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Principles of Foundation Engineering, SI Edition
8th Edition
ISBN: 9781305446298
Author: Braja M. Das
Publisher: Cengage Learning US
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Consider the forces acting on the handle of the wrench in Figure 1
The following table gives the variation of the field standard penetration number (№60) in a sand deposit:
Depth (m) N60
1.5
6
3.0
14
4.5
14
6.0
19
17
7.5
9.0
23
The groundwater table is located at a depth of 12 m. The dry unit weight of sand from 0 to a depth of 12 m is 17.6 kN/m³. Assume the mean grain size (D 50) of the sand deposit to be about 0.8 mm. Estimate the
variation of the relative density with depth for sand. Use the equation
N60 (0.23 +0.06/D50) 1.7
1
Dr (%)
=
9
σ'o/Pa
(Enter your answers to three significant figures.)
Depth (m)
N60
Dr (%)
1.5
6
3.0
14
4.5
14
6.0
19
7.5
17
9.0
23
0.5
(100)
00
N 50°
W
NAZ 310
Length &
plane
Survey!
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A (9
13
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- A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Cone resistance, qc (MN/m²) 2.0 3.5 5.0 6.5 8.0 4.02 5.15 6.04 10.13 13.10 The average unit weight of the sand is 16.5 kN/m³. Assume moderately compressible sand and hence Q Determine the relative density at each depth using the equation below. Dr = 1 305QOCR1.8 Яс Pa 0.5 0 Pa (Enter your answers to three significant figures.) Depth (m) Dr (%) 2.0 3.5 5.0 6.5 8.0 = 1.arrow_forwardA vane shear test was conducted in a saturated soft clay, using a 100 mm x 260 mm vane. When the vane was rotated at the standard rate of 0.1°/s, the torque measured in the torque meter increased to 60 N. m, and with further rotation reduced to 35 N. m. Determine the peak and residual undrained shear strengths of the clay. (Enter your answers to three significant figures.) Peak undrained shear strength Residual undrained shear strength = kN/m² kN/m²arrow_forwardFollowing is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) Neo N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 19 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 20.2 kN/m³. Estimate an average peak soil friction angle. Use the equation CN - [ 1 (o'o/Pa). 0.5 (Enter your answer to three significant figures.) $' =arrow_forward
- The beam shown in the figure below is typical for a floor system in an existing building.It needs to carry a uniform live load of 260 lb/ft and a uniform dead weight of 400 lb/ft,including its own weight. The owner wants to add a partition weighing 7 kip (live load) asshown. Assuming the added partition as live load, is the beam section adequate to safelycarry the extra live load? a. Determine the design moment capacity .b. Determine the factored applied bending moment. c. Is the beam safe and adequate for bending? Please explain your response.arrow_forwardFollowing is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 10 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 12.2 kN/m³. Use the relationship given in the equation C'N = 1 σo/Pa 0.5 to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) Depth (m) N60 (N1) 60 1.5 6 4.5 3 8 9 6 8 7.5 13 9 14arrow_forwardFollowing is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) 1.5 N60 5 3 6 4.5 9 6 7 7.5 9 10 11 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 18 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 20.2 kN/m³. Using the equation N60 0.5 - {11} Dr = determine the average relative density of sand. (Enter your answer to three significant figures.) Average D₁ = %arrow_forward
- The cantilever beam shown below supports a uniform service (unfactored) dead loadof 1.5 kip/ft plus its own self weight, plus two unknown concentrated service(unfactored) live loads, as shown. The concrete has f’c = 6,000 psi and the steel yieldstrength is 60 ksi.a. Determine the design mopment capacity .b. Set-up the factored applied bending moment equation. .c. Calculate maximum safe concentrated live load that the beam may carry.arrow_forwardA rectangular reinforced concrete beam 18 in. wide by 28 in. overall depth is to support a superimposed (additional to the self-weight) service dead load of 0.5 kip/ft and a service live load of 1.3 kip/ft. Reinforcing for positive moment is 60 ksi yield strength. f’c = 5,000 psi. Use 6#9 rebars a. Determine the design moment capacity . b. Set-up the factored applied bending moment . c. Determine the maximum simple span length on which this beam may be safely utilized.arrow_forward. . . . . . . . TUGAS-1 For a moist soil sample, the following are given: -Total Volume: V 1.2 m³ -Total mass: M = 2350 kg -Moisture Content: Wc = 8.6% -Spesific Gravity of Soil Solids : Gs = 2.71. Determine the following a. Moist Density (Y) b. Dry Density (yd) C. Void Ratio (e) e. f. g. Porosity (n) Degree of Saturation (Sr) Volume of water in the soil sample (Vw) Draw the three phase of the soil element complete with the number TUGAS-2 Mass (kg) Volum V Mac= V₁ = M = 2350 M₁ = ☐ Air Water Solid A saturated soil has a dry unit weight of 16.18 kN/m³. Its moisture content (WC) is 23%. Determine: a. Saturated unit weight, ysat b. Spesific gravity, Gs C. Void Ratio, e TUGAS-3 The dry density of a sand with a porosity of 0.387 is 1600 kg/m³. Determine the void ratio of the soil and the specific gravity of soil solids. POLIT V= POLITIarrow_forward
- 5. What is the lightest WT shape that would be adequate for tension yielding under a design tensile demand of 1,215 kips? Assume that geometric constraints within the structure require you to select a WT9 section.arrow_forwarda. Determine the effective area for the case shown in the figure below. Suppose that l = 6 in. For L5 × 5 × 5/8: A₁ = 5.90 in.², = 1.47 in. L5 × 5 × 5/8 Weld (Express your answer to three significant figures.) A₁ = in.² b. Determine the effective area for the case shown in the figure below. Suppose that l = 5 in. PL³/8 X 4 Weld (Express your answer to three significant figures.) Ae = in.2 2 c. Determine the effective area for the case shown in the figure below. -5" PL5/8 X 5 Weld (Express your answer to three significant figures.) Ae in.2 d. Determine the effective area for the case shown in the figure below. 2" 2" PL1/2 X 512 ос 3/4-in.-diam. bolts (Express your answer to three significant figures.) Ae = in.² e. Determine the effective area for the case shown in figure below. PL³/8 X 6 7/8-in.-diam. bolts (Express your answer to three significant figures.) Ae= in.2arrow_forwardA single-angle tension member of A36 steel must resist a dead load of 35 kips and a live load of 84 kips. The length of the member is 18 feet, and it will be connected with a single line of 1- inch-diameter bolts, as shown in the figure below. There will be four or more bolts in this line. For the steel Fy = 36 ksi and F₁ = 58 ksi. Try the tension members given in the table below. Tension member rz (in.) A, (in.²) L6 × 6 × 9.75 1.17 L 5 × 3 × 1/10 4.93 0.746 L5 × 3 × 16 5 2.56 0.758 L5 × 3 × 166 3.31 0.644 Bolt line a. Select a single-angle tension member to resist the loads. Use LRFD. A) L 6 × 6 × B) L 5 × 3 × C) L5 × 3 X D) L 5 × 3 × 6 -Select- V What is the required gross area? (Express your answer to three significant figures.) Ag = in.2 What is the required effective area? (Express your answer to three significant figures.) Ae = in.2 What is the minimum radius of gyration? (Express your answer to three significant figures.) 1min = in. b. Select a single-angle tension member to…arrow_forward
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