ok i think i now understand. Your bluntness is appreciated haha. I worked the question with both values and now see how designing to 80/1.5 delivers a greater diameter. I've attached the working if you could look over it and see if i have done the correct thing? thanks for your time once...
"calculate a suitable diameter to the nearest millimetre for the secondary drive shaft for a FOS of 1.5, if the shear stress in the shaft is not to exceed 80MPa when the shaft is under maximum load. (Hint – be careful about the point in your calculations where you apply the FOS. It isn’t just a...
ok well i suppose i need to be more specific to the factors of this question. It is a first year engineering question that i have been studying, i will post the details of the question for you guys to comment on. thanks for everyones time.
im working on a question that requires me to calculate the minimum shaft diameter for a maximum shear stress. the value is 80mpa. The factor of safety is 1.5, now I am aware that i can't simply increase the shaft diameter by 1.5 to do this correctly. Should I calculate a suitable diameter for...
so assuming that this is correct, i then attempted b.
it asks for max torque in shaft b, max torque occurs at max power.
max power occurs when 2444 rpm.
40/208 = .1923 rotations to every one of the small gear
x by 2444
/60
= 7.8 rots/sec of the large gear.
so
p=T.2(pi).f
28.5x10^3=T.2(pi).7.833...
so I am leaning towards the idea that power is simply .95 percent of the power supplied, meaning that the angular velocity and the torque are obviously different. am i on the right track of understanding this?
Homework Statement
The problem is attached, I am only looking for guidance on a, I am most probably just completely over complicating this.
Homework Equations
p=T.2(pi).f
t= torque
f=frequency/rot/sec
The Attempt at a Solution
im just stuck on wether the power in shaft b is...