300+ REAL TIME Strength of Materials Objective Questions & Answers 2023

250+ TOP MCQs on Mass Moment of Inertia and Answers

This set of Strength of Materials Multiple Choice Questions on “Mass Moment of Inertia”.

1. What is the product of the mass and the square of the distance of the center of gravity of the mass from an axis?
a) Moment of inertia
b) Mass moment of inertia
c) Center of gravity
d) Product of inertia
Answer: b
Clarification: The product of the mass and the square of the distance of the center of gravity of the mass from an axis is known as the mass moment of inertia about that axis.

2. What is the unit of mass moment of inertia?
a) m⁴
b) m⁶
c) N
d) m²
Answer: b
Clarification: The mass moment of inertia is the product of moment of inertia and area. So L⁴ x L² = L⁶. so its unit will be m⁶.

3. What is mass moment of inertia of circular plate?
a) Md²/3
b) Md²/12
c) Mr²/4
d) Mr²/3
Answer: c
Clarification: The mass moment of inertia of circular plate is Mr²/4.

4. What is the mass MOI of a rectangular plate about x-axis passing through the C.G of the plate if the y-axis is parallel to d and perpendicular to b?
a) Mb²/12
b) Md²/12
c) Md²/6
d) Mb²/6
Answer: b
Clarification: As the mass MOI is to be find along the x-axis, it would be Md²/12.

5. What is the mass MOI of right circular cone of radius R and height H about its axis?
a) 4MR²/10
b) MR²/10
c) 3MR²/10
d) MR²/12
Answer: c
Clarification: The mass MOI of right circular cone of radius R and height H about its axis is 3MR²/10.

6. What is the mass MOI of a hollow circular cylinder if R is the outer diameter and r is the inner diameter?
a) M(R + r)/4
b) M(R – r )/4
c) M(R+ r)/2
d) M(R – r)/2
Answer: a
Clarification: The mass MOI of a hollow circular cylinder is M(R + r)/4 where R is the outer diameter and r is the inner diameter.

7. What is the mass MOI of a rectangular plate about y-axis passing through the C.G of the plate if the y-axis is parallel to d and perpendicular to b?
a) Mb²/12
b) Md²/12
c) Md²/6
d) Mb²/6
Answer: a
Clarification: As the mass MOI is to be find along the y-axis, it would be Mb²/12.

8. The product of inertia at the principal axes is _____________
a) Minimum
b) Unit
c) Zero
d) Maximum
Answer: c
Clarification: The moment of inertia about x-axis and about y-axis, on the axis they are zero. So the product of inertia will be zero in the principal axis.

9. What is the unit of product of inertia?
a) mm⁴
b) mm²
c) mm
d) mm³
Answer: a
Clarification: The unit of product of inertia is same as that of moment of inertia I.e. mm⁴.

10. What is the product of inertia of the given following section?

a) 50mm⁴
b) 625mm⁴
c) 125mm⁴
d) 250mm⁴
Answer: b
Clarification: The product of inertia = area x points of C.G
= ( 10×5) x 5 x 2.5 = 625mm⁴.

11. What is the product of inertia of a circle of diameter 10mm?
a) 1862mm⁴
b) 1945mm⁴
c) 1963mm⁴
d) 2014mm⁴
Answer: c
Clarification: The product of inertia = area x C.G
= πx 10×10 / 4 x 5×5 = 1963mm².

250+ TOP MCQs on Maximum Shear Stress – 2 and Answers

This set of Strength of Materials Interview Questions and Answers for Experienced people on “Maximum Shear Stress – 2”.

1. Calculate the maximum shear force for square beam of side is 320 mm. If the shear force is 94kN.
a) 1.37N/mm²
b) 2.36N/mm²
c) 5.21N/mm²
d) 4.32N/mm²
Answer: a
Clarification: Maximum shear force is 3/2 × F/a×a ( a = side of square)
= 3/2 × 94×10³/320×320
= 1.3769 N/mm².

2. A simply supported beam of span 8 metres carries a udl of 16 kN/m at a point out of 60 kN acting at it’s centre. Calculate the maximum shear force.
a) 87kN
b) 45kN
c) 78kN
d) 94kN
Answer: d
Clarification: Maximum shear force is w×l/2
= 60+16×8 / 2
= 94 kN.

3. The ratio of creep strain to elastic strain is known as ___________
a) Creep factor
b) Creep postulate
c) Creep coefficient
d) Creep variable
Answer: c
Clarification: Creep is defined as plastic deformation under a constant load or stress the creep Coefficient which is defined as the ratio of ultimate creep strain to the elastic strain at various ages of loadings.

4. Poisson’s ratio for high strength concrete is __________
a) 0.049
b) 0.095
c) 0.1
d) 0.1111
Answer: c
Clarification: Poisons ratio varies between 0.1 for high strength concrete and 0.2 for weak concrete. Usually it is taken as 0.15 for strength design and 0.2 for serviceability conditions.

5. Partial safety factor for concrete is taken as ____________
a) 1.3
b) 1.2
c) 1.5
d) 1.6
Answer: c
Clarification: A higher value of partial safety factor for concrete 1.5 has been adopted because there are greater chances of variation of the strength of concrete due to improper compaction, inadequate curing, improper batching and mixing.

6. The design compressive strength of concrete is ___________ times of characteristic compressive strength of concrete.
a) 0.313
b) 0.253
c) 0.466
d) 0.411
Answer: c
Clarification: The compressive strength of concrete in the structure is assumed to be 0.67 times the characteristic strength of concrete. The partial safety factor equal to 1.5 is applied to the strength of concrete in addition to it therefore the design compressive strength of concrete is 0.67 fck / 1.5 equal to 0.446 fck. [fck = characteristic compressive strength].

7. In cantilever beams, the steel bars are placed at ___________
a) Bottom of the beam
b) Top of the Beam
c) Midspan of the Beam
d) Near supports
Answer: b
Clarification: In cantilever beams, steel bars are placed near the top of the beam to resist the tensile stresses developed in top layers due to the negative bending moment that is hogging bending moment.

8. Calculate the level arm factor of a section of M20 grade and if Fe 415 Steel. [Take critical neutral axis factor as 0.289].
a) 0.78
b) 0.9
c) 0.58
d) 0.73
Answer: b
Clarification: Lever arm factor (j) = 1-k/3
Where k= 0.289
j = 1-0.289/3
= 0.904~0.9.

9. Working stress method is based on elastic theory assumptions.
a) True
b) False
Answer: a
Clarification: Working stress method is based on elastic theory assuming reinforced concrete as elastic material. The stress strain curve of concrete is assumed as linear from zero at the neutral axis to a maximum value at the extreme fibre. In the working stress method, members are designed for working loads such that the stresses developed are within the allowable stress.

10. Modular ratio method is also known as ______
a) Ultimate stress method
b) Limit state method
c) Working stress method
d) Stress and strain method
Answer: c
Clarification: The stress in steel is linearly related to the stresses in adjoining concrete by constant factor called modular ratio (defined as the ratio of modulus of elasticity of steel to that of concrete) Working stress method is therefore also known as modular ratio method.

11. Find the moment of inertia about centroid axis of a triangular section are having base 100 mm and height 150 mm.

a) 9.21×10⁶mm⁴
b) 9.45×10⁶mm⁴
c) 9.37×10⁶ mm⁴
d) 8.51×10⁶ mm⁴
Answer: c
Clarification: b = 100mm
h = 150 mm
Moment of inertia about centroid Axis = bh³ / 36.
= 100 ×150³/ 36
= 9.37×10⁶mm⁴.

12. The stress corresponding to ______ of strain in the stress-strain curve of mild steel is known as proof stress.
a) 0.2%
b) 0.32%
c) 0.5%
d) 0.6%
Answer: a
Clarification: The stress corresponding to 0.2% of strain in the stress-strain curve of mild steel is known as proof stress. This is also taken as yield stress. The maximum stress is generally taken as yield stress.

13. __________ is the device used for measuring normal stresses on the surface of a stressed object.
a) Nephelometer
b) Straining appurtenances
c) Resistance strain gauge
d) Volt-Hypsometer gauge
Answer: c
Clarification: An electrical resistance strain gauge is a device for measuring normal strains on the surface of a stressed object. The gauges are small (less than half inch) made of wires that are bonded on surface of the object. We can use the transformation equations for plane strain to calculate the strains in various directions.

14. The compressive strength of brittle materials is ______ its tensile strength.
a) Less than
b) Greater than
c) Equal to
d) Depends on material
Answer: b
Clarification: The compressive strength of brittle materials is always greater than its tensile strength. In the same way, the tensile strength of ductile materials is greater than its compressive strength.

15. The breaking stress is _______ the ultimate stress.
a) Less than
b) Greater than
c) Depends on time
d) Equal to
Answer: a
Clarification: The stress corresponding to the ultimate load is known as ultimate stress and the stress corresponding to breaking point is known as breaking stress. In the stress strain curve, the ultimate stress is above the breaking stress. Hence the ultimate stress is greater than breaking stress.

Strength of Materials for Interviews,

250+ TOP MCQs on Analyse Propped Cantilever and Answers

This set of Strength of Materials Multiple Choice Questions on “Analyse Propped Cantilever”.

1. In cantilever beams, the extra support is known as ____________
a) Hinch
b) Prop
c) Cripple
d) Indeterminate end
Answer: b
Clarification: In case of cantilever beam, some support other than existing ones may be provided to reduce the amount of bending moment developed. The additional support is known as prop.

2. Prop reduces ___________ in the beam.
a) Deflection
b) Slope
c) Shear
d) Moment
Answer: a
Clarification: The extra support provided in case of cantilever beam excluding the existing ones is known as prop. It is provided in order to avoid excessive deflection caused due to unequal loading.

3. Which of the following is indeterminate structure?
a) Singly rereinforced beam
b) Propped cantilever beam
c) Over hanging beam
d) Simply supported beam
Answer: b
Clarification: The statically indeterminate structures are not capable of being analysed by using equation of statics. We need some more extra conditions for finding unknowns like €i and €y etc. A propped cantilever beam is an example of indeterminate structures.

4. ____________ is used to produce due to temperature variation in indeterminate structures.
a) Stresses
b) Strains
c) Deflections
d) Moment
Answer: a
Clarification: Statically indeterminate structures need some extra conditions for the further simplification. Normally stresses are produced due to variation in indeterminate beams.

5. In cantilever beams, the maximum deflection occurs at ___________
a) Fixed end
b) Free end
c) Through out
d) Point of loading
Answer: b
Clarification: The maximum deflection in cantilever beam occurs at free end. To resist that excessive deflection, the beam has to be supported by an extra support known as prop.

6. As per IRC, maximum width of lane considered as ____________
a) 2.44 m
b) 2.35 m
c) 3.5 m
d) 3.4 m
Answer: a
Clarification: As per IRC, the maximum width be 2.44 m. For a single lane, the width considered is 3.8 m. The pavement having two or more lanes the weight of 3.5 metre per lane is considered sufficient.

7. ______ is the area of land acquired and reserved for future development.
a) Right of pier
b) Carriage way
c) Right of way
d) Camber
Answer: c
Clarification: It is desirable to acquire more land because of the cost of adjoining land in variable increases after laying the road. The right of way is area of land acquired and reserved for future development.

8. Stability of high rise vehicles will be affected due to ____________
a) Camber
b) Gradient
c) Super elevation
d) Formation Width
Answer: a
Clarification: The rise given to the centre portion of the proposed carriageway with reference to its peripheral edge is called camber. Road users use more in the central portion of road and get worn out. The stability of high rise vehicles will be affected due to heavy camber.

9. The longitudinal rise or fall off road surface along its length is _________
a) Camber
b) Super elevation
c) Gradient
d) Carriage way
Answer: c
Clarification: The gradient is defined as the longitudinal rise or fall off road surface love its length is expressed as ratio 1 vertical: n horizontal or as a percentage.

10. Which of the following gradient is usually used in the construction of roads?
a) Exceptional gradient
b) Limiting gradient
c) Hydraulic gradient
d) Ruling gradient
Answer: d
Clarification: Ruling gradient may be used as isolated over in flat country roads carrying a large volume of slow moving traffic. Gradient up to the ruling gradient for different terrains. It is to be adopted in normal course of design.

11. According to IRC, the height of the object is taken to the height of ___________ mm.
a) 200 mm
b) 100 mm
c) 450 mm
d) 600 mm
Answer: b
Clarification: Sight distance is an important requirement for the safety of travel on highways the height of the object is taken to be at a depth of 100 mm above road.

12. What is the minimum shoulder width provided for village roads?
a) 1.25 m
b) 1.4 m
c) 0.5 m
d) 1 m
Answer: c
Clarification: The minimum shoulder width provided for village roads is 0.5m.

Class of road	Minimum shoulder width(Hilly terrain)
NH & SH ways	1.25 m
MD roads	0.5
Village roads	0.5

13. In case of vertical curves, the ____________ are taken above the road.
a) Gradient
b) Super elevation
c) Earth quantities
d) Summit
Answer: c
Clarification: The line of sight of a driver above the road is taken as 1.2m. The height of the object is taken to be height of 100 mm. Sight distance is an important resource requirement for the safety of travel. The designing layout plays a very vital role.

14. The time required for overtaking ___________ seconds.
a) 9 to 14
b) 8 to 10
c) 11 to 15
d) 14 to 19
Answer: a
Clarification: In the case of vertical curves, the sight distance is an important requirement for the safety of travel. It it is necessary that sight distance of adequate length should be available in different situations to permit driver enough time and distance to control their vehicles so that there are no unwarranted accidents.

15. _____ provide gradual introduction of super elevation.
a) Transition curves
b) Summit curves
c) Joint curves
d) Adjoining curves
Answer: a
Clarification: The transition curves are necessary for a vehicle to have smooth entry of straight section into circular curve. They provide aesthetic experience of the road. They provide a graduate introduction of super elevation.

250+ TOP MCQs on Thermal Stress and Answers

This set of Strength of Materials Multiple Choice Questions on “Thermal Stress”.

1. The length, Young’s modulus and coefficient of thermal expansion of bar P are twice that of bar Q. what will be the ration of stress developed in bar P to that in bar Q if the temperature of both bars is increased by the same amount?
a) 2
b) 8
c) 4
d) 16
Answer: c
Clarification: Temperature Stress = EαδT
Stress in bar P / Stress in bar Q = (E_P / E_Q) x (α_P / α_Q) = 2×2 = 4.

2. A steel bar 600mm long and having 30mm diameter, is turned down to 25mm diameter for one fourth of its length. It is heated at 30 C above room temperature, clamped at both ends and then allowed to cool to room temperature. If the distance between the clamps is unchanged, the maximum stress in the bar ( α = 12.5 x 10-6 per C and E = 200 GN/m²) is
a) 25 MN/m²
b) 40 MN/m²
c) 50 MN/m²
d) 75 MN/m²
Answer: d
Clarification: As temperature stress do not depend upon properties of cross section like length and area. They only depends upon properties of the material.
Therefore, σ=αEδT
= 12.5 x 10-6 x 200 x 10³ x 30
= 75 MN/m².

3. A cube having each side of length p, is constrained in all directions and is heated unigormly so that the temperature is raised to T.C. What will be the stress developed in the cube?
a) δET / γ
b) δTE / (1 – 2γ)
c) δTE / 2 γ
d) δTE / (1 + 2γ)
Answer: b
Clarification: δV/V = P / K = a³ (1 + aT)³ – a³) / a³
Or P / (E / 3(1-2γ)) = 3αT.

4. A steel rod 10mm in diameter and 1m long is heated from 20 to 100 degree celcius, E = 200 GPa and coefficient of thermal expansion is 12 x10^-6 per degree celcius. Calculate the thermal stress developed?
a) 192 MPa(tensile)
b) 212 MPa(tensile)
c) 192MPa(compressive)
d) 212 MPa(compressive)
Answer: c
Clarification: αEδT = (12 x 10^-6) ( 200 x 10³) (100-20) = 192MPa.

5. A cube with a side length of 1m is heated uniformly a degree celcius above the room temperature and all the sides are free to expand. What will be the increase in the volume of the cube? Consider the coefficient of thermal expansion as unity.
a) Zero
b) 1 m³
c) 2 m³
d) 3 m³
Answer: d
Clarification: Coefficient of thermal expansion = 3 x coefficient of volume expansion.

6. The thermal stress is a function of _____________
P. Coefficient of linear expansion
Q. Modulus of elasticity
R. Temperature rise
a) P and Q
b) Q and R
c) Only P
d) Only R
Answer: d
Clarification: Stress in the rod is only due to temperature rise.

7. A steel rod is heated from 25 to 250 degree celcius. Its coefficient of thermal expansion is 10^-5 and E = 100 GN/m². if the rod is free to expand, the thermal stress developed in it is:
a) 100 kN/m²
b) 240 kN/m²
c) Zero
d) Infinity
Answer: c
Clarification: Thermal stress will only develop if the body is restricted.

8. Which one of the following pairs is NOT correctly matched?
a) Temperature strain with permitted expansion – ( αTl – δ)
b) Temperature thrust – ( αTE)
c) Temperature stress – (αTEA)
d) Temperature stress with permitted expansion – E(αTl – δ) / l
Answer: a
Clarification: Dimension analysis gives Temperature strain with permitted expansion – ( αTl – δ)is wrong. In other options the dimensions are correctly matched.

9. A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of δT. The Youngs modulus is E and the coefficient of linear expansion is unity. The thermal stress in the rod is ____________
a) Zero
b) T
c) EδT
d) EδTL
Answer: c
Clarification: As α = δl / l δT
So, δl = l x 1 x δT
And temperature strain = δl / l = δT
As E = stress / strain
Stress = E δT.

10. A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stress are represented by σ_x and σ_z, then ___________
a) σ_x = 0, σ_y = 0
b) σ_x not equal to 0, σ_y = 0
c) σ_x = 0, σ_y not equal to 0
d) σ_x not equal to 0, σ_y not equal to 0
Answer: a
Clarification: We know that due to temperature changes, dimensions of the material change. If these changes in the dimensions are prevented partially or fully, stresses are generated in the material and if the changes in the dimensions are not prevented, there will be no stress set up. (Zero stresses).
Hence cylindrical rod Is allowed to expand or contract freely.
So, σ_x = 0 and σ_y = 0.

11. which one of the following are true for the thermal expansion coefficient?
a) α_aluminium > α_brass> α_copper > α_steel
b) α_brass > α_aluminium > α_copper > α_steel
c) α_copper > α_steel > α_aluminium > α_brass
d) α_steel > α_aluminium > α_brass > α_copper
Answer: a
Clarification: Aluminium has the largest value of thermal expansion coefficient, then brass and then copper. Steel among them has lowest value of thermal expansion coefficient.

12. The length, coefficient of thermal expansion and Youngs modulus of bar A are twice of bar B. If the temperature of both bars is increased by the same amount while preventing any expansion, then the ratio of stress developed in bar A to that in bar B will be ___________
a) 2
b) 4
c) 8
d) 16
Answer: b
Clarification: Temperature Stress = EαδT
So σ₁ / σ₂ = E₁α₁δT₁/E₂α₂δT₂
From question, α and E of bar A are double that of bar B.

250+ TOP MCQs on Types of Beams and Loads and Answers

This set of Strength of Materials Multiple Choice Questions on “Types of Beams and Loads”.

1. _______ is a horizontal structural member subjected to transverse loads perpendicular to its axis.
a) Strut
b) Column
c) Beam
d) Truss
Answer: c
Clarification: A beam is a horizontal structural member subjected to a transverse load perpendicular to its own axis. Beams are used to support weights of roof slabs, walls and staircases. The type of beam usually depends upon the span, type of load elasticity and type of structure.

2. Example for cantilever beam is ______
a) Portico slabs
b) Roof slab
c) Bridges
d) Railway sleepers
Answer: a
Clarification: A beam which is fixed at one end and is free at other end, it is called cantilever beam. The examples for it are portico slabs and sunshades.

3. The diagram depicts _______ kind of beam.

a) Cantilever
b) Continuous
c) Over hanging
d) Propped cantilever
Answer: d
Clarification: A beam which is fixed at one end and free at other end is called cantilever beam. In this case, some support other than the existing ones may be provided in order to avoid excessive deflection or to reduce the amount of bending moment, the additional support is known as a prop. The beam is known as a propped cantilever beam.

4. Fixed beam is also known as __________
a) Encastered beam
b) Built on beam
c) Rigid beam
d) Tye beam
Answer: a
Clarification: A beam which is fixed at both supports is called fixed beam or encastered beam. All framed structures are examples of fixed beams.

5. U.D.L stands for?
a) Uniformly diluted length
b) Uniformly developed loads
c) Uniaxial distributed load
d) Uniformly distributed loads
Answer: d
Clarification: These loads are uniformly spread over a portion or whole area. They are generally represented as rate of load that is Kilo Newton per meter length (KN/m).

6. Given below diagram is ______ load.

a) Uniformly distributed load
b) Uniformly varying load
c) Uniformly decess load
d) Point load
Answer: b
Clarification: A load which varies uniformly on each unit length is known as uniformly varying load. Sometimes the load is zero at one end and increases uniformly to the other forms of uniformly varying loads.

7. Moving train is an example of ____ load.
a) Point load
b) Cantered load
c) Rolling load
d) Uniformly varying load
Answer: c
Clarification: As train’s wheels (rolling stock) move in rolling way. The upcoming load will be considered as rolling load.

8. Continuous beams are _________
a) Statically determinate beams
b) Statically indeterminate beams
c) Statically gravity beams
d) Framed beams
Answer: b
Clarification: Fixed beams and continuous beams are statically indeterminate beams which cannot be analyzed only by using static equations.

9. A beam which extends beyond it supports can be termed as __________
a) Over hang beam
b) Over span beam
c) Isolated beams
d) Tee beams
Answer: a
Clarification: A Beam extended beyond its support. And the position of extension is called as over hung portion.

10. Units of U.D.L?
a) KN/m
b) KN-m
c) KN-m×m
d) KN
Answer: a
Clarification: As these loads distribute over span the units for this kind of loads will be load per meter length i.e KN/m. It is denoted by “w”.

250+ TOP MCQs on Combined Stress and Answers

This set of Strength of Materials Multiple Choice Questions on “Combined Stress”.

1. Bond stress for M20 grade concrete in tension is ____________
a) 1.4
b) 1.2
c) 1.5
d) 1.8
Answer: b
Clarification: Bond stress is the shear stress acting parallel to the bar on the interface between the reinforcing bar and the surrounding concrete. Hence it is the stress developed between the contact surface of Steel and concrete to keep them together. The value of M20 designs Bond stress is 1.2 in tension.

2. The formation of diagonal cracks at junctions is due to ________
a) Shear stress
b) Bond stress
c) Temperature stress
d) Lateral stress
Answer: a
Clarification: Bending is usually accompanied by shear. The combination of shear and bending stresses produces the principle stresses which causes diagonal tension in the beam section. This should be resisted by providing shear reinforcement in the form of vertical stirrups (or) bent up bars along with stirrups.

3. Calculate the factored bending moment of a rectangular reinforced concrete beam of effective span 4300 mm and load imposed 37.5 kN/m.
a) 100kNm
b) 127kNm
c) 130kNm
d) 145kNm
Answer: c
Clarification: Factored load (w) = 1.5×37.5 = 56.25 kN/m.
Factored bending moment for simply supported beam (M) = wl²/ 8. = 56.25×(4.3)²/ 8 = 130kNm.

4. Determine the limiting percentage of steel for singly reinforced sections of M20 grade & Fe415.
a) 0.68
b) 0.79
c) 0.96
d) 1.76
Answer: c
Clarification: The limiting percentage of steel for singly reinforced sections of M20 grade & Fe415 is 0.96.

Grade of concrete	Limiting percentage of tensile Steel for a Fe415
M15	0.72
M20	0.96
M25	1.19

5. Calculate the limiting depth of the neutral axis for mild steel of effective depth 400 mm.
a) 318mm
b) 212mm
c) 455mm
d) 656mm
Answer: b
Clarification: The limiting depth of neutral axis Fe 250 steel is
Xu (max) = 0.53 × d ( for Fe250)
= 0.53 × 400
= 212mm.

6. Lap splices should not be used for bars larger than _____ mm.
a) 45mm
b) 54mm
c) 36mm
d) 72mm
Answer: c
Clarification: Splices are provided when the length of the bar is less than that required. The splicing of reinforcement is provided either by lap joint or mechanical joint or welded Joint. Lap splices should not be used for bars larger than 36 mm for larger diameter, bars may be welded.

7. Anchorage value for “U” hook is ________
a) 16 × diameter of bar
b) 12 × diameter of bar
c) 10 × diameter of bar
d) 8 × diameter of bar
Answer: a
Clarification: Anchorage value for “U” hook is 16 × diameter of bar.

Type of Hook / Bend in degrees	Anchorage Value
U hook	16 × diameter of bar
45 bend	4 × diameter of bar
90 bend	8 × diameter of bar
135 bend	12 × diameter of bar

8. The standard __________ are provided in deformed bars.
a) Anglets
b) Bends
c) Fillets
d) Lugs
Answer: b
Clarification: In situations, where straight anchorage length cannot be provided due to lack of space. To improve the anchorage of bars, standard bends are provided in deformed bars.

9. Transverse bars are also called as _________
a) Main bars
b) Anchor bars
c) Distribution bars
d) Stirrups
Answer: c
Clarification: In addition to main bars, along the shorter direction provided at the bottom, minimum reinforcement along the longer span and are also provided on top of the main bars and at right angles to them. These are called distribution bars are transverse bars.

10. A slab supporting only in two edges opposite to each other is ______
a) Two way slab
b) One way slab
c) Continuous slab
d) Cantilever slab
Answer: b
Clarification: If the ratio of the longest span the shorter span is greater than 2 or A slab supporting only in two edges (opposite to each other) is called one way slab. This slab spans across shorter span practically.

11. Torsion reinforcement is provided in ___________ slab
a) One way slab
b) Two way slab
c) Simply supported slab
d) Cantilever slab
Answer: b
Clarification: A slab supporting on all four edges is known as two way slab. In this slab, the ratio of longest span to the shorter span is less than 2. It requires torsional reinforcement because there’s a chance of twisting at corners.

12. Generally in residential buildings, the width of stay is kept as ____________
a) 2m
b) 1m
c) 5m
d) 4m
Answer: b
Clarification: The stair consists of series of steps with landings at appropriate intervals. The width of stair depends upon the type of building in which it is provided. Generally, in residential buildings, the width of stair is 1 m.

13. As per IS 456:2000; the slope or pitch of stairs should be in between 25 ° to ___________
a) 45°
b) 90°
c) 40°
d) 120°
Answer: c
Clarification: Each step has one tread and one rise. As per IRC, the tread is in between 250mm to 300 mm. The slope or pitch of the stairs should be in between 25° to 40°.

14. When space is less, the ___________ staircases is much preferred.
a) Open well
b) Dog legged
c) Spiral stair
d) Circular
Answer: b
Clarification: The most common type of Stairs arranged with two adjacent flights running parallel with mid landing. Where the space is less, dog legged staircase is generally provided resulting in economical utilisation of available place.

15. The ______________ of a column is the distance between the points of zero bending moments.
a) Slenderness ratio
b) Eccentricity
c) Radius of gyration
d) Effective length
Answer: d
Clarification: Effective length of a column is the distance between the points of zero bending moments (point of contra flexure) of a buckled column the effective length of the column depends upon the unsupported length and the end conditions.