Answer :
he approximate induced stress on the rectangular section helical steel compression spring is approximately 1305.5556 psi (rounded to 4 significant figures).
To calculate the induced stress on a rectangular section helical steel compression spring, we can use the formula:
Stress = (F / A) + (M × c / I)
Where:
Stress is the induced stress on the spring
F is the applied force on the spring
A is the cross-sectional area of the rectangular section
M is the bending moment due to the rectangular section
c is the distance from the neutral axis to the extreme fiber of the rectangular section
I is the moment of inertia of the rectangular section
First, let's calculate the cross-sectional area (A) of the rectangular section:
A = t × b
Given that the thickness (t) is 0.3 in and the width (b) is 0.6 in:
A = 0.3 in × 0.6 in
A = 0.18 in²
Next, we need to calculate the moment of inertia (I) of the rectangular section:
I = (t × b^3) / 12
Given the values of t and b:
I = (0.3 in × (0.6 in)^3) / 12
I = 0.0054 in^4
The distance from the neutral axis to the extreme fiber (c) can be assumed to be half the thickness of the rectangular section:
c = t / 2
c = 0.3 in / 2
c = 0.15 in
Now, we can calculate the bending moment (M) due to the rectangular section:
M = F × (mean coil diameter / 2)
Given that the applied force (F) is 235 lb and the mean coil diameter is 1.8 in:
M = 235 lb × (1.8 in / 2)
M = 211.5 lb·in
Finally, we can calculate the induced stress (Stress):
Stress = (F / A) + (M × c / I)
Stress = (235 lb / 0.18 in²) + (211.5 lb·in × 0.15 in / 0.0054 in^4)
Stress ≈ 1305.5556 psi (rounded to 4 significant figures)
Therefore, the approximate induced stress on the rectangular section helical steel compression spring is approximately 1305.5556 psi (rounded to 4 significant figures).
To know more about moment of inertia :
brainly.com/question/14460640
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