Answer:
a) 2.452
b) 1.256
Explanation:
Stress due to dead weight. = 14 Ksi
Stress due to fully loaded tractor-trailer = 45Ksi
ultimate tensile strength of beam = 76 Ksi
yield strength = 50 Ksi
endurance limit = 38 Ksi
Determine the safety factor for an infinite fatigue life
a) If mean stress on fatigue strength is ignored
β = ( 45 - 14 ) / 2
= 15.5 Ksi
hence FOS ( factor of safety ) = endurance limit / β
= 38 / 15.5 = 2.452
b) When mean stress on fatigue strength is considered
β2 = 45 + 14 / 2
= 29.5 Ksi
Ratio = β / β2 = 15.5 / 29.5 = 0.5254
Next step: applying Goodman method
Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]
= 19.47 Ksi
hence the FOS ( factor of safety ) = Sa / β
= 19.47 / 15.5 = 1.256
Given:
(a)
According to the question,
→ [tex]\beta = \frac{45-14}{2}[/tex]
[tex]= \frac{31}{2}[/tex]
[tex]= 15.5 \ Ksi[/tex]
hence,
The safety factor will be:
= [tex]\frac{Endurance \ limit}{\beta}[/tex]
= [tex]\frac{38}{15.5}[/tex]
= [tex]2.452[/tex]
(b)
→ [tex]\beta_2 = \frac{45+14}{2}[/tex]
[tex]= \frac{59}{2}[/tex]
[tex]= 29.5 \ Ksi[/tex]
Ration will be:
= [tex]\frac{\beta}{\beta_2}[/tex]
= [tex]\frac{15.5}{29.5}[/tex]
= [tex]0.5254[/tex]
By applying Goodman method,
→ [tex]Sa = \frac{0.5254\times 38\times 76}{0.5254\times 76+38}[/tex]
[tex]= 19.47 \ Ksi[/tex]
hence,
The safety factor will be:
= [tex]\frac{Sa}{\beta}[/tex]
= [tex]\frac{19.47}{15.5}[/tex]
= [tex]1.256[/tex]
Thus the above response is correct.
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