

and z of w14x550 is = 1180 in3

b) plastic moment capapcity of W16x100 =743 kip-ft and ASD capapcity = 494 kip-ft

plastic moment capapcity of W14x550 =4430 kip-ft and ASD capapcity = 2940 kip-ft

c) plastic moment capapcity of w16x100 per formula = 0.9*50*200/12=750 kip-ft

ASD capacity of w16x100 = 50*200/(12*1.67)=499 kip-ft

plastic moment capapcity of w14x550 per formula = 0.9*50*1180/12=4425 kip-ft

ASD capacity of w16x100 = 50*1180/(12*1.67)=2944.1 kip-ft

  

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