The Architect does not like the standard tank profile and has asked what differe
ID: 1885086 • Letter: T
Question
The Architect does not like the standard tank profile and has asked what difference it would make, to the pressure on the soil under the tank, if the foundation for the tank was an L shaped slab as shown in the sketch below instead of the proposed circular footing.
Use a Newmark Chart to determine pressure profiles under each corner of the tank and plot the effective stress, and the “combined tank/base pressure + effective stress” profiles at each corner of the tank base.
Note:
(i) Your Newmark chart showing the squares overlaid is be included in your hand in, and all graphs are to be produced in Excel as in (a) above.
(ii) Plot all seven profiles on the same set of axes using an appropriate Excel graphing option
(b) On the same site a retaining wall 6m high, with a vertical back face is to be constructed using the blue-grey sand in Q1 (a) as backfill. If the Friction Angle f of the sand is measured at 300 determine, using Rankine’s formula, the maximum earth pressure against the wall and the thrust against the wall for the following cases:
Active earth Pressure and a backfill angle of 200
At rest Pressure for a level backfill surface
Show, by means of a sketch, the magnitude, location, and direction of the total thrust in each case above.
(c) If the first 8 metres of sloping ground is to be removed and a small retaining wall built at that point use Coulomb's graphical method to determine the thrust applied to the back of the 6 metre wall (assume that the friction angle with the wall d is 2/3f)
2. 200 m 2m Tv 2m Footing is 200mn wi der t han tank.Explanation / Answer
1. Since, in the first part of question, the depth at which pressure is to be caluculated is not mentioned, I am skiping through 1st Part.
2. Retaining wall height = 6m, coefficient of earth pressure in active state of failure = (1-sin (phi))/(1+sin(phi))
Phi = Angle of internal Friction, = 30 as in the problem mentioned,
Hence, coefficient of earth pressure in active state of failure of backfill = 0.33
Earth pressure coming onto Wall = coefficient of earth pressure * (0.5) *( Unit wt. of eath material) *(Height^2) = (0.33*0.5*18*6*6) ,
Note that Unit of earth material is assumed as 18 Kn/Cu.m.
Hence, Horizontal Thrust coming exerted on the wall per running mtr. of wall = 108 KN
3. If earth pressure at Rest is used then, Coefficient of earth pressure at rest = 1 - sin (phi) = 1 - sin 30 = 0.5
Hence, Force exerted on the wall at rest = 0.5 * 0.5 * 18 * 6 * 6 = 162 KN per running mtr. of wall
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