Question 16.3: A design for an industrial oven door has a 15-mm-diameter ho...

From the given data:
$φ = 110°$
$n_o = 5.0$ cycles/min
$D = 15$ mm
$L$ = Length of the bearing, in or mm
$F$ = (20.40 kN)/2 = 10.2 kN = 10 200 N on each hinge
$pV$ limit = 0.525 MPa . m/s
The allowable $pV$ value is

$(pV)_{all} = pV$ limit × 0.25 = 0.525 MPa . m/s × 0.25 = 0.1313 MPa . m/s
We can now compute the equivalent number of revolutions per minute from

$n_{eq} = n_o(2φ)/360$ = (5.0 cycles/min)(2)(110°)/360° = 3.06
Now the equivalent sliding velocity is
$V_e = πDn_{eq}/(60 000)$ m/s = V = π(15)(3.06)/(60 000) m/s = 0.00240 m/s
We can now solve for the limiting bearing pressure:

$p_{all} = (pV)_{all}/V_e$ = (0.1313 MPa . m/s)/(0.00240 m/s) = 54.69 MPa = 54.69 N/mm2
Let $p = F/LD$, we can then solve for the required length as follows:
$L = F/pD$ = (10 200 N)/(54.69 N/mm${}^2$)(15 mm) = 12.43 mm
A preferred value of $L = 16$ mm would be reasonable.