Question

Can you show step by step how to solve it, including the equations.

5. An 8.0 mm (0.31 in.) diameter cylindrical rod fabricated from a 7075-T6 alloy is subjected to reversed tension-compression load cycling along its axis. If the maximum tensile and compressive loads are +8400 N (1890 lby) and -8400 N (-1890 lb), respectively, determine its fatigue life.

Answer 1

Provided Informations in problem are as below

Cylindrical rod diameter = 8mm

Thus area A= π*8²/4= 50.27 mm^{​​​​​2}

Material used 7076-T6 Alloy which is having following properties ( from material standard)

Ultimate tensile strength = S_{​​​​​ut}​= 572 MPa

Tensile yield strength = S_{​​​​​Sy} = 503 Mpa

Fatigue strength = S_{​​​​​​e}​​​= 159 MPa

Maximum tensile L_{​​​​​​t} and L_c Compressive loads= +8400 N and -8400 N

Thus maximum tensile(S_{​​​​​​t}) and compressive (S_c)stresses = +(8400/50.27) and -(8400/50.27) = +167.1 N/mm^​​​2​​​​ and - 167.1 N/mm^{​​​​​2}

Find out fatigue life

Equations for fatigue life are as below

N = 10^-c/b*S_{​​​​​​a}^{​​​​​1/b}

b=( -1/3)* log (0.8*S_{​​​​​​ut}/S_{​​​​​e}​​​) = (-1/3)*log [(0.8*572/159)]= - 0.15393

c= log [(0.8*S_{​​​​​​ut})²/S_{​​​​​​e}​​​] = log [(0.8*572)²/159] = 3.12

Where N = Fatigue life in no of cycles

S_{​​​​​​a}​​​= Stress amplitude= [S_{​​​​​t}-(-S_{​​​​​c}​)]/2 = 167.1 N/ mm^​​2​

Substituting known values in above equation we can compute Fatigue life as below.

N= 10^{-(3.12/(-0.15393))}*167.2^1/(-0.15393)

= 10^20.79*167.2^-6.5= 2182550 Cycles