JNTU
JNTU B.Tech III DESIGN OF REINFORCED CONCRETE STRUCTURES Previous Question Paper (February - 2022 - R16)
Answer any five questions
All questions carry equal marks
Note: IS 456 & only the design charts for columns (SP 16) are allowed:
Main
Explain the terms balanced, over-reinforced and under-reinforced section in bending. Explain which of these should be recommended in design.
Answer
In reinforced concrete design, the behavior of a beam section depends on the relative strain in concrete and steel at the ultimate limit state:
- Balanced Section: A section where the concrete reaches its ultimate compressive strain (0.0035) and the steel reaches its yield strain (0.002 + 0.87fy/Es) simultaneously. It represents the transition point between ductile and brittle failure.
- Under-reinforced Section: A section where the steel reaches its yield strain before the concrete reaches its ultimate strain. This results in a ductile failure mode, providing ample warning through significant deflection and cracking before collapse.
- Over-reinforced Section: A section where the concrete reaches its ultimate strain before the steel yields. This leads to a sudden, brittle, and catastrophic failure without warning.
Designers should always recommend under-reinforced sections. They ensure ductile behavior, allowing the structure to absorb energy and provide sufficient warning before failure, which is a fundamental safety requirement in structural engineering.
A beam simply supported over an effective span of 7m carries alive load of 20 kN/m. Design the beam, using M20 concrete and HYSD bars of grade Fe415. keep the width equal to half the effective depth. Assume unit weight of concrete as 25 kN/m3. Sketch the reinforcement details.
Answer
1. Design Constants: For M20 and Fe415, the limiting depth factor is xu,max/d = 0.48. The lever arm factor is j = 1 - 0.42(xu,max/d) = 0.798. The moment of resistance factor R = 0.36 * (xu,max/d) * (1 - 0.42 * (xu,max/d)) * fck = 2.76 N/mm2.
2. Loads: Assume self-weight = 0.3 * 0.6 * 25 = 4.5 kN/m. Total load w = 20 + 4.5 = 24.5 kN/m. Factored load wu = 1.5 * 24.5 = 36.75 kN/m. Factored moment Mu = (wu * L2) / 8 = (36.75 * 72) / 8 = 225 kNm.
3. Dimensions: Mu = R * b * d2. Given b = 0.5d, then 225 * 106 = 2.76 * 0.5d * d2 = 1.38d3. Solving gives d = 555 mm. Provide d = 560 mm and b = 280 mm. Overall depth D = 610 mm.
4. Reinforcement: Ast = Mu / (0.87 * fy * (d - 0.42 * xu,max)) = 225 * 106 / (0.87 * 415 * 0.81 * 560) = 1375 mm2. Provide 3 bars of 25 mm diameter (1472 mm2).
5. Sketch: The reinforcement sketch should show a rectangular section 280 mm x 610 mm with 3-25 mm diameter bars at the bottom, 25 mm clear cover, and 8 mm diameter stirrups as shear reinforcement.
Explain the limiting moment of resistance and give the expression for the Fe 250 and Fe415 grade steel.
Answer
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Design a T-beam whose dimensions are bf= 700 mm and effective depth = 300 mm, width of web bw=200 mm, Df= 100 mm. Applied moment of 350 kNm. Use M20 grade concrete and Fe500 grade steel. Sketch the reinforcement details.
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Explain modes of failures in R.C beams with neat sketches.
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A simply supported beam 300 mm × 600 mm (effective) is reinforced with 5 bars of 25 mm diameter. It carries a uniformly distributed load of 80 kN/m (including its own weight) over an effective span is 6 m. out of 5 main bars can be bent up safely near the supports. Design the shear reinforcement for the beam. Use M20 grade of concrete and Fe415 steel.
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Explain IS456 method of design of reinforced concrete members subjected to torsional moment.
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A reinforced concrete beam of rectangular section with a width of 350 mm and overall depth of 800 mm is subjected to a factored bending moment of 215 kNm, ultimate torsional moment of 105 kNm an ultimate shear force of 150 kN. Using M-25 grade concrete and Fe 500 HYSD bars and side, top and bottom covers of 50 mm, design suitable reinforcement in the section.
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Explain the modes of failure of columns.
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Design a column of unsupported length 3m to carry an axial load of 2000 kN and a BM of 150 kNm at service conditions. Design the column as a short column. The column is subjected to severe exposure condition and grade of steel is Fe500. Provide equal reinforcement on all the faces. Use M30 concrete. Sketch reinforcement details.
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Derive the expression for the ultimate load for axially loaded short column.
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A braced reinforced concrete column of circular cross-section of 500mm diameter is to support a factored axial load of 2250 kN along with a factored moment of 160 kNm. The unsupported length of the column is 6.3m effective length of 5.5m. Design the column when it is to be provided with: Lateral ties and Spiral reinforcement. The M25 grade of concrete and HYSD steel bars of grade Fe500.
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What are the different types of foundations? Explain with figures.
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Design a rectangular isolated sloped footing for a column of size 250 mm × 750 mm carrying an axial load of 2600 kN. The S.B.C. of the soil is 300 kN/m2. Use M 25 grade concrete and Fe 500 grade steel.
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Sketch the reinforcement detailing of one way and two way slabs.
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Design a one−way slab, with a clear span of 4.0 m, simply supported on 230 mm thick masonry walls, and subjected to a live load of 4 kN/m2 and a surface finish of 1 kN/m2. Assume Fe 500 steel. Assume the beam is subjected to moderate exposure conditions.
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