Spring test

Aim: To find out spring constant and verify Hooks law for a simple

extension spring and 2 identical springs mounted in parallel.

Apparatus: spring, load, scale, mounting frame.

Theory:

Derivation of the Formula:

In order to derive a necessary formula which governs the behavior of springs, consider a closed coiled spring subjected to an axial load W.

Type equation here.

Fig: springs

W = axial load

D = mean coil diameter

d = diameter of spring wire

n = number of active coils

C = spring index = D / d For circular wires

l = length of spring wire

G = modulus of rigidity

x = deflection of spring

q = Angle of twist

When the spring is being subjected to an axial load to the wire of the spring gets be twisted like a shaft.

If q is the total angle of twist along the wire and x is the deflection of spring under the action of load W along the axis of the coil, so that

x = D / 2 q

Again l = p D n [ consider ,one half turn of a close coiled helical spring ]

Fig: springs

Assumptions:

(1) The Bending & shear effects may be neglected (2) (2) For the purpose of derivation of formula, the helix angle is

considered to be so small that it may be neglected.

Any one coil of a such a spring will be assumed to lie in a plane which is nearly perpendicular to the axis of the spring. This requires that adjoining coils be close together. With this limitation, a section taken

perpendicular to the axis the spring rod becomes nearly vertical. Hence to maintain equilibrium of a segment of the spring, only a shearing force V = F and Torque T = F. r are required at any X – section. In the analysis of springs it is customary to assume that the shearing stresses caused by the direct shear force is uniformly distributed and is negligible

So applying the torsion formula.

Using the torsion formula i.e.

𝑻

𝑱 =

𝝉

𝒓 =

𝑮𝜽

𝒍

As, we know that

T = W*d/2 ,

θ=𝟐∗𝒙

𝑫

W ∗ d/2

∏d4

32

=G2𝑥/D

∏D. n

𝑥 =8𝑊𝐷3n

𝐺𝑑4

As, 𝑘 = 𝑊

𝑥

𝐤 =𝐆𝐝𝟒

𝟖𝐧𝐃𝟑

Procedure:

Simple extension of springs

For simple extension spring, measure the thickness ,pull to pull length

values of spring by using Vernier caliper.

Mount the spring onto mounting frame and add weight in steps of 100

grams.

Tabulate the extension obtained from the scales and corresponding

weight.

Plot Force(N) versus extension(mm).

From graph, Slope would give value of K.

Verify experimental stiffness with the value of stiffness obtained from

theoretical expression of stiffness.

Spring in parallel

For identical springs in parallel, measure the thickness, full length

values of spring by using vernier caliper.

Mount the parallel spring’s setup and tabulates the extension obtained

from scales for corresponding weight which are added in set of 100

grams.

Repeat the same experiment with individual springs.

Plot Force(N) versus extension(mm) for parallel setup of spring.

Calculation:

Theoretical calculation:

As we know theoretically K (stiffness) can be calculated by

𝑘 = 𝑑4𝐺

8𝑛𝐷3

For single extension spring

n (no. of turns) = 79

𝑘 = 𝑑4𝐺

8𝑛𝐷3 = (1.08∗10−3)4∗(77∗109)

8∗79∗(12.74∗10−3)3 = 8.02*10−2 N/mm

For n = 57

𝑘 = 𝑑4𝐺

8𝑛𝐷3 = (1.12∗10−3)4∗(77∗109)

8∗57∗(12.68∗10−3)3 = 13.03*10−2 N/mm

For n = 52

𝑘 = 𝑑4𝐺

8𝑛𝐷3 = (1.12∗10−3)4∗(77∗109)

8∗52∗(12.68∗10−3)3 = 14.3*10−2 N/mm

For parallel combination

𝐾𝑓𝑖𝑛𝑎𝑙 = 𝐾1 + 𝐾2

𝐾𝑓𝑖𝑛𝑎𝑙 = 14.3 ∗ 10−2 + 13.03 ∗ 10−2

= 27 ∗ 10−2𝑁/𝑚𝑚

Graphical calculation:

For simple extension spring for n = 79

K =△𝐹

△𝑥=

7.16−6.16

44.5−32.5= 8.33 ∗ 10−2𝑁/𝑚𝑚

For simple extension spring for n = 52

K =△𝐹

△𝑥=

6.16−5.16

20−12= 12.5 ∗ 10−2𝑁/𝑚𝑚

For simple extension spring for n = 57

K =△𝐹

△𝑥=

7.16−6.16

58−45= 7.6 ∗ 10−2𝑁/𝑚𝑚

For parallel combination

K =△𝐹

△𝑥=

10.16−9.16

19−14.5= 22.22 ∗ 10−2𝑁/𝑚𝑚

Observation table:

For simple extension spring

D= 12.74mm, d = 1.08mm, n = 79, G = 77GPa

serial no

mass (kg)

force (N)

scale reading(mm) extension(mm)

1 0.016 0.16 179 0 2 0.116 1.16 180 1 3 0.216 2.16 180 1

4 0.316 3.16 180.5 1.5 5 0.416 4.16 189 10

6 0.516 5.16 200 21 7 0.616 6.16 211.5 32.5

8 0.716 7.16 223.5 44.5 9 0.816 8.16 235 56

For a individual spring which where mounted in parallel

D=12.68mm, d = 1.12mm, n1 = 52, n2 = 57, G = 77GPa

serial no

mass (kg)

force (N)

scale reading(mm) extension(mm)

n= 52 n = 57 n=52 n=57

1 0.016 0.16 152 153.5 0 0 2 0.116 1.16 152.5 153.5 0.5 0

3 0.216 2.16 153 153.5 1 0 4 0.316 3.16 153.5 159.5 1.5 6

5 0.416 4.16 156 172.5 4 19 6 0.516 5.16 164 185.5 12 32

7 0.616 6.16 172 198.5 20 45 8 0.716 7.16 179.5 211.5 27.5 58 9 0.816 8.16 187 225 35 71.5

For a spring which where mounted in parallel

serial no

mass (kg)

force (N)

scale reading(mm) extension(mm)

1 0.016 0.16 199 0

2 0.116 1.16 199 0 3 0.216 2.16 199 0

4 0.316 3.16 199.5 0.5 5 0.416 4.16 199.5 0.5 6 0.516 5.16 199.5 0.5

7 0.616 6.16 201 2 8 0.716 7.16 203.5 4.5

9 0.816 8.16 209 10 10 0.916 9.16 213.5 14.5

11 1.016 10.16 218 19

Graph:

For simple extension spring n = 79

For simple extension spring n = 57

0

1

2

3

4

5

6

7

8

9

0 10 20 30 40 50 60

forc

e i

n n

ew

ton

extension in mm

force vs extension for a spring

force (N)

0

1

2

3

4

5

6

7

8

9

-10 0 10 20 30 40 50 60 70 80

forc

e (

N)

extension (mm)

force vs extension for n = 57

For simple extension spring n = 52

For parallel combination:

0

1

2

3

4

5

6

7

8

9

0 5 10 15 20 25 30 35 40

forc

e (

N)

extension (mm)

force vs extension for n = 52

0

2

4

6

8

10

12

-5 0 5 10 15 20

forc

e (

N)

extension (mm)

force vs parallel springs

force (N)

Thank you