In: Civil Engineering
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