Answer:
Strain on bolt= 159.15MPa
Strain on sleeve = 39.79MPa
Stress on bolt= 0.00227mm/km
Stress on sleeve= 0.000884mm/mm
Explanation: Given : do= 20mm , di= 12mm , p= 8 × 10^3N, Eol= 70Gpa , Emg = 45Gpa
Normal stress = P/Ab
Ab= pi/4 ×d^2=3.142/4 × (8 × 10^-3)^2
Ab = 5.024× 10^-5m^2
Stress of bolt= 8×10^3/5.024×10^5=159.15Mpa
Eb= stress on bolt= 159.15×10^6/70×10^9
Eb= 0.00227mm/mm
For the sleeve
Normal stress= P/As As= pi×d^2/4 =3.142/4× ((20×10^-3)-(12×10^-3))^2
As = 5.024× 10^-5m
Stress on sleeve = 8×10^3/5.024×10^-5)
= 39.79MPa
Es= 39.79/ 45×10^9 =0.000884mm/mm
The strains will be "0.00227 mm/mm" and "0.00084 mm/mm".
According to the question,
The normal stress will be:
→ [tex]\sigma_b = \frac{P}{A_b}[/tex]
[tex]= \frac{8(10^3)}{\frac{\pi}{4} (0.008^2)}[/tex]
[tex]= 159.15 \ MPa[/tex]
→ [tex]\sigma_3 = \frac{P}{A_I}[/tex]
[tex]= \frac{8(10^3)}{\frac{\pi}{4} (0.02^2-0.012^2)}[/tex]
[tex]= 39.79 \ MPa[/tex]
By applying the Hooke's Law,
The normal strain will be:
→ [tex]\varepsilon_b =\frac{\sigma_b}{E_{aI}}[/tex]
[tex]= \frac{159.15(10^6)}{70(10^9)}[/tex]
[tex]= 0.00227 \ mm/mm[/tex]
→ [tex]\varepsilon_3= \frac{\sigma_3}{E_{mg}}[/tex]
[tex]= \frac{39.79(10^6)}{45(10^9)}[/tex]
[tex]= 0.00084 \ mm/mm[/tex]
Thus the above approach is correct.
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