This is problem 4.4-1 from William T. Segui\'s 5th Edition Steel Design textbook

Question

This is problem 4.4-1 from William T. Segui's 5th Edition Steel Design textbook.

An HSS 10 x 8 x 3/16 is used as a compression member with one end pinned and one end fixed against rotation but free to translate. The length is 12 feet. Compute the nominal compressive strength for A500 Grade B steel (F_y = 46 ksi). Note that this is a slender-element compressive member, and the equations AISC Section E7 must be used.

I have attached the work I have done so far on this problem. (I am not sure if all of it is correct though!)

The final answer should be 148 kips without iteration or 161 with iteration.

Given: HSS 10x3x to • L12 fact ASDD Grade 8 steel • one end pinned, Oter end fixed against rotation free to translate •F, 4 Ksi And Nominal compressive strength, P. Colutions Ky- Kyt 2,0 At 6.in b/- 4 3,0 = 3.88 in - 3.26 in W- 54.5 t= 0.74 KLY - ) = .23 (2012x12 74 K. Ly. (2.2H12 ) - . So y-axis controls 3.28) Shifened Cose 6 : A - E .000 = 5.15 1.40 43.0 + 3.— / > > slendor Slendor Stiffened to Q = aa (Q = .0) E7 .2-b – I-40 E. forff, no iteration use Equation E7-10 be = 1.92 tELL 0.38 E to 0.174 /) if LA to 3/16 =0.75 = .92(h6) , 1 - J46 43.0 b/t 43.0 7,033 quahon E7- 1c to Q. = O 3 2100 D3g 290Do be = A olts / 1%)42.1 b.

Explanation / Answer

Since we are designing this compression member so we should design it for the maximum of force which is possible on that member. And maximum force will be some factor*action load, where factor would be given in your standard codes. So t = 3/16 (max) as it is in inverse position, and b/t = 42.7 (min) to design safely. And for designing questions everybody gets different answers but we just have to show that our designed member is safe.

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