Mechanical Engineering - Workshop Technology - Discussion

Discussion Forum : Workshop Technology - Section 2 (Q.No. 10)
10.
Shrinkage allowance is made by adding to external dimensions and subtracting from internal dimensions.
Agree
Disagree
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Kanhai kumar said:   1 decade ago
Shrinkage allowance is provide for compensation of solid shrinkage by adding extra material for external dimension and subtracting from internal dimension.

Manish said:   10 years ago
Shrinkage allowance is always positive, irrespective of internal or external dimension.

So right answer is option "B".

Debasis said:   10 years ago
Shrinkage allowance is the extra dimension provided on the pattern for compensating solid Shrinkage taking place during cooling of the material from freezing temperature to room temp. Where shake allowance is consider as negative allowance as it is reducing the dimension of pattern. So answer is B.

Kukku said:   9 years ago
Shrinkage Allowance.

Generally, metals shrink in size during solidification and cooling in the mould. So casting becomes smaller than the pattern and the mould cavity.

Therefore, to compensate for this, mould and the pattern should be made larger than the casting of the amount of shrinkage .

Krish said:   9 years ago
Correct answer [A] because one need to provide extra material to compensate shrinkage.

For outer feature dimension added to provide extra material on pattern.

For inner feature dimension subtracted on pattern to facilitate extra molten metal in the mould cavity.

ATC said:   8 years ago
For internal features when shrinkage takes places the wall thickness reduces from inside to outside. So if there is a hole it enlarges. So dia should be reduced to compensate it.
(1)

TECH GURU said:   7 years ago
Yes right @Karish.
(1)

Shashank said:   6 years ago
Thanks all for explaining.

Varun Pratap said:   5 years ago
@Manish is right, it is added to both internal as well as external dimensions.
(1)

Salahuddin said:   5 years ago
I think given answer and the statement is correct. We have to subtract the internal dimensions.

For eg. If there is a hole (internal dimension) of dia 10 cm. After shrinkage it will expand to 12 cm hole.

Therefore at the first place, we have to SUBTRACT the dimension to 8 cm.
(1)

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