Question

Q.2 (12 Marks) An owner of a ceramic tile factory contacted a Geotechnical consultant to design foundation of the tall chimney to take care of flue gases formed during the tiles production. As per design of the Geotechnical Engineer, the chimney should have a square footing of 3.0 m side constructed at a depth of 2.50 m in the sub soil. It was found from the results of the laboratory test on the sample of sub-soil that the shear strength parameters in the form of angle of internal resistance and the cohesion intercept are 36° and 5 N/m² respectively. The moist unit weight is equal to 18.0 kN/m”. The chimney is carrying a concentrated load of 5000kN and two horizontal components of load of 200 kN acting at the height of 4.5 m perpendicular to length of the footing and 3.5 m perpendicular to the width of the footing measured from the base of footing. i. Suggest the most suitable analysis to be used in the above stated condition. Mention any two reasons to use the same. [2] Calculate the ultimate bearing capacity of the soil and factor of safety by using the above suggested analysis. [10] ii.

Answer 1

showed be Q=5000kN [* I think, C= 5KN/m not 5N/motherwises Ysat= 18 kN/m? ; C = Skov/m= (5x16**) /? $ = 360, (4.58200) l2 = M2 (3.5x200) voy Πας MI e, 5000 5000 0.14 m = 0.18m - B-221 = 2.64m 2.72 m Retrw=10kN/m² -> 3-0.1882) = Df= 2.5m L = 3-(0-14X2)-2-222 The most suitable analysis for the condition in the problem is Meryer hoof analysis of effective areas Considering general beasing capacity fromnla When the eccentercity is there, effective crea should be reduced to safe design the footing Also, when general becking opacity is considered shape, depth, inclination factors to be used for more safe design. qu= CNC Fs fed feit & Df Nq Fqs Fad Fait 815&B Nr Frs Frd Fri p=36 SN = (Nay coff No = = tan" (45+$Jettand' = 37.75 Ny= 2 (NqHD tand Df B are there = 50.58 NC = 56.3 2.5 2.64 = 0.95 21 C&scanned with CamScanner

B-2-2-64 m L' -L=2.72m shape factors Fes= HB = 1.72 Fas = 1+ B tand 43 = 1.71 Df В =0.95 Frs = 1-0.4 = 0.61 Depth factors Fdd = 1+2 tom & Cl-sin¢) D£ = 1.23 Fcd= Fqd- 1-Fad Nctand = 1.24 20.93 Erd=1 [po wort, vertical = 0 las Faiz Fali = (1-4)=(1-0)=1 Inclination factors Fri- A ) = (1-0) = 1 Tp = inclination of the load (vescical linclined applicó loud) no inclination of (and r=r= rsat-rw =(18-10) = 8 kN/m3 .-= (5x50-58% 1.72x1.24x 1)+(8*2.5 x 37-75X11718 1.23x2 + (0.5X 8 X 2.64 % 56.3 x 0.61* 1x1) taking, C=5KN/m? 2490.04 kN/m2 qu = CSScanned with CamScanner

Smax 1 = bei В. Lo [it Effective carea rit 2.64 -0.44 6 5000 6X0.18 e, 20118 <B LBS 2064x 2.72 2164 l2 = 0114 LB G 1981.15 kN/m2 L=2.72m 6e2 6x10.14 5000 1 + 1+ B'l (2.648 2.64 2672) 1917.85 kN/m2 max 2 = AFOS, qu 6max | mastallen 2490.04 = 2.54 981.15 2490.04 = 2171 917.85 FOS=2.71 FOS2 = qu omap2 And CS Scanned with CamScanner