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CE327 Old Final Exam

CE327 Old Final Exam - North Carolina State University...

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Unformatted text preview: North Carolina State University Reinforced Concrete Design CE 327 Spring 2003 Final Exam — May 9, 2003 [Closed Book and Notes. Only ACI 318-02 and one formula sheet allowed. Attach your formula sheet.] A reinforced concrete column with an isolated footing is shown. All loads shown in the figure are considered service hie loads. Assume concrete in the column and in the footing is of normal weight with a compressive strength of 4,000 psi, and steel is of Grade 60. Neglect self weight of the column. 0 ‘ 1. Find the maximum service lateral force (live load) that the column can resist. Remember to base your answer on the axial/flexural strength and the shear strength. Make sure you account for the area of concrete removed by the compression steel in your axial/flexural analysis. Cover to center of the bars in the column is 2.5 in. Do NOT use the refined equation for shear. 2. If the service lateral force (live load) of Part 1 is equal to 26,690 lb, find its corresponding immediate lateral deflection at the top of the column. Ignore any dead load effect. The formula for the lateral deflection of a cantilever column with a load at its top is given as follows: 3 A = ——QL , 3E1 3. If the factored lateral fOrce in Part 1 is equal to 42.71 k, check size and reinforcement in the footing. The column is an interior column in a building. Use a depth of 32 in for tension steel. Allowable soil pressure is given as 2.5 tons per square foot. Assume an average unit weight of 130 pcf for the soil and the concrete. Also, assume a surcharge of 100 psf. Make sure you check flexure, one-way (beam) shear and two-way (punching) shear. Properties of No. 3, No. 6, and No. 9 bars are given in Table 1. Table l. Diameters and Areas of Bars Used Cross-Sectional Area in2 PL = 353.1251]: « 4 Mn. 9 on Each side Axis of Bending 24” 13.!) ft 2.5 in mverto center 2.5 in mutt ta center .1 .-.-- KMW L -~ No. 3 fig}. “'1' cm W 12ft IL 13 No.fi@3“ca’uin but]: dirfictions MAL—4L. 1w. @PMLprstpsk— Q / L;- Po : 0.8513; (Ag—Aw)* A’f {‘7 = 0.35319 (LHL_3X|)+ SXUKOO = Z‘HLZ. L (FE = 0-<03x2t+n.7_ : ‘50:?- 1‘- Pa = 0.56 -) [MBA (above. balmud. 430mm) .4, ¢=o_b$‘ @ 4P0 Bu’f avumfl. bo‘Hfl Qctjlus 0*: §+<JL9 jm‘d P , Pu .-. 5(05— =8®1.7—5K ¢ 0.0; l)“: O‘gskfliba 1- A',(Q,_o.BSFé)_ A50) [email protected] 8mm ; ozsmxu‘xa + Lt (C00_o.85X*+) _ “coo => Q=\0-8\‘1"J ‘Rr REA Ks: —, [5,: 0.8: _, (L: .‘L =I?_.=i»7_z" C7D fin E.) g, d-(. 83c». , L\.$- ‘L.q7£ {3 " x0005 = 00°20? = ,__ :1 ’ (00 10 i c. \‘LJLZ £6 E,— quoo 30.00 7® 9-5 » C-‘Vgu ._ \7..=l»zg-2.s ‘ - .______.. o. =0-002H>& :9. v’ c. ”m?” )5 00.5 a 007.61. 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