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
[tex]\phi = 33^o[/tex]
saturated uit weight[tex] \gamma_{sat} = 115.0 lb/ft^3[/tex]
coeeficent of earth pressure [tex]K_o[/tex]
[tex]K_o = 1 -sin \phi[/tex]
= 1 - sin 33 = 0.455
for over consolidate
[tex]K_{con} = K_o \times OCR[/tex]
[tex] = 0.455 \times 1.5 = 0.683[/tex]
Pressure at bottom of wall is
[tex]P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H[/tex]
[tex]= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15[/tex]
P = 1474.88 lb/ft^3
Magnitude pf thrust is
[tex]F= \frac{1}{2} PH[/tex]
[tex]=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft[/tex]
the location must H/3 from bottom so
[tex]x = \frac{15}{3} = 5 ft[/tex]
