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JNTU B.Tech III Year DESIGN OF REINFORCED CONCRETE STRUCTURES Previous Question Paper (October - 2020 - R16)

B.Tech | III Year | I | R16 | October - 2020

Answer PaperPaper Code: 135AJMaximum Marks: 75Time: 2 hours

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1

Design the reinforcement of a concrete beam section subjected to a moment of 250 kNm. The overall size of the section is limited to 300 mm × 450 mm. Use M 25 concrete and Fe 415 steel reinforcement. Adopt limit state method of design.

15 marks

Answer

Given: M = 250 kNm, b = 300 mm, D = 450 mm, fck = 25 N/mm², fy = 415 N/mm². Assuming effective cover d' = 50 mm, effective depth d = 450 - 50 = 400 mm.

1. Check for limiting moment of resistance (Mu,lim):

Mu,lim = 0.138 × fck × b × d² = 0.138 × 25 × 300 × 400² = 165.6 kNm

Since Mu (250 kNm) > Mu,lim (165.6 kNm), the section must be designed as a doubly reinforced beam.

2. Calculate reinforcement for the singly reinforced portion (Mu1 = Mu,lim):

Ast1 = (0.5 × fck / fy) × [1 - sqrt(1 - 4.6 × Mu,lim / (fck × b × d²))] × b × d = 1385 mm²

3. Calculate reinforcement for the additional moment (Mu2 = Mu - Mu,lim = 250 - 165.6 = 84.4 kNm):

Ast2 = Mu2 / (fy × (d - d')) = 84.4 × 10⁶ / (415 × (400 - 50)) = 581 mm²

Total tension steel: Ast = Ast1 + Ast2 = 1385 + 581 = 1966 mm². Provide 4 bars of 25 mm diameter (1963 mm²).

4. Calculate compression reinforcement (Asc):

Assuming fsc ≈ 353 N/mm² for d'/d = 50/400 = 0.125:

Asc = Mu2 / (fsc × (d - d')) = 84.4 × 10⁶ / (353 × 350) = 683 mm²

Provide 2 bars of 22 mm diameter (760 mm²).

2

Explain about stress block parameters.

15 marks

Answer

In the limit state design of reinforced concrete, the stress-strain relationship of concrete is represented by a simplified rectangular-parabolic stress block. The parameters of this stress block are derived from the actual stress-strain curve to simplify the calculation of the ultimate moment of resistance.

The stress block parameters are defined as follows:

  • Compressive Stress Magnitude: The maximum compressive stress in the concrete is taken as 0.67fck, where fck is the characteristic compressive strength of concrete. This value accounts for the difference between the strength of concrete in a structure and the strength of a standard test cube, incorporating a partial safety factor of 1.5. Thus, the design stress is (0.67fck / 1.5) = 0.446fck.
  • Depth of Stress Block: The depth of the rectangular portion of the stress block is taken as 0.42xu, where xu is the depth of the neutral axis from the extreme compression fiber.
  • Total Compressive Force (C): The total compressive force acting on the concrete section is the product of the average stress, the width of the section (b), and the depth of the stress block. It is calculated as C = 0.36fck * b * xu.
  • Lever Arm (z): The resultant compressive force acts at the centroid of the stress block. The distance between the centroid of the compressive force and the centroid of the tensile force (in the steel) is the lever arm, given by z = d - 0.42xu, where d is the effective depth.

These parameters ensure that the internal moment of resistance calculated using the simplified block is equivalent to the moment calculated using the actual parabolic stress distribution, providing a reliable basis for the design of beams and slabs under flexure.

3

A simply supported RC beam of section 300 mm × 500 mm (effective depth) is subjected to an ultimate shear force of 150 kN at the supports. The beam is reinforced with 4 bars of 20 mm diameter. Design the shear reinforcement. Sketch the shear reinforcement details. Use M 20 concrete and Fe 415 steel.

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4

A rectangular reinforced concrete beam of size 300 mm × 600 mm overall depth is subjected to an ultimate moment of 75 kNm, a factored shear force of 90 kN and a factored torsional moment of 25 kNm. Design the longitudinal and transverse reinforcement. Sketch the reinforcement details. Use M 20 concrete and Fe 415 steel.

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5

Design a reinforced concrete column of effective length 3.6 m subjected to a factored axial load of 1250 kN and factored moments of 160 kNm and 125 kNm with respect to the major and minor centroidal axes respectively. Use M 25 grade concrete and Fe 415 steel.

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6

Design the reinforcement of an RCC column of size 300 mm × 350 mm subjected to a factored axial load of 1200 kN and factored moments of 40 kNm and 25 kNm with respect to the major and minor centroidal axes respectively at the top. The moments at the bottom end equal to 60 % of the corresponding values at the top. Assume the column is bent in double curvature and the effective length of the column is 6.0 m. Use M 25 grade concrete and Fe 415 steel.

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7

Design the footing for a rectangular column 300 mm × 450 mm subjected to an axial load of 900 kN. Assume the bearing capacity of soil is 180 kN/m2. Use M 20 concrete and Fe 415 steel.

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8

Design an RCC two-way slab of clear dimensions 4.5 m × 6 m with two adjacent edges continuous. The slab is subjected to live load of 3 kN/m2 and floor finish of 1 kN/m2. Assume the width of the supports is 300 mm. Use M 20 grade concrete and Fe 415 steel. Sketch the reinforcement details.

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This page contains the question paper of DESIGN OF REINFORCED CONCRETE STRUCTURES from JNTU for B.Tech CE III Year I, October - 2020, R16.

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