If a spring has plain ends then number of inactive coils is?
Explanation: There are no inactive coils in plain ends.
Spring having square ends has 1 inactive coil.
Explanation: There are 2 inactive coils.
The angle of twist for the equivalent bar to a spring is given by? (Symbols have their usual meaning)
Explanation: θ=Ml/GJ where M=PD/2, l=πDN and J=πd⁴/32.
Martin’s factor compensates for curvature effect in springs.
Explanation: Wahl’s factor accommodates curvature effect while designing spring.
The axial deflection of spring for the small angle of θ is given by?
Explanation: Deflection=θxD/2.
A spring of stiffness constant k is cut in two equal parts. The stiffness constant of new spring will be k/2.
Explanation: k=Gd⁴/8DᵌN, hence k is inversely proportional to number of coils. Thus result will be 2k.
For two spring connected in series, the force acting on each spring is same and equal to half of the external force.
Explanation: Te force on each spring is equal to the external force.
For two springs connected in series, the net deflection is equal to the sum of deflection in two springs.
Explanation: The net deflection is sum of the deflection of sprigs connected in series.
For two springs connected in parallel, net force is equal to the sum of force in each spring.
Explanation: Net force applied is distributed in the two springs.
Patenting is defined as the cooling below the freezing point of water.
Explanation: Patenting is heating steel above critical range followed by rapid cooling.
Find the Wahl’s factor if spring index is 6.
Explanation: K=[4C-1/4C-4]+0.615/C.
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.
Explanation: τ=K x 8PC/πd² where K=[4C-1/4C-4]+0.615/C.
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.
Explanation: D=Cd.
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.
Explanation: Deflection=8PDᵌN/Gd⁴ or N=4.4 or 5. Total coils=5+2(square grounded ends).
A railway wagon moving with a speed of 1.5m/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.
Explanation: mv²/2=Pxdeflection/2.
Doc | 8 Pages
Test | 15 questions | 30 min
Test | 10 questions | 20 min
Test | 10 questions | 30 min
Test | 8 questions | 30 min
Test | 10 questions | 20 min