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9781461432616 (2)_sm.pdf

Problem 7.1 Determine the soil pressure distribution under the footing. Elevation Plan M 180 e   1.5 ft P 120 (a) B= L= 8 ft L e  1.5 ft   1.33 ft 6 1 P 6 (P e) 180 6 (180) q2   2   3  4.92 kip/ft 2 BL BL

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Continuum Mechanics

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1 Problem 7.1 Determine the soil pressure distribution under the footing. Elevation Plan M 180e 1.5 ftP 120     (a)   B= L= 8 ft   Le 1.5 ft 1.33 ft6        2  22 2 321 2 3 6 (P e) 6 (180)P 180q 4.92 kip/ftB L B L 8(8) 86 P e 6 (180)P 180q 0.7 kip/ftB L B L 8(8) 8           (b)   L=10 ft B =5 ft Le 1.5 ft 1.66 ft6       22 2 221 2 2 6 (P e) 6 (180)P 180q 5.76 kip/ftB L B L 5(10) (5)(10)6 P e 6 (180)P 180q 1.42 kip/ftB L B L 5(10) (5)(10)            Problem 7.2 Determine the soil pressure distribution under the footing. Use a factor of 1.2 for DL and 1.6 for LL. Elevation Plan u D Lu D L P 1.2 P 1.6 P 1.2(200) 1.6(250) 640 kNM 1.2 M 1.6 M 0 1.6(64) 64 kN-m              3 uu M 64 Le 0.1 m 0.5 mP 640 6         2u uu1 2 32u uu2 2 3 P 6 (P e) 6 (64)640q 85.32 kN/mB L B L 3(3) 3P 6 P e 6 (64)640q 56.88 kN/mB L B L 3(3) 3            Problem 7.3 The effective soil pressure is 4  2 kip/ft . Determine the maximum allowable value for L. Elevation Plan M 48e .3478 ftP 138     Assume Le6    2 3 2 3 6 M 6 (48)P 1384L L L L       3req req 4L 138L 288 0 L 6.72 ft         4 Problem 7.4 The allowable soil pressure is  2allowable q 250 kN/m  , 3soil γ 18 kN/m  ,  3conc γ 24 kN/m  , P D  = 1000 kN and P L  = 1400 kN. Plan Elevation   D+L P 1000 1400 2400 kN      2effective allowable conc soil q q  γ t γ (h-t) 250 24(.4) 18(1 .4) 229.6 kN/m           u D L P 1.2 P 1.6 P 1.2(1000) 1.6(1400) 3440 kN        uu1 2 PqL L    1u 2 u22 u 1u L aV ( ) L q ( )2 2L q L aM ( )2 2 2  x x      
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