Nominal flexural strength of non-compact W, S, M, and C sections bend above major axis

All W, S, M, and C with F_y £ 50 ksi have compact flange (b/t) except 

W21x48, W14x99, W14x90, W12x65, W10x12, W8x31, W8x10, W6x15, W6x9, W6x8.5, M4x6.

All W, S, M, HP, C and MC sections in 13th edition with F_y £ 65 ksi have compact web (h/t_w).

In addition to requirements for compact sections, the following requirements for compression flange local buckling shall be applied

1. for section with non-compact flanges, l_p £ b/t £ l_r.

            M_n = {M_p - (M_p - 0.7 F_yS_x)[(l - l_p)/(l_r-l_p)]}           [AISC F3-1]       

2. for section with slender flanges, b/t > l_r.

            M_n = 0.9E k_c S_x/l²          [AISC F3-2]

Where l= b_f/2t_f, k_c = 0.35 £ 4/Ö(h/t_w) < 0.76.

I-shaped members and channels bend about their minor axis.

1. For compact sections

            M_n = M_p = F_yZ_y     £ 1.6 F_yS_y            [AISC F6-1]

2. For section with non-compact flange, l_p £ b/t £ l_r.

            M_n = {M_p - (M_p - 0.7 F_yS_y)[(l - l_p)/(l_r-l_p)]}           [AISC F6-2]

3. For section with slender flanges, b/t > l_r.

            M_n = F_cr S_y £ M_p       [AISC F6-3]

Where  F_cr = 0.69E/(b_f/2t_f)²         [AISC F6-3]

S_y for a channel shall be taken as the minimum section modulus.

Shear strength

Nominal shear strength, resistance factor f_vand safety factor W_v 

Nominal shear strength of beams, 

V_n = 0.6 F_y A_wC_v

Where A_w (= d t_w) is area of web, and C_v is determined as follows:

1. For all current ASTM W, S, and HP shapes except W44x230, W40x149, W36x135, W33x90, W24x55, W16x26, and W12x14 for F_y = 50 ksi steel.

f_v = 1.00, (LRFD)

W_b = 1.5  (ASD)

C_v = 1.0

2. For W44x230, W40x149, W36x135, W33x90, W24x55, W16x26, and W12x14 with F_y = 50 ksi and unstiffened web.

f_v = 0.9, (LRFD)

W_b = 1.67  (ASD)

C_v = 1

Example 2:

Situation: A structural steel beam with is supporting a roof as shown in the figure.  The beam is simply supported at each end.

Design Code: AISC LRFD 2nd edition

Roof live load: W_L = 12 psf

Roof dead load: W_D = 20 psf

Length of beam: L = 36 ft

Length of cantilever: a = 11 ft

Tributary width: TriB = 24 ft

Material: ASTM A992, yield strength, Fy = 50 ksi

Requirements:  Select a W18 beam

Solution:

Total load on beam: W = (1.2W_D+1.6W_L) TriB = 1037 lb/ft.

Maximum negative moment at cantilever end: M_uneg = W_u a²/2 = 62.7 kip-ft

Maximum unsupported length at cantilever, L_b = 11 ft

Try W18x35, From AISC Table, d = 17.7 in, t_w = 0.3 in

Elastic section modulus, S = 57.6 in³.

Plastic section modulus, Z = 66.5 in³.

Moment of inertia, I = 510 in⁴.

Check cantilever end:

From LRFD table, L_p = 4.3 ft < L_b = 11 ft < L_r = 11.5 ft

M_p = F_yZ = 277.1 ft-kip

F_L=Fy-10 ksi = 40 ksi

M_r = F_y S = 192 ksi

Cb = 1, f = 0.9

Maximum deflection at cantilever end, E = 29000 ksi

Service load, W = (W_D+W_L)TriB = 768 kip-ft

From structural analysis,

D = (Wa/24EI)(4a²L-L³+3a³) = -1.04 in               D/2a = 1/254    O.K.

Live load deflection, DL=D(12/32) = 0.39 in          D/2a = 1/678    O.K.

Check interior span:

Maximum negative moment is the same as cantilever end. 

Unsupported length is less.  O.K. by inspection.

Maximum positive moment:

M_upos = (W_u/8L²)(L+a)²(L-a)² = 138 ft-kip

The span is fully supported.

Calculate Moment strength: fM_n = fF_yZ_x =249.5 kip-ft  < 1.5fF_yS_x = 324 kip-ft O.K.

Check deflection:

Maximum deflection at x = (L/2)[1-(a/L)²] = 16.3 ft

Deflection D = (Wx/24E I L)(L⁴-2L²x²+Lx³-2a²L²+2a²x²)=1.52 in

D/L = 1/284   O.K.

Live load deflection D_L = (12/32)D = 0.57 in         DL/L= 1/758      O.K.

Check shear stress:

Shear force at cantilever end, V_u1=W_u a = 11. kip

Shear force at simply supported end, V_u2 = W_u (L+a)²/2L = 31.8 kips

Shear strength, fV_n = f0.6 Fy t_wd = 143 kip-ft     O.K.