Question

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the solid cylindrical shaft BC. Design specifications require that the displacement of D should not exceed 15 mm from the time the punch first touches the plastic sheet to the time it actually penetrates it. Determine the required diameter of shaft BC if the shaft is made of a steel with G.

Answer 1

Answer:

hello your question lacks some data and required diagram

G = 77 GPa,  т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P )  = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅  ,   where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅  of  CD on the shaft BC

Torque = 660 * 0.3 cos∅

             = 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

Answer 2

The required diameter of shaft BC will be 252.65 ×10⁻³A shaft's diameter is the line that goes through its center and splits it into two equal halves.

What is the diameter?

A diameter is a line that travels through the center of a sphere and intersects the circumference at opposing ends. It measures twice as long as the sphere's radius

The given data in the problem is;

P is the force = 660 N

d is the displacement = 15 mm

G is the constant= 77 GPa

τ is the shear stress= 80 MPa

From the torsional equation;

$\frac{\tau}{R} =\frac{T}{j} \\\\ \frac{\pi}{32} \times d^4=\\\\ \tau =\frac{T \times R}{J} \\\\\ \tau =\frac{T \times (\frac{d}{2} )}{J} \\\\ 80= \frac{32 \times 190 \times \frac{d}{2} }{\pi} \\\\\\d=252.65 \times 10^{-3}$

Hence the required diameter of shaft BC will be 252.65 ×10⁻³.

To learn more about the diameter refer to the link;

https://brainly.com/question/5501950