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A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety?
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A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile lo...
Calculation of Factor of Safety for a Flat Bar
To calculate the factor of safety for the given flat bar, we need to determine the maximum stress experienced by the bar and compare it with the yield point stress of the material. The factor of safety is defined as the ratio of the yield point stress to the maximum stress.
Given data:
Width of the flat bar (b) = 30 mm
Thickness of the flat bar (t) = 15 mm
Tensile load (P) = 80 kN = 80,000 N
Yield point stress (σ_yield) = 315 MPa = 315,000,000 Pa
Calculating the Maximum Stress:
The maximum stress (σ_max) in the flat bar can be calculated using the formula:
σ_max = P / (b * t)
Substituting the given values:
σ_max = 80,000 N / (30 mm * 15 mm)
= 177.78 MPa = 177,780,000 Pa
Calculating the Factor of Safety:
The factor of safety (FoS) is calculated as the ratio of yield point stress to the maximum stress:
FoS = σ_yield / σ_max
Substituting the given values:
FoS = 315,000,000 Pa / 177,780,000 Pa
= 1.77
Interpretation:
The factor of safety for the flat bar is 1.77. This means that the bar is designed to withstand a load that is 1.77 times greater than the maximum load it will experience before it reaches its yield point. A factor of safety greater than 1 indicates that the bar is designed to have a safety margin, providing a level of protection against failure and ensuring that it can handle unexpected loads or variations in loading conditions without experiencing permanent deformation or fracture.
To calculate the factor of safety for the given flat bar, we need to determine the maximum stress experienced by the bar and compare it with the yield point stress of the material. The factor of safety is defined as the ratio of the yield point stress to the maximum stress.
Given data:
Width of the flat bar (b) = 30 mm
Thickness of the flat bar (t) = 15 mm
Tensile load (P) = 80 kN = 80,000 N
Yield point stress (σ_yield) = 315 MPa = 315,000,000 Pa
Calculating the Maximum Stress:
The maximum stress (σ_max) in the flat bar can be calculated using the formula:
σ_max = P / (b * t)
Substituting the given values:
σ_max = 80,000 N / (30 mm * 15 mm)
= 177.78 MPa = 177,780,000 Pa
Calculating the Factor of Safety:
The factor of safety (FoS) is calculated as the ratio of yield point stress to the maximum stress:
FoS = σ_yield / σ_max
Substituting the given values:
FoS = 315,000,000 Pa / 177,780,000 Pa
= 1.77
Interpretation:
The factor of safety for the flat bar is 1.77. This means that the bar is designed to withstand a load that is 1.77 times greater than the maximum load it will experience before it reaches its yield point. A factor of safety greater than 1 indicates that the bar is designed to have a safety margin, providing a level of protection against failure and ensuring that it can handle unexpected loads or variations in loading conditions without experiencing permanent deformation or fracture.
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A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety?
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A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety? 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 flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety?.
A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety? 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 flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A flat bar 30 mm wide and 15 mm thick is loaded by a steady tensile load of 80 kN the material is mild steel with yield point stress of 315MPa find the factor of safety?.
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