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A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)
- a)56.57 mm
- b)63.4 mm
- c)28.18 mm
- d)46.37 mm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A steel plate is subjected to a completely reversed axial load of 40 ...
P = 土40 kN
Pa = 40 kN
kf = 1 + q(kt - 1)
= 1 + 0.82(2.5 - 1)
se= KaKbKcKdSe’
= 0.65 x 0.75 x 0.789 x 0.45 x (0.5 x 400)
= 34.5 N/mm2
For axial load (Se)a = 0.82 Se
= 28.28 N/mm2
kf = 1 + q(kt - 1)
= 1 + 0.82(2.5 - 1)
For safe loading
⇒ t = 56.57 mm
Most Upvoted Answer
A steel plate is subjected to a completely reversed axial load of 40 ...
Given data:
- Completely reversed axial load = 40 kN
- Diameter of hole = 10 mm
- Width of plate = 60 mm
- Surface finish factor (ka) = 0.65
- Size factor (kb) = 0.75
- Reliability factor (kc) = 0.789
- Notch sensitivity (q) = 0.82
- Theoretical stress concentration factor (kt) = 2.5
- Ultimate tensile strength (Sut) = 400 MPa
- Factor of safety (FoS) = 2
- Endurance limit = 50% of ultimate tensile strength
- Correction factor for axial load = 0.82
To find:
Thickness of the plate for infinite life
Solution:
1. Calculate the corrected endurance strength (Se) of the material
- Endurance limit = 0.5 * Sut = 0.5 * 400 = 200 MPa
- Correction factor for axial load = 0.82
- Se = Endurance limit * ka * kb * kc * q * kt * axial load correction factor
- Se = 200 * 0.65 * 0.75 * 0.789 * 0.82 * 2.5 * 0.82 = 76.08 MPa
2. Calculate the maximum alternating stress (Sa) in the plate
- Sa = FoS * (Load on the plate / Area of the plate)
- Load on the plate = Completely reversed axial load = 40 kN
- Area of the plate = (Width of the plate * Thickness of the plate) - Area of the hole
- Area of the hole = (π/4) * (Diameter of the hole)^2 = (π/4) * (10)^2 = 78.54 mm^2
- Area of the plate = (60 * Thickness) - 78.54
- Sa = 2 * (40 * 10^3) / ((60 * Thickness) - 78.54)
3. Calculate the mean stress (Sm) in the plate
- Sm = 0
4. Calculate the alternating stress amplitude (Sa') in the plate
- Sa' = (Sa^2 + 4*Sm^2)^0.5 = Sa
5. Calculate the stress concentration factor (Kf) due to the hole
- Kf = 3 + (Diameter of the hole / Width of the plate)
- Kf = 3 + (10 / 60) = 3.167
6. Calculate the fatigue strength reduction factor (Kf') due to the hole
- Kf' = (4 - Kf) / 2 = 0.917
7. Calculate the fatigue strength of the plate
- Sf = Se * Kf * Kf'
- Sf = 76.08 * 3.167 * 0.917 = 208.81 MPa
8. Calculate the required thickness of the plate for infinite life
- Sa' = Sf / FoS
- (Sa^2 + 4*Sm^2)^0.5 = Sf / FoS
- (Sa^2 + 4*0^2)^0.5 = (208.81 / 2) =
- Completely reversed axial load = 40 kN
- Diameter of hole = 10 mm
- Width of plate = 60 mm
- Surface finish factor (ka) = 0.65
- Size factor (kb) = 0.75
- Reliability factor (kc) = 0.789
- Notch sensitivity (q) = 0.82
- Theoretical stress concentration factor (kt) = 2.5
- Ultimate tensile strength (Sut) = 400 MPa
- Factor of safety (FoS) = 2
- Endurance limit = 50% of ultimate tensile strength
- Correction factor for axial load = 0.82
To find:
Thickness of the plate for infinite life
Solution:
1. Calculate the corrected endurance strength (Se) of the material
- Endurance limit = 0.5 * Sut = 0.5 * 400 = 200 MPa
- Correction factor for axial load = 0.82
- Se = Endurance limit * ka * kb * kc * q * kt * axial load correction factor
- Se = 200 * 0.65 * 0.75 * 0.789 * 0.82 * 2.5 * 0.82 = 76.08 MPa
2. Calculate the maximum alternating stress (Sa) in the plate
- Sa = FoS * (Load on the plate / Area of the plate)
- Load on the plate = Completely reversed axial load = 40 kN
- Area of the plate = (Width of the plate * Thickness of the plate) - Area of the hole
- Area of the hole = (π/4) * (Diameter of the hole)^2 = (π/4) * (10)^2 = 78.54 mm^2
- Area of the plate = (60 * Thickness) - 78.54
- Sa = 2 * (40 * 10^3) / ((60 * Thickness) - 78.54)
3. Calculate the mean stress (Sm) in the plate
- Sm = 0
4. Calculate the alternating stress amplitude (Sa') in the plate
- Sa' = (Sa^2 + 4*Sm^2)^0.5 = Sa
5. Calculate the stress concentration factor (Kf) due to the hole
- Kf = 3 + (Diameter of the hole / Width of the plate)
- Kf = 3 + (10 / 60) = 3.167
6. Calculate the fatigue strength reduction factor (Kf') due to the hole
- Kf' = (4 - Kf) / 2 = 0.917
7. Calculate the fatigue strength of the plate
- Sf = Se * Kf * Kf'
- Sf = 76.08 * 3.167 * 0.917 = 208.81 MPa
8. Calculate the required thickness of the plate for infinite life
- Sa' = Sf / FoS
- (Sa^2 + 4*Sm^2)^0.5 = Sf / FoS
- (Sa^2 + 4*0^2)^0.5 = (208.81 / 2) =
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A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer?
Question Description
A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. 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 steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. 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 steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer?.
A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. 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 steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. 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 steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A steel plate is subjected to a completely reversed axial load of 40 kN and it has 10 mm diameter hole. Width of the plate is 60 mm. If surface finish factor (ka) = 0.65 size factor (kb) = 0.75 Reliability factor (kc) = 0.789, Notch sensitivity (q) =.0.82 and theoretical stress concentration factor (kt) = 2.5. The ultimate tensile strength of the material (Sut) is 400 MPa. If the factor of safety is 2. What will be the thickness of the plate for infinite life (Take endurance limit as 50% of ultimate tensile strength and correction factor for axial load is 0.82)a) 56.57 mmb) 63.4 mmc) 28.18 mmd) 46.37 mmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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