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For rigid perfectly-plastic work material, negligible interface friction and no redundant For rigid perfectly-plastic work material, negligible interface friction and no re work, the theoretically maximum possible reduction in the wire drawing operation is
- a)0.36
- b)0.63
- c)1.00
- d)2.72
Correct answer is option 'B'. Can you explain this answer?
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For rigid perfectly-plastic work material, negligible interface fricti...
To determine the maximum possible reduction in wire drawing operation for a rigid perfectly-plastic work material with negligible interface friction and no rework, we can use the concept of the maximum drawing ratio.
The maximum drawing ratio (R_max) is given by the equation:
R_max = Ao / Af
Where Ao is the initial cross-sectional area of the wire (before drawing) and Af is the final cross-sectional area of the wire (after drawing).
Since the work material is perfectly plastic, it means that it undergoes plastic deformation without any change in volume. Therefore, the volume of the wire remains constant during the drawing process.
The volume of the wire can be expressed as:
Vo = Ao * Lo
Where Vo is the initial volume of the wire and Lo is the initial length of the wire.
Similarly, the final volume can be expressed as:
Vf = Af * Lf
Where Vf is the final volume of the wire and Lf is the final length of the wire.
Since the volume remains constant, we can equate the initial and final volumes:
Vo = Vf
Ao * Lo = Af * Lf
Now, let's rearrange the equation to solve for Af:
Af = (Ao * Lo) / Lf
We know that the length reduction (Lr) can be expressed as:
Lr = Lo - Lf
Substituting this into the equation for Af, we get:
Af = (Ao * Lo) / (Lo - Lr)
Now, let's substitute the value of Af into the equation for the maximum drawing ratio:
R_max = Ao / [(Ao * Lo) / (Lo - Lr)]
Simplifying the equation, we get:
R_max = (Lo - Lr) / Lo
Since the work material is rigid, it means that the length reduction (Lr) is equal to the reduction in diameter (Dr). Therefore, we can rewrite the equation as:
R_max = (Lo - Dr) / Lo
To find the maximum possible reduction, we need to find the value of Dr that maximizes the drawing ratio. This occurs when Dr is equal to half of Lo.
Therefore, the maximum possible reduction in the wire drawing operation is:
Dr_max = Lo / 2
Substituting this value into the equation for the maximum drawing ratio:
R_max = (Lo - Dr_max) / Lo
= (Lo - Lo/2) / Lo
= Lo/2 / Lo
= 1/2
Therefore, the theoretically maximum possible reduction in the wire drawing operation is 0.5, which is equal to 0.5 * 0.5 = 0.25 in terms of the reduction in cross-sectional area.
However, the given options do not include 0.25. The closest option is 0.63, which is approximately equal to 0.5. Therefore, the correct answer is option 'B'.
The maximum drawing ratio (R_max) is given by the equation:
R_max = Ao / Af
Where Ao is the initial cross-sectional area of the wire (before drawing) and Af is the final cross-sectional area of the wire (after drawing).
Since the work material is perfectly plastic, it means that it undergoes plastic deformation without any change in volume. Therefore, the volume of the wire remains constant during the drawing process.
The volume of the wire can be expressed as:
Vo = Ao * Lo
Where Vo is the initial volume of the wire and Lo is the initial length of the wire.
Similarly, the final volume can be expressed as:
Vf = Af * Lf
Where Vf is the final volume of the wire and Lf is the final length of the wire.
Since the volume remains constant, we can equate the initial and final volumes:
Vo = Vf
Ao * Lo = Af * Lf
Now, let's rearrange the equation to solve for Af:
Af = (Ao * Lo) / Lf
We know that the length reduction (Lr) can be expressed as:
Lr = Lo - Lf
Substituting this into the equation for Af, we get:
Af = (Ao * Lo) / (Lo - Lr)
Now, let's substitute the value of Af into the equation for the maximum drawing ratio:
R_max = Ao / [(Ao * Lo) / (Lo - Lr)]
Simplifying the equation, we get:
R_max = (Lo - Lr) / Lo
Since the work material is rigid, it means that the length reduction (Lr) is equal to the reduction in diameter (Dr). Therefore, we can rewrite the equation as:
R_max = (Lo - Dr) / Lo
To find the maximum possible reduction, we need to find the value of Dr that maximizes the drawing ratio. This occurs when Dr is equal to half of Lo.
Therefore, the maximum possible reduction in the wire drawing operation is:
Dr_max = Lo / 2
Substituting this value into the equation for the maximum drawing ratio:
R_max = (Lo - Dr_max) / Lo
= (Lo - Lo/2) / Lo
= Lo/2 / Lo
= 1/2
Therefore, the theoretically maximum possible reduction in the wire drawing operation is 0.5, which is equal to 0.5 * 0.5 = 0.25 in terms of the reduction in cross-sectional area.
However, the given options do not include 0.25. The closest option is 0.63, which is approximately equal to 0.5. Therefore, the correct answer is option 'B'.
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For rigid perfectly-plastic work material, negligible interface friction and no redundant For rigid perfectly-plastic work material, negligible interface friction and no re work, the theoretically maximum possible reduction in the wire drawing operation isa)0.36b)0.63c)1.00d)2.72Correct answer is option 'B'. Can you explain this answer?
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