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    Question is:

    A nylon cylinder of diameter 69.9mm is placed inside a steel tube having a slightly larger inner diameter. The nylon cylinder is then compressed by an axial force of 80 kN.

    What shall be the minimal value of the inner diameter of the steel tube to avoid lateral contact stresses? Compression on inner walls of steel tube due to lateral expansion of nylon bar must be avoided.

    Take E = 3.1GPa and Poisson’s ratio 0.4 for nylon. Required accuracy 0.01 mm.

    My working so far, taking L as 1m:

    Axial Strain = P/EA
    = 80kN/(3.1GPa*0.25π0.0699)
    = 6.72x10-4

    ΔL=Axial Strain*L
    = 6.72x10-4*1
    = 6.72x10-4

    To find increase in diameter:
    Δd=v*d*Strain
    =0.4*0.0669*6.72x10-4
    =1.798x10-5
    =0.01798mm

    Is this correct? :confused:
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    I also cant figure out how I should be working out:

    A steel bar of a square cross-section 67 mm on a side, is subject to an axial tensile force of 113 kN.

    Young's modulus (modulus of elasticity) is 200 GPa for the steel.

    Determine the strain in the bar with the required accuracy of 5%.
 
 
 
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