Physics > Elasticity > 10.0 Analogy of Rod as a Spring
Elasticity
1.0 Basic Definitions
2.0 Stress
3.0 Strain
4.0 Hooke’s Law and Modulus of Elasticity
5.0 Types of Modulus of Elasticity
6.0 Elongation of Rod Under its Self Weight
7.0 Bulk Modulus of Elasticity
8.0 Modulus of Rigidity
9.0 The Stress – Strain Curve
10.0 Analogy of Rod as a Spring
11.0 Potential Energy in a Stretched Wire
12.0 Thermal Stress & Strains
13.0 Poisson’s Ratio
14.0 Relation Between $Y$, $B$, $\eta$ and $\sigma$
10.1 Types of Arrangement
(a) Series Combination
The arrangement of rod shown in Fig. E. 19. (a) can be resolved into series spring combination as shown in Fig. E. 19. (b).
Now we know how to find the equivalent spring constant as we have learned in chapter Simple Harmonic Motion.
Therefore,
$$\begin{equation} \begin{aligned} \frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \\ {k_{eq}} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} = \frac{{\frac{{{A_1}{Y_1}}}{{{l_1}}}\frac{{{A_2}{Y_2}}}{{{l_2}}}}}{{\frac{{{A_1}{Y_1}}}{{{l_1}}} + \frac{{{A_2}{Y_2}}}{{{l_2}}}}} \\ {k_{eq}} = \frac{{{A_1}{A_2}{Y_1}{Y_2}}}{{{A_1}{Y_1}{l_2} + {A_2}{Y_2}{l_1}}} \\\end{aligned} \end{equation} $$
(b) Parallel Combination
The arrangement of rod shown in Fig. E. 20. (a) can be resolved into parallel spring combination as shown in Fig. E. 20. (b).
Now we know how to find the equivalent spring constant as we have learned in chapter Simple Harmonic Motion.
Therefore,
$$\begin{equation} \begin{aligned} {k_{eq}} = {k_1} + {k_2} \\ {k_{eq}} = \frac{{{A_1}{Y_1}}}{{{l_1}}} + \frac{{{A_2}{Y_2}}}{{{l_2}}} \\ {k_{eq}} = \frac{{{A_1}{Y_1}{l_2} + {A_2}{Y_2}{l_2}}}{{{l_1}{l_2}}} \\\end{aligned} \end{equation} $$
Example 6. Find the time period of the arrangement as shown in the Fig. E. 21.
Solution: As we know that the above arrangement of bars can be easily resolved in the combination of spring as shown in the Fig. E. 22 (a).
Now, we have to solve this combination of spring. Spring $1$ & $2$ are in series, therefore this combination is now resolved as in Fig. E. 22. (b).
$$\begin{equation} \begin{aligned} \frac{1}{{{k_{12}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \\ {k_{12}} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} \\\end{aligned} \end{equation} $$
Now spring $k_12$ & $k_3$ are in parallel arrangement therefore,
$$\begin{equation} \begin{aligned} {k_{eq}} = {k_{12}} + {k_3} \\ {k_{eq}} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} + {k_3} \\\end{aligned} \end{equation} $$
So, as we know the time period of the combination can be given by, $$T = 2\pi \sqrt {\frac{m}{{{k_{eq}}}}} $$
where as
$${k_{eq}} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} + {k_3}\ and\ {k_1} = \frac{{{A_1}{Y_1}}}{{{l_1}}},{k_2} = \frac{{{A_2}{Y_2}}}{{{l_2}}},{k_3} = \frac{{{A_3}{Y_3}}}{{{l_3}}}$$
