We are determined to provide the latest solutions related to all subjects FREE of charge!
Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.
Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program
Advertise your business, and reach millions of students around the world.
All the data tables that you may search for.
For Arabic Users, find a teacher/tutor in your City or country in the Middle East.
Find the Source, Textbook, Solution Manual that you are looking for in 1 click.
Need Help? We got you covered.
Fig. Ex. 20.18a gives the section of a long braced cut. The sides are supported by steel sheet pile walls with struts and wales. The soil excavated at the site is stiff clay with the following properties
c=800 lb / ft ^{2}, \phi=0, \gamma=115 lb / ft ^{3}Determine: (a) The earth pressure distribution envelope.
(b) Strut loads.
(c) The maximum moment of the sheet pile section.
The struts are placed 12 ft apart center to center horizontally.
(a) The stability number N_{s} from Eq. (20.57a) is
N_{s}=\frac{\gamma H}{c} \leq 4 (20.57a)
N_{s}=\frac{\gamma H}{c}=\frac{115 \times 25}{800}=3.6<4
The soil is stiff fissured clay. As such the pressure envelope shown in Fig. 20.28(c) is applicable. Assume p_{a}=0.3 \gamma H
p_{a}=0.3 \times 115 \times 25=863 lb / ft ^{2}
The pressure envelope is drawn as shown in Fig. Ex. 20.18(b).
(b) Strut loads
Taking moments about the strut head B_{1}(B)
\begin{aligned}&R_{A} \times 7.5=\frac{1}{2} 863 \times 6.25\left(\frac{6.25}{3}+6.25\right)+863 \times \frac{(6.25)^{2}}{2} \\&=22.47 \times 10^{3}+16.85 \times 10^{3}=39.32 \times 10^{3} \\&R_{A}=5243 lb / ft \\&R_{ B 1}=\frac{1}{2} \times 863 \times 6.25+863 \times 6.25-5243=2848 lb / ft\end{aligned}
Due to symmetry
\begin{aligned}&R_{A}=R_{C}=5243 lb / ft \\&R_{B 2}=R_{B 1}=2848 lb / ft\end{aligned}
Strut loads are:
\begin{aligned}&P_{A}=5243 \times 12=62,916 lb =62.92 kips \\&P_{B}=2 \times 2848 \times 12=68,352 lb =68.35 kips \\&P_{C}=62.92 kips\end{aligned}
(c) Moments
The shear force diagram is shown in Fig. 20.18c for sections D B_{1} \text { and } B_{2} E
Moment at A=\frac{1}{2} \times 5 \times 690 \times \frac{5}{3}=2,875 Ib-ft/ft of wall
Moment at m=2848 \times 3.3-863 \times 3.3 \times \frac{3.3}{2}=4699 Ib-ft/ft
Because of symmetrical loading
Moment at A = Moment at C = 2875 Ib-ft/ft of wall
Moment at m = Moment at n = 4699 Ib-ft/ft of wall
Hence, the maximum moment = 4699 Ib-ft/ft of wall.
The section modulus and the required sheet pile section can be determined in the usual way.