We appreciate your visit to A steel torsion spring has a wire diameter of 0 125 in a spring index of 6 and 10 coils It is loaded with a. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
A steel torsion spring has a wire diameter of 0.125 in, spring index of 6, and number of coils of 10. It is loaded with a bending moment of 114 lb-in and is inserted in an assembly.
To calculate the energy stored in the steel torsion spring, we can use the formula for torsional energy:
U = (1/2) × k × theta^2
Where:
U is the torsional energy stored in the spring
k is the torsional spring constant
theta is the angular deflection in radians
The torsional spring constant (k) can be calculated using the formula:
k = (G × J) / L
Where:
G is the shear modulus of the material
J is the torsion constant (also known as the polar moment of inertia)
L is the length of the spring
The torsion constant (J) can be calculated using the formula:
J = (π/32) × (D^4 - d^4)
Where:
π is a mathematical constant approximately equal to 3.14159
D is the outer diameter of the spring
d is the wire diameter of the spring
First, let's calculate the torsion constant (J) using the given values:
D = wire diameter + (2 × spring index)
= 0.125 in + (2 × 6)
= 0.125 in + 12
= 12.125 in
d = wire diameter
= 0.125 in
J = (π/32) × ((12.125^4) - (0.125^4))
= (π/32) × (23603.12207 - 0.00293)
= (π/32) × (23603.11914)
= 7367.97371 in^4
Next, we can calculate the torsional spring constant (k) using the given values:
L = number of coils × wire diameter
= 10 * 0.125 in
= 1.25 in
G is the shear modulus of the material, which depends on the specific type of steel used in the spring. Without that information, we cannot calculate the exact value of G. However, we can estimate the torsional spring constant (k) using a typical value for G of around 11.7 x 10^6 psi (80 GPa) for steel.
k = (G × J) / L
= (11.7 x 10^6 psi * 7367.97371 in^4) / 1.25 in
≈ 69.59766 x 10^6 lb-in/rad (rounded to 4 significant figures)
Now, we can calculate the energy stored in the spring (U) using the given bending moment (M) and the torsional spring constant (k):
U = (1/2) ×k × theta^2
Given bending moment (M) = 114 lb-in
We know that the relationship between bending moment (M) and angular deflection (theta) is:
M = k × theta
Rearranging the equation:
theta = M / k
theta = 114 lb-in / 69.59766 x 10^6 lb-in/rad
≈ 0.000001636 rad (rounded to 4 significant figures)
Now, substitute the values into the energy equation:
U = (1/2) × k × theta^2
= (1/2) × 69.59766 x 10^6 lb-in/rad × (0.000001636 rad)^2
≈ 0.000000000114 ft-lb (rounded to 4 significant figures)
Therefore, the energy stored in the steel torsion spring is approximately 0.000000000114 ft-lb (rounded to 4 significant figures).
To know more about torsion constant :
brainly.com/question/14378814
#SPJ11
Thanks for taking the time to read A steel torsion spring has a wire diameter of 0 125 in a spring index of 6 and 10 coils It is loaded with a. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
