Respuesta :
Answer:
a. [tex]\tau=51.55 MPa[/tex]
b.[tex]\tau=42.95MPa[/tex]
c.[tex]\theta=7.67\times 10^{-3}[/tex] rad.
Explanation:
Given: [tex]D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa [/tex]
We know that
[tex]\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}[/tex]
J for hollow shaft [tex]J=\dfrac{\pi (D_o^4-D_i^4)}{64}[/tex]
(a)
Maximum shear stress [tex]\tau =\dfrac{16T}{\pi Do^3(1-K^4)}[/tex]
[tex]K=\dfrac{D_i}{D_o}[/tex]⇒K=0.83
[tex]\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}[/tex]
[tex]\tau=51.55 MPa[/tex]
(b)
We know that [tex]\tau \alpha r[/tex]
So [tex]\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}[/tex]
[tex]\dfrac{51.55}{\tau}=\dfrac{210}{175}[/tex]
[tex]\tau=42.95MPa[/tex]
(c)
[tex]\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}[/tex]
[tex]\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}[/tex]
[tex]\theta=7.67\times 10^{-3}[/tex] rad.
