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A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :
(a) 176m * m ^ 2
(b) 225m * m ^ 2
(c) 364m * m ^ 2
(d) 505m * m ^ 2?
(a) 176m * m ^ 2
(b) 225m * m ^ 2
(c) 364m * m ^ 2
(d) 505m * m ^ 2?
Most Upvoted Answer
A rectangular beam cross-section having width as mm and effective dept...
Given Data
- Width of beam (b) = unspecified mm
- Effective depth (d) = 500 mm
- Design torsional moment (T) = 150 kNm = 150 × 10³ Nm
- Transverse shear force (V) = 80 kN = 80 × 10³ N
- Spacing of shear reinforcement = 150 mm
- Permissible shear stress (τ_c) = 0.36 N/mm²
- Grade of steel = Fe 415
Calculating the Design Shear Force
- The design shear force \( V_d \) is taken directly as 80 kN.
Calculating the Shear Stress
- The area of the cross-section \( A \) is calculated as:
\[ A = b \times d = b \times 500 \]
- The shear stress \( \tau \) is given by:
\[ \tau = \frac{V_d}{A} = \frac{80 \times 10^3}{b \times 500} \]
Permissible Shear Stress Check
- To ensure safety, the actual shear stress should not exceed the permissible shear:
\[ \frac{80 \times 10^3}{b \times 500} \leq 0.36 \]
Calculating Minimum Area of Shear Reinforcement
- The area of shear reinforcement \( A_{sv} \) can be calculated using:
\[ A_{sv} = \frac{V_d}{0.87 \times f_y} \]
- Where \( f_y = 415 \) N/mm² (for Fe 415).
- Substituting the values:
\[ A_{sv} = \frac{80 \times 10^3}{0.87 \times 415} \approx 225.52 \text{ mm}^2 \]
Conclusion
- The minimum area of total transverse reinforcement should be approximately:
- (b) 225 mm²
This meets the design requirements under the given conditions.
- Width of beam (b) = unspecified mm
- Effective depth (d) = 500 mm
- Design torsional moment (T) = 150 kNm = 150 × 10³ Nm
- Transverse shear force (V) = 80 kN = 80 × 10³ N
- Spacing of shear reinforcement = 150 mm
- Permissible shear stress (τ_c) = 0.36 N/mm²
- Grade of steel = Fe 415
Calculating the Design Shear Force
- The design shear force \( V_d \) is taken directly as 80 kN.
Calculating the Shear Stress
- The area of the cross-section \( A \) is calculated as:
\[ A = b \times d = b \times 500 \]
- The shear stress \( \tau \) is given by:
\[ \tau = \frac{V_d}{A} = \frac{80 \times 10^3}{b \times 500} \]
Permissible Shear Stress Check
- To ensure safety, the actual shear stress should not exceed the permissible shear:
\[ \frac{80 \times 10^3}{b \times 500} \leq 0.36 \]
Calculating Minimum Area of Shear Reinforcement
- The area of shear reinforcement \( A_{sv} \) can be calculated using:
\[ A_{sv} = \frac{V_d}{0.87 \times f_y} \]
- Where \( f_y = 415 \) N/mm² (for Fe 415).
- Substituting the values:
\[ A_{sv} = \frac{80 \times 10^3}{0.87 \times 415} \approx 225.52 \text{ mm}^2 \]
Conclusion
- The minimum area of total transverse reinforcement should be approximately:
- (b) 225 mm²
This meets the design requirements under the given conditions.
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A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2?
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A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2?.
A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2?.
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A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2?, a detailed solution for A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2? has been provided alongside types of A rectangular beam cross-section having width as mm and effective depth as 500 mm is subjected to a design torsional moment of 150 kNm and transverse shear force of 80 kN. If the spacing of the shear reinforcement provided is 150 mm, permissible shear is 0.36N / m * m ^ 2 and grade of steel is Fe 415, the minimum area of total transverse reinforcement should be :(a) 176m * m ^ 2(b) 225m * m ^ 2(c) 364m * m ^ 2(d) 505m * m ^ 2? theory, EduRev gives you an
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