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A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 is
- a)120/π
- b)160/π
- c)100/π
- d)80/π
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A close-coiled helical spring has 100 mm mean diameter and is made of ...
Closed coiled helical spring subjected to axial load (W) means that every section is subjected to torsion of WR where R is the radius of spring. From torsion formula,
d= diameter of wire
d = 10mm
W = 100 N
Shear stress due to load,
d= diameter of wire
d = 10mm
W = 100 N
Shear stress due to load,
Most Upvoted Answer
A close-coiled helical spring has 100 mm mean diameter and is made of ...
First, we need to find the shear modulus (G) of the steel wire using the given modulus of rigidity (G) and Poisson's ratio (v). We can assume v to be 0.3 for steel.
G = E/2(1+v) = 84 GPa/2(1+0.3) = 42 GPa
Next, we need to find the shear stress (τ) developed in the spring.
τ = (16/π) * (F/D^3) * N
where F is the axial load, D is the mean diameter, and N is the number of turns.
F = 100 N
D = 100 mm = 0.1 m
N = 20
d = 10 mm = 0.01 m
Substituting the values, we get:
τ = (16/π) * (100/0.1^3) * 20/(π*0.01^2/4) = 120 N/mm^2
Therefore, the shearing stress developed in the spring is 120 N/mm^2. Answer: \boxed{120}.
G = E/2(1+v) = 84 GPa/2(1+0.3) = 42 GPa
Next, we need to find the shear stress (τ) developed in the spring.
τ = (16/π) * (F/D^3) * N
where F is the axial load, D is the mean diameter, and N is the number of turns.
F = 100 N
D = 100 mm = 0.1 m
N = 20
d = 10 mm = 0.01 m
Substituting the values, we get:
τ = (16/π) * (100/0.1^3) * 20/(π*0.01^2/4) = 120 N/mm^2
Therefore, the shearing stress developed in the spring is 120 N/mm^2. Answer: \boxed{120}.
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A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer?
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A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer?.
A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 isa)120/πb)160/πc)100/πd)80/πCorrect answer is option 'D'. Can you explain this answer?.
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