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Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal to
- a)P/2
- b)2P/3
- c)P/9
- d)8P/9
Correct answer is option 'D'. Can you explain this answer?
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Two helical springs of the same material and of equal circular cross-s...
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Two helical springs of the same material and of equal circular cross-s...
Explanation:
To find the load carried by the inner spring, we can consider the system as two springs in series.
Concept:
When two springs are connected in series, the total spring constant is given by the reciprocal of the sum of the reciprocals of the individual spring constants.
Step 1: Calculating the spring constants:
The spring constant of a helical spring is given by the formula:
k = (Gd^4) / (8nD^3)
where,
k = spring constant
G = shear modulus of the material
d = wire diameter
n = number of turns
D = mean diameter of the spring (2 * radius)
For the given springs, both have the same material, circular cross-section, length, and number of turns. The only difference is in their radii.
Let's assume the spring constants of the inner and outer springs are k1 and k2 respectively.
Using the formula for spring constant, we can write:
k1 = (Gd^4) / (8nD1^3)
k2 = (Gd^4) / (8nD2^3)
Since the wire diameter (d), number of turns (n), and shear modulus (G) are the same for both springs, we can simplify the equations to:
k1 ∝ 1 / D1^3
k2 ∝ 1 / D2^3
Step 2: Calculating the total spring constant:
When the springs are connected in series, the total spring constant (K) is given by:
1 / K = 1 / k1 + 1 / k2
Substituting the expressions for k1 and k2, we get:
1 / K = 1 / (1 / D1^3) + 1 / (1 / D2^3)
1 / K = D1^3 + D2^3
Step 3: Calculating the load carried by the inner spring:
When the springs are compressed between two parallel planes with a load P, the load is shared by the springs in proportion to their spring constants.
Let's assume the load carried by the inner spring is P1 and by the outer spring is P2.
The load shared by the springs is given by:
P1 / P2 = k1 / k2
Substituting the expressions for k1 and k2, we get:
P1 / P2 = (1 / D1^3) / (1 / D2^3)
P1 / P2 = D2^3 / D1^3
Since the total load is P, we have:
P = P1 + P2
Substituting the expression for P1 / P2, we get:
P = P2 * (1 + D2^3 / D1^3)
Rearranging the equation, we get:
P2 = P / (1 + D2^3 / D1^3)
Hence, the load carried by the inner spring is:
P1 = P - P2
P1 = P - P / (1 + D2^3 / D1^3)
P1 = P * (1 - 1 / (1 + D2^3 / D1^3))
P1 = P * (
To find the load carried by the inner spring, we can consider the system as two springs in series.
Concept:
When two springs are connected in series, the total spring constant is given by the reciprocal of the sum of the reciprocals of the individual spring constants.
Step 1: Calculating the spring constants:
The spring constant of a helical spring is given by the formula:
k = (Gd^4) / (8nD^3)
where,
k = spring constant
G = shear modulus of the material
d = wire diameter
n = number of turns
D = mean diameter of the spring (2 * radius)
For the given springs, both have the same material, circular cross-section, length, and number of turns. The only difference is in their radii.
Let's assume the spring constants of the inner and outer springs are k1 and k2 respectively.
Using the formula for spring constant, we can write:
k1 = (Gd^4) / (8nD1^3)
k2 = (Gd^4) / (8nD2^3)
Since the wire diameter (d), number of turns (n), and shear modulus (G) are the same for both springs, we can simplify the equations to:
k1 ∝ 1 / D1^3
k2 ∝ 1 / D2^3
Step 2: Calculating the total spring constant:
When the springs are connected in series, the total spring constant (K) is given by:
1 / K = 1 / k1 + 1 / k2
Substituting the expressions for k1 and k2, we get:
1 / K = 1 / (1 / D1^3) + 1 / (1 / D2^3)
1 / K = D1^3 + D2^3
Step 3: Calculating the load carried by the inner spring:
When the springs are compressed between two parallel planes with a load P, the load is shared by the springs in proportion to their spring constants.
Let's assume the load carried by the inner spring is P1 and by the outer spring is P2.
The load shared by the springs is given by:
P1 / P2 = k1 / k2
Substituting the expressions for k1 and k2, we get:
P1 / P2 = (1 / D1^3) / (1 / D2^3)
P1 / P2 = D2^3 / D1^3
Since the total load is P, we have:
P = P1 + P2
Substituting the expression for P1 / P2, we get:
P = P2 * (1 + D2^3 / D1^3)
Rearranging the equation, we get:
P2 = P / (1 + D2^3 / D1^3)
Hence, the load carried by the inner spring is:
P1 = P - P2
P1 = P - P / (1 + D2^3 / D1^3)
P1 = P * (1 - 1 / (1 + D2^3 / D1^3))
P1 = P * (
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Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. Can you explain this answer?
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Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. 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 Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. 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 Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. Can you explain this answer?.
Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. 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 Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. 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 Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal toa)P/2b)2P/3c)P/9d)8P/9Correct answer is option 'D'. Can you explain this answer?.
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