Question

A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing stress of the shaft and key are 30 MPa and 240 MPa, respectively what is the length of the rectangular key if the width is one fourth that of the shaft diameter?

Answer 1

The length of the rectangular key is 0.4244 m

Step-by-step explanation:

Given that,

Power = 7.46 kW

Speed = 1200 rpm

Shearing stress of shaft = 30 MPa

Mini shearing stress of key = 240 MPa

We need to calculate the torque

Using formula of power

`P=(2\pi NT)/(60)`

Where, P = power

N = number of turns

Put the value into the formula

`7.46*10^(3)=(2\pi*1200* T)/(60)`

`T=(7.46*10^(3)*60)/(2\pi*1200)`

`T=59.36\ N-m`

We need to calculate the distance

`\tau_(max)=(16T)/(\pi d^3)`

`d^3=(16*59.36)/(\pi*30)`

`d=(10.077)^{(1)/(3)}`

`d=2.159\ m`

Width of key is one fourth of the shaft diameter

`W=(1)/(4)*2.159`

`W=0.53975\ m`

The shear stress induced in key

`\tau_(max)=(F)/(Wl)`

`\tau_(max)=((T)/((d)/(2)))/(wl)`

`\tau_(max)=(2T)/(dWl)`

`240=(2*59.36)/(2.159*0.53975* l)`

`l=(2*59.36)/(2.159*0.53975*240)`

`l=0.4244\ m`

Hence, The length of the rectangular key is 0.4244 m

Answer 2

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft,
`\tau _(shaft)` = 30 MPa

Shearing stress of key,
`\tau _(key)` = 240 MPa

width of key, w =
`(d)/(4)`

d is shaft diameter

Solution:

Torque, T =
`(P)/(\omega )`

where,

`\omega = (2\pi  N)/(60)`

`T = (7460)/((2\pi  (1200 ))/(60))` = 59.365 N-m

Now,

`\tau _(shaft) = \tau _(max) = (2T)/(\pi ((d)/(2))^(3))`

`30* 10^(6) = (2* 59.365)/(\pi ((d)/(2))^(3))`

d = 0.0216 m

Now,

w =
`(d)/(4)` =
`(0.02116)/(4)` = 5.4 mm

Now, for shear stress in key

`\tau _(key)` =
`(F)/(wl)`

we know that

T =
`F * r` = F.
`(d)/(2)`

⇒
`\tau _(key)` =
`((T)/((d)/(2)))/(wl)`

⇒
`240* 10^(6)` =
`((59.365)/((0.0216)/(2)))/(0.054l)`

length of the rectangular key, l = 4.078 mm