Answer :
Answer:
a) the lowest critical speed of the shaft is 880.57 rad/s
b) the new diameter when the critical speed is doubled is 0.05 m
c) the critical speed is 1761.14 rad/s and is doubled.
Explanation:
a) The moment of inertia is:
[tex]I=\frac{\pi d^{4} }{64}[/tex] (eq. 1)
The area is:
[tex]A=\frac{\pi d^{2} }{4}[/tex]
The speed of the shaft is:
[tex]w=(\frac{\pi }{L} )^{2} \sqrt{\frac{gEI}{A\gamma } }[/tex] (eq. 3)
Where
γ = weight density of the material = 76.5 kN/m³
E = modulus of elasticity = 207x10⁶kN/m²
L = length of the shaft = 0.6 m
d = 0.025 m
Replacing eq. 1, eq. 2 in eq. 3:
[tex]w=(\frac{\pi }{L} )^{2} (\frac{d}{4} )\sqrt{\frac{gE}{\gamma } }[/tex] (eq. 4)
[tex]w=(\frac{\pi }{0.6} )^{2} (\frac{0.025}{4} )\sqrt{\frac{9.8*207x10^{6} }{76.5} } =880.57rad/s[/tex]
b) The diameter of the shaft when the critical speed is doubled is equal to:
[tex]2w=(\frac{\pi }{L} )^{2} (\frac{d}{4} )\sqrt{\frac{gE}{\gamma } }\\d=\frac{4L^{2}*2w }{\pi ^{2} } \sqrt{\frac{\gamma }{gE} } \\d=\frac{4*0.6^{2}*2*880.57 }{\pi ^{2} } \sqrt{\frac{76.5}{9.8*207x10^{6} } }=0.05m[/tex]
c) Transforming equation 4 into:
[tex]\frac{L^{2}w }{d} =(\frac{\pi ^{2} }{4} )\sqrt{\frac{gE}{\gamma } }\\\sqrt{\frac{gE}{\gamma } }is-constant\\\frac{L_{1} ^{2}w_{1} }{d_{1} }=\frac{L_{2} ^{2}w_{2} }{d_{2} }[/tex]
If: L₂ = L/2 and d₂ = d/2, then:
[tex]\frac{L ^{2}w }{d }=(\frac{w_{2}(\frac{L}{2})^{2} }{\frac{d}{2} } )\\w_{2} =2w[/tex]
Replacing:
w₂ = 2 * 880.57 = 1761.14 rad/s
The critical speed is doubled.