DL CMU wall = 51.0 (lb/ft 2 ) 0.7 (ft) DL beam = 2.5 (lb/ft 2 ) 18.0 (ft) 5

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Isabella West
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1 SUJECT: HEADER EAM SELECTION SHEET 108 of 131 INTERIOR HEADER EAM SELECTION - ay length = 36 ft. (stairwell) INTERIOR HEADER EAM Header eam 1 2 Total ay Length = 36 (ft) Total ay Width = 10 (ft) 20.5 Fill eam Fy = 50 (ksi) Tributary width DL CMU wall = 51.0 (lb/ft 2 ) 0.7 (ft) DL beam = 2.5 (lb/ft 2 ) 18.0 (ft) 5 LL stairs = (lb/ft 2 ) 10.3 (ft) 10 DL joist = 56.5 (lb/ft 2 ) LL joist = 80.0 (lb/ft 2 ) WDL joist = (lb/ft) 15.5 WLL joist = 400 (lb/ft) Stair Opening CMU Wall DL joist = (lb) 36 LL joist = 3100 (lb) W DL(CMU) = 34.0 (lb/ft) W DL(beam) = 45.0 (lb/ft) Loading Diagram W LL(stair) = (lb/ft) Wu 3.33 Factored Load = Wu = (lb/ft) Factored Max. Moment (Mu) = 21.7 (k-ft) distributed loads factored load = = (lb) Factored Max. Moment (Mu) = 29.4 (k-ft) 2 equal, concentrated TOTAL Factored Max. Moment (Mu) = 51.1 (k-ft) symmetrically placed CHECK LOCAL CONDITIONS CHECK WE LRFD p Choose W12x14 eam Weight (k/ft) <than 85 plf assmd, sheet6 Mu = 51.1 (k-ft) φmp = 65 (k-ft) b f / 2 t f (LRFD p1-29) = 8.82 λp = 65 / Fy.5 = 9.2 OK λp > b f /2t f? h / tw (LRFD p1-29) = 54.3 λp = 640 / Fy.5 = 90.5 OK λp > h/tw?

2 SUJECT: FILL EAM SELECTION SHEET 109 of 131 INTERIOR HEADER EAM SELECTION - (continued) CHECK SHEAR CAACITY Vu = Wu * L / 2 + = 12.5 φvn = φ v (0.6 F yw ) A w (F yw = yield strength of web) = 64.3 (φ v =.9) A w = d * t w (LRFD p1-28) = 2.4 d = 12, tw = 0.2 OK φvn > Vu? (wu) = (5*w LL *L 4 )/(384*E*I) = () = ((*a)/(24*e*i))/(3*l 2-4*a 2 )= = L / 360 = (in) - LIMIT (in) - LL I (moment of inertia about x-x axis, LRFD)= 88.6 (in 4 ) NOTE -- The header beam has a combination of 4 distributed loads and 2 point loads. The distributed loads are the CMU dead load, the floor dead load (same for floor and stairwell - conservative), the stair live load and floor live load. Since there are only 2 joists, these are treated as point loads. These point loads are determined by a simple analysis of the loading on each joist and determining their reactions. In this case, deflection controls Cost er Ton 1950 ($) Total Cost $136.50

3 SUJECT: FILL EAM SELECTION SHEET 110 of 131 INTERIOR FLOOR FILL EAM SELECTION - ay length = 36 ft. (stairwell) INTERIOR FLOOR FILL EAM Interior Floor Fill eam C Note: Shear studs along the beams that are part of the lateral load-resisting truss are necessary to transfer the lateral loads from the floor diaphragms to the resisting members. This is outside the scope of this case study. 1 2 Total ay Length = 36 (ft) Total ay Width = 30 (ft) Fill eam Fy = 50 (ksi) Tributary width DL CMU wall = 51.0 (lb/ft 2 ) 0.7 (ft) 3 DL beam1 = 59.0 (lb/ft 2 ) 4.0 (ft) DL beam2 = 59.0 (lb/ft 2 ) 1.5 (ft) LL stairs = (lb/ft 2 ) 5.0 (ft) 5 LL office1 = 80.0 (lb/ft 2 ) 1.5 (ft) 10 LL office2 = 80.0 (lb/ft 2 ) 4.0 (ft) WDL CMU wall = 34 (lb/ft) Stair Opening WDL beam1 = (lb/ft) WDL beam2 = 88.5 (lb/ft) CMU Walls WLL stairs = 500 (lb/ft) Elevator Shaft WLL office1 = 120 (lb/ft) WLL office2 = 320 (lb/ft) DL header beam = (lb) LL header beam = (lb) Wu1 = (k/ft) 36 Wu2 = 0.8 (k/ft) 20.5 = 16.3 (k) Loading Diagrams Wu1 Wu2 = Wu1 + Wu R1 R Moment by superposition (k-ft) Shear by Superposition (k) at 14.7' at 18' at 20.5' at 23.8' R1 R2 Wu Wu Wu Wu Total Total Mmax = (K-ft)

4 SUJECT: FILL EAM SELECTION SHEET 111 of 131 INTERIOR FLOOR FILL EAM (continued) LRFD p Choose W24x62 eam Weight (k/ft) <than 85 plf assmd, sheet6 Mu = (k-ft) φmp = 578 (k-ft) CHECK LOCAL CONDITIONS b f / 2 t f (LRFD p1-17) = 5.97 λp = 65 / Fy.5 = 9.2 OK λp > b f /2t f? CHECK WE h / tw (LRFD p1-17) = 49.7 λp = 640 / Fy.5 = 90.5 OK λp > h/tw? CHECK SHEAR CAACITY Vu = Wu * L / 2 *a/l = 26.4 φvn = φ v (0.6 F yw ) A w (F yw = yield strength of web) = (φ v =.9) A w = d * t w (LRFD p1-16) = 8.4 d = 24, tw = 0.4 OK φvn > Vu? max= L / 360 = (in) - LIMIT Deflection at Mmax : ( ) = (*b*x)*(l^2 - b^2 - x^2) / (6*E*I*L) = ( Wu1) = W*a^2 *(L-x)*(4*L*x - 2*x^2 - a^2) / (24* E * I * L) = ( Wu2) = W*a^2 *(L-x)*(4*L*x - 2*x^2 - a^2) / (24* E * I * L) = (in) - LL I (moment of inertia about x-x axis, LRFD)= 1560 (in 4 ) NOTE -- The fill-beam has a combination of several uniform and partially distributed loads and 2 concentrated loads. The dead load for the floor and stair is uniformly distributed as well as the floor live load. The CMU dead load is only particially distrubuted along the length of the member LUS 1/2 of the distance along the header (the fill beam picks up half this load). This fill-beam also picks up 2 partially distributed live loads for the floor and stair. There is one concentrated dead and live load at the point where the header frames into the fill beam. Cost er Ton 1950 ($) Total Cost $2, In this case, deflection controls

5 SUJECT: FILL EAM SELECTION SHEET 112 of 131 INTERIOR FLOOR FILL EAM SELECTION - ay length = 36 ft. (stairwell) INTERIOR FLOOR FILL EAM Interior Floor Fill eam C 1 2 Total ay Length = 36 (ft) Total ay Width = 30 (ft) Fill eam Fy = 50 (ksi) ributary width DL CMU wall = 51.0 (lb/ft 2 ) 0.7 (ft) DL beam1 = 59.0 (lb/ft 2 ) 5.0 (ft) DL beam2 = 59.0 (lb/ft 2 ) 2.5 (ft) LL stairs = (lb/ft 2 ) 5.0 (ft) LL office1 = 80.0 (lb/ft 2 ) 2.5 (ft) 10 LL office2 = 80.0 (lb/ft 2 ) 5.0 (ft) 5 WDL CMU wall = 34 (lb/ft) Stair Opening WDL beam1 = (lb/ft) 5 WDL beam2 = (lb/ft) CMU Walls WLL stairs = 500 (lb/ft) Elevator Shaft WLL office1 = 200 (lb/ft) WLL office2 = 400 (lb/ft) DL header beam = (lb) LL header beam = (lb) Wu1 = (k/ft) 36 Wu2 = 1.0 (k/ft) 20.5 = 16.3 (k) Loading Diagrams Wu1 Wu2 = Wu1 + Wu R1 R Moment by superposition (k-ft) Shear by Superposition (k) at 14.7' at 18' at 20.5' at 23.8' R1 R2 Wu Wu Wu Wu Total Total Mmax = (K-ft)

6 SUJECT: FILL EAM SELECTION SHEET 113 of 131 INTERIOR FLOOR FILL EAM (continued) LRFD p Choose W21x83 eam Weight (k/ft) <than 85 plf assmd, sheet6 Mu = (k-ft) φmp = 735 (k-ft) CHECK LOCAL CONDITIONS b f / 2 t f (LRFD p1-17) = 5 λp = 65 / Fy.5 = 9.2 OK λp > b f /2t f? CHECK WE h / tw (LRFD p1-17) = 36.4 λp = 640 / Fy.5 = 90.5 OK λp > h/tw? CHECK SHEAR CAACITY Vu = Wu * L / 2 *a/l = 29.9 φvn = φ v (0.6 F yw ) A w (F yw = yield strength of web) = (φ v =.9) A w = d * t w (LRFD p1-16) = 8.4 d = 21, tw = 0.5 OK φvn > Vu? max= L / 360 = (in) - LIMIT Deflection at Mmax : ( ) = (*b*x)*(l^2 - b^2 - x^2) / (6*E*I*L) = ( Wu1) = W*a^2 *(L-x)*(4*L*x - 2*x^2 - a^2) / (24* E * I * L) = ( Wu2) = W*a^2 *(L-x)*(4*L*x - 2*x^2 - a^2) / (24* E * I * L) = (in) - LL I (moment of inertia about x-x axis, LRFD)= 1830 (in 4 ) NOTE -- The fill-beam has a combination of several uniform and partially distributed loads and 2 concentrated loads. The dead load for the floor and stair is uniformly distributed as well as the floor live load. The CMU dead load is only particially distrubuted along the length of the member LUS 1/2 of the distance along the header (the fill beam picks up half this load). This fill-beam also picks up 2 partially distributed live loads for the floor and stair. There is one concentrated dead and live load at the point where the header frames into the fill beam. Cost er Ton 1950 ($) Total Cost $2, In this case, deflection controls

7 SUJECT: FILL EAM SELECTION SHEET 114 of 131 INTERIOR FLOOR FILL EAM SELECTION - ay length = 36 ft. (stairwell) INTERIOR FLOOR FILL EAM 1 2 Interior Floor Fill eam 3 (races bottom flange of girder in lateral resisting truss) 10 5 C Stair Opening Elevator Shaft 20.5 Loading Diagrams 36 5 CMU Walls Spacing = 5 (ft) DL = 61.0 (lb/ft 2 ) LL = 80.0 (lb/ft 2 ) W DL = (lb/ft) W LL = (lb/ft) Applied Load - Wu = (lb/ft) LRFD p Choose W18x35 eam Weight (k/ft) > 11 plf assumed, add Wt. Mu = (k-ft) φmp = 249 (k-ft) = L / 360 = (in) - LIMIT = (5*w LL *L 4 )/(384*E*I) = (in) - LL I (moment of inertia about x-x axis, LRFD p.5-48)= 510 (in 4 ) Note: Fill beam for this bay will be the same on the 2nd and 3rd floors. On the 4th and roof use interior fill beams determined previously for this bracing beam. The beam is necessary to laterally brace the W 24 x 68 framing beam. The beam is part of the North/South lateral load-resisting truss. A plate is added to brace the bottom flange against buckling out-of-plane.

Beam Design and Deflections

Beam Design and Deflections tation: a = name for width dimension A = name for area Areq d-adj = area required at allowable stress when shear is adjusted to include self weight Aweb = area of the web of