Question

In: Civil Engineering

steel structure Q1. The column carries 60 kips dead load and 40 kips live load, its...

steel structure
Q1. The column carries 60 kips dead load and 40 kips live load, its length is 20 ft, and the steel is A992. Assume the member as pinned at the top and bottom.

(a) Compute Pa;

(b) Determine the effective length (KL);

(c) Use ASD method to select the lightest W section;

(d) Check the local buckling slenderness limits of the selected section.

Solutions

Expert Solution

Given: A992 (fy=50ksi ,fu=65ksi)

L=20ft

Dead load DL=60kips

Live load LL = 40kips

Answer a) Factore load Pu or Pa = 1.2*DL+1.6*LL

For ASD we calculate Pa=1.2*60+1.6*40

Pa=136k

Answer b) Effective Length (KL)

K=1 (pinned top and bottom)

L=20ft

KL=20ft

Answer c) selection of the lightest section

Based on ASD method

Assuming Kl/r =125 and using fy=50ksi

from Aisc table 4-22 we get =9.62ksi

Area Required =

=136/9.62

=14.13in2

lets try W14x53 (A=15.6in2,ry= 1.92in)

Kl/r = (12x20)/1.92

Kl/r=125

=9.62 x 15.6

=150.072K >136 K

Use W14x53

Answer d) Check for local buckling slenderness limits of W14x53

W14x53 (d=13.9in, bf=8.06in, tw=0.37in, tf=0.66in, k=1.25)

Check for slenderness of the flange element

b/tf = 8.06/(2 x 0.66) = 6.10 < 0.56

0.56 =0.56 x =13.49

b/tf < 0.56Non slender unstiffened flange element

Check for slenderness of the web element

h/tw = (d-2k)/tw

=  (13.9-2.50)/0.37

=30.81

1.49 =1.49 x =35.88

h/tw <1.49 Non slender stiffened web element


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