Question

An industrial machine requires a solid, round piston connecting rod 200 mm long, between pin ends that is subjected to a maximum compression force of 80,000 N. Using a factor of safety of 2.5, what diameter is required if aluminum is used with properties Sy = 496 MPa and E = 71 GPa?

Answer 1

diameter is 13.46 mm

Step-by-step explanation:

length of rod = 200 mm = 0.2 m

compression force = 80,000 N

factor of safety = 2.5

Sy = 496 MPa

E = 71 GPa

to find out

diameter

solution

first we calculate the allowable stress i.e. = Sy/factor of safety

allowable stress = 496/ 2.5= 198.4 MPa 198.4 ×
`10^(6)` Pa

now we calculate the diameter d by the Euler's equation i.e.

critical load =
`\pi ^(2)` E × moment of inertia / ( K × length )² ..........1

now we calculate the critical load i.e. allowable stress × area

here area =
`\pi` /4 × d²

so critical load = 198.4 ×
`\pi` /4 × d²

and K = 1 for pin ends

and moment of inertia is =
`\pi` / 64 ×
`d^(4)`

put all value in equation 1 and we get d

198.4 ×
`10^(6)` ×
`\pi` /4 × d² =
`\pi ^(2)` 71 ×
`10^(9)` ×
`\pi` / 64 ×
`d^(4)` / ( 1 × 0.2 )²

155.8229×
`10^(6)` × d² = 700.741912×
`10^(9)`× 0.049087×
`d^(4)` / 0.04

d=0.01346118 m

d = 13.46 mm

diameter is 13.46 mm