JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III DESIGN OF REINFORCED CONCRETE STRUCTURES Previous Question Paper (September - 2021 - R16)

B.Tech | III | I | R16 | September - 2021

Answer PaperPaper Code: 135AJMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

IS456 and design charts for columns (SP-16) are allowed.

Main

115 marks
a

With neat diagrams, derive the stress block parameters from first principles.

8 marks

Answer

The stress block parameters are derived based on the assumption that the strain distribution is linear across the depth of the beam. The concrete stress block is parabolic up to a strain of 0.002 and rectangular from 0.002 to the ultimate strain of 0.0035.

  1. Area of Stress Block: The area is the integral of the stress-strain curve. For the parabolic part (0 to 0.002) and rectangular part (0.002 to 0.0035), the total area is calculated as 0.36 fck xu.
  2. Resultant Force (C): The total compressive force C is the product of the average stress and the area of the concrete zone: C = 0.36 fck b xu.
  3. Depth of Resultant (ybar): By taking moments of the stress block area about the top fiber, the centroid of the compressive force is located at a distance of 0.42 xu from the extreme compression fiber.

These parameters ensure the internal compression force balances the tension in the steel, allowing for the calculation of the ultimate moment of resistance.

b

A simply supported reinforced concrete beam is 250 mm wide and 500mm effective depth and is reinforced with 4 bars of 20 mm diameter as tensile steel. If the beam is subjected to a factored shear of 65 KN at the support. Find the nominal shear stress at the support. Use M20 concrete and Fe 250 steel.

7 marks

Answer

To find the nominal shear stress (τv) at the support, we use the formula:

τv = Vu / (b × d)

Given data:

  • Factored shear force (Vu) = 65 kN = 65,000 N
  • Width of beam (b) = 250 mm
  • Effective depth (d) = 500 mm

Substituting the values into the formula:

τv = 65,000 / (250 × 500)
τv = 65,000 / 125,000
τv = 0.52 N/mm²

The nominal shear stress at the support is 0.52 N/mm².

215 marks
a

Draw the cross-section of singly reinforced rectangular beam and show the strain and stress diagrams.

8 marks

Answer

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b

Discuss the design criteria for single reinforced and doubly reinforced beam sections with neat sketches.

7 marks

Answer

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315 marks
a

Explain how the longitudinal reinforcement bent up nearer to the supports contribute to the shear resistance of RC beams?

7 marks

Answer

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b

Design the shear reinforcement for a simply supported RC beam of effective span 5m with width 300mm and effective depth 400mm and carrying a superimposed load of 10 kN/m. The beam is reinforced with 3 bars of 20 mm diameter. Use M20 concrete and Fe 415 grade steel.

8 marks

Answer

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415 marks
a

Explain the formulae with figures for the design of shear reinforcements in the following cases: (i) Series of bent-up bars or inclined stirrups at different sections. (ii) When a single bar or a group of bars are bent-up at the same cross-section.

8 marks

Answer

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b

Give neat sketches for the reinforcement details for simply supported and continuous beams with IS code provisions.

7 marks

Answer

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515 marks
a

Derive the equation of safety for uncracked section of an eccentrically loaded column.

7 marks

Answer

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b

Design an axially loaded tied column with an unsupported length of 3 m. The column is fixed at one end and pinned at the other end. The column has to carry a factored load of 2000 kN. Use M 25 Grade Concrete and Fe 415 grade steel.

8 marks

Answer

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615 marks
a

Explain the construction and use of interaction curves for uniaxial and biaxial Bending in columns.

7 marks

Answer

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b

Design a circular column of diameter 300mm to carry a factored axial load of 800 kN by using helical reinforcement. Use M20 Concrete and Fe415 steel.

8 marks

Answer

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715 marks
a

Explain how the one way and two-way shear is being handled in the design of footing with suitable IS code provisions and neat sketches.

8 marks

Answer

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b

Design an isolated footing for a column of size 300mm × 600mm carrying a factored axial load of 2500 kN. Assume a safe bearing capacity = 350 kN/m2, M20 grade concrete and Fe 415 grade steel.

7 marks

Answer

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815 marks
a

Write the IS codal provisions made for calculation and measurement of deflections and cracking in RCC beams.

7 marks

Answer

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b

A rectangular section beam with dimensions 200 mm × 450 mm (overall depth) is reinforced with 3 bars of 16 mm diameter at an effective depth of 420 mm. Two hanger bars of 12 mm diameter are provided near the compression face. The effective span of the beam is 5 m and it supports a service load of 10 kN /m. Compute the short-term deflection according to IS: 456-2000 code specifications. Adopt limit state method of design. Take fck = 20 MPa and fy = 415 MPa.

8 marks

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This page contains the solved question paper with answers of DESIGN OF REINFORCED CONCRETE STRUCTURES from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech CE III I, September - 2021, R16.

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