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

Design requirements 1. Maximum bending stress, fb must not exceed allowable stress, Fb. 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, fb. 3. Select a trial beam size and calculate allowable bending stress, Fb (see below) 4. Calculate deflection and check with allowable deflection ratio. 5. Calculate design shear and shear stress, fv. 6. Calculate allowable shear stress, Fv.

Determine bending stress and shear stress Bending 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 Channel Determine allowable bending stress Fb (ASD) W, I shape and channel hot-roll section bending about its major axis or shear center 1. Compact section: allowable bending stress, Fb = 0.66 Fy if Lb £ Lc where Fy 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. 2. Non-compact section: allowable bending stress if Lb £ 76 bf/Fy Fb = 0.60 Fy 3. Compact or non-compact section, Fb is the larger of the following

When

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Design of steel beam 89ed

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[ASD F1-6]

When [ASD F1-7] For any value of l/rT. [ASD F1-8] Note: for channels bent, allowable stress is determined from [F1-8]. Where rT (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 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, Fb = 0.75 Fy 2. Non-compact section: allowable bending stress Fb = 0.60 Fy Maximum width-thickness for compression flange for W, I and Channel section a. Compact section: bf/t £ 65/ÖFy. b. Non-compact secton: bf/t £ 95/ÖFy.

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 Lb 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.

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Design of steel beam 89ed

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

Acceptable deflection in most building codes Deflection limits listed in International Building Code 2003 Table 1604.3are Construction L S or W Roof members Supporting plaster ceiling L/360 L/360 Supporting non-plaster ceiling L/240 L/240 Not supporting ceiling L/180 L/180 Floor members L/360 Exterior walls and interior partitions With brittle finishes L/240 With flexible finishes L/120

D+L L/240 L/180 L/120 L/240 -

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 or Lc =20000/[(d/Af)Fy] = 20000 /[23.75/ 3.5) 36 ] = 6.9 ft > Lb = 5 ft Allowable stress: Fb = 0.66 Fy = 24 ksi < Fb = 24 ksi O.K. Bending stress: fb = M/S = 18.1 ksi Check deflection: Elastic modulus, E = 29000 ksi Total deflection, D = 5 W L4/(388 E I ) = 0.97 in

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

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 or Lc =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.

23/04/2015 14:39

Design of steel beam 89ed

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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 DL = (12/32)D = 0.54 in DL/L= 1/802 Check shear stress: Shear force at cantilever end, V1=W a = 9.3 kip

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O.K.

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.

Related Topics Design of metal roof Design of steel beams (13th ed) Design of steel columns (13th ed) Design of steel column (89 ed) Home | About | Calendar | Contact | Employees | Employment | FAQ | Information Links | News | Photo Gallery | Press | Products | Promotions | Services | Site Map Disclaimer: The content of this website was obtained and edited from various resources. The editor made reasonable effort of editing. Nevertheless, the editor does not warrant, and assume no liability for accuracy and completeness of its content. The viewer shall use his/her professional knowledge and judgment in use of the web content. Webmaster: www.ce-ref.com Copyright: www.ce-ref.com. All Rights Reserved.

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

Design requirements 1. Maximum bending stress, fb must not exceed allowable stress, Fb. 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, fb. 3. Select a trial beam size and calculate allowable bending stress, Fb (see below) 4. Calculate deflection and check with allowable deflection ratio. 5. Calculate design shear and shear stress, fv. 6. Calculate allowable shear stress, Fv.

Determine bending stress and shear stress Bending 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 Channel Determine allowable bending stress Fb (ASD) W, I shape and channel hot-roll section bending about its major axis or shear center 1. Compact section: allowable bending stress, Fb = 0.66 Fy if Lb £ Lc where Fy 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. 2. Non-compact section: allowable bending stress if Lb £ 76 bf/Fy Fb = 0.60 Fy 3. Compact or non-compact section, Fb is the larger of the following

When

23/04/2015 14:39

Design of steel beam 89ed

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[ASD F1-6]

When [ASD F1-7] For any value of l/rT. [ASD F1-8] Note: for channels bent, allowable stress is determined from [F1-8]. Where rT (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 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, Fb = 0.75 Fy 2. Non-compact section: allowable bending stress Fb = 0.60 Fy Maximum width-thickness for compression flange for W, I and Channel section a. Compact section: bf/t £ 65/ÖFy. b. Non-compact secton: bf/t £ 95/ÖFy.

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 Lb 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.

23/04/2015 14:39

Design of steel beam 89ed

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

Acceptable deflection in most building codes Deflection limits listed in International Building Code 2003 Table 1604.3are Construction L S or W Roof members Supporting plaster ceiling L/360 L/360 Supporting non-plaster ceiling L/240 L/240 Not supporting ceiling L/180 L/180 Floor members L/360 Exterior walls and interior partitions With brittle finishes L/240 With flexible finishes L/120

D+L L/240 L/180 L/120 L/240 -

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 or Lc =20000/[(d/Af)Fy] = 20000 /[23.75/ 3.5) 36 ] = 6.9 ft > Lb = 5 ft Allowable stress: Fb = 0.66 Fy = 24 ksi < Fb = 24 ksi O.K. Bending stress: fb = M/S = 18.1 ksi Check deflection: Elastic modulus, E = 29000 ksi Total deflection, D = 5 W L4/(388 E I ) = 0.97 in

23/04/2015 14:39

Design of steel beam 89ed

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

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 or Lc =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.

23/04/2015 14:39

Design of steel beam 89ed

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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 DL = (12/32)D = 0.54 in DL/L= 1/802 Check shear stress: Shear force at cantilever end, V1=W a = 9.3 kip

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O.K.

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.

Related Topics Design of metal roof Design of steel beams (13th ed) Design of steel columns (13th ed) Design of steel column (89 ed) Home | About | Calendar | Contact | Employees | Employment | FAQ | Information Links | News | Photo Gallery | Press | Products | Promotions | Services | Site Map Disclaimer: The content of this website was obtained and edited from various resources. The editor made reasonable effort of editing. Nevertheless, the editor does not warrant, and assume no liability for accuracy and completeness of its content. The viewer shall use his/her professional knowledge and judgment in use of the web content. Webmaster: www.ce-ref.com Copyright: www.ce-ref.com. All Rights Reserved.

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