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statika dan probabilitas, Exercises of Statistics

contoh soal soal statika dan probabilitas

Typology: Exercises

2021/2022

Uploaded on 09/27/2023

illona-putri
illona-putri 🇮🇩

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PROBLEMS 8
8.1 Refer to the constant-head permeability test arrangement in a two-layered soil as shown in
Figure 8.2. During the test, it was seen that when a constant head of h1 = 200 mm was
maintained, the magnitude of h2!was 80 mm. If k1!is 0.004 cm/sec, determine the value of
k2!given H1!= 100 mm and H2!= 150 mm.
Figure 8.2
Step1/4
Given that,
Head difference between top of the layer number 1 and bottom of the layer number 2!h1!= 200
mm,
Magnitude of!h2!= 80 mm,
Hydraulic conductivity of soil layer number 1 is!k1!= 0.004 cm/sec,
Height of the soil layer number 1 is!H1!= 100 mm,
Height of the soil layer number 2 is!H2!= 150 mm.
Step2/4
The constant head permeability test arrangement is shown in the figure.
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PROBLEMS 8

8.1 Refer to the constant-head permeability test arrangement in a two-layered soil as shown in Figure 8.2. During the test, it was seen that when a constant head of h1 = 200 mm was maintained, the magnitude of h 2 was 80 mm. If k 1 is 0.004 cm/sec, determine the value of k 2 given H 1 = 100 mm and H 2 = 150 mm. Figure 8. Step 1/ Given that, Head difference between top of the layer number 1 and bottom of the layer number 2 h 1 = 200 mm, Magnitude of h 2 = 80 mm, Hydraulic conductivity of soil layer number 1 is k 1 = 0.004 cm/sec, Height of the soil layer number 1 is H 1 = 100 mm, Height of the soil layer number 2 is H 2 = 150 mm. Step 2/ The constant head permeability test arrangement is shown in the figure.

Step 3/ In order to compute the hydraulic conductivity of soil layer number-2, we have the expression for as. Where Height of the soil layer number-1. Height of the soil layer number-2. Hydraulic conductivity of soil layer number-1. Hydraulic conductivity of soil layer number-2. Step 4/ On substitution, we get

Step 2/

Determine the head difference. Here, head difference is , head at upstream side of flow is , and head at downstream side of flow is. Substitute for , and for in formula. Thus, the head difference is.

Step 3/

Determine the seepage loss per meter length of sheet pile. Here, seepage loss is , hydraulic conductivity is , number of flow channels is , number of potential drops is , and head difference is. Substitute for , for , for , and for in formula.

Therefore, the seepage loss per meter length of sheet pile is.

Determine the head difference. Here, H is the head difference between upstream and downstream sides, H 1 is the head at the upstream side of the flow, and H 2 is the head at the downstream side of the flow. Substitute 3 m for H 1 and 0.5 m for H 2.

Step 4/

Determine the seepage loss per meter length of sheet pile. Here, q is the Seepage loss per meter length of the sheet pile, k is the hydraulic conductivity, H is the head difference between upstream and downstream sides, N f is the number of flow channels in the flow net, and N d is the number of potential drops. Substitute 3 for , 6 for , for , for , and for H.

Step 5/

Convert q from to.

Therefore, the seepage loss per meter length of sheet pile is. 8.6 Refer to Problem 8.5. Using the flow net drawn, calculate the hydraulic uplift force at the base of the hydraulic structure per meter length (measured along the axis of the structure). Figure 8.

Step 1/

The combination of equipotential lines and flow lines are collectively called as flow net. The process of drawing the flow net diagram is only by trial and error method.

Step 2/

Points to remember for drawing a flow net diagram is as follows:

  • The equipotential line and flow line must intersect at right angle.
  • The flow element after the intersection of equipotential line and flow line must be approximate square.

Step 3/

As it is trial and error method, so number of numbers of equipotential drop and number of flow channels may vary. Procedure of drawing a flow net as follows:

  • Make a two-dimensional scale drawing of the system.

Substitute 12 for , 5 for , for , and for in formula. Therefore, the seepage loss per meter length of sheet pile is.

Step 6/

Draw the flow net.

Step 7/

Determine the head difference. Here, number of potential drops is , and head difference is. Substitute for , and for in formula.

Thus, the head difference is. Determine pressure head at D. Here, head difference is , head at upstream side of flow is , and head at downstream side of flow is. Substitute for , and for in formula. Thus, the pressure head at D is.

Step 8/

Determine pressure head at E. Thus, the pressure head at E is. Determine pressure head at F. Thus, the pressure head at F is.

Step 9/

Therefore, the uplift force is. 8.9 Problem An earth dam is shown in Figure 8.27. Determine the seepage rate, q , in m^3 /day/m length. Given: α 1 = 35°, α 2 = 40°, L 1 = 5 m, H = 7 m, H 1 = 10 m, and k = 3 × 10−4^ cm/sec. Use Schaffernak’s solution. Figure 8.

Step 1/

Draw the seepage is shown in the figure.

Step 2/ Determine the value. Here, the height of the dam is. Substitute for and for. Step 3/ Determine the value of. Here, the height of water is and the top width of the dam is. Substitute 10 m for , for , for , 5 m for , 9.99 m for, and for. Step 4/ Determine the value of. Substitute 24.19 m for d , for , and for.

8.7 Draw a flow net for the weir shown in Figure 8.20. Calculate the rate of seepage under the weir.

Step 1/

Draw the flow net diagram for the hydraulic structure.

Step 2/

Calculate the rate of seepage under the weir by using the relation, …… (1) Here, is the number of flow channels in the flow net and is the number of potential drops and is hydraulic conductivity and for total head. Calculate the value of total head.

Step 3/

Substitute 4 for , 14 for , for k, and for in equation (1). Therefore, the rate of seepage under the weir is.