A solid cylindrical part with dimensions of 25 mm in diameter and 15 mm in height is to be produced from fine loose copper powder and is to have a tensile strength of 175 MPa. Assuming that the sintered density is equal to the compacting density, determine a) the compacting pressure, b) the press force and c) the initial volume of powder needed.

(a) 210 MPa (b) 377.97 KN (c) 7363.108 mm³

Explanation:

Solution

Given that

Diameter (d) = 25 mm

Height = 15 mm

Tensile Strength = 175 MPa

(a) We find the compacting pressure which is given below:

(Pe) = 2Sth/d

Here

St =175 MPa

h = 15 mm

d = 25 mm

Pe = 2 * 175 *15/25

=210 MPa

Pe = 210 MPa

(b) We calculate for the press force (Fp) shown below:

Fp = A *St

A = 2πr (r+h)

= 2π * 12.5 (12.5 + 5)

A =2159.84 mm²

So,

Fp = 2159.84 * 175

Fp = 377.97 KN

(c) We calculate for the initial volume of powder required which is given as:

Vp = πr² h

=π (12.5)² * 15 mm³

Vp = 7363.108 mm³

more information

explanation:

unlike the threaded fasteners which are used to clamp the machine members, power screws are used to transmit power. common applications of power screws are: screw jacks: to lift weight. lead screw of lathe: for axial movement of tool and its precise positioning.