Question 10.1: (a) Two parallel steel cylinders of radii 50mm and lOOmm are...
(a) Two parallel steel cylinders of radii 50 mm and l00 mm are brought into contact under a load of 2 kN. If the cylinders have a common length of 150 mm and elastic constants of E = 208 GN/m^{3} and \nu = 0.3 determine the value of the maximum contact pressure. What will then be the magnitude and position of the maximum shear stress?
(b) How would the values change if the larger cylinder were replaced by a flat surface?
For contacting parallel cylinders eqn. (10.9) gives the value of the maximum contact pressure (or compressive stress) as
\sigma _{c}=-p_{0} =-0.591\sqrt{\frac{P}{L\Delta } (\frac{1}{R_{1}}+\frac{1}{R_{2}} )} (10.9)
where
\Delta =\frac{1}{E_{1}} [1-\nu ^{2}_{1} ]+\frac{1}{E_{2}} [1-\nu ^{2}_{2} ]
=\frac{2}{E} [1-\nu ^{2} ] for similar materials
=\frac{2\times 0.91}{208\times 10^{9}}
Max. contact pressure
p_{0}=0.591\sqrt{\frac{2\times 10^{3}\times 208\times 10^{9}}{150\times 10^{-3}\times 2\times 0.91}(\frac{1}{50}+\frac{1}{100} )10^{3}}
=0.591\times 21.38\times 10^{7}
=126.4 MN/m^{2}
Maximum shear stress =0.295 p_{0}=37.3 MN/m^{2}
occurring at a depth d=0.786 b
with (from eqn. (10.8))
b=1.076\sqrt{\frac{P\Delta }{L(\frac{1}{R_{1}}+\frac{1}{R_{2}})} } (10.8)
=1.076\sqrt{\frac{2\times 10^{3}\times 2\times 0.91}{150\times 10^{-3}208\times 10^{9}\times 30} }
=1.076\times 0.624\times 10^{-4}
=0.067 mm
Depth of max shear stress =0.786\times 0.067=0.053 mm
(b) Replacing the 100 mm cylinder by a flat surface makes \frac{1}{R_{2}}=0 and
contact pressure p_{0}=0.591\sqrt{\frac{2\times 10^{3}\times 208\times 10^{9}}{150\times 10^{-3}\times 2\times 0.91}(\frac{1}{50} )10^{3}}
=0.591\times 17.48\times 10^{7}
=103.2 MN/m^{2}
with max shear stress =0.295\times 103.2=30.4 MN/n^{2}
and b=0.082 mm
Depth of max shear stress =0.786\times 0.082=0.064 mm