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A link in a mechanism is to be subjected to a tensile force that varies from 3500 to 500 N in a cyclical fashion as the mechanism runs. It has been decided to use SAE 1040 cold-drawn steel. Complete the design of the link, specifying a suitable cross-sectional shape and dimensions.

Respuesta :

Answer:

Design explained below

Explanation:

We are designing a link which is able to sustain 3500 N force because it will automatically sustain any lower force.

 

The tensile strength of AISI 1040 steel is : 585 MPa

Shear Strength = 338.15 MPa from data book

Now, 585 * 10^6 = 3500 / [Area]

(from pressure = force / area)

So, area = 5.98*10^-6= pi / 4 * d^2

So, diameter = 2.76 mm

Also, 338.15 *10^6 = 3500 / (pi * d * L)

So, L = 37.5 mm

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