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A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter?
Most Upvoted Answer
A hydraulic press exerts a total load of 3.5mega Newton this load is c...
Solution:
Given data:
Total load (F) = 3.5 mega Newton
Number of steel rods (n) = 2
Safe stress (σ) = 85 mega Pascal
Young's modulus (E) = 210 kilo Newton /mm^2
Length of each rod (L) = 2.5 meter
Formulae used:
Stress (σ) = Force (F) / Cross-sectional area (A)
Cross-sectional area (A) = πd^2/4
Elongation (δL) = FL/EA
Calculations:
Cross-sectional area of each rod:
A = πd^2/4
Where,
d = Diameter of each rod
Let's assume the diameter of each rod as 'd'.
Therefore,
A = πd^2/4
Stress in each rod:
σ = F/A
Therefore,
σ = F/(πd^2/4)
Elongation in each rod:
δL = FL/EA
Therefore,
δL = FL/(E(πd^2/4))
Substituting the given values in the above formulae, we get:
Cross-sectional area of each rod:
A = πd^2/4
A = π(∅/2)^2/4
A = π∅^2/16
Stress in each rod:
σ = F/A
σ = (3.5x10^6)/(π∅^2/16x10^6)
σ = 56.548/∅^2
As the safe stress is given as 85 mega Pascal, we can equate the above equation to get the diameter of the rod:
56.548/∅^2 = 85
∅^2 = 56.548/85
∅ = 0.345 meters (diameter of each rod)
Elongation in each rod:
δL = FL/(E(πd^2/4))
δL = (3.5x10^6x2.5)/(210x10^3xπx0.345^2/4)
δL = 2.172 mm
Therefore, the extension of each rod is 2.172 mm.
Conclusion:
The extension of each rod is 2.172 mm.
Given data:
Total load (F) = 3.5 mega Newton
Number of steel rods (n) = 2
Safe stress (σ) = 85 mega Pascal
Young's modulus (E) = 210 kilo Newton /mm^2
Length of each rod (L) = 2.5 meter
Formulae used:
Stress (σ) = Force (F) / Cross-sectional area (A)
Cross-sectional area (A) = πd^2/4
Elongation (δL) = FL/EA
Calculations:
Cross-sectional area of each rod:
A = πd^2/4
Where,
d = Diameter of each rod
Let's assume the diameter of each rod as 'd'.
Therefore,
A = πd^2/4
Stress in each rod:
σ = F/A
Therefore,
σ = F/(πd^2/4)
Elongation in each rod:
δL = FL/EA
Therefore,
δL = FL/(E(πd^2/4))
Substituting the given values in the above formulae, we get:
Cross-sectional area of each rod:
A = πd^2/4
A = π(∅/2)^2/4
A = π∅^2/16
Stress in each rod:
σ = F/A
σ = (3.5x10^6)/(π∅^2/16x10^6)
σ = 56.548/∅^2
As the safe stress is given as 85 mega Pascal, we can equate the above equation to get the diameter of the rod:
56.548/∅^2 = 85
∅^2 = 56.548/85
∅ = 0.345 meters (diameter of each rod)
Elongation in each rod:
δL = FL/(E(πd^2/4))
δL = (3.5x10^6x2.5)/(210x10^3xπx0.345^2/4)
δL = 2.172 mm
Therefore, the extension of each rod is 2.172 mm.
Conclusion:
The extension of each rod is 2.172 mm.
Community Answer
A hydraulic press exerts a total load of 3.5mega Newton this load is c...
Given:
P =3.5 MN = 3.5 � 106 N
stress
t
= 85
MPa
= 85N/mm
2
E =210
kN
/mm
2
=210 � 10
3
N/mm
2
l =2.5 m = 2.5 � 10
3
mm
1.Diameter of the rods
Let d= Diameter of the rods in mm.
Area=
pi*
d2/4= 0.8754 d2
Sincethe load
P
iscarried by two rods, therefore load carried by each rod is,
P
1= P/2 = 3.5 x10
6
/2 =1.75 � 10
6
N
Itis known that load carried by each rod (
P
1),
Stress=P/A rearranging for P
1.75 � 10
6
=
sigma
t
.
A
=85 � 0.7854
d
2
=66.76
d
2
d
2
= 1.75 � 106/66.76 = 26 213
d
=162 mm
2.Extension in each rod
Let:
d
l
=Extension in each rod.
It isknown that Young's modulus (E),
d
=212.5 � 10
3
/(210 � 10
3
) = 1.012
mm
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A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter?
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A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter? 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 hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter?.
A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter? 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 hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter?.
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A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter?, a detailed solution for A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter? has been provided alongside types of A hydraulic press exerts a total load of 3.5mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 mega Pascal And young's modulus e=210 kilo Newton /mm^2 find the extension of each rod I a length of 2.5 meter? theory, EduRev gives you an
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