Question

A solid 0.75 in diameter steel shaft transmits 7 hp at 3,200 rpm. Determine the maximum shear stress magnitude produced in the shaft. Hint: Use P=Tω and convert hp to ft-lbf/s. Find τ by using Tc/J. Recall max shear stress will be on the outer most surface.

Answer 1

Maximum shear stress is 11.47 MPa.

Step-by-step explanation:

Given:

D=.75 in⇒D=19.05 mm

P=7 hp⇒ P=5219.9 W

N=3200 rpm

We know that

P=T
`* \omega`

Where T is the torque and
`\omega` is the speed of shaft.

P=
`(2\pi N* T)/(60)`

So 5219.9=
`(2\pi * 3200* T)/(60)`

T=15.57 N-m

We know that maximum shear stress in shaft

`\tau _(max)=(16T)/(\pi * D^3)`

`\tau _(max)=(16* 15.57)/(\pi * 0.01905^3)`

`\tau _(max)`=11.47 MPa

So maximum shear stress is 11.47 MPa.

Answer 2

Answer:
`\tau _\left ( max\right )`=11.468MPa

Step-by-step explanation:

Given data

`power`
`\left ( P\right )`=7 hp=5220 W

N=3200rpm

`\omega`=
`(2\pi* N)/(60)`=335.14 rad/s

diameter
`\left ( d\right )`=0.75in=19.05mm

we know

P=
`Torque\left ( T\right )\omega`

5220=
`T* 335.14`

T=15.57 N-m

And

`\tau _\left ( max\right )`=
`(T)/(Polar\ modulus)`

`\tau _\left ( max\right )`=
`(T)/(Z_(P))`

`\tau _\left ( max\right )`=
`(16* T)/(\pi d^(3))`

`\tau _\left ( max\right )`=11.468MPa