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TIME SAVING DESIGN AID Page 1 of 23 Torsion Design The following example illustrates the design methods presented in the PCA book “Simplified Design Reinforced Concrete Buildings of Moderate Size and Height” third edition. Unless otherwise noted, all referenced table, figure, and equation numbers are from that book. Example Building Below is a partial plan of a typical floor in a cast-in-place reinforced building. In this example, beam CD is designed and detailed for the combined effects of f

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    TIME SAVING DESIGN AID Page 1 of 23 Torsion Design The following example illustrates the design methods presented in the PCA book “Simplified Design - Reinforced Concrete Buildings of Moderate Size and Height” third edition. Unless otherwise noted, all referenced table, figure, and equation numbers are from that book. Example Building Below is a partial plan of a typical floor in a cast-in-place reinforced building. In this example, beam CD isdesigned and detailed for the combined effects of flexure, shear, and torsion according to ACI 318-05. Design DataMaterials    Concrete: normal weight (150 pcf), f’ c = 4,000 psi    Mild reinforcing steel: Grade 60 (f  y = 60,000 psi) Loads    Dead load of joists = 77 psf    TIME SAVING DESIGN AID Page 2 of 23 Torsion Design    Superimposed dead load = 30 psf    Live load = 50 psf Additional data    Typical bay size = 24 × 32 ft (5 bays in N-S direction, 3 bays in E-W direction)    Typical story height = 12 ft Factored Torsional Moment u T    Since beam CD is part of an indeterminate framing system in which redistribution of internal forces canoccur following torsional cracking, the maximum factored torsional moment u T at the critical sectionlocated at a distance d from the face of the support can be determined from the following: ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ φ= cp2cpcu pAf4T  '  This type of torsion is referred to as compatibility torsion, the magnitude of which is greater than thefactored torsional moment min,u T  below which torsional effects can be neglected, where ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ φ= cp2cp'c,minu pAfT Since the beams and joists are cast monolithically, and for beam CD can include a portion of theadjoining slab. The effective width of the overhanging flange must conform to 13.2.4: cp A cp p e b e b = = 20.5 – 4.5 = 16.0 in. (governs) f hh − = = = 18.0 in. f h4 544 . × cp A = = 564.0 in. 2   ( ) ( 20.5×24+16.0×4.5 ) )) cp p = = 121 in. ( ) 220.5+24+16.0 cp2cp pA / = 2,629 in. 3  The torsional properties of the beam ignoring the overhanging flange are the following: cp A = = 492 in. 2   ( 24520 × . cp p = = 89 in. ( 245202 + . cp2cp pA / = 2,720 in.3 > 2,629 in. 3  Therefore, ignore flange per 11.6.1.1.720,2000,4475.0T  u ××= = 516,084 in.-lbs = 43.0 ft-kipsIt is assumed that the torsional loading on beam CD is uniformly distributed along the span.    TIME SAVING DESIGN AID Page 3 of 23 Torsion Design Shear Forces and Bending Moments in Beam CD ( ) ( ) 00012323077150 14424520w D ,//. ⎥⎦⎤⎢⎣⎡×++⎟ ⎠ ⎞⎜⎝ ⎛ ××= = 2.2 kips/ft ( ) 000123250w L ,// ×= = 0.8 kips/ft LDu w61w21w .. += = 3.9 kips/ftThis framing system satisfies the conditions for analysis by the coefficients of 8.3.3, as illustrated inFigure 2-2. Thus, use Figures 2-3, 2-4, and 2-7 to determine the maximum positive moment, negativemoment, and shear force in an interior span: =⎟ ⎠ ⎞⎜⎝ ⎛ −×== + 161222249.3 16wM 22nuu  l   119.8 ft-kips =×== − 11168.11911wM 2nuu  l   174.3 ft-kips == 2wV nuu  l   43.2 kips Adequacy of Cross-sectional Dimensions For solid sections: ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ +φ≤⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ +⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛  ' . cwc22ohhu2wu f8dbVA71pT dbV Using an average d = 20.5 − 2.5 = 18.0 in., the factored shear force at the critical section located ata distance d from the face of the support is = 37.4 kips. Also, the nominal shear strength provided bythe concrete is u Vdbf2V wcc ' = .Assuming a 1.5-in. clear cover to No. 4 closed stirrups: ( ) [ ] ( ) [ ] 2oh in5348505122450512520A ....... =−×−×−×−=   ( ) [ ] ( ) [ ] { } ...... in755051224505125202p h =−×−+−×−×=  Therefore, ( ) psi474000,48275.0psi207 5.3487.1 75000,120.43 1824400,37 222 =+×<= ⎟ ⎠ ⎞⎜⎝ ⎛ ×××+⎟ ⎠ ⎞⎜⎝ ⎛ × , OK. Transverse Reinforcement Required for Torsion θφ= cotfA2 T sA  yvout      TIME SAVING DESIGN AID Page 4 of 23 Torsion Design where = 0.85 o A 2ho .in2.2965.34885.0A =×=   °=θ 45 Therefore, leg.in/.in 019.045cot000,602.296275.0 000,120.43 sA 2t =°×××× ×=   Transverse Reinforcement Required for Shear ( ) 018000,6075.0 983,40400,37 18000,6075.0 18244000275.0400,37 dfVVsA  yvcuv <××−=×××××− =φφ−=  No transverse reinforcement for shear is required because cu VV φ< . However, since cu V5.0V φ> ,minimum area of transverse reinforcement shall be provided per ACI 11.5.6.1. Total Required Transverse Reinforcement For 4,000-psi concrete, minimum transverse reinforcement for shear and torsion= 019.0000,6024000,475.0 fbf75.0  yvw'c == in. 2 /in.= = 0.020 in. 2 /in. (governs) 000,60/2450f/b50  yvw ×= Required transverse reinforcement =+ sA2sA tv leg.in/.in 019.0legs20 2 ×+ = 0.038 in. 2 /in. > 0.020 in. 2 /in. OK.Maximum spacing of transverse reinforcement= = 75 / 8 = 9.4 in. < 12.0 in. 8/p h = = 18 / 2 = 9.0 in. (governs) 2/d Assuming No. 4 closed stirrups (area per leg = 0.2 in. 2 ), the required spacing s at the critical section =0.2/0.019 = 10.5 in. < 9.0 in.Provide No. 4 closed stirrups spaced at 9 in. on center at the critical section. In view of the shear andtorsion distribution along the span length, this same reinforcement and spacing can be provided from theface of the support to a distance + d = 24 + 18 = 42 in. = 3.5 ft past the location where it is no longerrequired. t b Longitudinal Reinforcement Required for Torsion 222 y yvht .in43.145cot 606075019.0cot ffpsAA =°×⎟ ⎠ ⎞⎜⎝ ⎛ ××=θ ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ ⎟ ⎠ ⎞⎜⎝ ⎛ =  l   l