The document summarizes an experiment to determine the spring constant of simple extension springs and springs connected in parallel using Hooke's law. Key points: 1) Springs were loaded with weights in increments and the extension was measured to calculate spring constant from the slope of force vs. extension graphs. 2) Theoretical spring constants were also calculated using the springs' material properties and dimensions. 3) For springs in parallel, the total spring constant was calculated as the sum of the individual spring constants, matching the experimental results.

1. Spring test

2. 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

3. 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

4. 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 =
𝐆𝐝 𝟒
𝟖𝐧𝐃 𝟑

5. 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.

6. Calculation:
Theoretical calculation:
As we know theoretically K (stiffness) can be calculated by
𝑘 =
𝑑4 𝐺
8𝑛𝐷3
 For single extension spring
n (no. of turns) = 79
𝑘 =
𝑑4 𝐺
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 =
𝑑4 𝐺
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 =
𝑑4 𝐺
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 ∗ 10−2
𝑁/𝑚𝑚

7. Graphical calculation:
For simple extension spring for n = 79
K =
△𝐹
△𝑥
=
7.16−6.16
44.5−32.5
= 8.33 ∗ 10−2
𝑁/𝑚𝑚
For simple extension spring for n = 52
K =
△𝐹
△𝑥
=
6.16−5.16
20−12
= 12.5 ∗ 10−2
𝑁/𝑚𝑚
For simple extension spring for n = 57
K =
△𝐹
△𝑥
=
7.16−6.16
58−45
= 7.6 ∗ 10−2
𝑁/𝑚𝑚
For parallel combination
K =
△𝐹
△𝑥
=
10.16−9.16
19−14.5
= 22.22 ∗ 10−2
𝑁/𝑚𝑚

8. Observationtable:
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

9. 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

10. 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
forceinnewton
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
force(N)
extension (mm)
force vs extension for n = 57

11. 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
force(N)
extension (mm)
force vs extension for n = 52
0
2
4
6
8
10
12
-5 0 5 10 15 20
force(N)
extension (mm)
force vs parallel springs
force (N)

12. Thank you