Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 7 (Q.No. 25)
25.
A footing 2 m x 1 m exerts a uniform pressure of 150 kN/m2 on the soil. Assuming a load dispersion of 2 vertical to 1 horizontal the average vertical stress (kN/m2) at 1.0 m below the footing is
Discussion:
6 comments Page 1 of 1.
Rakesh karn said:
3 years ago
How to calculate vertical stress for piles?
Please tell me.
Please tell me.
Renu said:
5 years ago
F=150*2*1 = 300kN.
Footing-2m * 1m.
Depth - 1.0m.
(150*2*1) ÷ [(2+1.0) * (1+1.0)] = 50 kN/m^2.
Footing-2m * 1m.
Depth - 1.0m.
(150*2*1) ÷ [(2+1.0) * (1+1.0)] = 50 kN/m^2.
Nirlipta Mohapatra said:
6 years ago
How is (4+2)? here L =1m, B=2m.
(2)
Babatunde said:
6 years ago
F = 150 * 2 * 1 (rectangular base) = 300kN.
Load disperses under the footing in a trapezoidal pattern, with 2V:1H.
Area of trapezium= 1/2(4+2)2 = 6m^2.
Ave vertical stress at 1m below = 300/6 = 50 kN/m^2.
Load disperses under the footing in a trapezoidal pattern, with 2V:1H.
Area of trapezium= 1/2(4+2)2 = 6m^2.
Ave vertical stress at 1m below = 300/6 = 50 kN/m^2.
(1)
Arnab said:
8 years ago
F= 150 *2*1 KN =300 KN,
Area of dispersion = (2+.5+.5)*(1+.5+.5)=6 m2,
Vertical stress = 300/6 = 50 KN/m2.
Area of dispersion = (2+.5+.5)*(1+.5+.5)=6 m2,
Vertical stress = 300/6 = 50 KN/m2.
Abirami said:
8 years ago
How to solve this?
Explain.
Explain.
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