Design of steel beam (ASD, Allowable Stress design)

Design code: AISC Allowable Stress Design 9^th edition, 1989.

Design requirements

1. Maximum bending stress, f_b must not exceed allowable stress, F_b.

2. Deflection should not exceed allowable limit.

3. Maximum shear stress, fv shall not exceed allowable shear stress.

Design procedure:

1. Calculate design load.

2. Calculate design moment, M and bending stress, f_b.

3. Select a trial beam size and calculate allowable bending stress, F_b (see below)

4. Calculate deflection and check with allowable deflection ratio.

5. Calculate design shear and shear stress, f_v.

6. Calculate allowable shear stress, F_v.

Determine bending stress and shear stress

Bending stress shall be determined as

f_b= M/S           where M is design moment, S is section modulus.

Design of steel beam with W, I shape or Channel

Determine allowable bending stress F_b (ASD)

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

1. Compact section: allowable bending stress,

F_b = 0.66 F_y                 if          L_b £ L_c

where

F_y is yield strength of steel members.

L_b is laterally unsupported length of the compression flanges,

L_c is the smaller of 76 b_f/ÖF_y or L_c = 20,000/[(d/A_f)F_y]

b_f is the width of the flange, A_f is area of the flange.

2. Non-compact section: allowable bending stress

F_b = 0.60 F_y                 if          L_b £ 76 b_f/F_y

3. Compact or non-compact section, F_b is the larger of the following

When                

                           [ASD F1-6]

When                                           

                                         [ASD F1-7]

For any value of l/r_T.

                                                     [ASD F1-8]

Note: for channels bent, allowable stress is determined from [F1-8].

Where

r_T (inch) is radius of gyration of a section comprising the compression flange plus 1/3 of the compression web area, taken about an axis in the plan of web. (note: r_T is available in AISC steel table for most W and I section)

A_f is area of flange,

C_b = 1.75+1.05(M₁/M₂)+0.3(M₁/M₂) but not more than 2.3.

M₁ and M₂ are the smaller and the larger applied moments

M₁/M₂ is positive if M₁ and M₂ have the same sign.

W and I shape hot-roll section bending about its minor axis,

1. Compact section: allowable bending stress,

F_b = 0.75 F_y

2. Non-compact section: allowable bending stress

F_b = 0.60 F_y

Maximum width-thickness for compression flange for W, I and Channel section

1. Compact section: b_f/t  £ 65/ÖF_y.

2. Non-compact secton: b_f/t £ 95/ÖF_y.

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

Shear stress, f_v =V/(t_wd) £ Allowable shear stress, F_v = 0.4 F_y

Where V is applied shear force, d is the depth of beam, t_w is thickness 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 as shown in the figure.  The beam is simply supported at each end.

Design Code: AISC ASD 9th 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, Fy = 36 ksi

Requirements:  Select a W24 beam

Solution:

Total load on beam: W = (W_L+W_D) TriB = 1120 lb/ft.

Maximum moment: M = W L²/8 = 171.5 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 = 114 in³.

Moment of inertia, I = 1350 in⁴.

Calculate compact length L_c = 76 bf/ÖFy = 76 (7) / Ö36 = 7.4 ft or

Lc =20000/[(d/Af)Fy] = 20000 /[23.75/ 3.5) 36 ] = 6.9 ft   > L_b = 5 ft

Allowable stress: F_b = 0.66 F_y = 24 ksi

Bending stress: f_b = M/S = 18.1 ksi       < F_b = 24 ksi   O.K.

Check deflection: Elastic modulus, E = 29000 ksi

Total deflection, D = 5 W L⁴/(388 E I ) = 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 stress,

Shear force, V = WL/2 = 19.6 kips

Shear stress, fv = V/(t_wd) = 2.7 ksi

Allowable shear stress, F_v = 0.4 F_y = 14.4 ksi

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 ASD 9th 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 = 12 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 = (W_L+W_D) TriB = 768 lb/ft.

Maximum negative moment at cantilever end: M_neg = W a²/2 = 55.3 kip-ft

Maximum unsupported length at cantilever, L_b = 12 ft

Try W18x35, From AISC Table, d = 17.7 in, b_f = 6 in, t_f =7/16 in, A_f = b_f t_f=2.62 in²,

t_w = 0.3 in, r_T = 1.49 in

Section modulus, S = 57.6 in³.

Moment of inertia, I = 510 in⁴.

Check cantilever end:

Calculate compact length L_c = 76 bf/ÖFy = 76 (6) / Ö50 = 5.3 ft or

Lc =20000/[(d/Af)Fy] = 20000 /[17.7/ 2.62) 50 ] = 4.9 ft   < L_b = 12 ft

Calculate a/r_T = 96

Calculate C_b, M₁ = 0, M₂ = 69.1 kip-ft, C_b = 1.75

Allowable stress from Eq. F1-6:

Allowable stress from Eq. F1-8:

Maximum bending stress:

Bending stress: f_b = M/S = 11.5 ksi       < F_b = 24.6 ksi   O.K.

Maximum deflection at cantilever end, E = 29000 ksi

From structural analysis,

D = (Wa/24EI)(4a²L-L³+3a³) = -0.93 in               D/2a = 1/248    O.K.

Live load deflection, DL=D(12/32) = 0.35 in          D/2a = 1/825    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_pos = (W/8L²)(L+a)²(L-a)² = 122.8 ft-kip

Bending stress: f_b = M_pos/S = 25.6 ksi

The span is fully supported. Allowable stress: F_b=0.66F_y=33 ksi     O.K.

Check deflection:

Maximum deflection at x = (L/2)[1-(a/L)²] = 16 ft

Deflection D = (Wx/24E I L)(L⁴-2L²x²+Lx³-2a²L²+2a²x²)=1.8 in

D/L = 1/310   O.K.

Live load deflection D_L = (12/32)D = 0.54 in         DL/L= 1/802      O.K.

Check shear stress:

Shear force at cantilever end, V₁=W a = 9.3 kip

Shear force at simply supported end, V₂ = W (L+a)²/2L =24.6 kips

Shear stress fv = V₂/t_wd = 4.6 ksi

Allowable shear stress, Fv = 0.4 Fy = 20 ksi        O.K.