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Design of steel beam (ASD, Allowable Stress design)Design code: AISC Allowable Stress Design 9th edition, 1989.

Design requirementsMaximum bending stress, fb must not exceed allowable stress, Fb.1.Deflection should not exceed allowable limit.2.Maximum shear stress, fv shall not exceed allowable shear stress.3.

Design procedure:

Calculate design load.1.Calculate design moment, M and bending stress, fb.2.Select a trial beam size and calculate allowable bending stress, Fb (see below)3.Calculate deflection and check with allowable deflection ratio.4.Calculate design shear and shear stress, fv.5.Calculate allowable shear stress, Fv.6.

Determine bending stress and shear stressBending stress shall be determined as

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

Design of steel beam with W, I shape or ChannelDetermine allowable bending stress Fb (ASD)

W, I shape and channel hot-roll section bending about its major axis or shear centerCompact section: allowable bending stress,1.

Fb = 0.66 Fy if Lb £ LcwhereFy is yield strength of steel members.Lb is laterally unsupported length of the compression flanges,

Lc is the smaller of 76 bf/ÖFy or Lc = 20,000/[(d/Af)Fy]bf is the width of the flange, Af is area of the flange.

Non-compact section: allowable bending stress2.

Fb = 0.60 Fy if Lb £ 76 bf/FyCompact or non-compact section, Fb is the larger of the following3.

When

[ASD F1-6]

When

[ASD F1-7]For any value of l/rT.

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

WhererT (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: rT is available in AISC steel table for most W and I section)Af is area of flange,Cb = 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 momentsM1/M2 is positive if M1 and M2 have the same sign.

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

Compact section: allowable bending stress,1.

Fb = 0.75 FyNon-compact section: allowable bending stress2.

Fb = 0.60 Fy Maximum width-thickness for compression flange for W, I and Channel section

Compact section: bf/t £ 65/ÖFy.a.

Non-compact secton: bf/t £ 95/ÖFy.b.

Determine unsupported length Simply supported beams

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 fullysupported. The unsupported length Lb is 0.

1.

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 thespacing of the joists or other beams.

2.

Cantilever beams: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 thecantilever beam.

1.

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

Continuous beams: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 simplysupported beam.

1.

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

2.

Check shear stressShear stress, fv =V/(twd) £ Allowable shear stress, Fv = 0.4 FyWhere V is applied shear force, d is the depth of beam, tw is thickness of web.

Acceptable deflection in most building codesDeflection limits listed in International Building Code 2003 Table 1604.3are

Construction L S or W D+L

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Roof membersSupporting plaster ceilingSupporting non-plaster ceilingNot supporting ceiling

L/360L/240L/180

L/360L/240L/180

L/240L/180L/120

Floor members L/360 - L/240Exterior walls and interior partitionsWith brittle finishesWith flexible finishes

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L/240L/120

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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: WL = 12 psf

Roof dead load: WD = 20 psf

Length of beam: L = 35 ft

Unsupported length (Joist spacing): Lb = 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 = (WL+WD) TriB = 1120 lb/ft.

Maximum moment: M = W L2/8 = 171.5 kip-ft

Maximum unsupported length, Lb = 5 ft

Try W24x55, From AISC steel Table, d = 23.57 in, bf = 7 in, tf =0.5 in, Af = bf tf=3.5 in2, tw = 5/16 in

Section modulus, S = 114 in3.

Moment of inertia, I = 1350 in4.

Calculate compact length Lc = 76 bf/ÖFy = 76 (7) / Ö36 = 7.4 ft orLc =20000/[(d/Af)Fy] = 20000 /[23.75/ 3.5) 36 ] = 6.9 ft > Lb = 5 ft

Allowable stress: Fb = 0.66 Fy = 24 ksi

Bending stress: fb = M/S = 18.1 ksi < Fb = 24 ksi O.K.

Check deflection: Elastic modulus, E = 29000 ksi

Total deflection, D = 5 W L4/(388 E I ) = 0.97 inDeflection ratio, L/D = 1/435 < 1/ 240 O.KLive load deflection, DL = D (12/32) = 0.36 in,

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

Check shear stress,

Shear force, V = WL/2 = 19.6 kips

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

Allowable shear stress, Fv = 0.4 Fy = 14.4 ksi

Example 2:

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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: WL = 12 psf

Roof dead load: WD = 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 = (WL+WD) TriB = 768 lb/ft.

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

Maximum unsupported length at cantilever, Lb = 12 ft

Try W18x35, From AISC Table, d = 17.7 in, bf = 6 in, tf =7/16 in, Af = bf tf=2.62 in2,

tw = 0.3 in, rT = 1.49 in

Section modulus, S = 57.6 in3.

Moment of inertia, I = 510 in4.

Check cantilever end:

Calculate compact length Lc = 76 bf/ÖFy = 76 (6) / Ö50 = 5.3 ft orLc =20000/[(d/Af)Fy] = 20000 /[17.7/ 2.62) 50 ] = 4.9 ft < Lb = 12 ft

Calculate a/rT = 96

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

Allowable stress from Eq. F1-6:

Allowable stress from Eq. F1-8:

Maximum bending stress:

Bending stress: fb = M/S = 11.5 ksi < Fb = 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:

Mpos = (W/8L2)(L+a)

2(L-a)

2 = 122.8 ft-kip

Bending stress: fb = Mpos/S = 25.6 ksi

The span is fully supported. Allowable stress: Fb=0.66Fy=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 inD/L = 1/310 O.K.Live load deflection DL = (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/twd = 4.6 ksi

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

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