Question

a sheet of aluminum of thickness 20 mm is cut with vertical machine that have blade of length 10 cm and width 0.2 cm while cutting the blades each exert a force of 30 N on the sheet . the length of each blade that makes contact with the sheet is approximately 0.5 mm calculate the shear stress on the sheet???

Answer 1

Task:

A sheet of aluminum of thickness $20\mathrm{mm}$ is cut with vertical machine that have blade of length $10\mathrm{cm}$ and width $0.2\mathrm{cm}$ while cutting the blades each exert a force of $30\mathrm{N}$ on the sheet. The length of each blade that makes contact with the sheet is approximately $0.5\mathrm{mm}$. Calculate the shear stress on the sheet.

Solution:

SHEET METAL WORK

Operations

**Cutting operations:** Piercing Blanking, Punching, Lancing, Parting of, cutting off, Notching, Shaving, Trimming.

**Bending operations:** Edge bending or wiping, U-bending, v-bending,

**Forming operations:** embossing, Flanging, Curling

**Drawing operations:** Drawing, deep drawing.

**Spinning, coining**

**Piercing:** It is an operation of making a hole in sheet. E.g. central hole in washer. Punch is of correct size and clearance (+ve) provided on die.

**Blanking:** It is an operation of cutting a required component from sheet metal. e.g. out side dia. Of washer, Hawai slipper. Die opening is of correct size and (-ve) clearance is provided on the punch.

**Lancing:** A combination cutting and bending operations. eg : Tortoise coil stand

**Parting off:** separating a blank from sheet metal by cutting along two lines.

**Cutting off:** separating a blank from sheet metal by cutting along one line.

**Notching:** Cutting along edges of the blank.

**Shaving:** Clearing and squaring of a blank or pierced hole.

**Trimming:** Correcting the edges a drawn product.

**Force required in cutting operation** $F_{sh} = \sigma_{sh} L t$

$L =$ length of cut $t =$ thickness of sheet $\sigma_{sh} =$ shear stress of sheet.

$$\sigma_{sh} = \frac{F_{sh}}{Lt} = \frac{30\,N}{20\,mm \cdot L} = \frac{1.5\,\frac{N}{mm}}{L}$$

$L$ is not totally clear from your description. Try to draw it. Then $L$ will become obvious.

Answer:

$$\sigma_{sh} = \frac{1.5\,\frac{N}{mm}}{L}$$