Design of steel beam (LRFD, Load and Resistance Factor Design)

Design code: AISC Load and Resistance Factor Design 2nd edition, 1994.

Design requirements

1. Maximum factored moment, M_u shall not exceed flexural bending strength, fM_n.
2. Deflection should not exceed allowable limit.
3. Maximum factored shear force, Vu shall not exceed shear strength, fV_u.

Design procedure:

1. Calculate factored design load.
2. Calculate factored moment, M_u.
3. Select a trial beam size and calculate flexural moment strength, fM_n (see below)
4. Calculate deflection and check with allowable deflection ratio.
5. Calculate factor design shear force, V_u.
6. Calculate shear strength, fV_n.

Design of steel beam with W, I shape or Channel

Determine factor moment, M_u

Load factors:

1.4D                                                                            [LRFD A4-1]

1.2D+1.6L+0.5(L_r or S or R)                                           [LRFD A4-2]

1.2D+1.6(L_r or S or R)+(0.5L or 0.8W)                             [LRFD A4-3]*

1.2D+1.3W+0.5L+0.5(L_r or S or R)                                   [LRFD A4-4]*

1.2D±1.0E+0.5L+0.2S                                                      [LRFD A4-5]*

0.9D±(1.3W or 1.0E)                                                      [LRFD A4-6]

where D is dead load, L is live load, Lr is roof live load, S is snow load, R is rain load, W is wind load, E is seismic load.

*Load factor for L is 1 for garage, public assembly and area over 100 psf live load.

Determine flexural strength, fM_n (LRFD)

W, I shape and channel hot-roll section bending about its major axis or shear center

1. Unbraced length, L_b £ L_p

f = 0.9

M_n = M_p = F_yZ £ 1.5 F_yS

Where Mp is plastic moment, Fy is yield strength of steel, Z is plastic section modulus, S is elastic section modulus, and

Lp = 300 r_y/ÖFy (value listed in part 4 of LRFD beam selection table)          [LRFD F1-4]

            Where r_y is radius of gyration about minor axis

2. L_p £ L_b £ L_r

f = 0.9

                        [LRFD f1-2]

where

M_r = F_L S_x, and F_L = F_y-10 ksi

C_b = 12.5M_max/(2.5M_max+3M_A+4M_B+3M_C)

M_max, M_A,M_B, M_C are absolute moment at maximum, 1/4, 1/2, and ¾ point of unbranced length, Cb can be conservatively taken as 1.0.

Lr = (r_yX₁/F_L)[1+(1+X₂F_L²)^1/2]^1/2 (value listed in part 4 of LRFD beam selection table)

X₁=(p/S_x)(EGJA/2)^1/2 (value listed in part I of LRFD property table)

X₂=(4C_w/I_y)(S_x/GJ)² (value listed in part I of LRFD property table)

S_x = section modulus in x direction, in³.

E = elastic modulus (29000 ksi)

G = shear modulous (11200 ksi)

I_y = moment of inernia in y axis, in⁴.

C_w = warping constant, in⁶.

3. L_b > L_r

            M_n = M_cr £ M_p       [AISC F1-12]

Where

            M_cr = C_b (p/L_b)[EI_yGJ+(pE/L_b)²I_yC_w]                         [AISC F1-13]

                    = [C_bS_xX₁Ö2/(L_b/r_y)]Ö[1+X1²X₂/[2(L_b/r_y)²]

Determine unsupported length

Simply supported beams

1. For simply supported beam, the top flange is in compression. If the beam is directly attached to roof deck or floor slab, the compression flange is fully supported.  The unsupported length L_b is 0.

2. When the beam supporting joists or other beams, and its flange is directly attached to the supported joists or other beams, the unsupported length is the spacing of the joists or other beams.

Cantilever beams:

1. For cantilever beam, the compression flange is at the bottom of the beam.  If the bottom flange is unbraced, the unsupported length is the length of the cantilever beam. 

2. If bracing is provided at the bottom flange, the unsupported length is the spacing between bracings.

Continuous beams:

1. For the positive moment portion of the beam, the compression flange is at the top of the beam.  The unsupported length is determined as a simply supported beam.

2. For the negative moment portion of the beam, the compression flange is at the bottom of the beam.  The unsupported length is determined as a cantilever beam.

Check shear stress

1. Factor shear strength,V_u £ shear strength, fV_n = f(0.6 F_y A_w)

Where f = 0.9, A_w is area of web.

Acceptable deflection in most building codes

Deflection limits listed in International Building Code 2003 Table 1604.3are

For cantilever beam, L is 2 time the length of cantilever.

Example 1:

Situation: A structural steel beam is supporting a roof.  

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 = 35 ft

Unsupported length (Joist spacing): L_b = 5 ft

Tributary width: TriB = 35 ft

Material: ASTM A36, yield strength, F_y = 36 ksi

Requirements:  Select a W24 beam

Solution:

Total factor load on beam: W_u= (1.2W_D+1.6W_L) TriB = 1512  lb/ft.

Maximum factored moment: M_u = W_u L²/8 = 231.5 kip-ft

Maximum unsupported length, L_b = 5 ft

Try W24x55, From AISC steel Table, 

d = 23.57 in, b_f = 7 in, t_f =0.5 in, A_f = b_f t_f=3.5 in², t_w = 5/16 in

Section modulus, S_x = 114 in³.

Moment of inertia, I_x = 1350 in⁴.

From LRFD table L_p = 5.6 ft > L_b = 5 ft

Calculate Moment strength: 

fM_n = fF_yZ_x = 0.9 (36)(134) =362 kip-ft  < 1.5fF_yS_x = 462 kip-ft O.K.

Check deflection: Elastic modulus, E = 29000 ksi

Service load, W = (W_D+W_L) TriB = 1120 lb/ft

Total deflection, D = 5 W L⁴/(388 E I_x ) = 0.97 in

Deflection ratio, L/D = 1/435   < 1/ 240 O.K

Live load deflection, D_L = D (12/32) = 0.36 in,

Live load deflection ratio, D_L /L = 1/1160 < 1/360 O.K.

Check shear force,

Factor shear force, V_u = W_uL/2 = 29.6 kips

Allowable shear stress, fV_n = f0.6 F_y t_wd = 143 kip

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.