A (9’x9’x32”) square footing is supporting 230 kips dead load and 130 kips live load from 18-in square column. The allo...
2. Design a square spread footing for the following conditions. Service dead load is 350 kips, service live load is 275 kips. Soil density is 130 lb/ftº. Allowable soil pressure is 4500 psf. Column is 18 in. square. Geotechnical analyses have determined the width to be 13 ft. fe' = 3500 psi normal weight concrete, and fy= 60 ksi. The base of the footing is placed 5 ft below the ground surface. Note: you don't need to design the bar...
S. Al-Martini, PhD, P. Eng. Question No. 5 (35minutes) 19 marks) As shown in the figure, a 9-ft 6-in square footing supporting an 18-in. -square tied concrete column and it is subjected to service dead load 225 kips, and service live load 175 kips. column is a typical interior column in a building. Assume normal-weight concrete.. soil pressure (ga) 5000 psf (5 ksf), f'e for column 4000 psi, f's for the footing lsteel 60, 000 psi. The density of the...
S. Al-Martini, PhD, P. Eng. Question No. 5 (35minutes) 19 marks) As shown in the figure, a 9-ft 6-in square footing supporting an 18-in. -square tied concrete column and it is subjected to service dead load 225 kips, and service live load 175 kips. column is a typical interior column in a building. Assume normal-weight concrete.. soil pressure (ga) 5000 psf (5 ksf), f'e for column 4000 psi, f's for the footing lsteel 60, 000 psi. The density of the...
Design a square single column footing to support an 18 inch square interior column reinforced with 8-#10 bars. The column carries an unfactored dead load of 245 Kips that includes weight of soil and weight of footing and an unfactored live load of 20 feet below finished grade. The footing is 10 feet by 10 feet and is 30 inches thick. The allowable soil pressure is 5000 psf, fe' 4000 psi and fy 60000 psi. Concrete used shall be sand...
1. Design a wall footing for the following conditions. Service dead load is 6 kips/ft, service live load is 8 kips/ft. Soil unit weight is 120 lb/ft?. Allowable soil pressure, qa is 4000 psf. The base of the footing is placed 3 ft below the ground surface. Wall is 12 in, thick. Geotechnical analyses have determined the width to be 4 ft. fe' = 3500 psi normal weight concrete, and fy = 60 ksi. Note: you don't need to design...
CVE 313- 1- Design a square footing for the interior column shown in the figure. The column carries 2 k 't of uniform live load and 1.5 k/t of uniform dead load (including the self weiht of the structure), the base of the footing is 5 ft below grade, the soil weight is 100 lb/ft', and q," 5000 psf. and concrete unit weight is 150 lb/ (25 points) NOTE: Your design should only include the following (one-way shear, two-way shear,...
15.3. psi. Design a single-column footing (including dowels) to support an 11 in. square column reinforced with eight No. 9 (No. 29) bars centered 2.5 in. from the column faces (equal number of bars on each face). The unfactored axial dead load = 135 kips, and the unfactored axial live load=125 kips. For the column, f=4000 psi and f=60,000 psi. The base of the footing will be 3 ft below grade. The allowable soil bearing pressure is 3000 lb/ft2. Material...
Design a square footing to support an 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 kips and an axial live load of 200 kips. The base of the footing is 4.0ft. below final grade and allowable soil pressure is 5 k/ft2 Use ??′ = 3.0 ksi and fy = 60.0 ksi
Design a square tied column to carry axial service loads of 320 kips dead load and 190 kips live load. There is no identified applied moment. Assume that the column is short. Use f’c =4000 psi and fy = 60,000 psi. Also, draw the flexural and shear reinforcement on a sketch. Case 3: Design of Short Columns - Small Eccentricity Design a square tied column to carry axial service loads of 320 kips dead load and 190 kips live load....
Complefe the following: problem Parameters/notes Change the load: Dead -10 psf. Live - 30 psf hint: first trace the load W to the ends of the rods, then draw a FBD of the beam and rod and solve for reactions at C and B. Use this info to find the force in the rod. Your answer should be the diameter of a round rod rounded to 1/8" increment). Change the load to 400,000 Ibs See Appendix table A3 for Steel...