JNTU
JNTU B.Tech III REINFORCED CONCRETE STRUCTURES DESIGN AND DRAWING Previous Question Paper (October - 2020 - R13)
Answer any five questions
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Main
Design the reinforcement in a rectangular beam of section 300 mm × 450 mm subjected to an ultimate twisting moment of 40 kN-m combined with an ultimate bending moment of 120 kN-m and ultimate shear force of 85kN. Assume M30 concrete and Fe415 steel.
Answer
To design the reinforcement for a beam subjected to combined bending, shear, and torsion, we follow the limit state method as per IS 456:2000.
1. Design Parameters:
Width (b) = 300 mm, Effective depth (d) = 400 mm (assumed cover 50 mm), fck = 30 N/mm², fy = 415 N/mm², Mu = 120 kNm, Vu = 85 kN, Tu = 40 kNm.
2. Equivalent Bending Moment (Me):
The equivalent bending moment is calculated as:
3. Longitudinal Reinforcement:
Calculate Ast for Me using the formula:
4. Equivalent Shear Force (Ve):
The equivalent shear force is calculated as:
5. Transverse Reinforcement:
Calculate the shear stress τve = Ve / (b * d) = 298.33 * 1000 / (300 * 400) = 2.48 N/mm². Since τve < τc,max (3.5 N/mm² for M30), the section is safe. Provide two-legged stirrups to resist torsion and shear. The spacing (sv) is determined by:
A reinforced concrete slab of 150mm thick is supported by reinforced concrete beams of width 300 mm and spaced at 3.3m c/c. The beams have a clear span of 6.6m and supported on 300mm wide supports. The beams are cast monolithic with the slab. The live load acting on the floor is 4 kN/m2 and the floor finish is 0.65 kN/m2. Design the T-beam section using M20 grade concrete and Fe 415 steel.
Answer
To design the T-beam, we first determine the effective span and loading. The effective span (L) is the lesser of clear span + effective depth (6.6 + 0.5 = 7.1m) or center-to-center of supports (6.6 + 0.3 = 6.9m). Adopting L = 6.9m. Assuming an overall beam depth of 600mm, the effective depth (d) is 550mm.
1. Effective Flange Width (bf): Using IS 456:2000, bf = (Lo / 6) + bw + 6Df. With Lo = 0.7 × 6.9 = 4.83m, bf = (4830 / 6) + 300 + 6(150) = 805 + 300 + 900 = 2005mm. Also, bf ≤ spacing = 3300mm. Thus, bf = 2005mm.
2. Loads: Slab self-weight = 0.15 × 25 = 3.75 kN/m2. Total dead load = 3.75 + 0.65 = 4.4 kN/m2. Total load on beam = (4.4 + 4.0) × 3.3 = 27.72 kN/m. Adding beam rib self-weight (0.3 × 0.45 × 25 = 3.375 kN/m), total factored load (wu) = 1.5 × (27.72 + 3.375) = 46.64 kN/m.
3. Bending Moment: Mu = (wu × L2) / 8 = (46.64 × 6.92) / 8 = 277.5 kNm.
4. Reinforcement: Check if neutral axis (xu) lies in the flange. Mu,lim for flange = 0.36 × fck × bf × Df × (d - 0.42 × Df) = 0.36 × 20 × 2005 × 150 × (550 - 0.42 × 150) = 100.9 × 106 Nmm = 100.9 kNm. Since Mu > Mu,lim, the neutral axis lies in the web. Calculating required steel area (Ast) using the T-beam formula: Mu = 0.87 × fy × Ast × (d - 0.42 × xu), assuming xu ≈ 1.5 × Df, Ast ≈ 1650 mm2. Provide 4 bars of 25mm diameter at the bottom.
Design the reinforcement of a cross section of a simply supported beam of span 5.2m is subjected to uniformly distributed live load of 50 kN/m. The size of the section is limited to 230mm × 400mm. Adopt using M20 grade concrete and Fe 415 steel. Sketch the reinforcement details.
Answer
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What is a bond stress? Explain what will happen if bond stress exceeds the allowable limit? What are the steps taken to keep the bond stress within allowable limits? If the beam is simply supported at what section is the bond stress maximum?
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Explain the procedure for serviceability requirement of limit state of deflection for the calculation of short term deflection.
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A simply supported two way slab with effective spans of 3.5 m and 4.5 m is reinforced with 10 mm bars @ 250 mm c/c along short direction and 10 mm bars @ 350 mm c/c along long direction. The total depth is 100 mm with floor finish of 1 kN/m2. Determine the maximum imposed load the slab can support.
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Design the reinforced concrete slab of clear dimensions 4.5 m × 6 m with two adjacent edges continuous. The slab is subjected to a live load of 3.5 kN/m2 and the floor finish is 1 kN/m2. Assume width of the supports is 300mm. Adopt using M20 grade concrete and Fe 415 steel. Assume mild exposure condition.
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A corner column 275 mm × 600 mm located in a multistory of a system of a braced frame is subjected to factored loads Pu=2000 kN, Mux=150 kN-m and Muy= 75 kN-m. The unsupported length of a column is 3.0m. Design the reinforcement in the column assuming M30 grade of concrete and Fe 415 HYSD steel.
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The details of a doglegged staircase of a multi-storied building are given below:
Number of steps in the flight = 10
Tread, T = 300 mm, Rise, R = 150 mm
Width of landing slab on either side = 1.2m
Use M-20 grade concrete and Fe-415 HYSD bars.
Design and detail any one of the flights of the staircase.
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