Closed Coil Helical Spring Under Axial Pull MCQ Quiz in मराठी - Objective Question with Answer for Closed Coil Helical Spring Under Axial Pull - मोफत PDF डाउनलोड करा

Concept:

Stiffness of the springs:

\(k = \frac{W}{δ}\),  \(\delta = \frac{{64\;W{R^3}n}}{{C{d^4}}}\)

\(k = \frac{W}{\delta } = \frac{W}{{\frac{{64\;W{R^3}n}}{{C{d^4}}}}} = \frac{{C{d^4}}}{{64{R^3}n}}\)

where,

k = stiffness of spring, W = axial load on spring, n = number of coils or turns, δ = deflection of the coil under load W, R = radius of coil, C = Modulus of Rigidity

d = diameter of the wire of the coil, θ = angle of twist, l = Length of wire = 2π Rn

Calculation:

Given:

Since spring is the same properties and diameter will not change, only the length of spring is reduced.

Initial length = l₁ , Final length = l₂ = \(\frac{{{l_1}}}{2}\)

\(l = 2\pi Rn\) ⇒ \(l \propto {n}\)

\(k = \frac{{C{d^4}}}{{64{R^3}n}} ⇒ k \propto \frac{1}{n}\) ⇒ \(K \propto \frac{1}{l}\)

\(\frac{{{k_1}}}{{{k_2}}} = \frac{{{l_2}}}{{{l_1}}} = \frac{{{l_1}/2}}{{{l_1}}}\)

\(\frac{{{k_1}}}{{{k_2}}} = \frac{{{1}}}{{{2}}}\) ⇒ k₂ = 2k₁

Hence the stiffness of the resulting spring will be double of initial.

Concept:

Shear stress developed in the spring is:

\({\tau _{max}} = \frac{{8PD}}{{\pi {d^3}}}\)

Deflection of spring

\(\delta = \frac{{8P{D^3}N}}{{G{d^4}}}\)

here d is wire diameter of spring, D is mean coil diameter, P is axial spring force, N is the number of active coils.

Calculation:

Given, 

d = 10 mm, N = 10, D =120 mm, P =200 N, G = 80 GPa = 80 × 10³ MPa

\({\tau _{max}} = \frac{{8PD}}{{\pi {d^3}}}= \frac{{8\times 200\times 120}}{{\pi {10^3}}}=61.11 \;Mpa\)

\(\delta = \frac{{8P{D^3}N}}{{G{d^4}}}= \frac{{8\times200\times{120^3}\times10}}{{80\times10^3\times{10^4}}}=34.56\;mm\)

Explanation:

For a helical spring,

\({\rm{Deflection\;\Delta }} = \frac{{8P{D^3}n}}{{G{d^4}}}\)

So,

\({\rm{\Delta }} \propto \frac{1}{{{d^4}}}\)

\(\frac{{{{\rm{\Delta }}_2}}}{{{{\rm{\Delta }}_1}}} = {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^4}\)

\(\frac{{{{\rm{\Delta }}_2}}}{{{{\rm{\Delta }}_1}}} = {\left( {\frac{{{d_1}}}{{2{d_1}}}} \right)^4}\)

[∵ Given d₂ = 2d₁]

\(\therefore \frac{{{{\rm{\Delta }}_2}}}{{{{\rm{\Delta }}_1}}} = \frac{1}{{16}}\)

Explanation:

Helical spring: Helical springs are designed to offer resistance against the linear compressing force applied along their axis. This is also called open-coiled helical spring.

If an open-coiled helical spring of mean diameter D, number of coils N and wire diameter d is subjected to an axial force P.

The wire of the spring will be subject to combined shear, bending and twisting. the following stresses also act on the wire:

1. Direct transverse shear stress 
2. Torsional shear stress
3. Tensile Stress

For a helical spring, when torsional shear stress, direct shear stress, and curvature stress is taken into consideration, then equivalent shear stress is given by-

\(τ=K\left ( \frac{8PD}{\pi d^3} \right )= K\left ( \frac{8PC}{\pi d^2} \right )\)

\(K = \frac{{4C - 1}}{{4C - 4}} + \frac{{0.615}}{C}\)

where, K = Wahl's factor, P = axial load on spring, D = mean coil diameter of spring, d = diameter of spring core or wire, C = spring index

As shear stress (τ) inversely related to the diameter of spring core or wire (d), shear stress increases with a decrease in the diameter of the spring core.

Concept:

The compression (deflection) of a helical spring is given by:

\(\delta = \frac{{8W{D^3}N}}{{G{d^4}}} \Rightarrow W = \frac{{\delta G{d^4}}}{{8{D^3}N}}\)

W = Axial load on the spring, D = Mean diameter of the spring coil, d = Diameter of the spring wire, N = Number of active coils

And equating the energy supplied by impact load to stored energy:

\(P\left( {h + \delta } \right) = \frac{1}{2}W\delta \)

Calculation:

d = 10 mm, D = 100 mm, N = 20, P = 100 N, δ = 60 mm

\(W = \frac{{\delta G{d^4}}}{{8{D^3}N}} = \frac{{60 \times 85000 \times {{10}^4}}}{{8 \times {{100}^3} \times 20}} = 318.75\;N\)

Thus:

\(100\left( {h + 60} \right) = \frac{1}{2} \times 318.75 \times 60 \Rightarrow h = 35.63\;mm\)

Concept:

The kinetic energy of the moving mass is converted into the potential energy of the spring.

Kinetic energy = \(\frac{1}{2}m{v^2}\;\)

Potential energy of spring = \(\frac{1}{2}k{x^2}\)

Stiffness: The force required to produce unit deflection in the spring is called as stiffness of the spring.

Calculation: 

Deflection (x) = 150 mm = 0.15 m 

\(\begin{array}{l} \frac{1}{2}k{x^2} = \frac{1}{2}m{v^2}\\ \Rightarrow k = m{\left( {\frac{v}{x}} \right)^2} = 2000 × {\left( {\frac{2}{{0.15}}} \right)^2} = 355.556\ kN/m \end{array}\)

Stiffness of one spring (k) \(= \frac{{355.556}}{2} = 177.78\ kN/m\)

Maximum force = k × x \(= 177.78 × {10^3} × 0.15 = 26666.67\ N\)

Explanation:

Given:

D = 150 mm, d = 20 mm, τ_per = 400 N / mm²

Now,

Spring index (C) is:

\(C = \frac{D}{d} = \frac{{150}}{{20}}\)

C = 7.5

\({\rm{Wahl\;factor\;}}\left( {{k_w}} \right) = \frac{{4c - 1}}{{4c - 4}} + \frac{{0.615}}{c}\)

\({k_w} = \frac{{4\left( {7.5} \right) - 1}}{{4\left( {7.5} \right) - 4}} + \frac{{0.615}}{{7.5}}\)

∴ k_w = 1.197

Now,

Permissible shear stress,

\(\tau = {k_w}\left( {\frac{{8PC}}{{\pi {d^2}}}} \right)\)

\(400 = 1.197\left( {\frac{{8P \times 7.5}}{{\pi \times {{20}^2}}}} \right)\)

P = 6998.81 N

Concept:

For closed coil helical spring,

\(Deflection\;under\;load,\;\delta = \frac{{4W{R^3}n}}{{G{r^4}}}\)

Where, R = radius of coil of spring, n = number of turns, G = modulus of elasticity and r = radius of wire of coils

From the above formula, it is clear that the deflection is directly proportional to the number of turns keeping all other parameters same.

Calculation:

Given,

Number of turns in spring A = ½ time the number of turns in B

\(\frac{{{\delta _A}}}{{{\delta _B}}} = \frac{{{n_A}}}{{{n_B}}}\)

We know, \({n_A} = \frac{1}{2}{n_B}\)

\(\therefore \frac{{{\delta _A}}}{{{\delta _B}}} = \frac{1}{2} \Rightarrow {\delta _A} = \frac{1}{2}{\delta _B}\)

Calculation:

The free body diagram of bar is

F is the force in spring

Taking moment about A

F × 1 = 3 × 3

F = 9 kN

Deflection at spring \(\delta = \frac{9}{1} = 9mm\)

∴ Deflection at B

\({\delta _B} = 3\delta = 27mm\)

Concept:

Helical spring subjected to Axial load:

D = Mean coil diameter, d = Wire diameter, n = Number of coils

l = Length of the wire, δ = Axial deflection, α = Angle of helix

The formula for deflection in helical spring is given as

 \(δ = \frac{{8P{D^3}N}}{{G{d^4}}}\)

\(δ \propto \frac{{1}}{{{d^4}}}\) 

where  P = load applied, D = Spring diameter, d = wire diameter, G = Shear modulus, N = no. of turns

Calculation:

Given:

P = 80 N, D = 16 mm, G = 80 GPa, d = 2 mm, N = 8

δ = \(\frac{8~\times~80~\times~16^3~ \times~8}{80~\times~1000~\times~2^4}\) = 16.384 mm