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Reinforced Concrete Design, 9th Edition Solution Manual

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    Instructor’s Solutions Manual To accompany Reinforced Concrete Design Ninth Edition Abi O. Aghayere Drexel University

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    Prob 1 - 11 Part (a): Dead Loads Slab weight (8”/12)(150pcf) = 100psf ½” light wt. floor finish = 4psf susp. ceiling = 2psf M/E (industrial) = 20psf Partitions = 20psf _______________________________________________  DL = 146psf (with partitions)  DL = 126psf (without partitions) Part (b): Live Load = 100psf 1.4D = (1.4)(126) = 176.4psf 1.2D + 1.6L = (1.2)(126) + (1.6)(100)= 311ps f  controls Part (c): Load on spandrel beam Trib. width = (18 ft/2) + 1.0 ft = 10 ft Cladding: Block: (55psf)(10 ft. – 2’ - 2”) = 431 plf Brick: (40psf)(10 ft.) = 400 plf Beam Stem: (8”/12)(12”/12)(150pcf) = 225 plf Service Dead Load: (126psf)(10 ft ) + 431plf + 400 + 225 plf = 2316 plf = 2.32 k/ft Service Live Load: (100psf)(10 ft) = 1000 plf = 1.0 k/ft Part (d): Factored Load on spandrel beam: 1.2D + 1.6L = (1.2)(2.32) + (1.6)(1.0)= 4.4plf 1 - 5

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    Part (e): Load on interior beam Trib. width = (18 ft/2) + (18 ft/2) + 1.0 ft = 19 ft Beam Stem: (8”/12)(12”/12)(150pcf) = 225 plf Service Dead Load: (126psf)(19 ft) + 225 plf = 2619 plf = 2.62 k/ft Service Live Load: (100psf)(19 ft) = 1000 plf = 1.9 k/ft Part (f): Factored Load on interior beam: 1.2D + 1.6L = (1.2)(2.62) + (1.6)(1.9)= 6.2k/ft Part (g): Spandrel Beam: 2 ) 36 )( 4 . 4 ( 2 L w V u u = = = 79.2 k 8 ) 36 )( 4 . 4 ( 8 L w M 2 2 u u = = = 713 k - ft 1 - 6

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    Prob 1 - 12 Roof Loads Typical Floor Loads Dead Load: 7” slab (7”/12 x 150 pcf) = 87.5 psf 5 ply + gravel = 6.5 psf Insulation = 3.5 psf M & E = 10 psf Suspended ceiling = 2 psf Total roof dead load, D = 109.5 psf Roof live load (snow) S = 35 psf Dead Load: 7” slab (7”/12 x 150 pcf) = 87.5 psf 0.5” light wt floor finish = 4 psf Partitions = 20 psf M & E = 10 psf Suspended ceiling = 2 psf Total roof dead load, D = 123.5 psf Floor live load (office) L = 50 psf A. Roof: w DL slab = 109.5 psf; w LL slab = 35 psf Using the load combinations, we find the maximum factored load to be, w u slab = 1.2 (109.5psf) + 1.6 (35 psf) = 188 psf Floor: w DL slab = 123.5 psf; w LL slab = 50 psf Using the load combinations, we find the maximum factored load to be, w u slab = 1.2 (123.5 psf) + 1.6 (50 psf) = 229 psf B. (i) TYPICAL INTERIOR FLOOR BEAM Tributary width of beam = 16 ft Beam stem weight = (16”/12) [(26” – 7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ft w DL beam = 123.5 psf x 16 ft + 317 Ib/ft = 2293 Ib/ft = 2.3 kip/ft w LL beam = 50 psf x 16 ft = 800 Ib/ft = 0.8 kip/ft w s beam = w DL beam + w LL beam = 3.1 kip/ft w u beam = w u slab x beam trib width + 1.2 x beam stem weight = 229 psf x 16 ft + 1.2 (317 Ib/ft) = 4045 Ib/ft = 4.1 kip/ft For a beam subjected to a uniformly distributed load, the beam reactions, assuming simple supports, are: R DL = w DL beam x L/2 = 2.3 kip/ft x 32 ft /2 = 37 kip R LL = w LL beam x L/2 = 0.8 kip/ft x 32 ft /2 = 12.8 kip R u = w u beam x L/2 or 1.2 (R DL ) + 1.6 (R LL ) = 4.1 kip/ft x 32 ft/2 = 66 kip or 1.2 (37 kip) + 1.6 (12.8 kip) = 66 kip (same!) 1 - 7

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    (ii) TYPICAL INTERIOR ROOF BEAM Tributary width of beam = 16 ft Beam stem weight = (16”/12) [(26” – 7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ft w DL beam = 109.5 psf x 16 ft + 317 Ib/ft = 2070 Ib/ft = 2.1 kip/ft w LL beam = 35 psf x 16 ft = 560 Ib/ft = 0.56 kip/ft w s beam = w DL beam + w LL beam = 2.7 kip/ft w u beam = wu slab x beam trib width + 1.2 x beam stem weight = 188 psf x 16 ft + 1.2 (317 Ib/ft) = 3389 Ib/ft = 3.4 kip/ft For a beam subjected to a uniformly distributed load, the beam reactions, assuming simple supports, are: R DL = w DL beam x L/2 = 2.1 kip/ft x 32 ft /2 = 34 kip R LL = w LL beam x L/2 = 0.56 kip/ft x 32 ft /2 = 9 kip R u = w u beam x L/2 or 1.2 (R DL ) + 1.6 (R LL ) = 3.4 kip/ft x 32 ft/2 = 55 kip or 1.2 (34 kip) + 1.6 (9 kip) = 55 kip (same!) (iii) TYPICAL FLOOR SPANDREL BEAM Tributary width of beam = (16 ft / 2 ) + [(16”/12)/2] = 8.67 ft Beam stem weight = (16”/12) [(26” – 7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ft Exterior Cladding load = 55 psf block x 7.83’ height + 40 psf brick x 10’ height = 0.83 kip/ft w DL beam = 123.5 psf x 8.67 ft + 317 Ib/ft + 830 Ib/ft = 2218 Ib/ft = 2.2 kip/ft w LL beam = 50 psf x 8.67 ft = 434 Ib/ft = 0.44 kip/ft w s beam = wDL beam + wLL beam = 2.7 kip/ft w u beam = w u slab x beam trib width + 1.2 x beam stem weight = 229 psf x 8.67 ft + 1.2 (317 Ib/ft + 830 Ib/ft) = 3362 Ib/ft = 3.4 kip/ft For a beam subjected to a uniformly distributed load, the beam reactions, assuming simple supports, are: R DL = wDL beam x L/2 = 2.2 kip/ft x 32 ft /2 = 36 kip R LL = wLL beam x L/2 = 0.44 kip/ft x 32 ft /2 = 7 kip R u = w u beam x L/2 or 1.2 (R DL ) + 1.6 (R LL ) = 3.4 kip/ft x 32 ft/2 = 55 kip or 1.2 (36 kip) + 1.6 (7 kip) = 55 kip (same!) 1 - 8

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    (iv) TYPICAL ROOF SPANDREL BEAM Tributary width of beam = (16 ft / 2 ) + [(16”/12)/2] = 8.67 ft Beam stem weight = (16”/12) [(26” – 7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ft On the roof beam and girder, there is usually a one to two feet high parapet. However, if we assume NO PARAPET in this example, the exterior cladding load on the roof spandrel beam will be: Exterior Cladding load = 55 psf block x 0’ height + 40 psf brick x 0’ height = 0 kip/ft w DL beam = 109.5 psf x 8.67 ft + 317 Ib/ft + 0 Ib/ft = 1267 Ib/ft = 1.3 kip/ft w LL beam = 35 psf x 8.67 ft = 303 Ib/ft = 0.3 kip/ft w s beam = w DL beam + wLL beam = 1.6 kip/ft w u beam = w u slab x beam trib width + 1.2 x beam stem weight = 188 psf x 8.67 ft + 1.2 (317 Ib/ft + 0 Ib/ft) = 2010 Ib/ft = 2.0 kip/ft For a beam subjected to a uniformly distributed load, the beam reactions, assuming simple supports, are: R DL = w DL beam x L/2 = 1.3 kip/ft x 32 ft /2 = 21 kip R LL = w LL beam x L/2 = 0.3 kip/ft x 32 ft /2 = 5 kip R u = w u beam x L/2 or 1.2 (R DL ) + 1.6 (R LL ) = 2.0 kip/ft x 32 ft/2 = 32 kip or 1.2 (21 kip) + 1.6 (5 kip) = 33 kip (same!) C. (i) TYPICAL INTERIOR FLOOR GIRDER Girder stem weight = (18”/12) [(26” – 7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ft This girder has TWO interior beams framing into it at the MIDSPA, in addition to a uniformly distributed loading (UDL) due to the girder stem weight. Therefore, the total concentrated loads acting on the girder are: P DL girder = 2 beams x R DL beam = 2 x 37 kip = 74 kip (acts at midspan) P LL girder = 2 x R LL beam = 2 x 12.8 kip = 26 kip (acts at midspan) P s girder = P DL girder + P LL girder = 74 + 26 = 100 kip (acts at midspan) P u girder = 1.2 P DL girder + 1.6 P LL girder = 1.2(74) + 1.6(26) = 131 kip (acts at midspan) 1 - 9

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    The uniformly distributed load (UDL) on this girder is due to the girder stem weight. Thus, w DL girder = girder stem weight = 357 Ib/ft w LL girder = 0 The total UDL on the girder are: w s girder = w DL girder + w LL girder = 357 Ib/ft w u girder = 1.2 w DL girder + 1.6 w LL girder = 1.2 (357 Ib/ft) + 1.6 (0) = 430 Ib/ft = 0.43 kip/ft The REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition to concentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simple supports, are: R DL = w DL girder x L/2 + P DLgirder /2 = (0.36 kip/ft x 32 ft /2) + (74 kip /2 ) = 43 kip R LL = w LL girder x L/2 + P LLgirder /2 = (0 kip/ft x 32 ft /2) + (26 kip /2 ) = 13 kip R u = w u girder x L/2 + P u girder /2 = (0.43 kip/ft x 32 ft /2) + (131 kip /2 ) = 73 kip (ii) TYPICAL INTERIOR ROOF GIRDER Girder stem weight = (18”/12) [(26” – 7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ft This girder has TWO interior beams framing into it at the MIDSPAN, in addition to a uniformly distributed loading (UDL) due t o the girder stem weight. Therefore, the total concentrated loads acting on the girder are: P DL girder = 2 beams x R DL beam = 2 x 34 kip = 68 kip (acts at midspan) P LL girder = 2 x R LL beam = 2 x 9 kip = 18 kip (acts at midspan) P s girder = P DL girder + P LL girder = 68 + 18 = 86 kip (acts at midspan) P u girder = 1.2 P DL girder + 1.6 P LL girder = 1.2(68) + 1.6(18) = 111 kip (acts at midspan) The uniformly distributed load (UDL) on this girder is due to the girder stem weight. Thus, w DL girder = girder stem weight = 357 Ib/ft; w LL girder = 0 The total UDL on the girder are: w s girder = w DL girder + w LL girder = 357 Ib/ft w u girder = 1.2 w DL girder + 1.6 w LL girder = 1.2 (357 Ib/ft) + 1.6 (0) = 430 Ib/ft = 0.43 kip/ft The REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition to concentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simple supports, are: 1 - 10

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    R DL = w DL girder x L/2 + P DLgirder /2 = (0.36 kip/ft x 32 ft /2) + (68 kip /2 ) = 40 kip R LL = w LL girder x L/2 + P LLgirder /2 = (0 kip/ft x 32 ft /2) + (18kip /2 ) = 9 kip R u = w u girder x L/2 + P ugirder /2 = (0.43 kip/ft x 32 ft /2) + (111 kip /2 ) = 63 kip (iii) TYPICAL FLOOR SPANDREL GIRDER Girder stem weight = (18”/12) [(26” – 7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ft Exterior Cladding load = 55 psf block x 7.83’ height + 40 psf brick x 10’ height = 0.83 kip/ft This girder has ONE interior beam framing into it at the MIDSPAN, in addition to a uniformly distributed loading (UDL) due to the girder stem weight, the girder edge distance, AND the exterior cladding. Therefore, the total concentrated loads acting on the girder are: P DL girder = 1 beam x R DL beam = 1x 37 kip = 37kip (acts at midspan) P LL girder = 1 x R LL beam = 1 x 12.8 kip = 13 kip (acts at midspan) P s girder = P DL girder + P LL girder = 37 + 13 = 50 kip (acts at midspan) P u girder = 1.2 P DL girder + 1.6 P LL girder = 1.2(37) + 1.6(13) = 65 kip (acts at midspan) Uniform floor load on Girder edge distance: w DL edge distance = (9”/12) x 123.5 psf = 93 Ib/ft w LL edge distance = 0 ( Because the WHOLE girder edge distance is occupied by the block wall cladding and thus there can be no live load in that edge distance ) The uniformly distributed load (UDL) on this girder is due to the girder stem weight, the load on the girder edge distance, AND the exterior cladding. Thus, w DLgirder = girder stem weight + girder edge distance + exterior cladding = 357 Ib/ft + 93 + 830 Ib/ft = 1280 Ib/ft ; w LL girder = 0 The total uniformly distributed load (UDL) on the girder is: w s girder = w DLgirder + w LLgirder = 1280 Ib/ft w u girder = 1.2 w DL girder + 1.6 w LL girder = 1.2 (1280 Ib/ft) + 1.6 (0) = 1536 Ib/ft = 1.54 kip/ft The REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition to concentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simple supports, are: R DL = w DL girder x L/2 + P DLgirder /2 = (1.28 kip/ft x 32 ft /2) + (37 kip /2 ) = 39 kip R LL = w LL girder x L/2 + P LLgirder /2 = (0 kip/ft x 32 ft /2) + (13 kip /2 ) = 6.5 kip R u = w u girder x L/2 + P ugirder /2 = (1.54 kip/ft x 32 ft /2) + (65 kip /2 ) = 57 kip 1 - 11

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    (iv) TYPICAL ROOF SPANDREL GIRDER Girder stem weight = (18”/12) [(26” – 7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ft (no parapet) This girder has ONE interior beam framing into it at the MIDSPAN, in addition to a uniformly distributed loading (UDL) due to the girder stem weight, the load on the girder edge distance, AND the exterior cladding. Therefore, the total concentrated loads acting on the girder are: P DL girder = 1 beam x R DL beam = 1x 34 kip = 34kip (acts at midspan) P LL girder = 1 x R LL beam = 1 x 9 kip = 9 kip (acts at midspan) P s girder = P DL girder + P LL girder = 34 + 9 = 43 kip (acts at midspan) P u girder = 1.2 P DL girder + 1.6 P LL girder = 1.2(34) + 1.6(9) = 55 kip (acts at midspan) Uniform roof load on Girder edge distance: w DL edge distance = (9”/12) x 109.5 psf = 82 Ib/ft w LL edge distance = (9”/12) x 35 psf = 26 Ib/ft The uniformly distributed load (UDL) on this girder is due to the girder stem weight, the load on the girder edge distance, AND the exterior cladding. Thus, w DLgirder = girder stem weight + girder edge distance + exterior cladding/parapet = 357 Ib/ft + 82 Ib/ft + 0 Ib/ft = 439 Ib/ft (assuming no parapet) w LL girder = 26 Ib/ft The total UDL on the girder are: w s girder = w DLgirder + w LLgirder = 465 Ib/ft w u girder = 1.2 w DL girder + 1.6 w LL girder = 1.2 (439 Ib/ft) + 1.6 (26 Ib/ft) = 568 Ib/ft = 0.57 kip/ft The REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition to concentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simple supports, are: R DL = w DL girder x L/2 + P DLgirder /2 = (0.44 kip/ft x 32 ft /2) + (34 kip /2 ) = 24 kip R LL = wLL girder x L/2 + P LLgirder /2 = (0.026 kip/ft x 32 ft /2) + (9 kip /2 ) = 4.9 kip R u = wu girder x L/2 + P ugirder /2 = (0.57 kip/ft x 32 ft /2) + (55 kip /2 ) = 36.6 kip 1 - 12

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    P rob 2 - 33 f c ’ = 3000 psi; f y = 60,000 psi; span, L = 14 ft; bar diameter = 5/8 = 0.625”; slab thickness, h =7” (a) Selfweight = (7”/12)(150 pcf) = 87.5 psf Effective depth, d = 7” – 1.5” cover – (0.625”/2) = 5.18” A s for #5 @ 14” o.c. = 0.265 in 2 /ft ( )( )( ) 2 2 (0.265)(60) 0.52" 0.85 (0.85)(3)(12) 0.52 0.9 0.265 60 5.18 2 2 5.89 . / . 8 8(5.89) 0.239 (14) 1.2 1.6 239 1.2(50 87.5) 1.6 , 46.7 ( ) s y c n s y u u u s A f a f b a M A f d ft k ft width M w ksf w D L psf L Therefore L psf b A   = = =      = − = −         = − = = = = + = + + = 2 ,min max 0.0018(12)(7") 0.15 / 5 5(7") 35" 18" # 4@15" . . # 4@14" . . ( ) in ft width s h or o c Use o c matches spacing of main rebar = = = = = Note : Check c/d t and the assumed strength reduction factor,  .  1 = 0.85 for f c ’ = 3000 psi c = (0.52”)/0.85 = 0.61” c/d t = 0.61/5.18 = 0.118 < 0.375, therefore,  = 0.9 as assumed. Prob 2 - 34 f c ’ = 4000 psi; f y = 60,000 psi 2 - 16

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