A steel shaft that makes up a drive svstem is shown in the f

A steel shaft that makes up a drive svstem is shown in the figure below, The three segments are connected by fillet welds of radius r = 3 mm. Under the current configuration. a motor supplies a torque T = 100 N.m to the center segment (40 mm dia. 75 mm long). Two pulleys that take up torques of 50 N.m each are connected to the segments (20 mm dia., 150 mm long) on either side. The material of the shaft is ASTMA36. Due to changes outside your control, the torque supplied by the motor is increased to 125 N.m. and the pulleys each take up 62.5 N.m. Suggest design changes that would accommodate this increase in torque, and perform the necessary analysis to show that the factor of safety is not less than that for the original design. Assume the yield strengths of the weld and that of the base material are the same, and, unless otherwise specified, assume the yield strength in shear is 0.577 of the yield strength in tension.

Solution

a) The torque acting on the shaft induces a shear stress in the which twists the shaft. The factor of safety is found by the ratio of ultimate loading to the elastic loading.

The A36 grade steel has the following properties,

Tensile strenght of the material (ultimate) = 552 MPa

Tensile yield = 221 MPa

Assuming that the ultimate tensile strenght of the shafts is 0.577 of the ultimate tensile strength,

Shear strength ultimate = 0.577 x 552 = 318.5 MPa

Now tha actual shear in the shafts is found using the relation,

tau = T x r / J

where J = pi x D^4 / 32 for solid shafts.

J for shaft 1 (D=0.04) = 3.14 x 0.04^4 / 32 = 0.0000002512 m^4

J for shaft 2 (D=0.02) = 3.14 x 0.02^4 / 32 = 0.0000000157 m^4

The shear on shaft 1, tau = 100 x 0.04/2 / 0.0000002512 = 7.9 MPa

The shear on pulley, tau = 100 x 0.02/2 / 0.0000000157 = 63.69 MPa

SOF for shaft  = ultimate shear / aplied shear = 318.5 / 7.9 = 40

SOF for pulley = 318.5 / 63.69 = 5

b) The increase in torque to 125Nm on the middle shaft has to be managed by increasing the dia of the material at work.

Considering the SOF for the shaft as 40. The applied shear for th shaft remains the same tau = 7.9 MPa

Now using the torque relation,

tau = Tx r / J

7.9 x 10 ^6 = 125 x (d/2) / (3.14 x d^4 / 32)

Solving for the dia we get d = 0.043 m = 43 mm

Similarly for the pulley,

63.69 x 10^6 = 62.5 x (d/2) / (3.14 x d^4 / 32)

Solve for the dia gives the value d = 0.017 m or 17 mm