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A mild steel tube 4m long, 30mm, internal diameter and 4mm thick is used as a strut with both ends hinged. Collapsing load is:-
Take E = 2.1 × 105 N/mm2
Take E = 2.1 × 105 N/mm2
- a)8.1 kN
- b)10 kN
- c)2.3kN
- d)16 kN
Correct answer is 'A'. Can you explain this answer?
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A mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused...
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A mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused...
To find the collapsing load of the mild steel tube, we can use Euler's formula for buckling load:
P = (π^2 * E * I) / (l^2)
Where:
P = Buckling load
E = Young's modulus of the material (2.1 x 10^11 Pa for mild steel)
I = Moment of inertia of the cross-sectional area
l = Length of the strut
First, let's calculate the moment of inertia (I) for the cross-sectional area of the tube. The moment of inertia for a hollow circular section can be calculated using the formula:
I = (π/64) * (D^4 - d^4)
Where:
D = External diameter of the tube
d = Internal diameter of the tube
Given:
External diameter (D) = 30 mm + 2 * 4 mm = 38 mm = 0.038 m
Internal diameter (d) = 30 mm = 0.03 m
Thickness (t) = (D - d)/2 = (0.038 - 0.03)/2 = 0.004 m
Substituting the values into the formula, we get:
I = (π/64) * ((0.038)^4 - (0.03)^4) = 9.61 x 10^-9 m^4
Next, substitute the values of E, I, and l into the buckling load formula:
P = (π^2 * E * I) / (l^2)
P = (π^2 * 2.1 x 10^11 Pa * 9.61 x 10^-9 m^4) / (4^2)
P = 1.47 x 10^6 N
Therefore, the collapsing load of the mild steel tube is approximately 1.47 x 10^6 Newtons.
P = (π^2 * E * I) / (l^2)
Where:
P = Buckling load
E = Young's modulus of the material (2.1 x 10^11 Pa for mild steel)
I = Moment of inertia of the cross-sectional area
l = Length of the strut
First, let's calculate the moment of inertia (I) for the cross-sectional area of the tube. The moment of inertia for a hollow circular section can be calculated using the formula:
I = (π/64) * (D^4 - d^4)
Where:
D = External diameter of the tube
d = Internal diameter of the tube
Given:
External diameter (D) = 30 mm + 2 * 4 mm = 38 mm = 0.038 m
Internal diameter (d) = 30 mm = 0.03 m
Thickness (t) = (D - d)/2 = (0.038 - 0.03)/2 = 0.004 m
Substituting the values into the formula, we get:
I = (π/64) * ((0.038)^4 - (0.03)^4) = 9.61 x 10^-9 m^4
Next, substitute the values of E, I, and l into the buckling load formula:
P = (π^2 * E * I) / (l^2)
P = (π^2 * 2.1 x 10^11 Pa * 9.61 x 10^-9 m^4) / (4^2)
P = 1.47 x 10^6 N
Therefore, the collapsing load of the mild steel tube is approximately 1.47 x 10^6 Newtons.
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Question Description
A mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. 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 mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. 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 mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. Can you explain this answer?.
A mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. 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 mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. 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 mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. Can you explain this answer?.
Solutions for A mild steel tube 4m long, 30mm,internal diameter and 4mm thick isused as a strut with both endshinged. Collapsing load is:-Take E = 2.1 × 105 N/mm2a)8.1 kNb)10 kNc)2.3kNd)16 kNCorrect answer is 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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