---
title: "Pattern allowances and pattern size"
subject: "Manufacturing and Production Processes"
chapter: 4
key: "semester_4/manufacturing_and_production_processes:pattern-allowances-numerical"
source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/pattern-allowances-numerical
site: BCT Engineering
---

# Pattern allowances and pattern size

Manufacturing and Production Processes · Chapter 4: Solidification Process and Powder Metallurgy

- Asked in: year not given, 5 marks
- Video: [Pattern Allowances in Casting | Types & Importance | Manufacturing Process](https://www.youtube.com/watch?v=wIfV0U4q_kw) — Tech Research Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/pattern-allowances-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-4

## Question

List the allowances provided on a pattern and state the reason for each. A cast iron casting has a finished length of 450 mm between two faces that are to be machined, with 3 mm machining allowance on each face. The shrinkage allowance for cast iron is 1% (10 mm per metre). Find the length of the pattern. Also find the diameter of the core for a finished bore of 80 mm diameter which is cast and then bored with a radial machining allowance of 2.5 mm.

## Short answer

Pattern allowances: shrinkage, machining, draft, shake, and distortion. To find pattern length, add machining allowances (456 mm) then apply 1% shrinkage → 460.56 mm. For core diameter, subtract radial bore allowance from finished bore to get cast diameter 75 mm, then apply shrinkage → 75.75 mm.

## Answer

#### Pattern allowances

| Allowance | Reason |
|---|---|
| Shrinkage | Metal contracts while cooling from solidification to room temperature, so the pattern is made larger |
| Machining (finish) | Extra metal on surfaces to be machined, so that the rough skin can be removed |
| Draft (taper) | Taper of 0.5-3 degree on vertical faces so the pattern can be withdrawn without damaging the mould |
| Shake (rapping) | The pattern is rapped before withdrawal, which enlarges the cavity slightly, so the pattern is made smaller |
| Distortion | Allows for warping of shapes such as U-sections or long thin castings |

#### Pattern length

Length of the casting, including the machining allowance on both faces:

$$
L_{c} = 450 + 2\times 3 = 456\ \text{mm}
$$

The pattern must be larger by the shrinkage allowance (1 percent):

$$
L_p = 456\,(1+0.01) = 460.56\ \text{mm}
$$

#### Core diameter

For an internal surface, the machining allowance makes the cast hole smaller, because boring will enlarge it. With 2.5 mm removed from the radius (5 mm from the diameter), the cast hole diameter is

$$
d_{cast} = 80 - 2\times 2.5 = 75\ \text{mm}
$$

As the casting cools, the hole contracts along with the rest of the metal, so the core must be made larger by the shrinkage allowance:

$$
d_{core} = 75\,(1+0.01) = 75.75\ \text{mm}
$$

**Answer:** pattern length $= 460.56$ mm (about 460.6 mm); core diameter $= 75.75$ mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the shake (rapping) allowance affect pattern size?
   A. Pattern is made larger
   B. Pattern is made smaller, since rapping enlarges the cavity
   C. No change
   D. It only applies to cores
   Answer: B. Rapping before withdrawal enlarges the cavity, so the pattern is made slightly smaller. (Key point: Shake allowance)

2. The purpose of draft (taper) on vertical faces is to
   A. Compensate metal contraction
   B. Allow the pattern to be withdrawn without damaging the mould
   C. Provide extra metal for machining
   D. Prevent warping
   Answer: B. Draft of 0.5-3 degree lets the pattern lift out cleanly. (Key point: Draft allowance)

3. To find the pattern length, the answer first adds machining allowance to both faces then
   A. Subtracts shrinkage
   B. Multiplies by (1 + 0.01) for shrinkage
   C. Adds 10 mm
   D. Divides by 1.01
   Answer: B. 456 mm x 1.01 = 460.56 mm, because the pattern must be larger by the shrinkage allowance. (Key point: Pattern length method)

4. Finished 450 mm, 3 mm machining on each face, 1% shrinkage. Pattern length is
   A. 450 mm
   B. 456 mm
   C. 460.5 mm (about 460.56 mm)
   D. 466 mm
   Answer: C. (450 + 2 x 3) x 1.01 = 460.56 mm. (Key point: Pattern length value)

## Related in Manufacturing and Production Processes

- [Solidification, shrinkage and casting defects](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Asolidification-shrinkage-casting-defects) — Manufacturing and Production Processes, chapter 4
- [Sheet bending allowance, force, springback](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Abending-allowance-force-springback-numerical) — Manufacturing and Production Processes, chapter 6

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