Question

$\text{The bending force required for a 1.5 mm thick steel (having ultimate tensile strength 800 MPa) sheet of 1 m length to be bent in a wiping die (K = 0.33) is kN. The die radius used is 3 mm. (By using the governing equation = bending force,}{F}_b=\frac{kL\sigma t^2}{W})\text{Every symbol has their using meaning.}$

• A. 79.2

Answer 1

The correct option is A 79.2
Die opening W $=1.5+3+3=7.5mm$

thickness of sheet, t $=1.5mm$

For a wiping die, K = 0.33

Length of the bent part, $L=1000mm$

$\text{Bending force,}{F}_b=\frac{kL\sigma t^2}{W}=\frac{0.33\times 1000\times 800\times (1.5{)}^2}{7.5}=79.2kN$