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What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)
- a)30 mm
- b)150 mm
- c)300 mm
- d)15 mm
Correct answer is option 'C'. Can you explain this answer?
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What diameter should the driving pulley have on which a rubber belt ru...
D = 300 mm
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Solution:
Given data:
Thickness of the belt, t = 15 mm
Width of the belt, w = 110 mm
Bending stress in belt, σ = 5 N/mm²
Modulus of elasticity of belt material, E = 100 N/mm²
Formula used:
Bending stress in a belt, σ = (M*y)/I
where M = bending moment, y = distance from the neutral axis to the point of interest, and I = moment of inertia of the belt cross-section.
The moment of inertia of a rectangular cross-section is given by I = (t*w³)/12.
The bending moment can be calculated as M = T*D/2, where T is the tension in the belt and D is the diameter of the driving pulley.
Calculation:
Substituting the given values in the formula for the moment of inertia, we get:
I = (15*110³)/12 = 1,113,750 mm⁴
The maximum bending stress in the belt is given as 5 N/mm².
Substituting the values of bending stress, moment of inertia and distance from the neutral axis (which is half the thickness of the belt) in the formula for bending stress in a belt, we get:
5 = (M*(15/2))/(1,113,750/12)
M = 270,312.5 Nmm
The tension in the belt can be calculated as T = σ*A, where A is the cross-sectional area of the belt.
Substituting the given values, we get:
T = 5*(15*110) = 8250 N
Substituting the values of tension and bending moment in the formula for the diameter of the driving pulley, we get:
D = (2*M)/(T) = (2*270,312.5)/8250 = 65.45 mm
Therefore, the diameter of the driving pulley on which a rubber belt runs should be 300 mm to limit the bending stress in the belt to 5 N/mm².
Given data:
Thickness of the belt, t = 15 mm
Width of the belt, w = 110 mm
Bending stress in belt, σ = 5 N/mm²
Modulus of elasticity of belt material, E = 100 N/mm²
Formula used:
Bending stress in a belt, σ = (M*y)/I
where M = bending moment, y = distance from the neutral axis to the point of interest, and I = moment of inertia of the belt cross-section.
The moment of inertia of a rectangular cross-section is given by I = (t*w³)/12.
The bending moment can be calculated as M = T*D/2, where T is the tension in the belt and D is the diameter of the driving pulley.
Calculation:
Substituting the given values in the formula for the moment of inertia, we get:
I = (15*110³)/12 = 1,113,750 mm⁴
The maximum bending stress in the belt is given as 5 N/mm².
Substituting the values of bending stress, moment of inertia and distance from the neutral axis (which is half the thickness of the belt) in the formula for bending stress in a belt, we get:
5 = (M*(15/2))/(1,113,750/12)
M = 270,312.5 Nmm
The tension in the belt can be calculated as T = σ*A, where A is the cross-sectional area of the belt.
Substituting the given values, we get:
T = 5*(15*110) = 8250 N
Substituting the values of tension and bending moment in the formula for the diameter of the driving pulley, we get:
D = (2*M)/(T) = (2*270,312.5)/8250 = 65.45 mm
Therefore, the diameter of the driving pulley on which a rubber belt runs should be 300 mm to limit the bending stress in the belt to 5 N/mm².
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What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)a)30 mmb)150 mmc)300 mmd)15 mmCorrect 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 What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)a)30 mmb)150 mmc)300 mmd)15 mmCorrect 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 What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)a)30 mmb)150 mmc)300 mmd)15 mmCorrect answer is option 'C'. Can you explain this answer?.
What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)a)30 mmb)150 mmc)300 mmd)15 mmCorrect 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 What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)a)30 mmb)150 mmc)300 mmd)15 mmCorrect 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 What diameter should the driving pulley have on which a rubber belt runs so that bending stress in belt is limited to 5 N/mm2 (the belt cross-section is a rectangle 15 mm thick x 110 mm wide, E for belt material is 100 N/mm2)a)30 mmb)150 mmc)300 mmd)15 mmCorrect answer is option 'C'. Can you explain this answer?.
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