Question

10. After the footing in Problem 7.9 was built, the groundwater table rose to a depth of 0.5 m below the ground surface. Compute the new factor of safety against a bearing capacity failure. Com- pare it with the original design value of 2 and explain why it is different. (use Terzaghi’s formula to solve)

7.9) A bearing wall carries a total unfactored load 220 kN/m. It is to be supported on a 400 mm deep continuous footing. The underlying soils are medium sands with c′ = 0, f′ = 37°, and g = 19.2 kN/m3. The groundwater table is at a great depth. Using ASD, compute the minimum footing width required to maintain a factor of safety of at least 2 against a bearing capacity failure. Express your answer to the nearest 100 mm.

Answer 1

Geven that Toral mtactored load=220 kN/mm. Depth (d )= Hoomm = 001 m. Q=37º, q=1 qol khel mosos i diso conseder Terzaghe's method, the ul Hemate bearing capacety for strep footing Que=CNCF Toold. No to.58 for 0=37, Nc = 70.1, Na = 638 , N = 684. No toolbi wedlh of foundation Bing YO

X'= 8-tw/2-10w-*)] 33 8'= 19.2-404117–10:15.) =)!= 96396-0.939 Enims. trzard Arc = 1962X Dode-o = 742 7768ballon from 0 yhn an@.39b-0.939 = 0X01 +7668* X 5308 +0.6 x 0.57 X68.IX 19 lo 5x0.57x6• 1x [9.396-01929) =) Quot = 413 41846 + 182-2256 - 180224 . 54© Pressure = 120 kNm Total lead=220 (CNing. Apply factor of safety ay pressure on wall footing = 2X220 = 440 kn dom 26 Equate ☺ q® by 413. 8 imb + 18202045b-180444 = 440

0et 5 b=0.700mm) Nowo general water table Re at a depon of hoomm., then po b qa39x0.7-00939 Quet = Znc+ entroidNq +0.5 di bną =Qut= 7868X5388 + 0.5X BrOH8 X04x68.|| Put= 800 balon) presure = 220 - 22.= 3(18-29 . New Fo Defa bon • 6 # १ Lut= GOD BiH:29 -110925 ] The nalauated frool re dearealed from 2 to 10925. due to rise en level of ground water table. a Scanned with CamScanner