3 A solid circular shaft with a radius of 15 mm is rigidly s

3. A solid circular shaft with a radius of 15 mm is rigidly supportod at one end and is suhjected to an axial loe N and que T as shown in ligue 3. At a pointon the surface ol the shal two stiin gauges are mounted; one in the direction ol the axis of the shalt and the other at 45 Uner lading the lirst strain gage gives a reading 4 and the sccond \' =-60x106 If the shaft is made out of steel with clastic modulus 210 GPa and Poisson\'s ratio is 0.29, a) calculate the axial oe and the tonqueT b) detemine the maximum shear stress and the maximum shear strain

Solution

GIVEN:-

r = 15mm, E = 210GPa, u = 0.29, €x = 140x10^-6, €a = -60x10^-6

TO FIND:-

Axial Force N =?, Torque T =?, Maximum shear stress Tmax=? and Maximum shear strain €max =?

SOLUTION:-

Sx = E x €x = 210000 x 140x10^-6 = 29.4 N/mm2

Also cross-section area of shaft A = 3.142 x r2 = 3.142 x 225 = 706.95 mm2

Axial force N = Snormal x A = 29.4 x 706.95 = 20784.33 N

Axial Force N = 29784.33 N (ANSWER)

Now €y = €a x sin(45) = -60 x 10^-6 x sin45 = 42.43 x 10^-6

G = E/2(1+u) = 210000 / 2(1+0.29) = 81395.35 MPa

Shear stress Ss = G x €y = 81395.35 x 42.43 x 10^-6 = 3.45 N/mm2

Now Maximum shear stress Tmax = [ (Sx/2)^2 + (Ss)^2]1/2 = [ (29.4/2)^2 + (3.45)^2]1/2 = 15.1 N/mm2

Maximum shear stress Tmax = 15.1 N/mm2 (ANSWER)

Torque T = (Tmax x 3.142 x 30^3) / 16 = 80062.1 Nmm (ANSWER)

Maximum shear strain €max = Tmax / G = 15.1 / 81395.35 = 1.85 x 10^-4 (ANSWER)