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An RCC beam of cross-section 250 mm x 400 mm is subjected to a factored bending moment of 150 kNm and a factored torsional moment of 150 kNm. The longitudinal reinforcement provided on the compression face will be designed for the moment of______ kNm (round up to 2 decimal places)(take cover to both the reinforcements be 50 mm)
    Correct answer is between '79,80'. Can you explain this answer?
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    An RCC beam of cross-section 250 mm x 400 mm is subjected to a factore...
    Given Data:

    Cross-section of RCC beam: 250 mm x 400 mm
    Factored bending moment: 150 kNm
    Factored torsional moment: 150 kNm

    Solution:

    To design the longitudinal reinforcement on the compression face, we need to calculate the moment for which the reinforcement will be designed.

    Step 1: Calculate the effective depth (d)

    The effective depth (d) of the beam is given by the formula:
    d = Overall depth of the beam - Clear cover - Diameter of reinforcement

    Given that the clear cover to both the reinforcements is 50 mm and assuming the diameter of reinforcement to be 10 mm, we can calculate the effective depth as follows:
    d = 400 mm - 2(50 mm) - 10 mm = 290 mm

    Step 2: Calculate the area of steel required for bending moment

    To calculate the area of steel required for bending moment, we use the formula:
    Ast = (M / (0.87 * fyd * d))

    Where,
    Ast = Area of steel required
    M = Factored bending moment
    fyd = Design yield strength of steel
    d = Effective depth

    Assuming the design yield strength of steel to be 415 MPa, we can calculate the area of steel required as follows:
    Ast = (150 * 10^6) / (0.87 * 415 * 10^6 * 290) = 0.0116 m^2

    Step 3: Calculate the area of steel required for torsional moment

    To calculate the area of steel required for torsional moment, we use the formula:
    Ast_torsion = (Mt / (0.87 * fyd * d))

    Where,
    Ast_torsion = Area of steel required for torsional moment
    Mt = Factored torsional moment
    fyd = Design yield strength of steel
    d = Effective depth

    Assuming the design yield strength of steel to be 415 MPa, we can calculate the area of steel required for torsional moment as follows:
    Ast_torsion = (150 * 10^6) / (0.87 * 415 * 10^6 * 290) = 0.0116 m^2

    Step 4: Total area of steel required

    The total area of steel required is the sum of the areas of steel required for bending moment and torsional moment, which is:
    Total area of steel required = Ast + Ast_torsion = 0.0116 m^2 + 0.0116 m^2 = 0.0232 m^2

    Step 5: Calculate the moment for which the reinforcement will be designed

    To calculate the moment for which the reinforcement will be designed, we use the formula:
    M_design = (Ast / (0.87 * fyd * d)) * (0.36 * fck * b * d^2)

    Where,
    M_design = Moment for which reinforcement will be designed
    Ast = Total area of steel required
    fyd = Design yield strength of steel
    d = Effective depth
    fck = Characteristic compressive strength of concrete
    b = Width of the beam

    Assuming the characteristic compressive strength of concrete
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    Community Answer
    An RCC beam of cross-section 250 mm x 400 mm is subjected to a factore...
    Me2 = 229.41 - 150 = 79.411 kNm
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    An RCC beam of cross-section 250 mm x 400 mm is subjected to a factored bending moment of 150 kNm and a factored torsional moment of 150 kNm. The longitudinal reinforcement provided on the compression face will be designed for the moment of______ kNm (round up to 2 decimal places)(take cover to both the reinforcements be 50 mm)Correct answer is between '79,80'. Can you explain this answer?
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    An RCC beam of cross-section 250 mm x 400 mm is subjected to a factored bending moment of 150 kNm and a factored torsional moment of 150 kNm. The longitudinal reinforcement provided on the compression face will be designed for the moment of______ kNm (round up to 2 decimal places)(take cover to both the reinforcements be 50 mm)Correct answer is between '79,80'. 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 An RCC beam of cross-section 250 mm x 400 mm is subjected to a factored bending moment of 150 kNm and a factored torsional moment of 150 kNm. The longitudinal reinforcement provided on the compression face will be designed for the moment of______ kNm (round up to 2 decimal places)(take cover to both the reinforcements be 50 mm)Correct answer is between '79,80'. 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 An RCC beam of cross-section 250 mm x 400 mm is subjected to a factored bending moment of 150 kNm and a factored torsional moment of 150 kNm. The longitudinal reinforcement provided on the compression face will be designed for the moment of______ kNm (round up to 2 decimal places)(take cover to both the reinforcements be 50 mm)Correct answer is between '79,80'. Can you explain this answer?.
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