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A bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar is
- a)63 mm
- b)95.4 mm
- c)59.44 mm
- d)53.144 mm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A bar of circular cross-section is subjected to alternating tensile f...
= 350 kN
Goodman theory
d = 59.44 mm
Most Upvoted Answer
A bar of circular cross-section is subjected to alternating tensile f...
Given data:
Minimum alternating tensile force (Fmin) = 200 kN
Maximum alternating tensile force (Fmax) = 500 kN
Ultimate tensile strength (UTS) = 900 MPa
Endurance limit (EL) = 700 MPa
Safety factor related to UTS = 3.5
Safety factor related to EL = 4
Stress concentration factor (Kf) = 1.65
Approach:
We need to calculate the diameter of the circular bar that can withstand the given alternating tensile forces with the given safety factors and using the Goodman straight line as the basis for design.
Steps:
1. Calculate the mean alternating stress (σm) using the given formula:
σm = (Fmax + Fmin) / 2×(π/4)×d²
where d is the diameter of the circular bar
On substituting the given values, we get:
σm = 350000 / (π/4)×d²
2. Calculate the alternating stress amplitude (σa) using the given formula:
σa = (Fmax - Fmin) / 2×(π/4)×d²
On substituting the given values, we get:
σa = 150000 / (π/4)×d²
3. Calculate the permissible alternating stress amplitude (σa') using the given formula:
σa' = UTS / (safety factor related to UTS + Kf)
On substituting the given values, we get:
σa' = 900 / (3.5 + 1.65) = 153.846 MPa
4. Calculate the permissible mean stress (σm') using the given formula:
σm' = EL / (safety factor related to EL)
On substituting the given values, we get:
σm' = 700 / 4 = 175 MPa
5. Calculate the value of σm/σm' + σa/σa' and equate it to 1, to obtain the value of d.
(σm/σm') + (σa/σa') = 1
On substituting the values of σm, σa, σm', and σa', we get:
(350000 / (π/4)×d²) / 175 + (150000 / (π/4)×d²) / 153.846 = 1
On solving the above equation, we get:
d = 59.44 mm
Therefore, the diameter of the circular bar that can withstand the given alternating tensile forces with the given safety factors and using the Goodman straight line as the basis for design is 59.44 mm, which is option C.
Minimum alternating tensile force (Fmin) = 200 kN
Maximum alternating tensile force (Fmax) = 500 kN
Ultimate tensile strength (UTS) = 900 MPa
Endurance limit (EL) = 700 MPa
Safety factor related to UTS = 3.5
Safety factor related to EL = 4
Stress concentration factor (Kf) = 1.65
Approach:
We need to calculate the diameter of the circular bar that can withstand the given alternating tensile forces with the given safety factors and using the Goodman straight line as the basis for design.
Steps:
1. Calculate the mean alternating stress (σm) using the given formula:
σm = (Fmax + Fmin) / 2×(π/4)×d²
where d is the diameter of the circular bar
On substituting the given values, we get:
σm = 350000 / (π/4)×d²
2. Calculate the alternating stress amplitude (σa) using the given formula:
σa = (Fmax - Fmin) / 2×(π/4)×d²
On substituting the given values, we get:
σa = 150000 / (π/4)×d²
3. Calculate the permissible alternating stress amplitude (σa') using the given formula:
σa' = UTS / (safety factor related to UTS + Kf)
On substituting the given values, we get:
σa' = 900 / (3.5 + 1.65) = 153.846 MPa
4. Calculate the permissible mean stress (σm') using the given formula:
σm' = EL / (safety factor related to EL)
On substituting the given values, we get:
σm' = 700 / 4 = 175 MPa
5. Calculate the value of σm/σm' + σa/σa' and equate it to 1, to obtain the value of d.
(σm/σm') + (σa/σa') = 1
On substituting the values of σm, σa, σm', and σa', we get:
(350000 / (π/4)×d²) / 175 + (150000 / (π/4)×d²) / 153.846 = 1
On solving the above equation, we get:
d = 59.44 mm
Therefore, the diameter of the circular bar that can withstand the given alternating tensile forces with the given safety factors and using the Goodman straight line as the basis for design is 59.44 mm, which is option C.
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A bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. Can you explain this answer?
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A bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. 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 bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. 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 bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. Can you explain this answer?.
A bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. 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 bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. 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 bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A bar of circular cross-section is subjected to alternating tensile forces varying from a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limit and a stress concentration factor of 1.65 for fatigue load. Using Goodman straight line as basis for design, the diameter of circular bar isa) 63 mmb) 95.4 mmc) 59.44 mmd) 53.144 mmCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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