JEE Exam  >  JEE Questions  >  Design a helical compression spring for maxim... Start Learning for Free
Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm?
Most Upvoted Answer
Design a helical compression spring for maximum load of 1000N , for a ...
Designing a Helical Compression Spring for Maximum Load of 1000N

Introduction:
A helical compression spring is a type of spring that is designed to resist compressive forces. It is made by winding a wire around a cylindrical shape, with the turns of the wire touching each other. When a load is applied to the spring, the wire compresses, and the spring stores energy. Helical compression springs are used in a variety of applications, including automotive suspension systems, industrial machinery, and consumer products.

Design Parameters:
To design a helical compression spring, we need to know the following parameters:

1. Maximum Load: The maximum load that the spring will be subjected to.
2. Deflection: The amount of compression that the spring will undergo when the load is applied.
3. Spring Index: The ratio of the mean coil diameter to the wire diameter.

Design Process:

1. Determine the Wire Diameter:
The wire diameter is calculated using the following formula:

d = (K*F)/(G*sqrt(N))

Where,
d = Wire Diameter
K = Spring Constant
F = Maximum Load
G = Modulus of Rigidity
N = Number of Active Coils

Here, the spring index is 5, so we can use the formula:

K = (4*C-1)/(4*C-4) + 0.615/C

Where,
C = Spring Index

Substituting the values, we get:

C = 5
K = 1.25

Substituting the values in the formula for wire diameter, we get:

d = (1.25*1000)/(84*sqrt(6)) = 2.45 mm

2. Determine the Mean Coil Diameter:
The mean coil diameter is calculated using the following formula:

D = d*(N+2)/C

Where,
D = Mean Coil Diameter
d = Wire Diameter
N = Number of Active Coils
C = Spring Index

Substituting the values, we get:

D = 2.45*(6+2)/5 = 3.68 cm

3. Determine the Number of Active Coils:
The number of active coils is calculated using the following formula:

N = (F*G*(N+2)^3)/(8*d^3*max_shear_stress)

Where,
F = Maximum Load
G = Modulus of Rigidity
d = Wire Diameter
max_shear_stress = Maximum Permissible Shear Stress

Substituting the values, we get:

N = (1000*10^3*84*(6+2)^3)/(8*(2.45*10^-3)^3*420*10^6) = 9.48

Since the number of coils must be an integer, we can round up to 10.

4. Check the Spring for Solid Height and Buckling:
The solid height is the length of the spring when it is fully compressed. It is calculated using the following formula:

H = d*(N-1)

Substituting the values, we get:

H = 2.45*(10-1) = 22.05 mm

The buckling load is the load at which the spring will buckle due to its own weight. It is calculated using the following formula:

F_c = (pi^2*E*d^4)/(8*(N/d)^2)

Where,
E = Modulus of Elasticity
Explore Courses for JEE exam
Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm?
Question Description
Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm?.
Solutions for Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? defined & explained in the simplest way possible. Besides giving the explanation of Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm?, a detailed solution for Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? has been provided alongside types of Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? theory, EduRev gives you an ample number of questions to practice Design a helical compression spring for maximum load of 1000N , for a deflection of 25 mm , using the value of spring index 5 . The maximum permissible shear stress for spring wire is 420 MPa and modulus of rigidity 84 KN / sq mm? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev