Lab_7_sol - UNIVERSITY OF CALIFORNIA, BERKELEY Instructor:...

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Unformatted text preview: UNIVERSITY OF CALIFORNIA, BERKELEY Instructor: Khalid M. Mosalam Department of Civil and Environmental Engineering CE123: Design of RC Structures Structural Engineering, Mechanics and Materials (SEMM) Spring Semester, 2009 CEE 123 Laboratory Project No. 7 (Solution) Problem 3.11 f'’c=4000psi fy=60,000psi b=24in h=14in d=11.5in As=3#10+2#11=3.81+3.12=6.93in2 a) Neglecting compression reinforcement . . . 5.096 6 But, 6.93 60 11.5 5.096 2 3.722 310 2 . 0.003 0.00275 0.004 Section is not allowed by ACI code and thus we should consider compressive steel reinforcement 0.71 0.483 833 220 b) Assume As’ yields . . . 3.23 2 331 . 3.80 6.93 2.54 60 11.5 3.23 2 2.54 60 11.5 2.5 0.003 298 11.5 3.8 3.8 0.0061 0.005 0.9 c) Compression steel strength fs’ We need to find f’s and to do that we solve the equilibrium of forces equation, which is a quadratic equation with the unknown being c as follows: 0.85 c=4.56in Thus, 0.003 29000 0.85 4.56 3.88 2 327 . . . . 39 2.54 39 60 60 11.5 3.88 2 2.54 39 11.5 2.5 6.93 0.003 282 11.5 4.57 4.57 0.00455 0.483 83.3 0.00455 0.86 Problem 3.12 f'’c=4000psi fy=60,000psi Mu=5780kip-in hf=5in bw=10in d=20in Assume a<hf 2 5780 0.9 60 20 5 2 6.12 Check a: 0.85 6.12 60 0.85 4 20 5.4 Find As’ to balance flange 0.85 2970 2.83 2 Find As in stem. First find Mn2 carried by stem: 3452 Try a=6in 3.38 2 2 0.85 5.96 0.003 6 20 7.01 7.01 0.0056 0.005 7.01 0.9 ! As=3.38+2.83=6.21in2 Use 4#11(#36)=6.24in2 placed in 2 layers H=d+spacing between layers+bar diameter+stirrup+cover=20+1+1.4+0.37+1.5=25in Problem 4.1 f'’c=4000psi Vu=44kips d=2bw a) No web reinforcement 1 2 2 44000 2 0.5 0.75 2√4000 928 22 44 b) Minimum web reinforcement 2 44000 2 464 0.75 2√4000 c) 16 32 Web reinforcement provides shear strength of Vs=2Vc 3 32 44000 2 155 10 3 0.75 2√4000 20 Problem 4.2 f'’c=4000psi fy=60000psi d=22in b=12in L=20ft As=3#11=3*1.56=4.68in2 DL=1.63kip/ft LL=3.26kip/ft Av=? using #3 stirrups Using the more approximate approach (Eq4.12b) for calculating Vc the design is as follows: wu=1.2(1.63)+1.6(3.26)=7.2kip/ft Vu=7.2*20/2=72Kips Vu @daway=7.2-(7.2*22/12)=58.8Kips 2 25 The ACI code allows the design to start from the shear value at a distance d away from the face of the support, thus for the first stretch: 58.8 25 0.22 0.0342 6.44 0.75 60 22 0.0342 Check for smax: 62.6 Thus, smax=smaller of Thus, smax=11in Use s1=6in for first stretch until a distance of 3 ft from the support NOTE: Start providing shear reinforcements at a distance s/2=3in from end of beam 4 66.8 2√4000 22 12 33.4 a) Avfy/(0.75*sqrt(4000)*12)=23.2in b)24in c)d/2=22/2=11in At x=3ft from support Vu=72-(72-25)*3/6.53=50.4Kips 25 50.4 25 0.22 0.026 8.57 11 0.75 60 22 0.026 Use s2=8in until the location where only minimum web reinforcement is required (theoretically no shear reinforcement is needed) i.e. when Vu<phi*Vc : x=10(72-25)/72=6.53ft from support The location where no shear reinforcement is needed is when Vu<phi*Vc/2 : x=10(72-12.5)/72=8.26in Thus, provide web reinforcements at the maximum allowed spacing of 11in from a distance of 6.53ft from the support until a distance of 8.26ft from the support and after that point until midspan no shear reinforcements are needed. (NOTE: The reinforcement is symmetrical on the two sides of the beam) Thus, s3=11in The following is a spacing pattern following the aforementioned calculations: 1 space at 3in=3 in 6 spaces at 6in=36in 6 spaces at 8in=48in 2 spaces at max spacing (11in)=22in The following is a table showing the calculated values at 1ft intervals until midspan: Distance from support, ft 0 1 1.83 2 3 4 5 6 7 8 9 10 3.00 6.53 8.26 Mu, ftkips 0.00 68.40 119.90 129.60 183.60 230.40 270.00 302.40 327.60 345.60 356.40 360.00 183.60 316.65 349.10 Vu, kips 72.00 64.80 58.80 57.60 50.40 43.20 36.00 28.80 21.60 14.40 7.20 0.00 50.40 24.98 12.53 Vu shaded 58.80 58.80 58.80 57.60 50.40 43.20 36.00 28.80 21.60 14.40 7.20 0.00 50.40 24.98 12.53 Vc, kips 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 33.39 phi*Vc 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 25.05 phi*Vc/2 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 12.52 The following plot shows the variations of the quantities presented in the above table, thus making the geometric location of critical sections easier: 80.00 70.00 60.00 50.00 Shear, Kips 40.00 30.00 20.00 10.00 0.00 0 1 2 3 4 5 6 7 8 9 10 Distance from support, ft Vu Vu cut at d away from support phi*Vc phi*Vc/2 Problem 4.3 f'’c=4000psi fy=60000psi d=22in b=12in L=20ft As=3#11=3*1.56=4.68in2 DL=1.63kip/ft LL=3.26kip/ft Av=? using #3 stirrups Using the more elaborate approach (Eq4.12a) for calculating Vc the design is as follows: wu=1.2(1.63)+1.6(3.26)=7.2kip/ft Vu=7.2*20/2=72Kips Vu @daway=7.2-(7.2*22/12)=58.8Kips 1.9 2500 3.5 Using the more elaborate approach, the method of design does not change, however, the calculations become tedious because Vc is not constant anymore and it varies as we move along the beam. The following table shows the variation of Vc and all other quantities needed along the length of the beam (until midspan since the design is symmetric): Distance from support, ft 0 1 1.83 2 3 4 5 6 7 8 9 10 3.33 6.52 8.31 Mu, ft-kips 0.00 68.40 119.90 129.60 183.60 230.40 270.00 302.40 327.60 345.60 356.40 360.00 199.84 316.40 349.72 Vu, kips 72.00 64.80 58.80 57.60 50.40 43.20 36.00 28.80 21.60 14.40 7.20 0.00 48.02 25.06 12.17 Vu shaded 58.80 58.80 58.80 57.60 50.40 43.20 36.00 28.80 21.60 14.40 7.20 0.00 48.02 25.06 12.17 (Vud/Mu)calculated 999.00 1.74 0.90 0.81 0.50 0.34 0.24 0.17 0.12 0.08 0.04 0.00 0.44 0.15 0.06 (Vud/Mu)used 1.00 1.00 0.90 0.81 0.50 0.34 0.24 0.17 0.12 0.08 0.04 0.00 0.44 0.15 0.06 Vc, kips 43.41 43.41 42.23 41.24 37.60 35.74 34.58 33.76 33.14 32.62 32.16 31.72 36.87 33.42 32.47 phi*Vc 32.55 32.55 31.67 30.93 28.20 26.80 25.93 25.32 24.85 24.46 24.12 23.79 27.65 25.06 24.35 phi*Vc/2 16.28 16.28 15.84 15.47 14.10 13.40 12.97 12.66 12.43 12.23 12.06 11.90 13.83 12.53 12.18 The following plot presents the quantities presented in the above table graphically. 80.00 70.00 60.00 50.00 Shear, Kips 40.00 30.00 20.00 10.00 0.00 0 1 2 3 4 5 6 7 8 9 10 Distance from Support, ft Vu Vu cut at d away value phi*Vc phi*Vc/2 ACI allows the design to start by using the value of shear at a distance d away from the support, thus at x=d=1.83ft: 31.67 Vu=58.8Kips 58.8 31.67 0.22 0.027 8.15 11 0.75 60 22 0.027 Use s1=8in until roughly a distance of L/6=40in=3.33ft from face of support NOTE: start providing stirrups from a distance s/2=8/2=4in from end of beam At x=3.33ft 27.65 Vu=48.02Kips 48.02 27.65 0.22 0.0206 10.68 11 0.75 60 22 0.0206 Use s2=10in until the location where only minimum web reinforcement is required (theoretically no shear reinforcement is needed) i.e. when Vu<phi*Vc : x=6.52ft (from table results) The location where no shear reinforcement is needed is when Vu<phi*Vc/2 : x=8.31ft (from table results) Thus, provide web reinforcements at the maximum allowed spacing of 11in from a distance of 6.52ft from the support until a distance of 8.31ft from the support and after that point until midspan no shear reinforcements are needed. (NOTE: The reinforcement is symmetrical on the two sides of the beam) Thus, s3=11in The following is a spacing pattern following the aforementioned calculations: 1 space at 4in=4 in 6 spaces at 8in=48in 4 spaces at 10in=40in 2 spaces at max spacing (11in)=22in ...
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