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

Answer:

diameter is 13.46 mm

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