Answer:
The dimension of the square cross section is = 30mm * 30mm
Explanation:
Before proceeding with the calculations convert MPa to Kpsi
Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi
the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut
at 10^6 cycles" for 111.65 Kpsi = f ( fatigue strength factor) = 0.83
To calculate the endurance limit use Se' = 0.5 Sut since Sut < 1400 MPa
Se'( endurance limit ) = 0.5 * 770 = 385 Mpa
The surface condition modification factor
Ka = 57.7 ( Sut )^-0.718
Ka = 57.7 ( 770 ) ^-0.718
Ka = 0.488
Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one
The endurance limit at the critical location of a machine part can be expressed as :
Se = Ka*Kb*Se'
Se = 0.488 * 0.85 * 385 = 160 MPa
Next:
Calculating the constants to find the number of cycles
α = [tex]\frac{(fSut)^2}{Se}[/tex]
α =[tex]\frac{(0.83*770)^2}{160}[/tex] = 2553 MPa
b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]
b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex] = -0.2005
Next :
calculating the fatigue strength using the relation
Sf = αN^b
N = number of cycles
Sf = 2553 ( 10^4) ^ -0.2005
Sf = 403 MPa
Calculate the maximum moment of the beam
M = 2000 * 0.6 = 1200 N-m
calculating the maximum stress in the beam
∝ = ∝max = [tex]\frac{Mc}{I}[/tex]
Where c = b/2 and I = b(b^3) / 12
hence ∝max = [tex]\frac{6M}{b^3}[/tex] = 6(1200) / b^3 = 7200/ b^3 Pa
note: b is in (meters)
The expression for the factor of safety is written as
n = [tex]\frac{Sf}{\alpha max }[/tex]
Sf = 403, n = 1.5 and ∝max = 7200 / b^3
= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex] making b subject of the formula in other to get the value of b
b = 0.0299 m * 10^3 mm/m
b = 29.9 mm therefore b ≈ 30 mm
since the size factor assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2
the dimension = 30 mm by 30 mm
