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A hydraulic press exerts a total load of 4 MN. This load' is carried by two steel rods, supporting the upper head of the press. If the safe stress is 100 MPa and E =210 kN/mm2, what is the design diameter of supporting steel rods?
- a)160 mm
- b)165 mm
- c)170 mm
- d)175 mm
Correct answer is option 'A'. Can you explain this answer?
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A hydraulic press exerts a total load of 4 MN. This load' is carri...
Load carried by one rod,
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A hydraulic press exerts a total load of 4 MN. This load' is carri...
Given data:
Total load carried by the steel rods = 4 MN
Safe stress = 100 MPa
Elastic modulus of steel (E) = 210 kN/mm2
To find: Design diameter of supporting steel rods
We can use the following formula to calculate the stress:
Stress = Load/Area
Let's assume the diameter of each rod as 'd' and the cross-sectional area as 'A'.
The load carried by each rod can be calculated as half of the total load, since there are two rods.
Load carried by each rod = Total load/2 = 4 MN/2 = 2 MN
We can rearrange the formula to find the area:
Area = Load/Stress
Substituting the values, we get:
A = (2 MN)/(100 MPa) = (2 × 106 N)/(100 × 106 N/m2) = 20 × 10-3 m2
The area of a circle is given by the formula:
A = πd2/4
Rearranging the formula, we can find the diameter:
d = √(4A/π)
Substituting the value of area, we get:
d = √(4 × 20 × 10-3/π) = √(80 × 10-3/π) = √(80/π) × 10-3 ≈ 160 mm
Therefore, the design diameter of the supporting steel rods is approximately 160 mm.
Hence, option A (160 mm) is the correct answer.
Total load carried by the steel rods = 4 MN
Safe stress = 100 MPa
Elastic modulus of steel (E) = 210 kN/mm2
To find: Design diameter of supporting steel rods
We can use the following formula to calculate the stress:
Stress = Load/Area
Let's assume the diameter of each rod as 'd' and the cross-sectional area as 'A'.
The load carried by each rod can be calculated as half of the total load, since there are two rods.
Load carried by each rod = Total load/2 = 4 MN/2 = 2 MN
We can rearrange the formula to find the area:
Area = Load/Stress
Substituting the values, we get:
A = (2 MN)/(100 MPa) = (2 × 106 N)/(100 × 106 N/m2) = 20 × 10-3 m2
The area of a circle is given by the formula:
A = πd2/4
Rearranging the formula, we can find the diameter:
d = √(4A/π)
Substituting the value of area, we get:
d = √(4 × 20 × 10-3/π) = √(80 × 10-3/π) = √(80/π) × 10-3 ≈ 160 mm
Therefore, the design diameter of the supporting steel rods is approximately 160 mm.
Hence, option A (160 mm) is the correct answer.
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A hydraulic press exerts a total load of 4 MN. This load' is carried by two steel rods, supporting the upper head of the press. If the safe stress is 100 MPa and E =210 kN/mm2, what is the design diameter of supporting steel rods?a)160 mmb)165 mmc)170mmd)175 mmCorrect answer is option 'A'. Can you explain this answer?
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