Equivalent Stiffness MCQ Quiz in தமிழ் - Objective Question with Answer for Equivalent Stiffness - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

From the above figure we can see that spring (2) and spring (3) are connected in parallel

∴ k + k = 2k

Now,

Spring (1) is connected in series with spring (2) and (3)

Now,

This k_eq is connected in parallel with spring 4

Now,

Natural frequency is given by:

Concept:

Force in spring 1 is,

F₁ = W

Taking moment about the fixed end,

⇒ F1 × l1 = F2 × l2

Force in spring 2,

⇒ F₂ =   

Now, using the superposition principle (Taking one spring at a time to find the deflection of the mass m)

The direction of mass = Deflection of the mass m due spring 1 when spring 2 is not there + Deflection of the mass m due spring 2 when spring 1 is not there

The direction of mass = Deflection of the spring 1+  (Deflection of spring 2)

Now, the natural frequency of the vibration,

Calculation:

Given:  S1 = 100 N/m S2 = 100 N/m, l1 = 200 mm l2 = 160 mm, m = 10 kg

Substitution of the values,

⇒ w_n = 1.97 rad/s

Concept:

Springs in series:

Springs in parallel:

k_e = k₁ + k₂

Calculation:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k₁

k₁ = k + k

k₁ = 2k

Now k₁ and 2k are in series connection, let their equivalent spring constant is k_e

k_e = k

Explanation:

Displacing the rod by a small distance x.

Taking moment about '0'

Kx × 2L = mg × L

From similar triangle property

Using static deflection of the mass 'm'

Concept:

Make an equivalent system with a single spring and mass

Calculation:

The system can be replaced with a single spring as follows:

Equivalent stiffness (K_e)

      (1)

Calculation:

Here,

x_B = Lθ, x_C = 3Lθ, x_D = 4Lθ

Now,

Let PE be potential energy

PE_B + PE_C = PE_D

      (2)

Substituting (2) in (1) gives

     (3)

= 16.296 N/m   (option a)

Let the natural frequency be ω_n

     (4)

Given:

 m = 4.5 kg and k = 40 N/m

      (4)

ω_n = 1.9 rad/s    (option b)

Concept:

When two spring connected in parallel:

k_eq = k₁ + k₂

When two spring in series:

Calculation:

Given:

k₁ = k/2, k₂ = k/2, k₃ = k.

For the given spring-mass system (1) and (2) are in parallel whose equivalent stiffness is k₁₂. The overall equivalent stiffness will be in series between k₁₂ and k₃.

Between 1 and 2:

k₁₂ = k1 + k2

Between k₁₂ and k₃:

k₁₂ and k₃ are in series hence

Hence Natural frequency 

Concept:

When arrangement of springs are in parallel

k_e = k₁ + k₂ + k₃ + ....

When the arrangement of springs are in series

And the natural frequency is given as,  

Calculation:

Three springs of stiffness "k" are in parallel, so combined spring stiffness k_e1 = 3k

Now springs of stiffness k_e1 and k are in series, so equivalent spring stiffness

Equation of motion is:

Concept:

Springs in series:

Springs in parallel:

ke = k1 + k2

Calculation:

Given:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k1

k1 = k + k

k1 = 2k

Now k1, k and k are in series connection, let their equivalent spring constant is keq

k_eq = 0.4 k

Explanation:

• The stiffness of a material is its resistance to elastic deformation. It is a measure of the elastic modulus for the particular deformation mode

• Spring stiffness is the force or load (F) required to cause unit deflection in spring (Δ)

⇒ k = F/Δ