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The results of two plate load tests performed on a given location with two circular plates are given below:
1. Diameter = 750 mm, S = 15 mm, Q = 150 kN
2. Diameter = 300 mm, S = 15 mm, Q = 50 kN
Use Housel’s equation i.e. Q = Aq + Ps
A = contact area
q = bearing pressure beneath area A (constant)
P = perimeter of footing
s = Perimeter shear (constant)
Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.
- a)279.6
- b)402.18
- c)512.2
- d)558.8
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The results of two plate load tests performed on a given location wit...
⇒ 150 = 0.44q + 2.36s …(i)
Similarly
Q2 = 50 kN
⇒ 50 = 0.071q + 0.94s ...(ii)
Solving (i) and (ii)
S = 46.13 kN/m
q = 93.48 kN/m2
Now for footing
Q = 279.6 kN
Most Upvoted Answer
The results of two plate load tests performed on a given location wit...
To determine the load on a circular footing that will cause a settlement of 15 mm, we can use Housel's equation, which relates the load (Q) to the bearing pressure (q), contact area (A), and perimeter shear (P) of the footings.
Given data:
Footing 1: Diameter = 750 mm, S = 15 mm, Q = 150 kN
Footing 2: Diameter = 300 mm, S = 15 mm, Q = 50 kN
We need to calculate the load on a circular footing with a diameter of 1.2 m.
Let's calculate the contact area (A) for both footings:
- For footing 1:
Radius (r) = Diameter / 2 = 750 mm / 2 = 375 mm
Area (A1) = π * r^2 = 3.1416 * (375 mm)^2 = 441,786.065 mm^2
- For footing 2:
Radius (r) = Diameter / 2 = 300 mm / 2 = 150 mm
Area (A2) = π * r^2 = 3.1416 * (150 mm)^2 = 70,685.834 mm^2
Now, we can calculate the bearing pressure (q) for both footings:
- For footing 1:
q1 = Q1 / A1 = 150 kN / 441,786.065 mm^2 = 0.339 kN/mm^2
- For footing 2:
q2 = Q2 / A2 = 50 kN / 70,685.834 mm^2 = 0.708 kN/mm^2
Next, we need to calculate the perimeter shear (P) for both footings:
- For footing 1:
P1 = π * D1 = 3.1416 * 750 mm = 2,355.2 mm
- For footing 2:
P2 = π * D2 = 3.1416 * 300 mm = 942.48 mm
Now, we can rearrange Housel's equation to solve for the load (Q) for the desired settlement (S):
Q = Aq / Ps
Substituting the values, we have:
- For footing 1:
Q = (A1 * q1) / (P1 * S) = (441,786.065 mm^2 * 0.339 kN/mm^2) / (2,355.2 mm * 15 mm) = 279.6 kN
Therefore, the load on a circular footing with a diameter of 1.2 m that will cause a settlement of 15 mm is 279.6 kN, which corresponds to option A.
Given data:
Footing 1: Diameter = 750 mm, S = 15 mm, Q = 150 kN
Footing 2: Diameter = 300 mm, S = 15 mm, Q = 50 kN
We need to calculate the load on a circular footing with a diameter of 1.2 m.
Let's calculate the contact area (A) for both footings:
- For footing 1:
Radius (r) = Diameter / 2 = 750 mm / 2 = 375 mm
Area (A1) = π * r^2 = 3.1416 * (375 mm)^2 = 441,786.065 mm^2
- For footing 2:
Radius (r) = Diameter / 2 = 300 mm / 2 = 150 mm
Area (A2) = π * r^2 = 3.1416 * (150 mm)^2 = 70,685.834 mm^2
Now, we can calculate the bearing pressure (q) for both footings:
- For footing 1:
q1 = Q1 / A1 = 150 kN / 441,786.065 mm^2 = 0.339 kN/mm^2
- For footing 2:
q2 = Q2 / A2 = 50 kN / 70,685.834 mm^2 = 0.708 kN/mm^2
Next, we need to calculate the perimeter shear (P) for both footings:
- For footing 1:
P1 = π * D1 = 3.1416 * 750 mm = 2,355.2 mm
- For footing 2:
P2 = π * D2 = 3.1416 * 300 mm = 942.48 mm
Now, we can rearrange Housel's equation to solve for the load (Q) for the desired settlement (S):
Q = Aq / Ps
Substituting the values, we have:
- For footing 1:
Q = (A1 * q1) / (P1 * S) = (441,786.065 mm^2 * 0.339 kN/mm^2) / (2,355.2 mm * 15 mm) = 279.6 kN
Therefore, the load on a circular footing with a diameter of 1.2 m that will cause a settlement of 15 mm is 279.6 kN, which corresponds to option A.
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The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer?
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The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. 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 The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer?.
The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. 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 The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer?.
Solutions for The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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ample number of questions to practice The results of two plate load tests performed on a given location with two circular plates are given below:1. Diameter = 750 mm, S = 15 mm, Q = 150 kN2. Diameter = 300 mm, S = 15 mm, Q = 50 kNUse Housel’s equation i.e. Q = Aq + PsA = contact areaq = bearing pressure beneath area A (constant)P = perimeter of footings = Perimeter shear (constant)Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.a)279.6b)402.18c)512.2d)558.8Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice GATE tests.
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