Equivalent Stiffness MCQ Quiz in తెలుగు - Objective Question with Answer for Equivalent Stiffness - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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)

\(\begin{array}{l} \therefore \frac{1}{{{k_{eq}}}} = \frac{1}{k} + \frac{1}{{2k}}\\ \therefore {k_{eq}} = \frac{{2{k^2}}}{{2k + k\;}} = \frac{{2{k^2}}}{{3k}} = \frac{{2k}}{3} \end{array}\)

Now,

This k_eq is connected in parallel with spring 4

\(\therefore {k_{eq}} = \frac{{2k}}{3} + k = \frac{{5k}}{3}\)

Now,

Natural frequency is given by:

\({\omega _n} = \sqrt {\frac{{{k_{eq}}}}{m}}\)

\(\therefore {\omega _n} = \sqrt {\frac{{5k}}{{3m}}}\)

Concept:

Force in spring 1 is,

F₁ = W

Taking moment about the fixed end,

⇒ F1 × l1 = F2 × l2

Force in spring 2,

⇒ F₂ =  \(\frac{Wl_1}{l_2}\) 

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+ \(\frac{l_1}{l_2}\) (Deflection of spring 2)

\(\Delta = \frac{W}{{{S_1}}} + \frac{{{l_1}}}{{{l_2}}}\left( {\frac{{W{l_1}/{l_2}}}{{{S_2}}}} \right)\)

\(⇒ \Delta = W\left( {\frac{1}{{{S_1}}} + \frac{{{{\left( {{l_1}/{l_2}} \right)}^2}}}{{{S_2}}}} \right)\)

\(⇒ \Delta = mg\left( {\frac{{{S_2} + {S_1}{{\left( {{l_1}/{l_2}} \right)}^2}}}{{{S_1}{S_2}}}} \right)\)

Now, the natural frequency of the vibration,

\({w_n} = \sqrt {\frac{g}{\Delta }} = \sqrt {\frac{{{S_1}{S_2}}}{{\left( {{S_2} + {S_1}{{\left( {{l_1}/{l_2}} \right)}^2}} \right)m}}} \)

Calculation:

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

Substitution of the values,

\({w_n} = \sqrt {\frac{{100 \times 100 }}{{\left( {100 + 100{{\left( {\frac{{200}}{{160}}} \right)}^2}} \right) \times 10}}} \)

⇒ w_n = 1.97 rad/s

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\)

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

\(\frac{1}{{{k_e}}} = \frac{1}{{2k}} + \frac{1}{{2k}} = \frac{1}{k}\)

k_e = k

Explanation:

Displacing the rod by a small distance x.

Taking moment about '0'

Kx × 2L = mg × L

\(x = \frac{{mg}}{{2k}}\)

From similar triangle property

 \(\frac{x}{{2L}} = \frac{\delta }{L}\)

\(\delta = \frac{x}{2} = \frac{{mg}}{{4k}}\)

Using static deflection of the mass 'm'

\({\omega _n} = \sqrt {\frac{g}{\delta }} = \sqrt {\frac{{4K}}{m}} \)

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)

\({k_e} = \;\frac{{k\; \times \;k'}}{{k\; + \;k'}}\)      (1)

Calculation:

Here,

x_B = Lθ, x_C = 3Lθ, x_D = 4Lθ

Now,

Let PE be potential energy

PE_B + PE_C = PE_D

\(\frac{1}{2}2Kx_B^2 + \frac{1}{2}Kx_C^2 = \;\frac{1}{2}K'x_D^2\) 

\(\frac{1}{2}2K\;{\left( {L\theta } \right)^2} + \frac{1}{2}K{\left( {3L\theta } \right)^2} = \;\frac{1}{2}K'{\left( {4L\theta } \right)^2}\)

\(K' = \;\frac{{11K}}{{16}}\)      (2)

Substituting (2) in (1) gives

\({k_e} = \frac{{11k}}{{27}}\)     (3)

= 16.296 N/m   (option a)

Let the natural frequency be ω_n

\({\omega _n} = \;\sqrt {\frac{{{k_e}}}{m}}\)     (4)

Given:

 m = 4.5 kg and k = 40 N/m

\({\omega _n} = \;\sqrt {\frac{{11k}}{{27m}}}\)      (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:

\(\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\)

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

\({k_{12}} = \frac{k}{2} + \frac{k}{2} = k\)

Between k₁₂ and k₃:

k₁₂ and k₃ are in series hence

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_{12}}}} + \frac{1}{{{k_3}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}\)

\({k_{eq}} = \frac{k}{2}\)

Hence Natural frequency \({\omega _n} = \sqrt {\frac{{{k_{eq}}}}{m}} = \sqrt {\frac{k}{{2m}}} \)

Concept:

When arrangement of springs are in parallel

k_e = k₁ + k₂ + k₃ + ....

When the arrangement of springs are in series

\(\frac{1}{k_e}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+...\)

And the natural frequency is given as,  \(\omega _n=\frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{{{\rm{k}}_{{\rm{eq}}}}}}{{\rm{M}}}} \)

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

\({k_{e}} = \frac{{3{\rm{k}} \times {\rm{\;k}}}}{{3{\rm{k}} + {\rm{k}}}}\)

\({k_{e}} = \frac{{3{\rm{k}}}}{4}\)

\({k_{e1}} = {k_1} + {k_2}\) 

\(\frac{1}{{{k_{e2}}}} = \frac{1}{{{k_{e1}}}} + \frac{1}{{{k_3}}} \Rightarrow {k_{e2}} = \frac{{{k_3}\;{k_{e1}}}}{{{e_{e1}} + {k_3}}} = \frac{{\left( {{k_1} + {k_2}} \right){k_3}}}{{{k_1} + {k_2} + {k_3}}}\) 

Equation of motion is:

\({m_{eq}}\ddot x + {k_{eq}}x = 0\)

\(m\ddot x + \frac{{\left( {{k_1} + {k_2}} \right){k_3}}}{{{k_1} + {k_2} + {k_3}}}x = 0\)

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \)

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

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{k_e}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{2k}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{k_{eq}}=\frac{1+2+2}{2k}\)

\(\frac{1}{k_{eq}}=\frac{5}{2k}\)

\(k_{eq}=\frac{2k}{5}\)

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