# two springs are connected end to end,same load is used.you need to relate spring constant of one spring with other.what is the effective spring constants in terms of k1 and k2.

### Asked by archana | 18th Sep, 2015, 08:18: AM

Expert Answer:

### Springs connected in** series** (Figure B):
Two strings having individual spring stiffness of k_{1} and k_{2} units connected in series as shown in the figure.
Since each spring supports the load W, the total deflection of the spring will be the sum of the deflections of the individual springs.
Deflection of first spring, y_{1}=W/k_{1} and that of the second spring, y_{2}=W/k_{2}.
As the deflection of the spring with constant k is given by, y=y_{1}+y_{2}.
Therefore, W/k=W/k_{1}+W/k_{2 }or 1/k=1/k_{1}+1/k_{2} or effective of the spring, k=k_{1}k_{2}/(k_{1}+k_{2}).
Springs connected in **parallel** (Figure A):
Two strings having individual spring stiffness k_{1} and k_{2} are connected in parallel.
Since the load is shared by the two springs jointly the sum of the loads W_{1} and W_{2} taken by individual springs must be equal to the axial load W applied on the composite spring.
Thus the same deflection 'y' of the two springs the load taken by the individual springs will be:
Load taken by first spring, W_{1}=k_{1}y, and the load taken by the second spring, W_{2}=k_{2}y.
With k as the stiffness of the spring the load taken by the spring is W=ky.
W=W_{1}+W_{2}, we get, ky=k_{1}y+k_{2}y
The effective of the springs connected in parallel is given by, k=k_{1}+k_{2}.

**series**(Figure B):

_{1}and k

_{2}units connected in series as shown in the figure.

_{1}=W/k

_{1}and that of the second spring, y

_{2}=W/k

_{2}.

_{1}+y

_{2}.

_{1}+W/k

_{2 }or 1/k=1/k

_{1}+1/k

_{2}or effective of the spring, k=k

_{1}k

_{2}/(k

_{1}+k

_{2}).

**parallel**(Figure A):

_{1}and k

_{2}are connected in parallel.

_{1}and W

_{2}taken by individual springs must be equal to the axial load W applied on the composite spring.

_{1}=k

_{1}y, and the load taken by the second spring, W

_{2}=k

_{2}y.

_{1}+W

_{2}, we get, ky=k

_{1}y+k

_{2}y

_{1}+k

_{2}.

### Answered by Faiza Lambe | 18th Sep, 2015, 11:25: AM

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