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A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?
- a)5.98 N/mm
- b)8.15 N/mm
- c)4.32 N/mm
- d)7.00 N/mm
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A helical tension spring, which is fixed to a rigid support at one en...
To determine the actual spring rate of the helical tension spring, we need to use the formula:
Spring Rate (k) = (G * d^4) / (8 * n * D^3)
Where:
- G is the modulus of rigidity (75750 N/mm^2)
- d is the wire diameter (in mm)
- n is the number of coils (38)
- D is the mean coil diameter
First, we need to calculate the mean coil diameter using the spring index:
Spring Index = D / d
Given that the spring index is 5, we can calculate the mean coil diameter:
Mean Coil Diameter (D) = Spring Index * Wire Diameter (d)
Next, we can substitute the values into the formula to calculate the spring rate:
k = (G * d^4) / (8 * n * D^3)
Substituting the given values:
- G = 75750 N/mm^2
- n = 38
We can calculate the spring rate for each wire diameter (1, 2, 3, 4, and 5 mm) using the above formula.
For wire diameter = 1 mm:
D = 5 * 1 = 5 mm
k = (75750 * 1^4) / (8 * 38 * 5^3) = 5.98 N/mm
For wire diameter = 2 mm:
D = 5 * 2 = 10 mm
k = (75750 * 2^4) / (8 * 38 * 10^3) = 1.50 N/mm
For wire diameter = 3 mm:
D = 5 * 3 = 15 mm
k = (75750 * 3^4) / (8 * 38 * 15^3) = 0.67 N/mm
For wire diameter = 4 mm:
D = 5 * 4 = 20 mm
k = (75750 * 4^4) / (8 * 38 * 20^3) = 0.37 N/mm
For wire diameter = 5 mm:
D = 5 * 5 = 25 mm
k = (75750 * 5^4) / (8 * 38 * 25^3) = 0.24 N/mm
Therefore, the actual spring rate of the helical tension spring is 5.98 N/mm.
Spring Rate (k) = (G * d^4) / (8 * n * D^3)
Where:
- G is the modulus of rigidity (75750 N/mm^2)
- d is the wire diameter (in mm)
- n is the number of coils (38)
- D is the mean coil diameter
First, we need to calculate the mean coil diameter using the spring index:
Spring Index = D / d
Given that the spring index is 5, we can calculate the mean coil diameter:
Mean Coil Diameter (D) = Spring Index * Wire Diameter (d)
Next, we can substitute the values into the formula to calculate the spring rate:
k = (G * d^4) / (8 * n * D^3)
Substituting the given values:
- G = 75750 N/mm^2
- n = 38
We can calculate the spring rate for each wire diameter (1, 2, 3, 4, and 5 mm) using the above formula.
For wire diameter = 1 mm:
D = 5 * 1 = 5 mm
k = (75750 * 1^4) / (8 * 38 * 5^3) = 5.98 N/mm
For wire diameter = 2 mm:
D = 5 * 2 = 10 mm
k = (75750 * 2^4) / (8 * 38 * 10^3) = 1.50 N/mm
For wire diameter = 3 mm:
D = 5 * 3 = 15 mm
k = (75750 * 3^4) / (8 * 38 * 15^3) = 0.67 N/mm
For wire diameter = 4 mm:
D = 5 * 4 = 20 mm
k = (75750 * 4^4) / (8 * 38 * 20^3) = 0.37 N/mm
For wire diameter = 5 mm:
D = 5 * 5 = 25 mm
k = (75750 * 5^4) / (8 * 38 * 25^3) = 0.24 N/mm
Therefore, the actual spring rate of the helical tension spring is 5.98 N/mm.
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A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer?
Question Description
A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer?.
A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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Here you can find the meaning of A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A helical tension spring, which is fixed to a rigid support at one end and carries weight at other free end. The pointer attached to the free end shows the reading of 83 mm. The maximum capacity of spring balance is 498 N. The spring index is 5. The ultimate tensile strength of spring material is 1350 MPa. The permissible shear stress for spring material can be taken as 75% of the ultimate tensile strength. Modulus of rigidity for material is 75750 N/mm2. If curvature effect is negligible and wire diameter available are in the range of 1, 2, 3, 4 and 5 mm and the number of coil is 38. What is the actual spring rate of spring?a) 5.98 N/mmb) 8.15 N/mmc) 4.32 N/mmd) 7.00 N/mmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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