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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