Question

In: Mechanical Engineering

You are to make a bowling ball out of 2024-0 by forging in an impression die...

You are to make a bowling ball out of 2024-0 by forging in an impression die with flash. What force (tons) is needed? What will be the weight of the finished bowling ball?

Yield strength = 75 MPa = 11 ksi

Ultimate tensile strength = 190 MPa = 27 ksi

Max bowling ball diameter = 8.595 inches

Density = 0.1 lbs/in³

Expert Solution

according to the information provided,

we understand that there is negligiable friction acting at the die-workpiece interface.

hence, forging load = compressive force acting on the cross-section of workpiece

the compressive force, F = Y*A

where, Y= yield strength of the material or flow stress

and, A = cross section of bowling ball on which the impression force will be acting

thus, A = (4r²)/2 = 2r²

where r = radius of the bowling ball

A= 2* * (8.595*0.0254)² meter² = 0.3 meter²

and Y= 75 MPa

thus compressing force, F = (75 x 10⁶)* 0.3 = 22.5 x 10⁶ N = 2294.36 ton (metric)

Now to find the weight of the bowling ball,

m = volume*density

volume of ball = (4r³)/3 = 4**(0.109³)/3 = 0.005424 meter³

density of 2024-0 aluminium = 0.1 lbs/in³ = 0.1* 27679.905 kg/meter³ = 2767.99kg/ m³

thus weight of finished bowling ball = 2767.99 * 0.005424 kg = 15.01 kg

samet mamat answered 2 years ago

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