Respuesta :
The smallest allowable depth is [tex]d=16.04 \mathrm{mm}[/tex] for the milled portion of bar.
Explanation:
Given,
Magnitude of force,[tex]\mathbf{p}=18 \mathrm{kN}[/tex]
[tex]a=30 \mathrm{mm}[/tex]
[tex]=0.03 \mathrm{m}[/tex]
Allowable stress,[tex]\sigma_{a l l}=135 \mathrm{MPa}[/tex]
cross sectional area of bar,
[tex]A=a \times d[/tex]
[tex]A=a d[/tex]
e - eccentricity
[tex]e=\frac{a}{2}-\frac{d}{2}[/tex]
The internal forces in the cross section are equivalent to a centric force P and a bending couple M.
[tex]M=P e[/tex]
[tex]=P\left(\frac{a}{2}-\frac{d}{2}\right)[/tex]
[tex]=\frac{P(a-d)}{2}[/tex]
Allowable stress
[tex]\sigma=\frac{P}{A}+\frac{M c}{I}[/tex]
[tex]c=\frac{d}{2}[/tex]
Moment of Inertia,
[tex]I=\frac{b d^{3}}{12}[/tex]
[tex]=\frac{a d^{3}}{12}[/tex]
[tex]\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}[/tex]
[tex]\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\[/tex]
[tex]\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)[/tex]
[tex]\sigma\left(a d^{2}\right)=P d+3 P a-3 P d[/tex]
[tex]\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a[/tex]
[tex]\left(\sigma a d^{2}\right)=-2 P d+3 P a[/tex]
[tex]\sigma d^{2}=-\frac{2 P}{a} d+3 P[/tex]
By substituting values we get,
[tex]\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[/tex]
[tex]\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0[/tex]
On solving above equation we get,[tex]d=0.01604 \mathrm{m}\\[/tex]
[tex]d=16.04 \mathrm{mm}[/tex]
