Question

Geotechnical Engineering

-Draw in scale the distribution of the passive pressure

- Calculate the total forces and its components (in case there is)

-Locate the force in m with respect to the lower point

(Alpha is 90 degrees and Beta is 0, Delta is 10 degrees)

Layer	γ	Thickness	Φ
1	18	2	27
2	16	1	32
3	17	3	28
4	18.5	2.50	30
5	17.5	1.5	29

Answer 1

Solution:- the values given in the question are as follows:

Layer	Unit Weight() kN/m^3	Thickness m	Angle of Friction()
1	18	2	27
2	16	1	32
3	17	3	28
4	18.5	2.5	30
5	17.5	1.5	29

Calculating coefficient of passive earth pressure for all layers:-

Kp=(1+sin)/(1-sin)

Kp1=(1+sin1)/(1-sin1)

Kp1=(1+sin27)/(1-sin27)

Kp1=2.6629

Kp2=(1+sin32)/(1-sin32)

Kp2=3.25458

Kp3=(1+sin28)/(1-sin28)

Kp3=2.7698

Kp4=(1+sin30)/(1-sin30)

Kp4=3

Kp5=(1+sin29)/(1-sin29)

Kp5=2.882

Calculating Passive pressure of different layers:-

Pp1=0.5*Kp1*1*H1^2

Pp1=0.5*2.6629*18*2^2

Pp1=95.8658 KN/m

force Pp1 acting at distance 1 from base

1=(H5+H4+H3+H2+H1/3)

1=(1.5+2.5+3+1+2/3)

1=8.6667 m

Pp2=Kp2*1*H1*H2

Pp2=3.25458*18*2*1

Pp2=117.1648 kN/m

force Pp2 acting a distance 2 from base

2=(H5+H4+H3+H2)/2

2=(1.5+2.5+3+1)/2

2=4 m

Pp3=0.5*Kp2*2*H2^2

Pp3=0.5*3.25458*16*1^2

Pp3=26.03664 kN/m

force Pp3 acting at distance 3 from base

3=(H5+H4+H3+H2/3)

3=(1.5+2.5+3+1/3)

3=7.333 m

Pp4=Kp3*2*H2*H3

Pp4=2.7698*16*1*3

Pp4=132.9504 kN/m

force Pp4 acting at distance 4 from base

4=(H5+H4+H3)/2

4=(1.5+2.5+3)/2

4=3.5 m

Pp5=0.5*Kp3*3*H3^2

Pp5=0.5*2.7698*17*3^2

Pp5=211.8897

force Pp5 acting at distance 5 from base

5=(H5+H4+H3/3)

5=(1.5+2.5+3/3)

5=5 m

Pp6=Kp4*3*H3*H4

Pp6=3*17*3*2.5

Pp6=382.5 kN/m

force Pp6 acting at distance 6 from base

6=(H5+H4)/2

6=(1.5+2.5)/2

6=2 m

Pp7=0.5*Kp4*4*H4^2

Pp7=0.5*3*18.5*2.5^2

Pp7=173.4375 kN/m

force Pp7 acting at distance 7 from base

7=(H5+H4/3)

7=(1.5+2.5/3)

7=2.3333 m

Pp8=Kp5*4*H4*H5

Pp8=2.882*18.5*2.5*1.5

Pp8=199.9387 kN/m

force Pp8 acting at distance 8 from base

8=(H5)/2

8=(1.5)/2

8=0.75 m

Pp9=0.5*Kp5*5*H5^2

Pp9=0.5*2.882*17.5*1.5^2

Pp9=56.7393 kN/m

force Pp9 acting at distance 9 from base

9=H5/3

9=(1.5)/3

9=0.5 m

let total passive pressure is Pp and acting at distance from base

Total passive force(Pp)=Pp1+Pp2+Pp3+Pp4+Pp5+Pp6+Pp7+Pp8+Pp9

Totalpassiveforce(Pp)=95.8658+117.1648+26.03664+132.9504+211.8897+382.5+173.4375+199.9387+56.7393

Total passive force(Pp)=1396.5228 kN/m

=(Pp1*1+Pp2*2+Pp3*3+Pp4*4+Pp5*5+Pp6*6+Pp7*7+Pp8*8+Pp9*9)/(Pp1+Pp2+Pp3+Pp4+Pp5+Pp6+Pp7+Pp8+Pp9)

=(95.8658*8.666+117.1648*4+26.03664*7.333+132.9504*3.5+211.8897*5+382.5*2+173.4375*2.33+199.9387*0.85+56.7393*0.5)/(95.8658+117.1648+26.03664+132.9504+211.8897+382.5+173.4375+199.9387+56.7393)

=3.12428 m

total force Pp acting at distance(height) is 3.12428 m from base

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