I want the lightest possible polycarbonate push rod that wil

I want the lightest possible polycarbonate push rod that will have a circular cross section, be 50 mm long, and able to support 800 N. My vendor sells polycarbonate with a strength of 83 MPa and a Young\'s modulus of 2.6 GPa. State the diameter of your design as well as the slenderness ratio. Make sure it will not fail in either axial compression or buckling.

Solution

Given:

L = 50mm

P= 800 N

Ultimate tensile stress = 83 Mpa = 83 N/mm²

E = 2.6 Gpa = 2600 N/mm²

Assume F.O.S = 2

Max Cylinder Force, F_max

= working pressure x Cylinder area

= 800 x x d² / 4

= 628.31 x d² N

Moment of inertia , I = X d⁴/64

Critical load of Buckling, F_cr

= ² x E x I/ L²

= ³ x E x d⁴/64 xL²

= .503x d⁴ N

As F.O.S is normally assumed to be 2

Critical buckling = 2 x Working force

.503x d⁴ = 2 x 628.31 x d²

Solving above equation we get

‘d2 = 2497

So, d = 49.97 mm 50 mm   (cylinder will not fail in with this diameter size in buckling).

Now, Torque applied on cylinder is

= F.L

= 800 x 50 = 40000 N-mm

As Ultimate tensile stress = 83 Mpa = 83 N/mm²,

Shear stress () = Ultimate tensile stress /F.O.S

= 41.5 N/mm²

Polar moment of inertia, J

= x d⁴/32

=.1 x d⁴

We know that,

/r = T/J

41.5/(d/2) = 40000/(.1 x d⁴)

Solving above equation,

.1 x d⁴ / (d/2) = 40000/41.5

‘d³ = 963

‘d = 9.875 << 50 mm(diameter obtained while buckling calculation)

Moment of inertia,

I = X d⁴/64

= 306640 mm⁴

C/s Area , A = X d²/4

= 1963 mm²

Radius of gyration, r = Sqrt (I/A)

=12.5

Slenderness ratio = L/r

= 50/12.5

= 4