Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the diagram design to solve the above question.
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](https://tex.z-dn.net/?f=%5Cphi%20%3D%2033%5Eo)
saturated uit weight![\gamma_{sat} = 115.0 lb/ft^3](https://tex.z-dn.net/?f=%20%5Cgamma_%7Bsat%7D%20%3D%20115.0%20lb%2Fft%5E3)
coeeficent of earth pressure ![K_o](https://tex.z-dn.net/?f=K_o)
![K_o = 1 -sin \phi](https://tex.z-dn.net/?f=K_o%20%3D%201%20-sin%20%5Cphi)
= 1 - sin 33 = 0.455
for over consolidate
![K_{con} = K_o \times OCR](https://tex.z-dn.net/?f=K_%7Bcon%7D%20%3D%20K_o%20%5Ctimes%20OCR)
![= 0.455 \times 1.5 = 0.683](https://tex.z-dn.net/?f=%20%3D%200.455%20%5Ctimes%201.5%20%3D%200.683)
Pressure at bottom of wall is
![P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H](https://tex.z-dn.net/?f=P%20%3DK_%7Bcon%7D%20%5Ctimes%20%28%5Cgamma_%7Bsat%7D%20-%20%5Cgamma_%7Bw%7D%29%20%2B%20%5Cgamma_w%20%5Ctimes%20H)
![= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15](https://tex.z-dn.net/?f=%3D%200.683%20%5Ctimes%20%28115%20-%2062.4%29%20%5Ctimes%2015%20%2B%2062.4%20%5Ctimes%2015)
P = 1474.88 lb/ft^3
Magnitude pf thrust is
![F= \frac{1}{2} PH](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B1%7D%7B2%7D%20PH)
![=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%201474.88%5Ctimes%2015%20%3D%2011061.65%20lb%2Fft)
the location must H/3 from bottom so
![x = \frac{15}{3} = 5 ft](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B15%7D%7B3%7D%20%3D%205%20ft)