Question

a transmission shaft supporting two pulleys A AND B and mounted between two bearing C1 and C2 . power is transmitted from pulley A&B .the shaft is made of plain carbon steel 45c8 syt is 380 & sut is 600 the pulleys are keyed to the shaft . determine the shaft diameter the ASME code if Kb is 1.5 and Kt is 1.0 .also determine the shaft diameter on the basis of torsional rigidity if the permissible angle of twist between two pulleys is 0.5 degree and the modulus of rigidity is 79300N/mm^2. Forces are 2000 & 500N

Answer 1

We know that , through torque equation we can say that,

$\frac{T}{J}= \frac {G \theta}{l}$

Here , G= modulus of rigidity = 79300 $\frac {N}{mm^2}$ , angle of twist ,$\theta= 0.5 = 0.5 \times (\frac{\pi}{180})=0.0087$

Net force = $\sqrt{2000^2 + 500^2}= 2061.55 N$

T= F$\times \frac{d}{2}$ , J $=\frac{\pi d^4 }{32}$

$\frac {F\times d \times 32}{2 \times d^4} = \frac {G \theta}{l}$

$\frac {2061.55\times d \times 32}{2 \times d^4} = \frac {79300 \times0.0087}{1}$

$d^3= 4.78 \times 10^5$

d= 78.2 mm

and for above situation there FOS is missing so we can not apply soderberg method to find the diameter.