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Elena-2011 [213]
3 years ago
15

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

ed 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.
Engineering
1 answer:
monitta3 years ago
3 0

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

<u>Explanation:</u>

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 \times d

A=a d

e - eccentricity

e=\frac{a}{2}-\frac{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(\frac{a}{2}-\frac{d}{2}\right)

=\frac{P(a-d)}{2}

Allowable stress

\sigma=\frac{P}{A}+\frac{M c}{I}

c=\frac{d}{2}

Moment of Inertia,

I=\frac{b d^{3}}{12}

=\frac{a d^{3}}{12}

\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}

\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\

\sqrt{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}=-\frac{2 P}{a} d+3 P

By substituting values we get,

\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0

\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0

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

d=16.04 \mathrm{mm}

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