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A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearly
- a)20 N/mm2
- b)33.3 N/mm2
- c)55.6 N/mm2
- d)92.6 N/mm2
Correct answer is option 'C'. Can you explain this answer?
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A close-coiled helical spring is made of 5 mm diameter wire coiled to ...
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A close-coiled helical spring is made of 5 mm diameter wire coiled to ...
To determine the maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, we can use the concept of shear stress in a close-coiled helical spring.
Given data:
- Diameter of wire for the first spring (D1) = 5 mm
- Mean diameter of the first spring (d1) = 50 mm
- Maximum shear stress in the first spring (τ1) = 20 N/mm²
The formula for calculating maximum shear stress in a close-coiled helical spring is given as:
τ = (16 * W * R) / (π * d^3 * n)
where:
τ = maximum shear stress (N/mm²)
W = axial force acting on the spring (N)
R = mean radius of the spring (mm)
d = diameter of the wire (mm)
n = number of active coils in the spring
The number of active coils (n) can be calculated using the formula:
n = (L - C) / (π * d)
where:
L = length of the spring (mm)
C = length of one coil (mm)
Now, let's calculate the maximum shear stress in the second spring.
Step 1: Calculate the number of active coils (n1) for the first spring
Given: C1 = π * d1 (length of one coil for the first spring)
Using the formula: n1 = (L - C1) / (π * d1)
Since the length of the spring is not given, we can assume a value for L (let's say 100 mm)
n1 = (100 - π * 5) / (π * 5) = (100 - 15.7) / 15.7 ≈ 5.36
Step 2: Calculate the mean radius (R1) for the first spring
Given: d1 = 5 mm
Using the formula: R1 = d1 / 2 = 5 / 2 = 2.5 mm
Step 3: Calculate the axial force (W) for both springs
Since the axial force is given as the same for both springs (let's assume W = 100 N)
Step 4: Calculate the maximum shear stress (τ2) for the second spring
Given: d2 = 3 mm, R2 = d2 / 2 = 1.5 mm
Using the formula: τ2 = (16 * 100 * 1.5) / (π * 3^3 * n1)
τ2 = (2400) / (27.57 * 5.36)
τ2 ≈ 56.0 N/mm²
Therefore, the maximum shear stress in the spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force, will be approximately 56.0 N/mm² (option C).
Given data:
- Diameter of wire for the first spring (D1) = 5 mm
- Mean diameter of the first spring (d1) = 50 mm
- Maximum shear stress in the first spring (τ1) = 20 N/mm²
The formula for calculating maximum shear stress in a close-coiled helical spring is given as:
τ = (16 * W * R) / (π * d^3 * n)
where:
τ = maximum shear stress (N/mm²)
W = axial force acting on the spring (N)
R = mean radius of the spring (mm)
d = diameter of the wire (mm)
n = number of active coils in the spring
The number of active coils (n) can be calculated using the formula:
n = (L - C) / (π * d)
where:
L = length of the spring (mm)
C = length of one coil (mm)
Now, let's calculate the maximum shear stress in the second spring.
Step 1: Calculate the number of active coils (n1) for the first spring
Given: C1 = π * d1 (length of one coil for the first spring)
Using the formula: n1 = (L - C1) / (π * d1)
Since the length of the spring is not given, we can assume a value for L (let's say 100 mm)
n1 = (100 - π * 5) / (π * 5) = (100 - 15.7) / 15.7 ≈ 5.36
Step 2: Calculate the mean radius (R1) for the first spring
Given: d1 = 5 mm
Using the formula: R1 = d1 / 2 = 5 / 2 = 2.5 mm
Step 3: Calculate the axial force (W) for both springs
Since the axial force is given as the same for both springs (let's assume W = 100 N)
Step 4: Calculate the maximum shear stress (τ2) for the second spring
Given: d2 = 3 mm, R2 = d2 / 2 = 1.5 mm
Using the formula: τ2 = (16 * 100 * 1.5) / (π * 3^3 * n1)
τ2 = (2400) / (27.57 * 5.36)
τ2 ≈ 56.0 N/mm²
Therefore, the maximum shear stress in the spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force, will be approximately 56.0 N/mm² (option C).
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A close-coiled helical spring is made of 5 mm diameter wire coiled to ...
Maximum shear stress = 16 * force*mean radius of the spring/ pie * diameter of wire^3 max shear stress1 / max shear stress 2 = R1*d2^3/R2*d1^3 20/ max shear stress 2 = 25 * 3^3/15*5^3 max shear stress 2 = 15*5^3*20/25*3^3 max shear stress 2 = 37500/675 max shear stress 2 =55.56 N/mm^2
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A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer?.
A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice A close-coiled helical spring is made of 5 mm diameter wire coiled to 50 mm mean diameter. Maximum shear stress in the spring under the action of an axial force is 20 N/mm2. The maximum shear stress in a spring made of 3 mm diameter wire coiled to 30 mm mean diameter, under the action of the same force will be nearlya)20 N/mm2b)33.3 N/mm2c)55.6 N/mm2d)92.6 N/mm2Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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