JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III Year REINFORCED CONCRETE STRUCTURES DESIGN AND DRAWING Previous Question Paper (September - 2021 - R13)
Answer any five questions. All questions carry equal marks.
Main
With neat sketches represent Balanced section, over reinforced section and under reinforced section as per the IS code provisions.
Answer
The behavior of an RCC section under flexure is classified based on the strain distribution across the depth. In a Balanced section, the concrete reaches its ultimate strain (0.0035) simultaneously with the steel reaching its yield strain (0.002 + 0.87fy/Es). In an Under-reinforced section, the steel reaches the yield strain before the concrete reaches the ultimate strain, leading to a ductile failure. In an Over-reinforced section, the concrete reaches its ultimate strain before the steel yields, resulting in a brittle, sudden failure. IS 456:2000 restricts the design to under-reinforced sections to ensure safety.
Sketches should represent the stress-strain profiles: for a balanced section, the neutral axis depth (xu) equals the limiting depth (xu,max); for under-reinforced, xu < xu,max; and for over-reinforced, xu > xu,max.
Design the shear reinforcement for a RCC beam supported on wall of 250mm thick spaced at a clear span of 6.75m. The beam carries a super imposed UDL of 25kN/m. The size of the beam is restricted to 250 × 500mm. Use M20 grade concrete and Fe500 grade steel. The beam is exposed to mild environmental condition. Sketch the details.
Answer
1. Design Parameters: Clear span = 6.75m. Effective span (L) = min(6.75 + 0.25, 6.75 + 0.5) = 7.0m. Total load (w) = 25 + (0.25 × 0.5 × 25) = 28.125 kN/m. Factored load (wu) = 1.5 × 28.125 = 42.19 kN/m.
2. Shear Force Calculation: Maximum factored shear force (Vu) = (wu × L) / 2 = (42.19 × 7.0) / 2 = 147.67 kN.
3. Shear Stress: Effective depth (d) = 500 - 25 (cover) - 10 (bar dia) = 465mm. Nominal shear stress (τv) = Vu / (b × d) = 147670 / (250 × 465) = 1.27 N/mm². Design shear strength of concrete (τc) depends on percentage of steel (Pt). Assuming 1% steel, τc ≈ 0.67 N/mm². Since τv > τc, shear reinforcement is required.
4. Shear Reinforcement: Shear to be resisted by stirrups (Vus) = Vu - (τc × b × d) = 147670 - (0.67 × 250 × 465) = 69,582.5 N. Using 8mm diameter 2-legged stirrups (Asv = 100.5 mm²), spacing (Sv) = (0.87 × fy × Asv × d) / Vus = (0.87 × 500 × 100.5 × 465) / 69582.5 = 292mm. Provide 8mm 2-legged stirrups at 250mm c/c spacing throughout the beam.
5. Detailing: Provide 2 bars of 12mm diameter as hanger bars and 3 bars of 20mm diameter at the bottom, with stirrups spaced at 250mm c/c.
Simply supported beam of C/S 350×700mm is provided with (5#25 + 5#20) on the tension side and 4#25 on the compression side. Calculate moment of resistance. Grade of concrete is M25 with severe exposure. Grade of reinforcement is Fe550.
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Explain about limit state design for serviceability.
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Design a slab for an hall of internal dimensions 12 × 4m carrying a live load of 4kN/m². The slab is supported on the wall of thickness 230 mm. Adopt mild exposures with M20 grade concrete and Fe415 steel. Sketch the reinforcement details.
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Design the slab for a hall 7.5×27m. The slab is supported on RCC beams of 300mm wide and spaced at 3m c/c. The LL on the slab may be taken as 2.75kN/m². Account for floor finishes and light partitions. Use M30 concrete and Fe 500 steel. Assume mild condition and sketch the details.
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Design a short reinforced rectangular column of size 200mm × 400mm, to carry an ultimate axial load of 800kN and ultimate moment of 120kNm. With respect to major axis. Design the ties and sketch the reinforcement details. Use M20 grade of concrete and Fe415 grade of steel for mild exposure condition.
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Design a solid footing to take up dead load (excluding self weight of footing) of 1000kN and an imposed load of 400kN from a square column 400×400mm (with 16mm bars.) Assuming fy=415N/mm² and fck=20N/mm² and safe bearing capacity to be 200kN/m². Design the footing and sketch the reinforcement details?
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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 and Fe415 steel. Also sketch the reinforcement details.
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An engineer in charge has to inspect the safety of the beam with respect to Xu, Cu, Tu, lever arm, Mu. A rectangular simply supported beam has a cross section of 230×550mm overall consisting of 4#25 as tensile steel and 3#20 as compression steel. Also, determine safe UDL if it is provided as a simply supported beam with effective span as 7m. Grade of steel and concrete to be used are Fe500 and M30 respectively with moderate exposure.
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