Respuesta :
Answer:[tex]\tau _\left ( max\right )[/tex]=11.468MPa
Explanation:
Given data
[tex]power[/tex][tex]\left ( P\right )[/tex]=7 hp=5220 W
N=3200rpm
[tex]\omega [/tex]=[tex]\frac{2\pi\times N}{60}[/tex]=335.14 rad/s
diameter[tex]\left ( d\right )[/tex]=0.75in=19.05mm
we know
P=[tex]Torque\left ( T\right )\omega [/tex]
5220=[tex]T\times 335.14[/tex]
T=15.57 N-m
And
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Polar\ modulus}[/tex]
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Z_{P}}[/tex]
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{16\times T}{\pi d^{3}}[/tex]
[tex]\tau _\left ( max\right )[/tex]=11.468MPa
Answer:
Maximum shear stress is 11.47 MPa.
Explanation:
Given:
D=.75 in⇒D=19.05 mm
P=7 hp⇒ P=5219.9 W
N=3200 rpm
We know that
P=T[tex]\times \omega[/tex]
Where T is the torque and [tex]\omega[/tex] is the speed of shaft.
P=[tex]\frac{2\pi N\times T}{60}[/tex]
So 5219.9=[tex]\frac{2\pi \times 3200\times T}{60}[/tex]
T=15.57 N-m
We know that maximum shear stress in shaft
[tex]\tau _{max}=\dfrac{16T}{\pi \times D^3}[/tex]
[tex]\tau _{max}=\dfrac{16\times 15.57}{\pi \times 0.01905^3}[/tex]
[tex]\tau _{max}[/tex]=11.47 MPa
So maximum shear stress is 11.47 MPa.
