MEMBERS UNDER AXIAL COMPRESSION AND BENDING
P
P
x
My = Pex
x
ex
ey
y
y
<= =>
Mx = Pey
NSCP 2010
506.6
I-shaped Members and Channels Bent about the Minor Axis
The nominal flexural strength , Mn , shall be the lower value obtained according to
the limit states of yielding (plastic moment) and flange local buckling.
506.6.1 Yielding
Mn = Mp = Fy Zy ≤ 1.6 Fy Sy
(506.6.1)
506.6.2 Flange Local Buckling
1. For sections with compact flanges the limit state of yielding shall apply.
2. For sections with noncompact flanges
Mn = [Mp – (Mp – 0.7Fy Sy) (λ – λpf )/ (λrf – λpf ) ]
(506.6.2)
3. For sections with slender flanges
Mn = Fcr Sy
(506.6.3)
where
Fcr = (0.69E) / (bf / 2tf) 2
(506.6-4)
λ = width-thickness ration of unstiffened flange
λpf = λp = limiting slenderness for a compact flange
(Tables 502.4.1 and 502.4.2)
λrf = λr = limiting slenderness for a noncompact flange
(Tables 502.4.1 and 502.4.2)
Sy = for a channel, shall be taken as the minimum section modulus
Problem: A W14 x 120, A36 steel section is to carry an eccentric load of 800 kN
(300kN dead load and 500kN live load), with eccentricities ex = 150mm
and ey = 210mm . The unsupported length is 6m in both x and y axes, and
the ends of the member are both hinged. Fy = 248 Mpa and E = 200,000 Mpa .
C b = 1.0 .Determine the adequacy of the section.
Solution:
800 kN
y
Properties of W14x120:
d = 367.8 mm
bf = 372.6 mm
A = 22,774 mm2
Ix = 574,399,000 mm4
Sx = 3,123,431 mm3
rx = 158.81 mm
x
tw = 15.0 mm
tf = 23.9 mm
rT = 102.62 mm
Iy = 206,035,000 mm4
Sy = 1,105,931mm3
ry = 95.12 mm
ex
ey
Determining available compressive strength:
For the unstiffened flange element, λr = 0.56 E/Fy = 15.9
(½ bf )/tf = ½ (372.6)/ 23.9 = 7.79 < λr (= 15.9) compact/ not slender
For the stiffened web element , λr = 1.49 E/Fy
= 42.31
h = d – 2tf = 367.8– 2(23.9)= 320
h/tw = 320 / 15 = 21.33 < λr (= 42.31)
compact/not slender
therefore, the member is without slender elements and
Q = Qa Qs = 1.0
Effective length factor k = 1.0 in both x and y axes ( both ends hinged)
controlling kL/rmin = 1.0(6,000)/ 95.12 = 63.07
4.71 E/Fy = 133.75
For members without slender elements:
when KL/r ≤ 4.71 E/Fy
Fcr = [0.658 Fy/Fe ] Fy
Fe = π2 E / (KL/r )2 = π 2 ( 200,000)/(63.07)2 = 496.23 Mpa
Fcr = [0.658 248/496.23 ] 248 = 201.25 MPa
Pn = Fcr A g = 201.25 (22,774 ) = 4,583,267.5 N = 4583.26 kN.
Using ASD:
Pc = Pn / Ω = 4583.26 /1.67 = 2744.46 kN
Using LRFD
Pc = ф Pn = (0.9)4583.26 = 4124.934 kN
Determining available flexural strength in the x-axis:
Zx = 2 [ (bf )(tf ) (d/2 - tf /2) + (h/2)(tw )(h/4)]
= 2 [ (bf )(tf ) (d - tf 2)/2 + (h2 /8)(tw )]
= 2 [ (372.6)(23.9) (367.8 - 23.9)/2 + (3202 /8)( 15)]
= 3,446,477.65mm3
Lp = 1.76ry E/Fy = 1.76 (95.12 )
= 4,754mm = 4.754 m
200,000/248
Lr = 1.95 rts (E/0.7Fy )
1 + 1 + 6.76 [ (0.7Fy Sx ho )/ (EJc)] 2
Jc / (Sx ho )
ho = d – tf = 367.8 - 23.9 = 343.9 mm
Cw = warping constant = Iy ho2 /4 = (206,035,000)(343.92 )/4
= 6.09 x 10 12 mm6
r2 ts = [ Iy Cw ] / Sx =
(206,035,000)(6.09 x 10 12 ) / 3,123,431
= 11340.89
r ts = 106.49 mm
J = 1/3(∑bt3 ) = (1/3) [ 2((372.6)(23.9)3 + (320)( 15)3 ]
= 3751136.68 mm4
c = 1.0 for doubly symmetric I-section
Lr = 1.95 rts (E/0.7Fy )
Jc / (Sx ho )
1 + 1 + 6.76 [ (0.7Fy Sx ho )/ (EJc)] 2
1 + 6.76 [ (0.7Fy Sx ho )/ (EJc)] 2 = 1+ 6.76 0.79(248)(3,123,431)(343.9) 2
(200,000)(3751136.68)(1.0)
= 1.24
Jc / (Sx ho )
=
(3751136.68) (1.0)/ [3,123,431 (343.9)] = 0.059
Lr = [1.95(106.49)(200,000)/[(0.7)(248)]][0.059][
= 21,125mm = 21.125m
Lb = 6m ,
thus
1 + 1.24
Lp < Lb < Lr
Sec. 506.2. For Doubly Symmetric Compact I-shaped Members and Channels
Bent About their Major Axis, Mn shall be the lowest value obtained
according to the limit states of yielding and lateral-torsional buckling.
Sec. 506.2.1. Yielding
Mn = Mp = Fy Z x
where
Fy = specified minimum yield strength
Z x = plastic section modulus about x-axis
Mn = 248 MPa(3,446,477.65mm3)
= 854.726 kN-m.
Sec. 506.2.2 Lateral-Torsional Buckling
When Lp ≤ Lb ≤ Lr
Mn = Cb Mp - (Mp - 0.7Fy Sx) Lb – Lp
Lr – Lp
≤ Mp
= 1.0[854.726 – [854.726 – 0.7 (248)(3,123,431)(10-6][ 6- 4.754]/[21.125 – 4.754]
= 830.758 kN-m < Mp
Thus, Mnx = 830.758 kN-m
Using ASD:
Mcx = Mnx / Ωb = 830.758/1.67 = 497.46 kN-m.
Using LRFD:
Mcx = фbMnx = 0.9(830.758) = 747.68 kN-m.
Determining available flexural strength in the y-axis:
506.6
I-shaped Members and Channels Bent about the Minor Axis
The nominal flexural strength , Mn , shall be the lower value obtained according
to the limit states of yielding (plastic moment) and flange local buckling.
506.6.1 Yielding
Mn = Mp = Fy Zy ≤ 1.6 Fy Sy
506.6.2 Flange Local Buckling
1. For sections with compact flanges the limit state of yielding shall apply.
Zy = 2[ 2(bf tf /2)(bf / 4) + (htw /2)(tw /4)]
= (bf 2 tf )/2 + htw 2 / 4
= (372.62 )(23.9)/2 + (320)(152 )/4
= 1,677,027.58 mm3
Mn = Mp = 248 (1,677,027.58)
≤ 1.6 (248) (1,105,931)
= 415.9 kN-m
≤ 438.833 kN-m
thus, Mny = 415.9 kN-m
Using ASD:
Mcy = Mny / Ωb = 415.9 /1.67 = = 249.04 kN-m
Using LRFD
Mcy = ф Mny = (0.9)(415.9) = 374.31kN-m.
Check adequacy (ASD):
Pr = D + L = 300 + 500 kN = 800 kN
Mrx = Pr ey = 800(0.210) = 168 kN-m
Mry = Pr ex = 800(0.150) = 120 kN-m
Pr / P c = 800/ 2744.46 = 0.29 > 0.2
For Pr / P c > 0.2
(Pr / P c ) + 8/9 [ (M rx / M cx ) + (M ry / M cy )] ≤ 1.0
0.29 + (8/9)[ 168/ 497.46 + 120/ 249.04 ] ≤ 1.0
1.019 > 1.0 therefore NOT ADEQUATE!!!
Check adequacy (LRFD):
Pr = 1.2D + 1.6L = 1.2(300) + 1.6(500) kN = 1160kN
Mrx = Pr ey = 1160(0.210) = 243.6kN-m
Mry = Pr ex = 1160(0.150) = 174.0kN-m
Pr / P c = 1160/ 4124.934 = 0.28 > 0.2
For Pr / P c > 0.2
(Pr / P c ) + 8/9 [ (M rx / M cx ) + (M ry / M cy )] ≤ 1.0
0.28 + (8/9)[243.6/ 747.68 + 174.0/ 374.31 ] ≤ 1.0
0.984 < 1.0
Therefore, ADEQUATE!!!