JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III STRUCTURAL ENGINEERING - I (RCC) Previous Question Paper (September - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
A hall has C/C dimensions of 21000mm × 8000mm with beam spacing of 3500mm c/c. Overall dimensions of web is 300mm × 475mm with tension steel 4#20 for end L-section. Thickness of slab = 125 mm. Calculate xU, CU, TU, lever arm, MR at collapse and the safe superimposed load on the slab excluding the self weight of the slab. The beam is exposed to moderate condition. Grade of concrete and steel is M25 and Fe550 respectively.
Answer
Given: Effective span L = 8000 mm, Beam spacing = 3500 mm, Web width bw = 300 mm, Overall depth D = 475 mm, Slab thickness Df = 125 mm, Tension steel Ast = 4 × (π/4) × 202 = 1256.6 mm2, Concrete M25 (fck = 25 N/mm2), Steel Fe550 (fy = 550 N/mm2). Assuming effective cover = 50 mm, effective depth d = 425 mm.
1. Neutral Axis (xu): Assuming xu ≤ Df, 0.36 fck bf xu = 0.87 fy Ast. With effective flange width bf = L0/6 + bw + 6Df = 8000/6 + 300 + 6(125) = 2383 mm. xu = (0.87 × 550 × 1256.6) / (0.36 × 25 × 2383) = 63.5 mm. Since 63.5 mm ≤ 125 mm, the assumption holds.
2. Forces and Lever Arm: Cu = 0.36 fck bf xu = 0.36 × 25 × 2383 × 63.5 = 1360.7 kN. Tu = 0.87 fy Ast = 0.87 × 550 × 1256.6 = 601.1 kN. Lever arm z = d - 0.416 xu = 425 - 0.416(63.5) = 398.6 mm.
3. Moment of Resistance (Mu): Mu = Tu × z = 601.1 × 103 × 398.6 = 239.6 kNm.
4. Safe Superimposed Load: For a simply supported beam, Mu = (wu L2) / 8. wu = (8 × 239.6 × 106) / 80002 = 29.95 kN/m. Factored self-weight of beam = 1.5 × (0.3 × 0.475 × 25) = 5.34 kN/m. Factored self-weight of slab = 1.5 × (0.125 × 25 × 3.5) = 16.41 kN/m. Remaining load capacity wsuper = 29.95 - 5.34 - 16.41 = 8.2 kN/m. Safe superimposed load = 8.2 / 1.5 = 5.47 kN/m2.
Illustrate characteristic strength of a material as applied to limit state design with a normal distribution curve. Outline the reasons for adopting partial safety factors for materials and loads in limit state design.
Answer
In limit state design, the characteristic strength of a material is defined as that value of strength below which not more than 5% of the test results are expected to fall. It is statistically determined based on a normal distribution curve of material test results.
The distribution curve shows the frequency of test results plotted against strength. The characteristic strength (fk) is calculated as:
Where fm is the mean strength and σ is the standard deviation. By setting the threshold at 1.65 standard deviations below the mean, the design ensures a 95% confidence level in the material's structural capacity.
The adoption of partial safety factors is necessary to account for the uncertainties inherent in structural engineering, which are categorized into two main groups:
- Material uncertainties: These account for the difference between the strength of materials in the laboratory and those in the actual structure, variations in material quality, and the possibility of workmanship defects during construction.
- Load uncertainties: These account for the possibility of actual loads exceeding the design loads, inaccuracies in the assessment of load effects, and unforeseen variations in the distribution of loads within the structure.
By applying separate partial safety factors to materials (γm) and loads (γf), the limit state method provides a more rational and probabilistic approach to safety compared to earlier deterministic methods, ensuring that the structure remains safe throughout its intended service life.
At the limit state of collapse, an RC beam is subjected to a flexural moment of 150 kN-m, shear force of 20 kN and torque of 8 kNm. The beam is 230 mm × 450 mm. Determine equivalent shear force and equivalent flexural moment for designing longitudinal tension steel.
Answer
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Determine the minimum effective depth and the corresponding area of tension reinforcement for the rectangular beam having width 230mm to resist ultimate moment of resistance 200 kN-m using M25 grade concrete and Fe500 HYSD bars.
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Design a slab over a room of internal dimensions 4 m × 5 m supported on 230 mm thick wall. All the edges of the slab are simply supported. The corners are free to lift. Use live load 2 kN/m2, floor finish 1.5 kN/m2, and light partition 1.5 kN/m2. Adopt mild exposure with M20 grade concrete and Fe415 steel.
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Design a dog legged staircase for an office building, given the following data: Height between floor = 3.2m; Riser = 160mm, Tread = 270mm; Width of flight = 1.25m; Live load = 5kN/m2; Finishes load = 0.6kN/m2. Assume the stairs to be supported on 230mm thick masonry walls at the outer edges of the landing, parallel to the risers. Adopt mild exposure with M20 grade concrete & Fe415 steel. Also sketch the reinforcement details.
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A rectangular column 300 mm wide and 600 mm deep is subjected to an axial load of 1000 kN and moments along the major and minor and major axis are 130 kNm and 80 kNm respectively. Assume M30 concrete and Fe 500 steel. Assume severe condition and sketch the details.
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Design a reinforced concrete footing for a rectangular column of section 300 mm by 500 mm supporting an axial factored load of 1500 kN. The SBC of the soil at site is 185kN/m2. Adopt M20 grade concrete and Fe415 grade steel.
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