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A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18?
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A rectangular beam reinforced on both sides is 30cm wide and 55cm deep...
Given Data
- Width of the beam (b) = 30 cm
- Depth of the beam (d) = 55 cm
- Distance of steel from edges = 5 cm
- Limiting stress in concrete (f_c) = 5 N/mm²
- Limiting stress in steel (f_s) = 140 N/mm²
- Bending Moment (M) = 80 KN-m
- Modular Ratio (M) = 18
Effective Depth Calculation
- Effective depth (d) = Total depth - distance of steel from the edge
- d = 55 cm - 5 cm = 50 cm
Bending Moment Conversion
- Convert Bending Moment to N-mm:
- M = 80 KN-m = 80 × 10^3 × 1000 = 80,000,000 N-mm
Calculate Required Steel Area (A_s)
- Using the formula: M = f_s * A_s * (d - a/2)
- Where 'a' = depth of the neutral axis, for bending stress conditions, use:
- a = (f_c / (f_c + f_s/M)) * d
- Substituting values to find A_s, rearranging gives A_s = M / (f_s * (d - a/2))
Neutral Axis Depth Calculation
- For balanced sections, the neutral axis (a) can be approximated from:
- a = (f_c / (f_c + (f_s/M))) * d
- Using limiting stresses and effective depth, calculate 'a'.
Final Area of Steel Calculation
- Substitute back to find A_s:
- A_s = (80,000,000) / (140 * (50 - a/2))
Conclusion
- Calculate the value of A_s to determine the amount of steel required. Ensure that the calculated area meets the design requirements while adhering to the limiting stress values provided.
- Width of the beam (b) = 30 cm
- Depth of the beam (d) = 55 cm
- Distance of steel from edges = 5 cm
- Limiting stress in concrete (f_c) = 5 N/mm²
- Limiting stress in steel (f_s) = 140 N/mm²
- Bending Moment (M) = 80 KN-m
- Modular Ratio (M) = 18
Effective Depth Calculation
- Effective depth (d) = Total depth - distance of steel from the edge
- d = 55 cm - 5 cm = 50 cm
Bending Moment Conversion
- Convert Bending Moment to N-mm:
- M = 80 KN-m = 80 × 10^3 × 1000 = 80,000,000 N-mm
Calculate Required Steel Area (A_s)
- Using the formula: M = f_s * A_s * (d - a/2)
- Where 'a' = depth of the neutral axis, for bending stress conditions, use:
- a = (f_c / (f_c + f_s/M)) * d
- Substituting values to find A_s, rearranging gives A_s = M / (f_s * (d - a/2))
Neutral Axis Depth Calculation
- For balanced sections, the neutral axis (a) can be approximated from:
- a = (f_c / (f_c + (f_s/M))) * d
- Using limiting stresses and effective depth, calculate 'a'.
Final Area of Steel Calculation
- Substitute back to find A_s:
- A_s = (80,000,000) / (140 * (50 - a/2))
Conclusion
- Calculate the value of A_s to determine the amount of steel required. Ensure that the calculated area meets the design requirements while adhering to the limiting stress values provided.
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A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18?
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A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18? 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 reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18?.
A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18? 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 reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18?.
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A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18?, a detailed solution for A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18? has been provided alongside types of A rectangular beam reinforced on both sides is 30cm wide and 55cm deep the centres of steel are 5cm from the respective edges. If the limiting stresses in concrete and steel are 5N/mm² & 140N/mm² respectively determine the steel areas for a B.M of 80KN-m take M=18? theory, EduRev gives you an
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