JNTU
JNTU B.Tech III REINFORCED CONCRETE STRUCTURES DESIGN AND DRAWING Previous Question Paper (February - 2022 - R13)
Answer any five questions
All questions carry equal marks
Use of only IS 456:2000 and design charts for columns is allowedMain
Explain with figure balanced, under reinforced, over reinforced sections.
Answer
In limit state design, the behavior of a reinforced concrete section depends on the relative strain between concrete and steel at failure. A balanced section is one where the concrete reaches its ultimate compressive strain (0.0035) at the same time the tensile steel reaches its yield strain. An under-reinforced section has a lower percentage of steel than the balanced section, causing the steel to yield first, leading to a ductile failure with visible deflection and cracking. An over-reinforced section has more steel than the balanced section, causing the concrete to crush before the steel yields, resulting in a sudden, brittle failure.
The stress-strain diagrams show the neutral axis depth (xu) relative to the limiting depth (xu,max):
- Balanced: xu = xu,max
- Under-reinforced: xu < xu,max
- Over-reinforced: xu > xu,max
Determine the ultimate moment of resistance of a doubly reinforced beam of rectangular section having a width of 300 mm and reinforced with 5 bars of 25 mm diameter at an effective depth of 600 mm. The compression steel is made up of 2 bars of 25 mm diameter at an effective cover of 60 mm. Adopt M-20 grade concrete and Fe 500 HYSD bars.
Answer
Given: b = 300 mm, d = 600 mm, d' = 60 mm, M20 (fck = 20 N/mm²), Fe500 (fy = 500 N/mm²). Area of tension steel Ast = 5 × (π/4) × 25² = 2454.37 mm². Area of compression steel Asc = 2 × (π/4) × 25² = 981.75 mm².
1. Calculate limiting neutral axis depth: xu,max = 0.46 × 600 = 276 mm.
2. Calculate actual neutral axis depth (xu) using equilibrium of forces (C = T): 0.36 fck b xu + fsc Asc = 0.87 fy Ast. Assuming fsc ≈ fy = 434.78 N/mm² (for d'/d = 0.1), 0.36 × 20 × 300 × xu + 434.78 × 981.75 = 0.87 × 500 × 2454.37. Solving gives xu = 243.5 mm. Since xu < xu,max, the section is under-reinforced.
3. Calculate Ultimate Moment of Resistance (Mu): Mu = Mu1 + Mu2. Mu1 = 0.36 fck b xu (d - 0.42 xu) = 0.36 × 20 × 300 × 243.5 × (600 - 0.42 × 243.5) = 296.7 kNm. Mu2 = fsc Asc (d - d') = 434.78 × 981.75 × (600 - 60) = 230.4 kNm. Total Mu = 296.7 + 230.4 = 527.1 kNm.
What do you mean by neutral axis and leaver arm? Explain briefly with neat sketches.
Answer
Sign in to view the complete answer paper.
Find the moment of resistance of a T- beam having a web width of 240 mm, effective depth of 400 mm, flange width of 740 mm and flange thickness equal to 100 mm. The beam is reinforced with 5-16 mm diameter, Fe 500 bars. Use M 20 concrete.
Answer
Sign in to view the complete answer paper.
Explain the critical sections for design shear as per IS 456 with requisite sketches.
Answer
Sign in to view the complete answer paper.
A simply supported rectangular beam 300 × 500 mm of effective span 7.0 m, is reinforced with 5 bars of 20 mm diameter on tension side and 4 bars of 16 mm diameter on compression side. Check the deflection criteria. Use Fe415 steel.
Answer
Sign in to view the complete answer paper.
What is nominal shear reinforcement? Give the reasons for providing minimum shear reinforcement.
Answer
Sign in to view the complete answer paper.
A simply supported beam, 300 mm wide and 500 mm effective depth carries a uniformly distributed load of 50 kN/m, including its own weight over an effective span of 6 m. Design shear reinforcement in the form of vertical stirrups. Assume that the beam contains 0.75% reinforcement throughout the length. The concrete is of M 20 grade and steel for stirrups is of Fe 250 grade. Take width of supports as 40 mm.
Answer
Sign in to view the complete answer paper.
Explain the difference in the behavior of one-way and two-way slabs.
Answer
Sign in to view the complete answer paper.
Design a two way slab for an office floor size 3.5 m × 4.5 m discontinuous and simply supported edges on all the sides with the corners prevented from lifting and supporting a service live load of 4.4 kN/m2. Adopt M 20 grade and Fe 415 HYSD bars.
Answer
Sign in to view the complete answer paper.
Give neat sketches for the reinforcement details for one way simply supported and two way continuous slabs.
Answer
Sign in to view the complete answer paper.
Design continuous RC slab for a class room 7 m wide and 14 m long. The slab is supported on RCC beams spaced at 3.5 m intervals. The width of each beam is 230 mm. Design the slab for LL of 3 kN/m2. Assume weight of roof finishing as 1 kN/m2 use M20 grade concrete and Fe 415 grade steel.
Answer
Sign in to view the complete answer paper.
Outline the procedure for design of axially loaded reinforced concrete column.
Answer
Sign in to view the complete answer paper.
A braced reinforced concrete column of circular cross-section of 500 mm 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 6.3 m effective length of 5.5 m. 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 Fe 500.
Answer
Sign in to view the complete answer paper.
What are the rules for providing dowels connecting footing and column?
Answer
Sign in to view the complete answer paper.
A longitudinal type of staircase spans a distance of 3.75 m c/c beams, R=175 mm, G=250 mm, T=270 mm. The treads have 15 mm granolithic finish and consists of 50 steps. Assume M30 and Fe 500. Design a staircase for a LL of 5 kN/m2. Assume breadth of staircase of 1.5 m.
Answer
Sign in to view the complete answer paper.