Problem 12.7 For Problems 12.1 to l2.5, desigh Wall fooconcreteDistance from bottom ofWallthickness (in.)ProblemD (k/ft)L(k/ft)(ksi) ,(ksi)9. (ksf)footing to final grade (ft)no.12.11415124.603.12.2142120604.12.3151518605n12.413243046044.12.512108.4604.(Answer to Problem 12.1: 15-in. footing, 10 ft 3 in, wide with #7 bars @ 9-in, main steel)(Answer to Prablem 12.3: 15-in. footing, 7 ft 6 in. wide with #5 bars @ 6-in. main steel)(Answer to Problem 12.5; 10-in. footing, 4 ft 0 in. wide with #5 bars @ 15-in. main steelProblem 12.6Repeat Problem 12.1 if a masonry wall is used.Column FootingsFor Problems 12.7 to 12.12, design square single-column footings for the values given. All celumns are interior columns.Columnsize (in.)Distance from bottom offooting to final grade (ft)ProblemD (k)L(k)f(ksi),(ksi)9. (ksf)no.12.712 x 121001604.60412.812 x 12100803601512.915 x 151601504604.12.1015 x 15150120604.12.1116 x 161101006012.12Round, 18 dia.24014046038888 3

Following is the given data : Rebar = #7, CC = 3 in, d = thickness - 1.5db - CC Assume footing thickness equal to 19 in(1.583 ft), d = 18.00 inches. Estimated to c.g. of Top Layer of Flexural Steel. Calculate qe (effective bearing pressure capacity) as follows: qe = 5000 psf - 1.583ft(150 pcf) - (4ft - 1.583ft)(100pcf) = 4520.85 psf Calculate required area of footing as follows: Area = Total Service Loads/qe = (100k + 160k)/4520.85 psf = 57.51 ft2 Side of square footing: L = W = (57.51 ft2)1/2 = 7.584 ft = 8.0 ft Footing size is 8 ft x 8ft. Area of footing = 64ft2 Calculate the Bearing Pressure for Strength Design as follows: qu = [1.2(100k) + 1.6(160k)]/(64ft2) = [120 k + 256 k] / 64ft2 = 5875 psf Depth Required for Two-Way or Punching Shear b = column size + d b = 12 inch + 14.6875 inch = 26.6875 inches (2.224ft) bo = 4(26.6875 inches) = 106.75 inches Punching Area = 64 ft2 - (2.224ft)2 = 59.10 ft2 Vu2 = 59.10 ft2(5875 psf) = 347.212 k d = Vu2/phi(4(fc'0.5)(bo) = 347.212 k/0.75(4)(4000psi)1/2(106.75in) = 17.14 inches < 18.00 inches Hence, OK in Punching Shear.Depth Required for One-Way Shear b = (8ft/2) - (12inch/2) - 15.00 in = 17 inch(1.42ft) Vu1 = bLqu = 8ft(1.42ft)(5875 psf) = 66.740 kLd = d = Vu1/phi(2(fc')1/2(L) = 66.740 k/0.75(2)(4000psi)1/2(8ft) = 8.8 in < 14.6875 inches - OK! Hence assumed d and thickness is sufficient. Strength Design: b = (8ft/2) - (12inch/2) = 48 inches (4ft) Mu = 4ft(8ft)(5875 psf)(4ft/2) = 376.00ft-k Rn = Mu/phibd2 = 376.00ft-k/0.90(84in)(14.6875in)2 = 276.66 psi rho = 0.85fc'/fy(1 - (1 - 2Rn/0.85fc')1/2) = 0.002 rhomin = 200/fy = 200/60000psi = 0.00333 > 3(fc')0.5/fy = 0.00274, Thus, use rhomin = 0.00333>0.002 As = rho(bd) = 0.0033(84in)(14.6875in) = 4.11 in2 Ab = 0.25pi(0.875in)2 = 0.601in2 No. of rebars = 4.11/0.601 = 6.84 say 7 pcs (bothways) Summary: Use 8 ft x 8 ft x 20 inch thick footing reinforced with 7 - #7 rebars .

Engineering

Civil Engineering

Column Footings For Problems 12.7 to 12.12, design square single-column footings for the values given. All coiumns are interior columns. Column size (in.) Distance from bottom of footing to final grade (ft) Problem D (k) L(k) (ksi), (ksi) 9. (ksf) no. 12.7 12 x 12 100 160 60 4. 128 100 60

Column Footings For Problems 12.7 to 12.12, design square single-column footings for the values given. All coiumns are interior columns. Column size (in.) Distance from bottom of footing to final grade (ft) Problem D (k) L(k) (ksi), (ksi) 9. (ksf) no. 12.7 12 x 12 100 160 60 4. 128 100 60

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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