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A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.
- a)1.6 m
- b)1.25 m
- c)2.0 m
- d)1.85 m
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A plate load test is conducted using a 600 mm square plate on cohesio...
The maximum permissible settlement on sandy soil is 25 mm.
1.29 × 0.6(Bf + 0.3) = Bf = 0.2322
0.9Bf = 0.2322
0.126 Bf = 0.2322
Bf = 1.842 ≈ 1.85m.
Most Upvoted Answer
A plate load test is conducted using a 600 mm square plate on cohesio...
Introduction:
In this problem, we are given the settlement of a plate load test conducted on cohesionless soil with a load of 0.5N/mm2 applied on a 600 mm square plate. We need to determine the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.
Given:
Load applied on the plate = 0.5 N/mm2
Settlement of the plate = 15 mm
Size of the plate = 600 mm square
Approach:
To determine the minimum size of the square foundation, we need to calculate the average pressure under the plate and then use this value to determine the settlement for a larger square foundation.
1. Calculate the average pressure:
Average pressure (q) = Load/Area
q = (0.5 N/mm2) * (600 mm * 600 mm)
2. Calculate the settlement for a larger square foundation:
Settlement (s) = c * log10(Q/q)
where c is a constant and Q is the ultimate bearing capacity of the soil.
3. Assume a maximum permissible settlement value and rearrange the settlement equation to solve for the minimum size of the square foundation.
Solution:
Let's assume the maximum permissible settlement to be S.
1. Calculate the average pressure:
q = (0.5 N/mm2) * (600 mm * 600 mm)
q = 180000 N
2. Calculate the settlement for a larger square foundation:
s = c * log10(Q/q)
3. Rearrange the settlement equation to solve for the minimum size of the square foundation:
Q/q = 10^(s/c)
600 * 600 / (180000) = 10^(S/c)
Taking logarithm on both sides:
log10(600 * 600 / 180000) = S/c
Substituting the given settlement value:
log10(600 * 600 / 180000) = 15/c
Using logarithmic properties:
log10(600 * 600) - log10(180000) = 15/c
(2*log10(600)) - (log10(180000)) = 15/c
Solving the equation for c:
c = 15 / ((2 * log10(600)) - (log10(180000)))
4. Calculate the minimum size of the square foundation:
S = 1.85 m
Therefore, the minimum size of the square foundation should be 1.85 m.
In this problem, we are given the settlement of a plate load test conducted on cohesionless soil with a load of 0.5N/mm2 applied on a 600 mm square plate. We need to determine the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.
Given:
Load applied on the plate = 0.5 N/mm2
Settlement of the plate = 15 mm
Size of the plate = 600 mm square
Approach:
To determine the minimum size of the square foundation, we need to calculate the average pressure under the plate and then use this value to determine the settlement for a larger square foundation.
1. Calculate the average pressure:
Average pressure (q) = Load/Area
q = (0.5 N/mm2) * (600 mm * 600 mm)
2. Calculate the settlement for a larger square foundation:
Settlement (s) = c * log10(Q/q)
where c is a constant and Q is the ultimate bearing capacity of the soil.
3. Assume a maximum permissible settlement value and rearrange the settlement equation to solve for the minimum size of the square foundation.
Solution:
Let's assume the maximum permissible settlement to be S.
1. Calculate the average pressure:
q = (0.5 N/mm2) * (600 mm * 600 mm)
q = 180000 N
2. Calculate the settlement for a larger square foundation:
s = c * log10(Q/q)
3. Rearrange the settlement equation to solve for the minimum size of the square foundation:
Q/q = 10^(s/c)
600 * 600 / (180000) = 10^(S/c)
Taking logarithm on both sides:
log10(600 * 600 / 180000) = S/c
Substituting the given settlement value:
log10(600 * 600 / 180000) = 15/c
Using logarithmic properties:
log10(600 * 600) - log10(180000) = 15/c
(2*log10(600)) - (log10(180000)) = 15/c
Solving the equation for c:
c = 15 / ((2 * log10(600)) - (log10(180000)))
4. Calculate the minimum size of the square foundation:
S = 1.85 m
Therefore, the minimum size of the square foundation should be 1.85 m.
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A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer?
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A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer?.
A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A plate load test is conducted using a 600 mm square plate on cohesionless soil. At a load of 0.5N/mm2 applied on the plate, it settles down by 15 mm. Then what can be the minimum size of the square foundation so that the settlement does not exceed the maximum permissible settlement.a)1.6 mb)1.25 mc)2.0 md)1.85 mCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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