JNTU
JNTU B.Tech III DESIGN OF REINFORCED CONCRETE STRUCTURES Previous Question Paper (August - 2022 - R16)
Answer any five questions
All questions carry equal marks
Note: Use of IS 456:2000 and only design charts of SP16 is allowed.
Main
Derive the ‘stress block’ parameters from first principles.
Answer
The stress block parameters are derived based on the assumptions of Limit State Design for a rectangular section. The stress-strain curve for concrete is parabolic up to a strain of 0.002 and constant at 0.446fck up to the ultimate strain of 0.0035.
1. Area of stress block: The area is the integral of the stress distribution over the depth of the compression zone (xu). It consists of a parabolic part (0 to 0.002 strain) and a rectangular part (0.002 to 0.0035 strain). The total area is calculated as 0.36fck * xu.
2. Depth of centroid (y_bar): The centroid of the stress block is found by taking the moment of the stress area about the extreme compression fiber. For the parabolic and rectangular components, the resultant force acts at a distance of 0.42xu from the extreme compression fiber.
3. Resultant compressive force (C): The total compressive force is C = 0.36fck * b * xu, where b is the width of the section.
Calculate the ‘balanced percentage of steel’ for M30 concrete and Fe 500 grade steel from first principles, as per Limit State Design.
Answer
For M30 concrete (fck = 30 N/mm²) and Fe 500 steel (fy = 500 N/mm²), the balanced percentage of steel is calculated using the limiting depth of the neutral axis (xu,max).
1. Limiting depth of neutral axis: For Fe 500, xu,max/d = 0.46. Thus, xu,max = 0.46d.
2. Equating C and T: At balanced failure, C = T. Therefore, 0.36 * fck * b * xu,max = 0.87 * fy * Ast,lim.
3. Calculation: Substituting the values:
4. Percentage of steel (pt,lim):
The balanced percentage of steel for M30 and Fe 500 is approximately 1.14%.
Design and flexural reinforcement for an RC beam of size 300 mm wide and 450 mm deep to resist an ultimate moment of 400 kNm. Assume moderate exposure condition. Use M25 concrete and Fe 500 grade steel. Adopt Limit State method. Sketch the reinforcement details.
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With a neat sketch, explain the force components that participate in the shear transfer mechanism at a flexural-shear crack location in a reinforced concrete beam.
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A simply supported beam of 6.3 m effective span, is to carry a uniformly distributed load (dead load) of 20 kN/m including its self weight, and a live load of 30 kN/m. Design the beam for shear using Limit state method. Use M25 concrete and Fe 500 grade steel. Sketch the reinforcement details.
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List the main factors that influence bond strength of a reinforcing bar.
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Design the torsional reinforcement in a rectangular beam section 350 mm wide and 800 mm deep, subjected to an ultimate twisting moment of 180 kNm, an ultimate hogging bending moment of 300 kNm and an ultimate shear force of 150 kNm. Use M30 concrete and Fe 415 grade steel. Assume moderate exposure condition. Sketch the reinforcement details.
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What are the functions of transverse reinforcement in reinforced concrete column.
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A column 360 mm × 450 mm has an effective length of 3.4 m. It is subjected to an ultimate load of 1400 kN and an ultimate moment of 300 kNm about its major axis. Determine the longitudinal and transverse reinforcement. Use M30 concrete and Fe 415 grade steel. Assume moderate exposure condition. Sketch the cross-section showing reinforcement details.
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Derive the expression, from first principles, for ‘Ultimate Load Carrying Capacity’ of an axially loaded short RC column with lateral ties.
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Design longitudinal reinforcement and ties if an axially loaded column subjected to a working load of 1000 kN. The unsupported length of the column is 3.5 m. Use M25 concrete and Fe 500 grade steel. Sketch the cross-section showing reinforcement details.
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Design an isolated rectangular footing for a reinforced concrete column 300 mm × 550 mm reinforced with 6 reinforcing bars of 25 mm diameter, subjected to a factored axial load of 1200 kN and a factored uniaxial moment (about major axis), of 150 kNm at the column base. The safe bearing capacity of the soil is 220 kN/m2 at a depth of 1.5 m. Assume M25 concrete and Fe415 grade steel. Sketch the reinforcement details.
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A simply supported one-way slab 180 mm thick having an effective span of 4.5 m is reinforced with 10 mm diameter bars spaced at 125 mm c/c at an effective cover of 25 mm. The slab is subjected to a live load of 3 kN/m2 and a surface finish of 1 kN/m2. Use M25 concrete and Fe 500 grade steel. Assume ultimate shrinkage strain = 0.0003 and creep coefficient = 1.6. Estimate the only the long-term deflection.
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