Question

6. Steel Column Analysis Span B -- -- -- For the given axially loaded steel W-section, determine the maximum floor live load capacity, P LL. Assume the column is pinned top and bottom: K = 1.0, and there is no intermediate bracing. Use AISC-LRFD steel equations to determine phi Pn and the load. E = 29000 ksi. Span A -3- DATASET: 1 -2- W-section Fy Span A Span B Height L Floor Dead Load W10X39 50 KSI 36 FT 49 FT 18 FT 17 PSF A K=lo Score Your Response KIPS Correct Answer SUBMIT SUBMIT SUBMIT SUBMIT # 1 2 3 4 5 6 7 8 9 Question Total unfactored floor dead load on the column Controlling slenderness ratio Seperation slenderness value, 4.71(E/Fy)^.5 Euler stress, Fe Critical stress, For Nominal load, Pn Factored nominal load, phi Pn Factored live load on column Actual unfactored floor live load SUBMIT SUBMIT KSI KSI KIPS KIPS KIPS PSF SUBMIT SUBMIT SUBMIT Current Score: 0 / 45

Please answer questions 1 - 9.

Be sure to circle final answers.

Thank you!

Answer 1

(1)

The UDL acting on the span can be considered as a point load in the following way

D.L=17 psfx49 ftx36 ft

D.L=29.988 kips

(2)

KL/ry=18x12/=1.98=109

(3)

4.71(E/Fy)^0.5=4.71(29000/50)^0.5=113

(4)

Fe=$\pi$²E/(KL/r)²

F_e=$\pi$²x29000/(109)²

F_e=24.14 ksi

(5)

Fcr=(0.658^Fy/Fe)Fy

Fcr=0.658^50/24.14)50

Fcr=21.02 ksi

(6)

Pn=FcrAg

Pn=21.02x11.5

Pn=241 kips

(7)

$\phi$Pn=0.9x241

=217.5 kips

(8)

Pu=1.2(D.L)+1.6(L.L)

217.5=1.2(29.98)+1.6(L.L)

L.L=113.45 kips

1.6L.L=181.52 kips

(9)

Unfactored load=151.26x1000/(1.6x36x49)

=64.315 psf