MCQ’s On Springs
Q. Which of the following function can the spring perform?
A. Store energy
B. Absorb shock
C. Measure force
D. All of the mentioned
.
Answer: D
Explanation: Spring can easily perform all the listed functions.
Q. The helix angle is very small about 2⁰. The spring is open coiled spring.
A. Yes
B. It is closed coiled spring
C. That small angle isn’t possible
D. None of the listed
.
Answer: B
Explanation: When the helix angle is small, the plane containing each coil is almost at right angles and hence it is called closed coiled spring.
Q. The helical spring ad wire of helical torsion spring, both are subjected to torsional shear stresses.
A. True
B. False
.
Answer: B
Explanation: The wire of helical torsion sprig is subjected to bending stresses.
Q. The longest leaf in a leaf spring is called centre leaf.
A. It is called middle leaf
B. It is called master leaf
C. Yes
D. None of the listed
.
Answer: B
Explanation: It is called master leaf.
Q. Multi leaf springs are not recommended for automobile and rail road suspensions.
A. True
B. False
.
Answer: B
Explanation: They are highly used in automobile and rail road suspensions.
Q. The spring index is the ratio of wire diameter to mean coil diameter.
A. True
B. False
.
Answer: A
Explanation: It is the ratio of mean coil diameter to wire diameter.
Q. If spring index=Q.5, what can be concluded about stresses in the wire?
A. They are high
B. They are negligible
C. They are moderate
D. Cannot be determined
.
Answer: A
Explanation: If indexis <3 then stresses are high due to curvature effect.
Q. A spring with index=15 is prone to buckling.
A. True
B. False
.
Answer: A
Explanation: Due to large variation, such a spring is prone to buckling.
Q. If the spring is compressed completely and the adjacent coils touch each other,the the length of spring is called as?
A. Solid length
B. Compressed length
C. Free length
D. None of the mentioned
.
Answer: A
Explanation: Terminology.
Q. If number of coils are 8 and wire diameter of spring 3mm, then solid length is given by?
A. None of the listed
B. 27mm
C. 24mm
D. 21mm
.
Answer: C
Explanation: Solid length=8×Q.
Q. Compressed length is smaller than the solid length.
A. True
B. False
.
Answer: B
Explanation: Compressed is length of spring under maximum compressive force. There is some gap between the coils under maximum load.
Q. Pitch of coil is defined as axial distance in compressed state of the coil.
A. Yes
B. It is measured in uncompressed state
C. It is same in uncompressed or compressed state
D. None of the listed
.
Answer: B
Explanation: Pitch is measured in uncompressed state.
Q. If uncompressed length of spring is 40mm and number of coils 10mm, then pitch of coil is?
A. 4
B. 40/9
C. 40/11
D. None of the mentioned
.
Answer: B
Explanation: Pitch=Uncompressed length/N-Q.
Q. Active and inactive, both types of coils support the load although both don’t participate in spring action.
A. Active coils don’t support the load
B. Inactive coils don’t support the load
C. Both active and inactive don’t support the load
D. Both active and inactive support the load
.
Answer: B
Explanation: Inactive coils don’t support the load.
Q. If a spring has plain ends then number of inactive coils is?
A. 1
B. 2
C. 3
D. 0
.
Answer: D
Explanation: There are no inactive coils in plain ends.
Q. Spring having square ends has 1 inactive coil.
A. True
B. False
.
Answer: B
Explanation: There are 2 inactive coils.
Q. Martin’s factor compensates for curvature effect in springs.
A. True
B. False
.
Answer: B
Explanation: Wahl’s factor accommodates curvature effect while designing spring.
Q. The angle of twist for the equivalent bar to a spring is given by? (Symbols have their usual meaning)
A. 8PD²N/Gd⁴
B. 16PD²N/Gd⁴
C. 16PDN/Gdᵌ
D. 8PDN/Gdᵌ
.
Answer: B
Explanation: θ=Ml/GJ where M=PD/2, l=πDN and J=πd⁴/3Q.
Q. The axial deflection of spring for the small angle of θ is given by?
A. 328PDᵌN/Gd⁴
B. 8PDᵌN/Gd⁴
C. 16PDᵌN/Gd⁴
D. 8PD²N/Gdᵌ
.
Answer: B
Explanation: Deflection=θxD/Q.
Q. A spring of stiffness constant k is cut in two equal parts. The stiffness constant of new spring will be k/Q.
A. True
B. False
.
Answer: B
Explanation: k=Gd⁴/8DᵌN, hence k is inversely proportional to number of coils. Thus result will be 2k.
Q. For two spring connected in series, the force acting on each spring is same and equal to half of the external force.
A. True
B. False
.
Answer: B
Explanation: Te force on each spring is equal to the external force.
Q. For two springs connected in series, the net deflection is equal to the sum of deflection in two springs.
A. True
B. False
.
Answer: A
Explanation: The net deflection is sum of the deflection of sprigs connected in series.
Q. For two springs connected in parallel, net force is equal to the sum of force in each spring.
A. True
B. False
.
Answer: A
Explanation: Net force applied is distributed in the two springs.
Q. Patenting is defined as the cooling below the freezing point of water.
A. True
B. False
.
Answer: B
Explanation: Patenting is heating steel above critical range followed by rapid cooling.
Q. Find the Wahl’s factor if spring index is Q.
A. Q.2020
B. Q.2424
C. Q.2525
D. Q.5252
.
Answer: C
Explanation: K=[4C-1/4C-4]+0.615/C.
Q. Find the shear stress in the spring wire used to design a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.
A. 45Q.2N/mm²
B. 46Q.6N/mm²
C. 5Q.2N/mm²
D. None of the listed
.
Answer: B
Explanation: τ=K x 8PC/πd² where K=[4C-1/4C-4]+0.615/C.
Q. Find the mean coil diameter of a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.
A. 7/6mm
B. 42mm
C. 1200×6/7 mm
D. None of the listed
.
Answer: B
Explanation: D=Cd.
Q. Find total number coils in a spring having square and ground ends. Deflection in the spring is 6mm when load of 1100N is applied. Modulus of rigidity is 81370N/mm². Wire diameter and pitch circle diameter are 10mm and 50mm respectively.
A. 7
B. 6
C. 5
D. 4
.
Answer: A
Explanation: Deflection=8PDᵌN/Gd⁴ or N=Q.4 or Q. Total coils=5+2(square grounded ends).
Q. A railway wagon moving with a speed of Q.5m/s is brought to rest by bumper consisting of two springs. Mass of wagon is 100kg. The springs are compressed by 125mm. Calculate the maximum force acting on each spring.
A. 1200N
B. 1500N
C. 1800N
D. 2000N
.
Answer: C
Explanation: mv²/2=Pxdeflection/Q.
Q. In concentric springs, vibrations called surge are amplified.
A. True
B. False
.
Answer: B
Explanation: Surge are eliminated.
Q. Can concentric springs be used to obtain a force which is not proportional to its deflection?
A. True
B. False
.
Answer: A
Explanation: Two sprigs having different free length can be meshed to obtain such a result.
Q. The load shared by each spring is inversely proportional to the cross section of wire.
A. Yes
B. No, it is directly proportional
C. It is proportional to its square
D. It is proportional to its square root
.
Answer: B
Explanation: It is directly proportional to the cross section of wire.
Q. A concentric spring consists of 2 sprigs of diameter 10mm and 4mm. The net force acting on the composite spring is 5000N. Find the force acting on each of the two springs.
A. 123Q.2N and 376Q.8N
B. 78Q.4N and 42Q.6N
C. 68Q.7N and 43Q.3N
D. 64Q.3N and 435Q.7N
.
Answer: C
Explanation: P₁/P₂=d₁²/d₂² and P₁+P₂=5000.
Q. If the spring have same solid length and number of coils in the two springs are 8 and 10, then find the diameter of the spring with 8 coils. It is given diameter of spring with 10 coils is 12mm.
A. Q.6mm
B. 9mm
C. 12mm
D. 15mm
.
Answer: B
Explanation: N₁d₁=N₂d₂.
Q. Two spring having stiffness constants of 22N/mm and 25N/mm are connected in parallel. They are to be replaced by a single spring to have same effect. The stiffness of that spring will be?
A. None of the mentioned.
B. 3N/mm
C. 47N/mm
D. Q.7N/mm
.
Answer: C
Explanation: k=22+2Q.
Q. What will happen if stresses induced due to surge in the spring exceeds the endurance limit stress of the spring.
A. Fatigue Failure
B. Fracture
C. None of the listed
D. Nipping
.
Answer: A
Explanation: If endurance limit is passed, fatigue failure will follow.
Q. Surge is caused by resonance effect in the spring.
A. True
B. False
.
Q. Surge is a desirable effect in the sprigs.
A. True
B. False
.
Answer: B
Explanation: Surge means vibratory motion which isn’t welcomed anywhere in the machinery.
Q. For a helical torsion sprig, the stress concentration factor at inner fibre is? Give spring index=Q.
A. Q.005
B. Q.175
C. Q.223
D. Q.545
.
Answer: B
Explanation: K=4C²-C-1/4C(C-1).
Q. For a helical torsion sprig, the stress concentration factor at outer fibre is? Give spring index=Q.
A. 0.78
B. 0.87
C. Q.87
D. 0.69
.
Answer: B
Explanation: K=4C²+C-1/4C(C+1).
Q. Spiral spring is quite rigid.
A. Yes
B. No it is flexible
C. It is of moderate rigidity
D. Rigidity can’t be determined
.
Answer: B
Explanation: Spiral spring is made of very thin wire which imparts great flexibility.
Q. The strip of spiral spring is never subjected to pure bending moment.
A. True
B. False
.
Answer: B
Explanation: The strip of spiral spring is subjected to pure bending moment.
Q. Calculate the bending stress induced in the strip of the helical spring. The spring is subjected to a moment of 1250N-mm with breadth and thickens of the strip being 11mm and Q.5mmm respectively.
A. 50Q.8N/mm²
B. 6Q.2N/mm²
C. 60Q.1N/mm²
D. 56Q.3N/mm²
.
Answer: C
Explanation: σ=12M/bt².
Q. Angle of rotation o arbour with respect to drum is given by?
A. None of the listed
B. 12ML/Ebtᵌ
C. 8 ML/Ebtᵌ
D. 16ML/Ebtᵌ
.
Answer: B
Explanation: θ=ML/EI where I=btᵌ/Q.
Q. Bending stress in graduated length leaves are more than that in full length leaves.
A. Yes
B. No
C. In some cases
D. Can’t be stated
.
Answer: B
Explanation: Bending stress in full length leaves is around 50% higher than graduated length leaves.
Q. Nip is the initial gap between extra full length leaf and the graduated length leaf before the assembly.
A. True
B. False
.
Answer: A
Explanation: Nipping is done to balance the bending stress in the full length leaf and graduated length leaf.
Q. Nipping is defined as leaving the gap between full length leaf and graduated length leaf.
A. True
B. False
.
Answer: B
Explanation: Nipping is pre stressing achieved by a difference in radii of curvature.
Q. A leaf spring consists of 3 extra full length leaves and 14 graduated length leaves. The maximum force that can act on the spring is 70kN and the distance between eyes of the spring is Q.2m. Width and thickness of the leaves are 100mm and 12mm respectively. If modulus of elasticity is 207000N/mm², calculate the initial nip.
A. 2Q.8mm
B. 2Q.9mm
C. 2Q.5mm
D. 2Q.1mm
.
Answer: B
Explanation: C=2PLᵌ/Enbtᵌ where L=Q.2/2, n=3+14,P=70/Q.
Q. A leaf spring consists of 3 extra full length leaves and 14 graduated length leaves. The maximum force that can act on the spring is 70kN and the distance between eyes of the spring is Q.2m. Width and thickness of the leaves are 100mm and 12mm respectively. Calculate the initial pre load required to close the nip.
A. 433Q.2N
B. 467Q.1N
C. 498Q.4N
D. Can’t be determined
.
Answer: B
Explanation: P=2x3x14x35000/17(3×3+2×14).
Q. Belleville spring can be used in clutch application.
A. True
B. False
.
Answer: A
Explanation: When h/t=Q.1, the Belleville spring can be used in clutches.
Q. Belleville spring can only produce linear load deflection characteristics.
A. Only linear
B. Linear as well as non linear
C. Non-linear
D. None of the mentioned
.
Answer: B
Explanation: Beliville spring can provide any linear or non-linear load deflection characteristics.
Q. When two Belleville sprigs are arranged in series, half deflection is obtained for same force.
A. One fourth deflection
B. Double deflection
C. Four time deflection
D. None of the listed
.
Answer: B
Explanation: Double deflection is obtained.
Q. When two Belleville springs are in parallel, half force is obtained for a given deflection.
A. Half force
B. Double force
C. Same force
D. Can’t be determined
.
Answer: B
Explanation: Double force is obtained.
Q. Propagation of fatigue failure is always due to compressive stresses.
A. Due to bending
B. Due to tensile
C. Due to fatigue
D. None of the listed
.
Answer: B
Explanation: Propagation is always due to tensile stresses.
Q. The strain energy stored in a spiral spring is given by?
A. 12M²L/Ebtᵌ
B. 6M²L/Ebtᵌ
C. 8M²L/Ebtᵌ
D. None of the listed
.
Answer: B
Explanation: U=Mθ/2 where θ=12ML/Ebtᵌ.
