Question

A TQy250K IT QE 36 GWEN x=8 8 = 130pf &= 30 C-Loops FIND Tmax, mon. DRAW THE PRESSURE DIAGRAMS & 4x4 F K UNDER THE FOOTNG FinD FACTOR OF Superty Against OYERTURNING ABOUT K'.

Answer 1

Given data: y =130 pcf 0 = 30° C = 800 psf The active pressure coefficient is given by, K 1-sino * 1+sin Substitute the values, 1-sin 30° K, 1sin 30 The active pressure on soil is given by, 0,=yh 04 = K.5, -20.K. Calculate pressure at point A. 0.= 1300 = 0 =-923.76 psf

Calculate pressure at point B. 0, =130x4 = 520 psf =-750.427 psf The pressure diagram is shown below. 923.76 173.33 750.427

Calculate moment about point k. 1. For rectangle section. The height is 4 ft, so C.G. is 2 ft from point k. 756.427x4x2 = 6003.416 pounds which is clockwise 2. For triangular section, The height is 4 ft, so C.G. is =x4 from point k. 173,3x4x<<4) = 924.427 pounds which is clockwise 3. For 30 k load, 30x1000x(4+2) =180000 pounds which is clockwise 4. For 50 k load, 50x1000x2 = 100000 pounds which is anti-clockwise

The factor of safety is given by, FOS - Restoring moment Rotating moment Substitute the values 100000 FOS-a 6003.416+924.427+180000 =0.534 The value of c is given by, M -M C=- Σ ' Substitute the values 100000-6003.416-924.427-180000 50x1000 =-1.738 The eccentricity is given by, Substitute the values, e ==-(-1.738) =3.738

The maximum pressure is given by, Substitute the values, 50x1000, 6x3.738 02244 = 82586.5 psf The minimum pressure is given by, Substitute the values 0.501000[,- 623,738] =-57587.5 psf The pressure distribution under footing is given by,

57587.5 82.586.5