Question

A 15 ft high vertical wall retains an overconsolidated soil where OCR-1.5, c'-: O, ф , --33°, and 1 1 5.0 lb/ft3.

Answer 1

Answer:

magnitude of thrust uis  11061.65 lb/ft

location is 5 ft from bottom

Explanation:

Given data:

Height of vertical wall is 15 ft

OCR  is 1.5

$\phi = 33^o$

saturated uit weight$\gamma_{sat} = 115.0 lb/ft^3$

coeeficent of earth pressure $K_o$

$K_o = 1 -sin \phi$

        = 1 - sin 33 = 0.455

for over consolidate

$K_{con} = K_o \times OCR$

            $= 0.455 \times 1.5 = 0.683$

Pressure at bottom of wall is

$P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H$

   $= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15$

P = 1474.88 lb/ft^3

Magnitude pf thrust is

$F= \frac{1}{2} PH$

   $=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft$

the location must H/3 from bottom so

$x = \frac{15}{3} = 5 ft$