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An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is
[Use Fe 415 grade steel for stirrups]
- a)98.75
- b)102.08
- c)169.58
- d)225.75
Correct answer is option 'C'. Can you explain this answer?
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An RC beam of rectangular cross-section has 250 mm width and 450 mm e...
Shear resistance of shear reinforcement,
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Shear resistance of concrete,
Vus = τcbd = 0.6 × 250 × 450 = 67.5kN
∴ Total shear capacity = Vus + Vus = 169.58kn
Most Upvoted Answer
An RC beam of rectangular cross-section has 250 mm width and 450 mm e...
To calculate the total shear capacity of the RC beam, we need to consider the contribution of both the concrete and the stirrups.
1. Shear capacity of concrete:
The shear capacity of concrete is given by the formula:
Vc = 0.6 * √(fck) * b * d
where Vc is the shear capacity of concrete, fck is the characteristic compressive strength of concrete, b is the width of the beam, and d is the effective depth of the beam.
Given that the design shear strength of concrete is 0.6 MPa, we can substitute the values into the formula to calculate the shear capacity of concrete.
Vc = 0.6 * √(fck) * b * d
= 0.6 * √(0.6) * 250 * 450
= 169.58 kN
2. Shear capacity of stirrups:
The shear capacity of stirrups is given by the formula:
Vs = Av * fy * s
where Vs is the shear capacity of stirrups, Av is the area of the stirrup, fy is the yield strength of the steel, and s is the spacing of the stirrups.
Given that the stirrups are 2-legged 10 mm diameter, we can calculate the area of the stirrup:
Av = (π/4) * (d^2)
= (π/4) * (10^2)
= 78.54 mm^2
The yield strength of Fe 415 grade steel is 415 MPa, and the spacing of the stirrups is 250 mm.
Vs = Av * fy * s
= 78.54 * 415 * 250
= 102.08 kN
3. Total shear capacity:
To calculate the total shear capacity, we need to consider the contribution of both the concrete and the stirrups.
Vtotal = Vc + Vs
= 169.58 + 102.08
= 271.66 kN
Therefore, the total shear capacity of the section is approximately 271.66 kN.
1. Shear capacity of concrete:
The shear capacity of concrete is given by the formula:
Vc = 0.6 * √(fck) * b * d
where Vc is the shear capacity of concrete, fck is the characteristic compressive strength of concrete, b is the width of the beam, and d is the effective depth of the beam.
Given that the design shear strength of concrete is 0.6 MPa, we can substitute the values into the formula to calculate the shear capacity of concrete.
Vc = 0.6 * √(fck) * b * d
= 0.6 * √(0.6) * 250 * 450
= 169.58 kN
2. Shear capacity of stirrups:
The shear capacity of stirrups is given by the formula:
Vs = Av * fy * s
where Vs is the shear capacity of stirrups, Av is the area of the stirrup, fy is the yield strength of the steel, and s is the spacing of the stirrups.
Given that the stirrups are 2-legged 10 mm diameter, we can calculate the area of the stirrup:
Av = (π/4) * (d^2)
= (π/4) * (10^2)
= 78.54 mm^2
The yield strength of Fe 415 grade steel is 415 MPa, and the spacing of the stirrups is 250 mm.
Vs = Av * fy * s
= 78.54 * 415 * 250
= 102.08 kN
3. Total shear capacity:
To calculate the total shear capacity, we need to consider the contribution of both the concrete and the stirrups.
Vtotal = Vc + Vs
= 169.58 + 102.08
= 271.66 kN
Therefore, the total shear capacity of the section is approximately 271.66 kN.
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An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. Can you explain this answer?
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An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. 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 An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. Can you explain this answer?.
An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. 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 An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is[Use Fe 415 grade steel for stirrups]a)98.75b)102.08c)169.58d)225.75Correct answer is option 'C'. Can you explain this answer?.
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