The diameter of the shaft will be ; ≥ 52.13 mm
Using the given data :
Power ( p ) = 20 * 10³ W
N ( revolution ) = 200 r.p.m
W ( load weight ) = 900 Newtons
L = 3 m
Shear stress ( [tex]\frac{\pi }{16} * 42 *d^3[/tex] ) = 42 Mpa = 42 N/mm²
Tensile stress ( Τ ) = 56 Mpa = 56 N/mm²
lets assume Diameter of shaft = d
First step ; calculate the value of the torque transmitted
[tex]T = \frac{P *60}{2\pi N}[/tex] = [tex]\frac{20000 * 60 }{2*\pi *200}[/tex] = 955 N-m = 955 * 10³ N-mm
Step 2 ; determine the max bending moment
M = [tex]\frac{W*L}{4}[/tex] = [tex]\frac{900 *3}{4}[/tex] = 675 * 10³ N-mm
next ; Determine the equivalent twisting moment ( T )
[tex]T = \sqrt{M^2 + T^2}[/tex] = [tex]\sqrt{(675 *10^3 )^2 + ( 955*10^3))^2}[/tex] = 1169 * 10³ N-mm
Final step ; Determine the diameter of the shaft
note ; equivalent twisting moment can be expressed as
T = [tex]\frac{\pi }{16} * 42 * d^3[/tex]
∴ 1169 * 10³ = 0.1963 * 42 * d³
d³ = ( 1168 * 10³ ) / ( 0.1963 * 42 )
d = [tex]\sqrt[3]{141668.486}[/tex] = 52.13 mm
Given that the maximum tensile or compressive stress is ≤ 56 Mpa
Hence we can conclude that The max diameter of the shaft will be ; ≥ 52.13 mm
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