The 8 mm diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the materials at A is rigid. Eal = 70 GPa, Emg = 45 GPa.

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

friction

explanation:

they us to move forward and backwards