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A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)
Correct answer is between '100,102'. Can you explain this answer?
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A reinforced concrete beam with rectangular cross section (width = 300...
B = 300 mm
d = 580 mm
Vu = 440 kN
Concrete used is M30
Raft steel is Fe415
d = 580 mm
Vu = 440 kN
Concrete used is M30
Raft steel is Fe415
- Spacing of 2-legged shear reinforcement
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A reinforced concrete beam with rectangular cross section (width = 300...
Given Data:
- Width of the beam (b) = 300 mm
- Effective depth of the beam (d) = 580 mm
- Longitudinal tension reinforcement = 1% of the cross-sectional area
- Concrete grade (fck) = M30 (30 N/mm2)
- Steel grade (fy) = Fe 415 (415 N/mm2)
- Design shear strength of the beam (τc) = 0.66 N/mm2
- Factored shear force (Vu) = 440 kN
Solution:
Step 1: Calculation of Cross-sectional area of tension reinforcement (Ast)
Given that the longitudinal tension reinforcement is 1% of the cross-sectional area of the beam.
Cross-sectional area of the beam (A) = b × d
A = 300 mm × 580 mm = 174,000 mm2
Cross-sectional area of tension reinforcement (Ast) = 1% of A
Ast = 1% × 174,000 mm2
Ast = 1,740 mm2
Step 2: Calculation of Design shear strength (τv)
The design shear strength (τv) is given by the formula:
τv = 0.17 √(fck)
τv = 0.17 √(30 N/mm2)
τv = 0.17 × 5.48 N/mm2
τv = 0.93 N/mm2
Step 3: Calculation of Required shear reinforcement (Vus)
The required shear reinforcement (Vus) is given by the formula:
Vus = (Vu - 0.75 × τc × b × d) / (2 × d × τv)
Vus = (440 kN - 0.75 × 0.66 N/mm2 × 300 mm × 580 mm) / (2 × 580 mm × 0.93 N/mm2)
Vus = 440,000 N - 72,270 N / 1,076.4 N
Vus = 367.7 mm2
Step 4: Calculation of Spacing of vertical stirrups (s)
The spacing of vertical stirrups is given by the formula:
s = (0.87 × fy × Ast) / (0.4 × b × τv)
s = (0.87 × 415 N/mm2 × 1,740 mm2) / (0.4 × 300 mm × 0.93 N/mm2)
s = 614,145 N - 565,161 N / 111.6 N
s = 485.8 mm
To round off to the nearest integer, the spacing of the vertical stirrups is 486 mm.
Therefore, the spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is 486 mm.
- Width of the beam (b) = 300 mm
- Effective depth of the beam (d) = 580 mm
- Longitudinal tension reinforcement = 1% of the cross-sectional area
- Concrete grade (fck) = M30 (30 N/mm2)
- Steel grade (fy) = Fe 415 (415 N/mm2)
- Design shear strength of the beam (τc) = 0.66 N/mm2
- Factored shear force (Vu) = 440 kN
Solution:
Step 1: Calculation of Cross-sectional area of tension reinforcement (Ast)
Given that the longitudinal tension reinforcement is 1% of the cross-sectional area of the beam.
Cross-sectional area of the beam (A) = b × d
A = 300 mm × 580 mm = 174,000 mm2
Cross-sectional area of tension reinforcement (Ast) = 1% of A
Ast = 1% × 174,000 mm2
Ast = 1,740 mm2
Step 2: Calculation of Design shear strength (τv)
The design shear strength (τv) is given by the formula:
τv = 0.17 √(fck)
τv = 0.17 √(30 N/mm2)
τv = 0.17 × 5.48 N/mm2
τv = 0.93 N/mm2
Step 3: Calculation of Required shear reinforcement (Vus)
The required shear reinforcement (Vus) is given by the formula:
Vus = (Vu - 0.75 × τc × b × d) / (2 × d × τv)
Vus = (440 kN - 0.75 × 0.66 N/mm2 × 300 mm × 580 mm) / (2 × 580 mm × 0.93 N/mm2)
Vus = 440,000 N - 72,270 N / 1,076.4 N
Vus = 367.7 mm2
Step 4: Calculation of Spacing of vertical stirrups (s)
The spacing of vertical stirrups is given by the formula:
s = (0.87 × fy × Ast) / (0.4 × b × τv)
s = (0.87 × 415 N/mm2 × 1,740 mm2) / (0.4 × 300 mm × 0.93 N/mm2)
s = 614,145 N - 565,161 N / 111.6 N
s = 485.8 mm
To round off to the nearest integer, the spacing of the vertical stirrups is 486 mm.
Therefore, the spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is 486 mm.
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A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer?
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A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer?.
A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm2. The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is ______ mm. (round off to the nearest integer)Correct answer is between '100,102'. Can you explain this answer?.
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