Strength of Materials MCQ Set 1
1. A solid shaft of circular in section is subjected to torque which produces maximum shear stress in a shaft. Calculate the diameter of the shaft.
a) (16T/πf)3/2
b) (16f/πT)1/2
c) (16f/π) 1/2
d) (πT/16f) 1/2
Answer
Answer: a [Reason:] From torsional equation
T/J = f/R
T = f.Z
T = f×π/16d3.
D= (16T/πf) 3/2.
2. When two dissimilar shafts are connected together, then the shaft is __________
a) Integrated shafts
b) Composite shafts
c) Differential shafts
d) Combined shafts
Answer
Answer: b [Reason:] When two dissimilar shafts are connected together to form one shaft then the shaft can be termed as composite shaft.
3. __________ torque occurs along maximum shear stress due to combined bending and torsion.
a) Equipment
b) Coaxial
b) Biaxial
d) Lateral
Answer
Answer: a [Reason:] Equipment torque is the twisting moment which acts along producing a maximum shear stress due to the combined bending as well as torsion.
4. When a shaft is subjected to pure twisting then the type of stress developed is__
a) Bending
b) Axial
c) Shear
d) Normal
Answer
Answer: c [Reason:] Shear stress is produced when the shaft is subjected to pure twisting (torsion). The shear stress due to twisting moment is zero at the axis of the shaft.
5. The torque which produces unit twist per unit length is ________
a) Torsional rugosity
b) Torsional rigidity
c) Torsional viscosity
d) Torsional mean radius
Answer
Answer: a [Reason:] The product of shear modulus(C) and polar moment of inertia (J) is called torsional rigidity. Torsional rigidity produces a twist of 1 radian in a shaft of unit length.
6. The level of top of weir can be termed as __________
a) Talus
b) Curtain walls
c) Crest
d) Shutter
Answer
Answer: c [Reason:] The level of top of weir is known as crest. The shutters are provided on the crest and can be raised or laid flat during the time of floods.
7. ________ possesses less silting and scouring.
a) Weir
b) Barrage
c) Dams
d) Reservoir
Answer
Answer: b [Reason:] The barrage is a low obstructive barrier constructed across the river. There will be less silting and better control over the levels due to low set crest.
8. In __________there will be no means for silt disposal.
a) Weir
b) Barrage
c) Reservoir
d) Dams
Answer
Answer: a [Reason:] The weir may be defined as a solid obstruction/wall built across the river to raise the water level. Raised crest causes silting at upstream and there is no means silt disposal.
9. _________ is a pure water pressure.
a) Uplift
b) Percolation
c) Scour
d) Flood bank
Answer
Answer: a [Reason:] Uplift occurs when pore water pressure under a structure or a low permeability confining layer becomes larger than the mean overburden pressure.
10. __________ causes of uplift of structure.
a) Percolation
b) Scour
c) Critical Velocity
d) Slope Failure
Answer
Answer: a [Reason:] The effect of percolation on an irrigation structure like a weir to cause uplift pressure on the structures and topple the structure at any moment.
11. ________ protects the weir from erosive forces during floods.
a) Talus
b) Curtain walls
c) Shutter
d) Upstream solid apron
Answer
Answer: d [Reason:] Upstream solid apron is a concrete bed which is provided on the upstream side of weir to protect the weir from erosive forces during floods. The length of apron depends upon maximum discharge of the river.
12. Gross storage – Dead storage is _______
a) Live storage
b) Virtual storage
c) Excessive storage
d) Free storage
Answer
Answer: a [Reason:] It is also called as available or effective storage. It is the difference between the gross storage and dead storage. It is the amount of water available from FRL to the sill of lowest sluice.
13. Which of the following is not sound proof?
a) G I sheets
b) A C sheets
c) PVC sheets
d) Fabric sheets
Answer
Answer: a [Reason:] Galvanised iron sheets are commonly used as roofing material. These are very durable and fire proof. The main disadvantage is they are not sound proof.
14. Which of the following is not affected by temperature?
a) Fabric sheets
b) G I sheets
c) AC sheets
d) Flat roofs
Answer
Answer: c [Reason:] Asbestos cement sheets are cheaper in initial cost. They are fire resisting. They are heavy in weight and they are not affected by temperature.
15. Which of the following possess good insulation properties?
a) Battened roofs
b) Wooden roofs
c) Jack arch roofs
d) Flat roofs
Answer
Answer: d [Reason:] Flat roofs are easier in construction and maintenance. A flat roof is more stable against high wards. It has better architectural appearance and it has good insulation properties.
Strength of Materials MCQ Set 2
1. Bond stress for M20 grade concrete in tension is
a) 1.4
b) 1.2
c) 1.5
d) 1.8
Answer
Answer: b [Reason:] 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
Answer: a [Reason:] 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
Answer: c [Reason:] 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
Answer: c [Reason:] 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 neutral axis for mild steel of effective depth 400 mm
a) 318mm
b) 212mm
c) 455mm
d) 656mm
Answer
Answer: b [Reason:] 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
Answer: c [Reason:] Splices are provided when the length of 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
Answer: a [Reason:] 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
Answer: b [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: b [Reason:] 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
Answer: b [Reason:] 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
Answer: c [Reason:] 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
c) Circular
Answer
Answer: b [Reason:] 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
Answer: d [Reason:] 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.
Strength of Materials MCQ Set 3
1. In flitched beams, ______ remains same for both materials.
a) Stress
b) Strain
c) Section modulus
d) Young’s modulus
Answer
Answer: b [Reason:] Due to bending, the strain will be same in both the materials.
A timber beam strengthened by steel strips.
Where E of timber / E of steel = m
The equivalent width = b + 2mt.
2. What is the moment due to dead load in case of continuous beams at the middle of interior spans?
a) w L3 / 12
b) w L2/ 14
c) w 3/ 20
d) w L2 / 24
Answer
Answer: d [Reason:] The moment due to dead load in case of continuous beams at the middle of interior spans is w L2 / 24.
| Position | Moment due to dead load |
| Near middle of end span | W L2/ 12 |
| At the middle of interior span | W L2/ 24 |
| At the support next to and support | -W L2/ 22 |
.
3. A continuous beam is one which is _______
a) Infinitely long
b) Supported at two points
c) Supported it more than two supports
d) Supported by a prop
Answer
Answer: c [Reason:] A beam which is supported by more than two supports is known as continuous beam. In this beam, bending moment is low and hence the deflection in the beam is also comparatively less. This beam is stiffer when compared to the other traditional beams.
4. The effective length of column depends upon ________
a) the cross section of beam
b) end conditions
c) maximum bending moment
d) extreme fibres
Answer
Answer: b [Reason:] The effective length of column depends upon end conditions.
| End condition | Effective length |
| Both ends hinged | L |
| Both ends fixed | L/2 |
| One end is fixed and other end free | 2L |
5. The phenomenon under which the strain of material varies under constant stress is known as ________
a) Creep
b) Hysteresis
c) Viscoelasticity
d) Strain hardening
Answer
Answer: a [Reason:] A creep is a plastic deformation underweight the strain of material where is under constant stress this is one of the mechanical properties of the engineering materials. The best example is the failure of concrete.
6. Volumetric strain = 3× _____ strain.
a) Lateral
b) Linear
c) Composite
d) Yield
Answer
Answer: b [Reason:] eV (volumetric strain) = 3× linear strain
= 3×e
The volumetric strain is algebraic sum of all the linear(or) axial strain when a solid to be subjected to equal normal sources of the same type of all faces we will have €x , €y and €z equal in value. In this case the volumetric strain will be 3 times the linear strain in any of the three axes.
7. The stress corresponding to breaking point is known as _____________
a) yield stress
b) ultimate stress
c) breaking stress
d) normal stress
Answer
Answer: c [Reason:] After reaching ultimate stress, the stress strain curve suddenly falls with rapid increase in strain and specimen breaks. The stress corresponding to breaking point is known as breaking stress and it is denoted by G.
8. Determine the yield stress of a steel rod 20 mm diameter, if the yield load on the steel rod is 88kN.
a) 240.55 N/mm2
b) 280.25 N/mm2
c) 325 N/mm2
d) 290.45 N/mm2
Answer
Answer: b [Reason:] Initial area of the Steel rod of 20 mm = 314 mm2 [ area of circle ]
Yield stress = yield load/ Area
= 88 × 103/ 314
= 280.25 N/mm2.
9. What is the elongation percentage of a steel rod of 50 mm diameter if the total extension is is 54 mm and gauge length is 200 mm.
a) 27%
b) 23%
c) 43%
d) 35%
Answer
Answer: a [Reason:] Percentage elongation = Total extension / Gauge length × 100
= 54/200 × 100
= 27%.
10. __________ joints are provided when there is a break in the concreting operation.
a) transverse joints
b) longitudinal joints
c) construction joints
d) warpage joints
Answer
Answer: c [Reason:] The construction joints are provided when there is a break in a concreting operation. Although effort is always made to complete the concrete work in one day, sometimes it is not possible and therefore, construction joints are provided. For beams, the joints should be at the centre of the span or within the middle third.
Strength of Materials MCQ Set 4
1. Calculate that torque, if the diameter of the shaft is 50 mm and revolutions @ 130 rpm. The maximum shear stress is 62.5 N/mm2.
a) 1564 Nm
b) 1478 Nm
c) 1534 Nm
d) 1494 Nm
Answer
Answer: c [Reason:] Diameter of shaft = 50 mm
Revolutions of shaft = 130 rpm.
Maximum shear stress = f= 62.5 N/mm2.
T = f π D3/ 16
T = 62.5 ×503× π / 16.
T = 1534 Nm.
2. What is the example for centrifugal pump?
a) Reciprocating pump
b) Suction pump
c) Rotodynamic pump
d) Delivery pump
Answer
Answer: c [Reason:] Rotodynamic pumps have a rotating element through which as the liquid passes its angular momentum changes, due to which the pressure energy of the liquid is increased. The centrifugal pump is a rotodynamic pump.
3. Reciprocating pump is an example of ___________
a) Positive displacement pump
b) Delivery pump
c) Suction pump
d) Rotodynamic pump
Answer
Answer: a [Reason:] Positive displacement pumps are those pumps in which the liquid is sucked and then it is pushed to the thrust exerted on it by a piston. The most common example of the positive displacement pump is the reciprocating pump.
4. ____ is the difference between theoretical discharge and actual discharge of the pump.
a) Crank
b) Hook
c) Slip
d) Centile
Answer
Answer: c [Reason:] c [Reason:] Slip of a pump is defined as the difference between the theoretical discharge and actual discharge after pump.
5. _____ is a phenomenon by which the study and continuous flow of liquid is obstructed.
a) Slip
b) Separation
c) Air vessels
d) Knockage
Answer
Answer: b [Reason:] Separation of reciprocating pump is that phenomenon by which the steady and continuous flow of liquid is affected by the presence of air and dissolved gases.
6. Negative slip occurs when the______ is more than theoretical discharge.
a) Virtual discharge
b) Actual discharge
c) Mean discharge
d) Mode discharge
Answer
Answer: b [Reason:] When the delivery valve opens before the suction stroke is completed, the actual discharge is more than the theoretical discharge. In such cases, the slip of the pump is known as negative slip.
7. _____ slip occurs, when the delivery pipe is short and suction pipe is long.
a) Positive
b) Critical
c) Negative
d) Zero
Answer
Answer: b [Reason:] The slip occurs when the delivery pipe is short and suction pipe is long. The pump is running at high speeds as the delivery valve open before suction stroke is completed, the slip of the pump is known as negative slip.
8. ________ reduces the possibility of separation.
a) Air vessels
b) Casing
c) Impeller
d) Vortex
Answer
Answer: a [Reason:] An air vessel may be a closed chamber having the compressed air in a top portion and the water at the bottom. It reduces the possibility of separation and it ensures the pump to run at high speed.
9. If the absolute pressure falls below ___ m, the pump prone to separation.
a) 3 m
b) 2 m
c) 1.5 m
d) 2.5 m
Answer
Answer: d [Reason:] If the absolute pressure falls below 2.5 metres of water, the dissolved gases will be appearing in a liquid and continuous flow will be chocked. This phenomenon can be termed as separation.
10. The phenomenon of separation can also be known as ___
a) Cavitation
b) Priming
c) Positive head
d) Pulsate
Answer
Answer: a [Reason:] Separation is a phenomenon of obstructing the flow by the presence of dissolved gases when the absolute pressure falls below 2.5 metres of water. This phenomenon of separation can also be known as knocking (or) cavitation in the reciprocating pump.
11. The work done against friction is reduced due to _____
a) Impeller
b) Priming
c) Air vessel
d) Vortex
Answer
Answer: c [Reason:] An air vessel is fitted to the suction and delivery pipes at a point close to the cylinder of a single acting reciprocating. The pump increases the length of suction pipe and reduces the work done against friction.
12. Volute is a type of ________
a) Delivery pipe
b) Casing
c) Impeller
d) Suction pipe
Answer
Answer: b [Reason:] Casing is an airtight chamber covering the impeller. The different types of casing
1.Volute casing
2.Vortex casing
3.Casing with guide blades.
13. ______ pumps, the torque is uniform.
a) Reciprocating pump
b) Suction pump
c) Delivery pump
d) Centrifugal pump
Answer
Answer: d [Reason:] Centrifugal pump is used for lifting highly viscous liquids such as oils, muddy and sewage water, paper pulp etc. In centrifugal pump, torque is uniform and no air vessels are required.
14. What is the practical maximum suction lift in reciprocating pump?
a) 3.5 m
b) 4.5 m
c) 5 m
d) 6.5 m
Answer
Answer: d [Reason:] Reciprocating pump can handle only pure water or less viscous liquids free from impurities. It can be operated at low speeds only. The practical maximum section lift is 6.5 metres.
15. _____ pumps give larger discharge.
a) Suction
b) Reciprocating
c) Centrifugal
d) Positive displacement
Answer
Answer: c [Reason:] Centrifugal pump are example of rotodynamic pump the basic principle of centrifugal pump is that “when a certain mass of liquid is rotated by an external force, then centrifugal head is impressed which enables it to rise to a higher level”. A centrifugal pump discharges larger quantity when compared to other pumps.
Strength of Materials MCQ Set 5
1. The obstruction or a barrier built across the stream or river is called as____
a) Barrage
b) Weir
c) Dam
d) Reservoir
Answer
Answer: c [Reason:] A dam may be defined as an obstruction or a barrier built across the stream or river these are artificial storage works. It retains water to create an impounding reservoir.
2. FTL Stands for ________
a) Free tank level
b) Full tank level
c) Full top level
d) Fill toe level
Answer
Answer: b [Reason:] It is also called as full reservoir level (FRL). It is a level up to which the water stored obviously the crest of the spillway is fixed at this level.
3. _______ is openings extending from upstream to downstream of dam.
a) Guide banks
b) Divide voids
c) Sluices
d) Spillway
Answer
Answer: c [Reason:] Sluices are openings or conduits extending from upstream face of the dam to downstream face of the dam. They are used to clean the silt from the reservoir. They also the decrease the peak flood in the reservoir.
4. Water stored in dam is expressed in _______
a) Mega cumec metres
b) Million cubic metres
c) Metric cumec
d) Million cusec metres
Answer
Answer: b [Reason:] It is the total quantity of water stored up to FRL. It includes dead storage also. It is expressed generally in thousand hectare metre or million cubic metres (Mm3).
5. MDDL Stands for ________
a) Minimum draw down level
b) Maximum draw down level
c) Million drop down level
d) Mega drop down level
Answer
Answer: a [Reason:] It is the lowest level up to which the reservoir is depleted from the considerations of hydropower generation. So this level is known as minimum draw down level (MDDL).
6. _____ dam which resists are the external forces by virtue of its self weight.
a) Earthen dam
b) Storage dam
c) Detention dam
d) Gravity dam
Answer
Answer: d [Reason:] A gravity dam is that, which is stable against all the external forces achieved by the weight of the dam itself. This is the most permanent one and hence it is very commonly used. It may be constructed in all localities.
7. The factor of safety against overturning should not be less than ______
a) 1.8
b) 2.25
c) 1.5
d) 1.75
Answer
Answer: c [Reason:] In the dam section, the overturning takes place when a resultant force cuts the base of the dam downstream of the toe. The factor of safety against overturning is the ratio of the stabilizing moment to the overturning moments. The safety against overturning should not be less than 1.5.
8. In sliding failure, the co-efficient of friction varies from ________
a) 0.65 – 0.75
b) 0.8 – 0.9
c) 0.45 – 0.65
d) 0.85 – 1
Answer
Answer: a [Reason:] To avoid sliding, the factor of safety against the sliding should be greater than 1.
F.S = M(V-U) / €H > 1
Where M = Co-efficient of friction. It varies from 0.65 to 0.75
V = Total vertical force
U = Upward force.
9. Which of the following forces do not act on dam?
a) Silt pressure
b) Wave pressure
c) Creep pressure
d) Uplift
Answer
Answer: c [Reason:] Among the above forces, creep pressure does not act on dam. Generally on gravity dam number of forces such as water pressure, wave pressure, wind pressure, ice pressure etc. will be acting in horizontal direction. In the same way, uplift, self weight acts in vertical direction.
10. The elementary profile of a dam is generally a ________
a) Isosceles triangle
b) Right angled triangle
c) Scalene triangle
d) Equilateral triangle
Answer
Answer: b [Reason:] In the absence of any other forces, the forces due to water and self weight of the dam form an elementary profile which will be in triangular section having zero top width at water level, where the pressure is zero and maximum base width is at bottom where the maximum water pressure acts.
11. _____ acts as inspection chamber in Dams.
a) Spillway
b) Heel
c) Drainage gallery
d) Toe
Answer
Answer: c [Reason:] A drainage gallery is an opening in the body of a dam which runs longitudinally. It runs through the length of the dam. Generally it is rectangle shape with flat a semi-circular head usually 1.5 m wide and 2.5m height.
12. The minimum standard height for construction joint is about ________
a) 1.2 m
b) 1.5 m
c) 2.1 m
d) 2.3 m
Answer
Answer: b [Reason:] The joints which facilitate construction of the dam to proceed in small lifts. These joints are also known as horizontal joints. A lift may be defined as the vertical distance between two consecutive construction joints. The height is about 1.5 m each.
13. Cracks developed in the body of dam section can be avoided by ________
a) Construction joints
b) Contraction joints
c) Transverse joints
d) Longitudinal joints
Answer
Answer: b [Reason:] Due to variation in temperature it causes contraction and expansion in masonry or concrete of the dam. It will develop fine cracks in the body of the dam. By providing contraction joints, these cracks can be avoided.
14. ______ is the over flow section or portion of the dam.
a) Heel
b) Toe
c) Spillway
d) Gallery
Answer
Answer: c [Reason:] A spillway is the overflow section or portion of the dam over which surplus discharge flows from reservoir to downstream face. This structure is provided in the body of the dam or near the dam or on the periphery of the reservoir.
15. _______ is common type of spillway used in gravity dams.
a) Ogee spillway
b) Trough spillway
c) Side channel spillway
d) Emergency spillway
Answer
Answer: a [Reason:] An ogee spillway is very common type of spillway used in gravity dams. It consists of two parts namely ¡)ogee crest and ¡¡) a bucket. In this spillway water spills and flows over and ogee crest in the form of a rolling sheet of water. Due to this, development of negative pressures can be avoided.
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