Determine the adequacy of a cantilevered retaining wall The 20-ft high retaining wall, shown in Figure 10.45, has a 300 psf surcharge at the top of the wall and has 1 ft of soil over the toe of the footing. Determine the adequacy of the wall by calculating the factor of safety for sliding and

Question

Determine the adequacy of a cantilevered retaining wall
The 20-ft high retaining wall, shown in Figure 10.45, has a 300 psf surcharge at the top of the wall and has 1 ft of soil over the toe of the footing. Determine the adequacy of the wall by calculating the factor of safety for sliding and overturning, as well as calculating the bearing pressure at the toe and heel of the footing. Additional information is as follows:

Density of the soil, [latex]γ_s = 120\frac{\#}{ft^2}[/latex]
Weight of concrete, [latex]γ_c = 150\frac{\#}{ft^2}[/latex]
Angle of internal friction, ∅ = 35°
Coefficient of friction, µ = 0.4

Accepted Answer

It is typical to draw the retaining wall with the pressure distributions of the loads acting on the wall as shown in Figure 10.46, and locate the resultants of the lateral pressures.

It is important to note that the passive pressure is being considered for the full depth of the footing below the grade at the toe. The soil above the toes is often conservatively not considered due to the possible removal of this soil during the life of the structure.

To determine the lateral earth pressure, the Rankine lateral earth pressure coefficients are calculated as follows.

From Equation 10.15, the coefficient of active lateral earth pressure is calculated as follows:

[latex]K_o =1- \sin Φ [/latex] (10.15)

[latex]K_a=\frac{1-\sin ∅}{1+\sin ∅}[/latex]

[latex]k_a=\frac{1-35}{1+35}=0.271[/latex]

and similarly, from Equation 10.16, the coefficient of passive lateral earth pressure is as follows:

[latex]K_a =\frac{1-\sin Φ }{1+ \sin Φ} [/latex] (10.16)

[latex]K_p=\frac{1+\sin ∅}{1-\sin ∅}[/latex]

[latex]k_p=\frac{1+35}{1-35}=3.69[/latex]

Then, the lateral earth pressures can be calculated on a, per foot, linear length of wall.

The active pressure is calculated as follows:

[latex]P_a=K_a\gamma_s=0.271(120^{\#/ft^3})=\frac{32.52^{\#/ft^2}}{linear foot of wall}[/latex]

[latex]H_a=\frac{P_ah_T^2}{2}=\frac{\left(32.52^{\#/ft^2} ight)(21.5^\prime)^2}{2(1000^{\#/ft})}=\frac{7.52^k}{linear foot of wall}[/latex]

The lateral pressure due to the surcharge is calculated as follows:

[latex]P_L=K_aW_L=0.271(300^{\#/ft^2})=\frac{81.3^{\#/ft^2}}{linear foot of wall}[/latex]

[latex]H_L=P_L(h_T)=\frac{\left(81.3^{\#/ft^2} ight)21.5^\prime}{1000^{\#/ft}}=\frac{1.75^k}{linear foot of wall}[/latex]

Passive pressure (resistance) is calculated as follows:

[latex]P_p=K_p \gamma_s=3.69(120^{\#/ft^3})=\frac{442.8^{\#/ft^2}}{linear foot of wall}[/latex]

[latex]H_p=\frac{P_ph_k^2}{2}=\frac{\left(442.8^{\#/ft^2} ight)(5.5^\prime)^2}{2(1000^{\#/ft})}=\frac{6.7^k}{linear foot of wall}[/latex]

Now that we have the pressures and forces acting on the wall, we can now calculate the force to resist against sliding and the overturning moment.

Summing the horizontal forces which will cause the wall to slide is simply as follows:

[latex]H_{sliding} = H_a + H_L = 7.52^k + 1.75^k = 9.27^k[/latex]

and summing the moments which will cause the footing to overturn about the toe, is calculated as follows:

[latex]M_{overturning}= H_L(10.75^\prime)+H_a(7.17^\prime)=1.75^k(10.75^\prime)+7.52^k(7.17^\prime)=72.73^{k\cdot ft}[/latex]

The component weights of the wall are calculated to determine the resulting frictional force. The weights are calculated here and are performed in the same manner as the forces, that is, based on per foot of linear wall. Their locations are shown in Figure 10.47.

[latex]W_L = 0.3^{k/ft^2}(9^\prime)=2.7^k\W_{Sa} = 0.120^{k/ft^3}(20^\prime)(9^\prime)=21.6^k\W_{Sp} = 0.120^{k/ft^3}(3^\prime)(1^\prime)=0.36^k\W_w = 0.150^{k/ft^3}(23^\prime)(1.5^\prime)=5.18^k\W_b = 0.150^{k/ft^3}(13.5^\prime)(1.5^\prime)=3.04^k[/latex]

The frictional resistance is then calculated as follows:

[latex]F = µ(W_L +W_{Sa}+W_{Sp}+ W_w + W_b) =0.4(32.88^k)=13.15^k[/latex]

Total resistance, against sliding, is the sum of the frictional force and the passive pressure.

[latex]H_{resistance} = F + H_p = 13.15^k + 6.7^k = 19.85^k[/latex]

Then, the factor of safety for sliding is determined as follows:

Factor of safety, F.S. = [latex]\frac{H_{resistance}}{H_{sliding}}=\frac{19.85^k}{9.27^k}2.14 \gt 1.5[/latex], therefore OK

Next, calculate the resistance to overturning or resisting moment. The weight of the soil on top of the toe and the passive pressure is conservatively omitted in determining the overturning moment.

From Figure 10.47

[latex]+ ⤸∑M_A =(W_L+W_{Sa})(9^\prime) +W_w(3.75^\prime)+W_b(6.75^\prime)[/latex]

[latex]=(24.3^k)(9^\prime) +(5.51^k)(3.75^\prime)+(2.81^k)(6.75^\prime)[/latex]

[latex]=258.3^{k\cdot ft}[/latex]

Then, the factor of safety against overturning is determined as follows:
Factor of safety, F.S. = [latex]\frac{M_R}{M_o}=\frac{258.3^{k\cdot ft}}{72.73^{k\cdot ft}}=3.55\gt 2.0[/latex], therefore OK

Finally, to calculate the unfactored soil pressure at the base, as shown in Figure 10.48, we must determine the eccentricity on the footing/base.

Using the familiar eccentric footing procedure first, determine the kern of the footing.

[latex]e=\frac{L}{6}=\frac{13.5}{6}=2.25^\prime[/latex]

Set an equilibrium equation to determine the center of gravity.

[latex]R\bar{x} =M_R − M_0[/latex]

The resultant is the sum of the wall components.

[latex]R = W_L + W_w + W_{Sa} + W_b = 2.7^k + 5.51^k + 21.6^k + 2.81^k = 32.62^k[/latex]

Solve for [latex]\bar{x}[/latex]:

[latex]\bar{x}=\frac{M_R-M_o}{R}=\frac{258.3^{k\cdot ft}-72.73^{k\cdot ft}}{32.62^k}=5.68^\prime[/latex]

then,

[latex]e^\prime=\bar{x}[/latex]

and

[latex]e=6.75^\prime-5.68^\prime=1.07^\prime \lt \frac{L}{6}[/latex]

Thus, the footing/base has full bearing along the bottom of the footing and use Equation 10.6 to calculate the maximum and minimum unfactored bearing pressures.

[latex]q_{max/min}=\frac{R\left(1\pm \frac{6e}{L} ight)}{(L)(B)}=\frac{32.62^k\left(1\pm \frac{6(1.07^\prime)}{13.5^\prime} ight)}{(13.5^\prime)(1^\prime)}[/latex]
[latex]q_{max} = 3.56^k[/latex]/linear foot of wall
[latex]q_{min} = 1.27^k[/latex]/linear foot of wall

Question 10.12: Determine the adequacy of a cantilevered retaining wall The ...

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