Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Question
Chapter 16, Problem 16.9P
To determine
Find the gross allowable load
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Students have asked these similar questions
7. [Soil Bearing Capacity]
a gross load of 700 kN. Using a factor of safety 3, determine the required value of B. Assume
general shear failure. N. = 63. 53 , Ng = 47.16, N, = 54.36
»). A square footing is shown below. The footing will carry
%3D
Q = 700 kN
y = 17.4 kN/m
0 = 36°
C = 0
1.2 m
B
b) As shown in the figure below, a rectangular footing (B x L = 2m x 2.2m) is subjected
to a vertical load (400 kN) and moment (100 kN-m). The eccentricity is in the direction
of L. Determine the effective width B'
=
9max =
, and
400 kN
2.2 m
Rock
1
the effective length L'
100 kN-m
1. A circular footing 3 m in diameter is shown below.
[Soil Bearing Capacity] (
Assume the general shear failure and use a factor of safety 2.8. Determine the follow
N. = 25.13 , N, = 12.72, N, = 8. 34
a. The gross allowable bearing capacity.
b. Net Allowable bearing capacity
c. The safe load that the footing can carry.
Ground surface
Y = 18.5 kN/m³
C = 80 kPa
+ = 25°
1.1 m
D: = 1.8 m
Water table
Tse = 19.2 kN/m?
Diameter =
3 m
Chapter 16 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
Ch. 16 - A continuous footing is shown in Figure 16.17....Ch. 16 - Refer to Problem 16.1. If a square footing with...Ch. 16 - Redo Problem 16.1 with the following: = 115...Ch. 16 - Redo Problem 16.1 with the following: = 16.5...Ch. 16 - Redo Problem 16.1 using the modified general...Ch. 16 - Redo Problem 16.2 using the modified general...Ch. 16 - Redo Problem 16.3 using the modified general...Ch. 16 - Redo Problem 16.4 using the modified general...Ch. 16 - Prob. 16.9PCh. 16 - If the water table in Problem 16.9 drops down to...
Ch. 16 - Prob. 16.11PCh. 16 - A square footing is subjected to an inclined load...Ch. 16 - A square footing (B B) must carry a gross...Ch. 16 - Redo Problem 16.13 with the following data: gross...Ch. 16 - Refer to Problem 16.13. Design the size of the...Ch. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Refer to the footing in Problem 16.16. Determine...Ch. 16 - Figure 16.21 shows a continuous foundation with a...Ch. 16 - The following table shows the boring log at a site...
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Similar questions
- A square footing (1.5 m × 1.5 m) is located at a depth of 1.0 m. The footing is subjected to an eccentric load of 450 kN, with an eccentricity of 0.2 m along one of the symmetrical axes. Determine the factor of safety against bearing failure. U se Vesic's equation. Take y= 21 kN/m³, c = 100 kN/m², p= 0; N = 5.14; s = 1.14; s = 1; d. 1.27 ; d = 1; i = ig = 1. = q qarrow_forward1. A continuous footing is shown in Figure 16.19. Using Terzaghi's bearing capacity factors, determine the gross allowable load per unit area (4an) that the footing can carry. Assume general shear failure. Given: y = 115 lb/ft, c' = 600 lb/ft', d' = 25°, D, = 3.5 ft, B = 4 ft, and factor of safety = 3. qall Unit weight of soil = y Figure 16.19arrow_forward[Soil Bearing Capacity] ( a gross load of 700 kN. Using a factor of safety 3, determine the required value of B. Assume general shear failure. N. = 63. 53,Nq = 47.16, Ny = 54.36 ). A square footing is shown below. The footing will carry Q = 700 kN Y = 17.4 kN/m $ = 36° C = 0 1.2 marrow_forward
- Q2: A square footing (2x2 m) is constructed on a sandy soil as shown below. Find the maximum allowable load (Q) that the footing can carry. 1.5 m Im Assume general shear failure and the factor of safety is 3. E Y = 16 kN/m 7-18 kN/m 6349arrow_forwardA continuous footing is shown in Figure 16.17. Using Terzaghi’s bearing capacity factors, determine the gross allowable load per unit area (all ) that the footing can carry. Assume general shear failure. Given: γ = 19 kN/m3, c′ = 31kN/m2 , , Df = 1.5 m, B = 2 m, and factor of safety = 3.5.arrow_forwardDetermine the size of square footing to carry net allowable load of 295 KN. FS=3. Use Terzaghi equation assuming general shear failure. 295 kN + = 35° C = 0.0 Ya = 18.15 kN/m3 1 m $ = 25" C = 50 kN/m? Ya = 20 kN/m3arrow_forward
- A 1.57 -m. deep 2.5 m. x 3.3c m. footing has to be constructed as shown in Figure 1. The soil properties are: γ=18.7 kN/m3 , γsat=20.9 kN/m3 , angle of friction is 29 degrees, and D1 = 0.55 m. Using FS = 3.3, determine the value of ultimate bearing capacity and the ultimate load capacity (inclined) using β = 17 degrees . (use: γw=9.81 kN/m3)arrow_forwardA rectangular footing is constructed on saturated sand. This footing is placed under 1000 KN column load and 500 KN.m moment as shown in the image. Find the eccentricity in both directions and calculate the equivalent footing size. 3.0 M=500 KN.m 0.50 2.5 P1=1000 KN 0.50arrow_forwardb) As shown in the figure below, a rectangular footing (B x L = 2m x 2.2m) is subjected to a vertical load (400 kN) and moment (100 kN-m). The eccentricity is in the direction of L. Determine the effective width B': , and = 9max = — 400 KN t 2.2 m Y 100 kN-m Rock the effective length =arrow_forward
- A square footing 2 m x 2 m is located at a depth of 1.2 m below the ground surface. The soilproperties are cohesion, c = 10 kPa, φ = 15 degrees, γ = 16.7 kN/m3, and γsat = 20 kN/m3.Use Terzhagi’s bearing capacity equation: ?? = 1.3??? + ????? + 0.4????, bearingcapacity factors Nc = 12.9, Nq = 4.4, Nγ = 2.5, find qu under the following conditions:arrow_forward). A circular footing 3 m in diameter is shown below. 6. [Soil Bearing Capacity] (* Assume the general shear failure and use a factor of safety 2.8. Determine the follow N. = 25. 13 , Nq = 12.72, N, = 8.34 a. The gross allowable bearing capacity. b. Net Allowable bearing capacity c. The safe load that the footing can carry. Ground surface Y= 18.5 kN/m? C = 80 kPa = 25° 1.1 m D,- 1.8 m Water table Ye = 19.2 kN/m? Diameter = 3marrow_forwardAn rc column footing 2.26m in square shape is to rest 1.5m below level ground level is on cohesive soil. The unit weight is 17.6kN/m^3.What is the safe load of cohension is 30kN/m^3 Factor of safety 2.4 and an angle of internal fraction 33° by IS code. Take NC=38.6; Nq= 26.1 & Ny=29.3arrow_forward
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