Question

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are applied at the centers of the ends of the bar. Knowing that a = 30 mm and σall = 135 MPa, determine the smallest allowable depth d of the milled portion of the bar.

Answer 1

The smallest allowable depth is
`d=16.04 \mathrm{mm}` for the milled portion of bar.

Step-by-step explanation:

Given,

Magnitude of force,
`\mathbf{p}=18 \mathrm{kN}`

`a=30 \mathrm{mm}`

`=0.03 \mathrm{m}`

Allowable stress,
`\sigma_(a l l)=135 \mathrm{MPa}`

cross sectional area of bar,

`A=a * d`

`A=a d`

e - eccentricity

`e=(a)/(2)-(d)/(2)`

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

`M=P e`

`=P\left((a)/(2)-(d)/(2)\right)`

`=(P(a-d))/(2)`

Allowable stress

`\sigma=(P)/(A)+(M c)/(I)`

`c=(d)/(2)`

Moment of Inertia,

`I=(b d^(3))/(12)`

`=(a d^(3))/(12)`

`\therefore \sigma=(P)/(a d)+((P(a-d))/(2) * (d)/(2))/((a d^(3))/(12))`

`\sigma=(P)/(a d)+(3 P(a-d))/(a d^(2))\\`

`√(x) \sigma\left(a d^(2)\right)=P d+3 P(a-d)`

`\sigma\left(a d^(2)\right)=P d+3 P a-3 P d`

`\sigma\left(a d^(2)\right)=(P-3 P) d+3 P a`

`\left(\sigma a d^(2)\right)=-2 P d+3 P a`

`\sigma d^(2)=-(2 P)/(a) d+3 P`

By substituting values we get,

`\left(135 * 10^(6)\right) d^(2)+(2 * 18 * 10^(3))/(0.03) d-3\left(18 * 10^(3)\right)=0`

`\left(135 * 10^(6)\right) d^(2)+\left(12 * 10^(5)\right) d-54 * 10^(3)=0`

On solving above equation we get,
`d=0.01604 \mathrm{m}\\`

`d=16.04 \mathrm{mm}`