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Tuesday, November 4, 2025

1010235110 – Design Of Rcc Structures (Limit State Method)

 

1010235110 – Design Of Rcc Structures (Limit State Method)

 ATTEND 20 MARKS MCQ TEST MCQ TEST LINK

 

1. Curtailment of bars is permitted when:
a) Shear is minimum
b) Moment is reduced
c) Sufficient development length is available
 d) All of the above


2. In singly reinforced beam, steel is provided in the:
a) Compression zone only
 b) Tension zone only
c) Both zones
d) Web only


3. Effective span of a cantilever beam is:
a) Clear span
 b) Clear span + d
c) Clear span + 0.5d

d) Clear span + 2d


4. Which type of section occurs when xu < xu,lim in a beam?
 a) Under-reinforced
b) Balanced
c) Over-reinforced
d) Brittle


5. What is the main purpose of a landing in a staircase?
a) To reduce the number of steps
 b) To provide a resting place and change direction
c) To make the staircase look wider
d) To reduce the cost of construction


6. For a simply supported beam of length (l) under a UDL (w), the maximum shear force at the support is:
a) wl
 b) wl/2
c) wl/4
d) wl/8


7. The formula for the shear force resisted by vertical stirrups (Vus) is:
 a) 0.87 fy Asv d / Sv
b) 0.87 fy Asv Sv / d
c) τc b d
d) τv b d


8. The minimum width of a staircase for residential buildings should be:
a) 600 mm
b) 750 mm
c) 900 mm
d) 1200 mm


9. As per IS 456:2000, for continuous beams or slabs, if support width is wider than 1/12 of clear span (or) 600 mm whichever less, for end span with one end fixed and the other continuous or intermediate spans, the effective span is:
 b) Centre-to-centre distance
a) Average of support widths
c) Clear span between supports
d) Not defined


10. As per IS 456:2000, for continuous beams or slabs, if support width is wider than 1/12 of clear span (or) 600 mm whichever less, for end span with one end free and the other continuous, the effective span is:
a) Clear span + half the width of discontinuous support
b) Clear span + half the effective depth
 c) Greater of (A) & (B)
d) Lesser of (A) & (B)**


11. As per IS 456:2000, for continuous beams or slabs, if support width is less than 1/12 of clear span, the effective span is:
a) Clear span + Effective Depth
 b) Centre-to-centre of supports
c) Greater of (A) & (B)
d) Lesser of (A) & (B)


12. For imposed load (not fixed), the coefficient for span moment near middle of end span is:
 a) +1/16
b) -1/16
c) +1/10
d) -1/10


13. Effective span of a simply supported one-way slab is:
a) Clear span + effective depth
b) Clear span + width of support
 c) Lesser of (clear span + effective depth) or center-to-center of supports
d) Clear span only


14. The width of the middle strip in a two-way slab is taken as:
a) The full width of the slab
 b) Half of the slab width
c) 0.125 times the slab length
d) 0.75 times of slab length


15. In a two-way slab, torsional reinforcement is provided at:
a) Edges
b) Center
 c) Corners
d) Mid-span


16. The width of the edge strip in a two-way slab is taken as:
a) The full width of the slab
 b) One-half of the slab width
c) 0.125 times the slab length
d) 0.75 times slab length


17. Long columns fail due to:
 a) Buckling
b) Crushing
c) Bending
d) Torsion


18. The footing is considered safe when:
a) Shear > moment
 b) Soil pressure ≤ SBC
c) Load > area
d) Reinforcement is more


19. The minimum diameter of longitudinal bars used in columns shall be:
a) 8 mm
b) 10 mm
 c) 12 mm
d) 16 mm


20. Which footing is preferred for two columns close together?
a) Strip footing
 b) Combined footing
c) Sloped footing
d) Mat footing

 

1020235110 – ELEMENTS OF MACHINE DESIGN STUDY MATERIALS

 

20 MARKS MCQ TEST LINK MCQ TEST LINK

 

 

1. The preferred number series used in engineering design follows which
a) Arithmetic
 b) Geometric
c) Harmonic
d) Random

👉 Explanation: Engineering preferred number series (like R5, R10) are based on geometric progression for standardization of sizes.


2. The maximum distortion energy theory is also known as:
 a) Von Mises theory
b) Rankine theory
c) Tresca theory
d) Mohr's theory

👉 Explanation: Von Mises theory = Maximum distortion energy theory → used for ductile materials.


3. The factor of safety for brittle materials is usually ______ than for ductile materials.
 a) Higher
b) Lower
c) Equal
d) Unpredictable

👉 Explanation: Brittle materials fail suddenly → so higher safety factor is used.


4. The maximum distortion energy theory is best suited for:
a) Brittle materials
 b) Ductile materials
c) Composites
d) Ceramics

👉 Explanation: Von Mises (Distortion energy) applies to ductile materials like steel.


5. Flange couplings are classified as:
a) Flexible couplings
 b) Rigid couplings
c) Fluid couplings
d) Magnetic couplings

👉 Explanation: Flange coupling rigidly connects shafts — no flexibility for misalignment.


6. In a pin-type flexible coupling, flexibility is achieved by:
a) Increase speed
b) Reduce power loss
 c) Compensate misalignment
d) Lower cost

👉 Explanation: Pins with rubber bushes allow slight misalignment between shafts.


7. Which coupling type would be most suitable for connecting shafts with angular misalignment?
a) Rigid flange coupling
b) Marine coupling
 c) Universal coupling
d) Protected flange coupling

👉 Explanation: Universal joint allows angular motion and misalignment between shafts.


8. The main advantage of marine couplings is:
 a) Corrosion resistance
b) High speed capacity
c) Low maintenance
d) Light weight

👉 Explanation: Marine couplings are designed to resist sea water corrosion.


9. What does the designation 'B75' for a V-belt typically indicate?
a) Belt type 'B' and 75 cm inside length
 b) Belt type 'B' and 75 inches pitch length
c) Belt type 'B' and 75 mm width tension
d) Belt type 'B' and a maximum of 75 N

👉 Explanation: In V-belts, number = pitch length in inches.


10. When designing a V-belt drive using manufacturer’s catalogues, the first step is to determine the:
 a) Design power (Service factor × Power)
b) Center distance
c) Belt length
d) Number of belts

👉 Explanation: Design starts with design power, considering service conditions.


11. The 'angle of contact' for the smaller pulley in an open belt drive is:
a) Always 180°
 c) Always <180°
b) Always >180°
d) Independent of pulley diameters

👉 Explanation: In open belt drive, belt leaves pulley tangentially → less than 180°.


12. A designer needs a belt for 15 kW. Manufacturer’s rating per belt = 4.2 kW. The minimum number of belts is:
a) 3
 b) 5
c) 4
d) 7

👉 Explanation: 15 ÷ 4.2 = 3.57 → take next higher integer = 5 belts (for safety).


13. A bearing designed for axial forces is called:
 a) Thrust bearing
b) Radial bearing
c) Journal bearing
d) Roller bearing

👉 Explanation: Thrust bearing → supports axial (end) loads.


14. In ball bearing code “6205”, bore size is indicated by:
a) First digit
b) Middle digit
c) Prefix letters
 d) Last two digits

👉 Explanation: For bearing codes, bore = last two digits × 5 mm → 05 → 25 mm bore.


15. Journal bearing, D = 50 mm, L = 50 mm, W = 6 kN. Find bearing pressure.
 a) 2.55 MPa
b) 2.30 MPa
c) 2.0 MPa
d) 2.40 MPa

👉 Explanation:
Bearing pressure p=WL×D=60000.05×0.05=2.55 MPap = \frac{W}{L \times D} = \frac{6000}{0.05 \times 0.05} = 2.55 \text{ MPa}p=L×DW​=0.05×0.056000​=2.55 MPa


16. The coefficient of friction in properly lubricated journal bearings is:
 a) 0.001–0.01
b) 0.1–0.3
c) 0.5–0.7
d) 0.8–1.0

👉 Explanation: With lubrication, friction is very low (around 0.001–0.01).


17. The primary role of CAD in design is to:
a) Replace engineers
 b) Assist in design process
c) Manufacture products
d) Market products

👉 Explanation: CAD = Computer-Aided Design, helps engineers design accurately.


18. Value Engineering focuses on:
a) Increasing cost
b) Maintaining function
c) Reducing quality
d) Extending timeline

👉 Explanation: Value engineering = reduce cost without affecting function/quality.


19. The main advantage of solid modeling over wireframe is:
a) Lower cost
 b) Complete representation
c) Faster computation
d) Simpler algorithms

👉 Explanation: Solid models show volume, mass & surface, unlike wireframe.


20. Which modeling approach uses primitives and Boolean operations?
a) B-Rep
 b) CSG
c) Wireframe
d) Surface

👉 Explanation: Constructive Solid Geometry (CSG) uses Boolean ops (union, subtract, intersect) with primitives like cube, cylinder.


Final Answer Key Summary:
1-b | 2-a | 3-a | 4-b | 5-b | 6-c | 7-c | 8-a | 9-b | 10-a | 11-c | 12-b | 13-a | 14-d | 15-a | 16-a | 17-b | 18-b | 19-b | 20-b

 

1010235110 – Design Of Rcc Structures (Limit State Method)

  1010235110 – Design Of Rcc Structures (Limit State Method)  ATTEND 20 MARKS MCQ TEST MCQ TEST LINK   1. Curtailment of bars is permi...