The shaft with the dimensions given below is made of colddra

The shaft with the dimensions given below is made of cold-drawn AISI 1050 steel and is simply supported at B and C. The load F makes an angle of 20 degree with the z-direction and is pulsating between 0 and F_max. If the shaft diameter d= 20 mm, Find F_max that will cause static failure in the shaft. Find F_max that will cause fatigue failure in the shaft (Use Goodman criterion).

Solution

The Shaft is subjected to both Bending and Torsion. The torsion is same throughout the length of the shaft.

T= Torsion = (tangential component of F * radius) = F Cos20^o x (150/2) = F * 0.9397 * 75 = 70.477F Nmm.

Shaft Diameter = 20mm.

J= PI/32 x d^4 = 15707

_max = Maximum Shear stress due to the torsion = TD/2J = 0.0448F MPa

The Max bending moment occurs at a point on the support C. This is equal to F x 100 = 100F Nmm.

Section Modulus of shaft Z = PI /32 * d^3 = 785.4.

Corresponding Bending stress _max = Bmax/Z = 0.127F

Since the Load is pulsating from 0 to F_Max, Torsion T as well as Bending moment also will vary 0 to F_Max.

Therefore the mean bending stress is _m = F_Max/2 = 0.0635F MPa and

alternating stress stress : _a = _max /2=0.0635F MPa

Therefore the mean torsional stress = _m= _max /2 = 0.0224F MPa

and alternating stress stress = _a : _max /2=0.0224F MPa

We shall apply Von-Mises Stress theory for calculating the resultant stresses acting on the shaft.

For a single normal stress and shear stress as our case is,

\'_m= Mean stress = SQRT( _m² + 3 _m²) = 0.074F

_a= alternating bending stress = SQRT( _a² + 3 _a²) = 0.074F.

For cold drawn 1050 CD, Sut = 690MPa, Sy = 580MPa, S’e = 347 MPa & Se= 235 MPa.

Part 1: For the Case of Static Load:

Again apply Von-Mises Stress theory for calculating the resultant stresses acting on the shaft. Here the resulting Max stress = SQRT( _max² + 3 _max²)= 0.1488F MPa.

In static condition failure will occur at Max Stress = Sy.

0.1488F= 580. Therefore F=580/0.1488= 3897 N. this the max stress the shaft can take in static condition.

Part 2: For Pulsating stress:

\'_m= Mean stress = 0.074F

_a= alternating stress = 0.074F.

Draw Goodman diagram as shown below. Read Max Stress amplitude as: 0.08F ( approx.)

And corresponding load as 235/0.08= 2937 N. ( approx.)