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A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.
- a)14.80mm
- b)13.95mm
- c)16.5mm
- d)17.2mm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal ra...
Explanation: σ(b)=Mh/ AeR or σ(b)=1000x4Dx(4D-0.2D-3.5D)/ 0.8D²x0.02Dx3.5D , σ(b)=21428.6/D²
21428.6/D² = 110 or D=13.95mm.
21428.6/D² = 110 or D=13.95mm.
Most Upvoted Answer
A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal ra...
To calculate the stress in the curved beam, we can use the formula for stress:
σ = (M * y) / I
where σ is the stress, M is the moment, y is the distance from the centroid, and I is the moment of inertia.
First, let's calculate the moment of inertia for the cross-sectional area of the curved beam. The moment of inertia can be calculated as:
I = (π / 4) * (r_outer^4 - r_inner^4)
where r_outer is the outer radius and r_inner is the inner radius.
Given that the inner radius is 3.5D and the outer radius is 4.5D, we have:
r_outer = 4.5D
r_inner = 3.5D
Therefore, the moment of inertia is:
I = (π / 4) * ((4.5D)^4 - (3.5D)^4)
Next, let's calculate the moment applied to the curved beam. The moment can be calculated as:
M = F * e
where F is the applied force and e is the eccentricity.
Given that the applied force is 1kN and the eccentricity is 0.02D, we have:
F = 1kN
e = 0.02D
Therefore, the moment is:
M = 1kN * 0.02D
Now, let's calculate the distance y from the centroid. The distance y can be calculated as:
y = (r_outer + r_inner) / 2
Given that the outer radius is 4.5D and the inner radius is 3.5D, we have:
r_outer = 4.5D
r_inner = 3.5D
Therefore, the distance y is:
y = (4.5D + 3.5D) / 2
Finally, we can calculate the stress using the formula:
σ = (M * y) / I
Substituting the values we have calculated, we get:
σ = (1kN * 0.02D * (4.5D + 3.5D) / 2) / ((π / 4) * ((4.5D)^4 - (3.5D)^4))
Simplifying further, we have:
σ = (0.01kN * 8D) / ((π / 4) * (91D^4))
σ = (0.08kN * D) / (π * (91D^4))
Therefore, the stress in the curved beam is (0.08kN * D) / (π * (91D^4)).
σ = (M * y) / I
where σ is the stress, M is the moment, y is the distance from the centroid, and I is the moment of inertia.
First, let's calculate the moment of inertia for the cross-sectional area of the curved beam. The moment of inertia can be calculated as:
I = (π / 4) * (r_outer^4 - r_inner^4)
where r_outer is the outer radius and r_inner is the inner radius.
Given that the inner radius is 3.5D and the outer radius is 4.5D, we have:
r_outer = 4.5D
r_inner = 3.5D
Therefore, the moment of inertia is:
I = (π / 4) * ((4.5D)^4 - (3.5D)^4)
Next, let's calculate the moment applied to the curved beam. The moment can be calculated as:
M = F * e
where F is the applied force and e is the eccentricity.
Given that the applied force is 1kN and the eccentricity is 0.02D, we have:
F = 1kN
e = 0.02D
Therefore, the moment is:
M = 1kN * 0.02D
Now, let's calculate the distance y from the centroid. The distance y can be calculated as:
y = (r_outer + r_inner) / 2
Given that the outer radius is 4.5D and the inner radius is 3.5D, we have:
r_outer = 4.5D
r_inner = 3.5D
Therefore, the distance y is:
y = (4.5D + 3.5D) / 2
Finally, we can calculate the stress using the formula:
σ = (M * y) / I
Substituting the values we have calculated, we get:
σ = (1kN * 0.02D * (4.5D + 3.5D) / 2) / ((π / 4) * ((4.5D)^4 - (3.5D)^4))
Simplifying further, we have:
σ = (0.01kN * 8D) / ((π / 4) * (91D^4))
σ = (0.08kN * D) / (π * (91D^4))
Therefore, the stress in the curved beam is (0.08kN * D) / (π * (91D^4)).
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A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. Can you explain this answer?
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A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. 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 curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. 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 curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. Can you explain this answer?.
A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. 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 curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. 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 curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.a)14.80mmb)13.95mmc)16.5mmd)17.2mmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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