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A load of 2 kN is dropped axially on a close coiled helical spring from a height of 250 mm. The spring has 20 effective turns, and it is made of 25 mm diameter wire. The spring index is 8. The amounts of compression produced is 290 mm, the modulus of rigidity for the material of the spring is 84 kN/mm2. The maximum shear stress induced in the spring is __________ MPa.
(A) 285.5
(B) 286.5
Correct answer is between ','. Can you explain this answer?
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A load of 2 kN is dropped axially on a close coiled helical spring fr...
Given data:
- Load (F) = 2 kN
- Height (h) = 250 mm
- Number of effective turns (N) = 20
- Wire diameter (d) = 25 mm
- Spring index (C) = 8
- Compression (x) = 290 mm
- Modulus of rigidity (G) = 84 kN/mm^2
Calculation:
1. Calculate the spring constant (k):
- The spring constant can be calculated using the formula:
k = (Gd^4)/(8ND^3)
where G is the modulus of rigidity, d is the wire diameter, N is the number of turns, and D is the mean coil diameter.
- The mean coil diameter can be calculated using the spring index formula:
D = Cd
where C is the spring index and d is the wire diameter.
- Substitute the given values into the formula to calculate D:
D = 8 * 25 mm = 200 mm
- Substitute the values of G, d, N, and D into the formula to calculate k:
k = (84 kN/mm^2 * (25 mm)^4)/(8 * 20 * (200 mm)^3) = 0.2625 kN/mm
2. Calculate the potential energy (PE):
- The potential energy can be calculated using the formula:
PE = F * h
where F is the load and h is the height.
- Substitute the given values into the formula to calculate PE:
PE = 2 kN * 250 mm = 500 kNm
3. Calculate the maximum shear stress (τ):
- The maximum shear stress can be calculated using the formula:
τ = (8FxD)/(πd^3)
where F is the load, D is the mean coil diameter, and d is the wire diameter.
- Substitute the given values into the formula to calculate τ:
τ = (8 * 2 kN * 200 mm)/(π * (25 mm)^3) = 2.2857 MPa (approximately)
Answer:
The maximum shear stress induced in the spring is approximately 2.2857 MPa.
- Load (F) = 2 kN
- Height (h) = 250 mm
- Number of effective turns (N) = 20
- Wire diameter (d) = 25 mm
- Spring index (C) = 8
- Compression (x) = 290 mm
- Modulus of rigidity (G) = 84 kN/mm^2
Calculation:
1. Calculate the spring constant (k):
- The spring constant can be calculated using the formula:
k = (Gd^4)/(8ND^3)
where G is the modulus of rigidity, d is the wire diameter, N is the number of turns, and D is the mean coil diameter.
- The mean coil diameter can be calculated using the spring index formula:
D = Cd
where C is the spring index and d is the wire diameter.
- Substitute the given values into the formula to calculate D:
D = 8 * 25 mm = 200 mm
- Substitute the values of G, d, N, and D into the formula to calculate k:
k = (84 kN/mm^2 * (25 mm)^4)/(8 * 20 * (200 mm)^3) = 0.2625 kN/mm
2. Calculate the potential energy (PE):
- The potential energy can be calculated using the formula:
PE = F * h
where F is the load and h is the height.
- Substitute the given values into the formula to calculate PE:
PE = 2 kN * 250 mm = 500 kNm
3. Calculate the maximum shear stress (τ):
- The maximum shear stress can be calculated using the formula:
τ = (8FxD)/(πd^3)
where F is the load, D is the mean coil diameter, and d is the wire diameter.
- Substitute the given values into the formula to calculate τ:
τ = (8 * 2 kN * 200 mm)/(π * (25 mm)^3) = 2.2857 MPa (approximately)
Answer:
The maximum shear stress induced in the spring is approximately 2.2857 MPa.
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A load of 2 kN is dropped axially on a close coiled helical spring from a height of 250 mm. The spring has 20 effective turns, and it is made of 25 mm diameter wire. The spring index is 8. The amounts of compression produced is 290 mm, the modulus of rigidity for the material of the spring is 84 kN/mm2. The maximum shear stress induced in the spring is __________ MPa.(A) 285.5(B) 286.5Correct answer is between ','. Can you explain this answer?
Question Description
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