CEE 8102 Foundation Engineering: Design of Retaining Walls

This report details the design of temporary retaining walls, focusing on soldier beam and steel sheet pile methods. It includes calculations for earth pressure (active and passive), surcharge pressure, and bending moments under various conditions. The report presents three designs, each with specific parameters and calculations to determine the optimal length and bracing load for the retaining walls. The designs consider factors like soil properties, surcharge loads, and different earth pressure coefficients. The analysis leads to equations for determining the required depth of embedment for stability, ensuring the retaining walls can withstand the applied loads. Desklib offers a wealth of similar solved assignments and past papers to aid students in their studies.

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FOUNDATION ENGINEERING
DESIGN 1
Conditions
For stratum 1, Ka = 0.31 and Kp = 3.28
For stratum 2, Ka = 0.26 and Kp = 3.85
Earth pressure
Passive earth pressures Pp
Section 1 = (130-62.4) x 3 x D = 202.8D
Active earth pressure Pa
Section 2: (7 x 110 x 0.31) = 238.7
Section 3: (4 x 110 x 0.26 ) = 114.4
Section 4: (4 x 120 x 0.26) = 124.8
Section 5: (section 3 + 4) 114,.4 + 124.8 = 239.2
Section 6: (130-62.4) x 0.26 x D = 17.58D
Surcharge pressure
300 x 0.31 = 93
300 x 0.26 = 78

S.B Moments
1 0.5x202.8DxDxD/3 = +33.8D3
2 -0.5x7x238.7(D+4+7/3) = -835.45D -5288.4
3 -7X114.4X(Dx4/2) = -800.8D -1601.6
4 -0.5x7x124.8x(D +4/3) = -436.8D -582.4
5 -0.5xDXdx239.2 = -119.6D2
6 -0.5xDx17.58DxD/3 = -2.9D3
7 -7x9.3 x (d+4+7.2) = -651D -4882.5
8 -0.5x (D+4)2 x 78 = -39.5D2 -355D -987
Overall moment = +30.9D3 -159.1D2 -3079.05D -13341.9
Length of S.B
Given that Ax3+Bx2+Cx+D =0
The moments turn out to formation of the equation
= +30.9D3-159.1D2-3079.05D-13341.9 =0
The positive value of the 14.25’. Therefore the best Length is 14.25’
DESIGN 2
According to coulomb law
Pa = 0.5KaH2

Fb = 0.66x50 = 33
Earth pressure coefficients
For stratum 1, Ka = 0.31 and Kp = 3.28
For stratum 2, Ka = 0.26 and Kp = 3.85
Earth pressure
Passive earth pressures Pp
Section 1 = 0.5(130-62.4) x 0.26 x D = 8.8D
Active earth pressure Pa
Section 2: 0.5 (7 x 110 x 0.31) = 119.35
Section 3: 0.5 (4x 110 x 0.26 ) = 57.2
Section 4: 0.5(4x 120 x 0.26) = 62.4
Section 5: (section 3 + 4) 57.2+ 62.4= 119.6
Section 6: 0.5(130-62.4) x 0.26 x D = 1.48D
Surcharge pressure
300 x 0.31 = 93
300 x 0.26 = 78
S.B Moments
1 0.5x8.8DxDxD/3 = +1.48D3
2 -0.5x7x119.4(D+4+7/3) = - 417.9D - 2646.7
3 -7X57.2(D+4/2) = -400.4D -8008.6
4 -0.5x7x62.4(D +4/3) = -218.4D -291.2
5 -0.5xDXdx119.6 = -59.8D2
6 -0.5xDx8.8DxD/3 = -0.73D3
7 -7x9.3 x (d+4+7.2) = -651D -4882.5
8 -0.5x (D+4)2 x 78 = -39.5D2 -355D -987

Overall moment = +0.75D3 -99.3D2 -2042.7D - 16816
Length of S.B
Given that Ax3+Bx2+Cx+D =0
The moments turn out to formation of the equation
= +0.75D3-99.3D2-2042.7D-16816 =0
The positive value of the 151.37’. Therefore the best Length is 151.37’
Design 3
25’H braced steel sheet pile wall design
For stratum 1, Ka = 0.31 and Kp = 3.28
For stratum 2, Ka = 0.26 and Kp = 3.85
Earth pressure
Passive earth pressures Pp
Section 1 = 0.8(130-62.4) x 3 x D = 162.24D
Active earth pressure Pa
Section 2: 0.8 (7 x 110 x 0.31) = 191
Section 3: 0.8 (10x 110 x 0.26 ) = 228.8

Section 4: 0.8 (10x 120 x 0.26) = 249.6
Section 5: (section 3 + 4) 228.2 + 249.6 = 478.4
Section 6: 0.8 (130-62.4) x 0.26 x D = 14.06D
Surcharge pressure
300 x 0.31 = 93
300 x 0.26 = 78
Calculating the bracing load
Pa = 0.8KaH = 0.8x0.31x110x 25 = 682
Pp = 300x0.31 = 93
Sum of the moments
(0.8x25x682x25/2) + ( 93x25) = 17B
= 170,500+ 2325 = 17B
= B =6,913lf/wall
Lower resisting force and sizing of the
R = (0.8x25x682)+(93x25) – 6,913
R = 13640+ 2325– 6,913
R = 9,052 lf/wall
Moment = 0.5(8-7)2 x(682) + 0.5 x7x682x(1+7/3) + 0.5x93x8x8
= 341+ 7956.67+2976
= 11273.67 x 8
= 90189.36
Fb = 0.66x50 = 33x 1000
S = 90189.36x 11/33,000 = 60.13 inch3 use NP 15x45 to give s = 63.23 in3