PROBLEM 3. A POWER OF ONE THOUSAND KILOWATTS IS TO BE...

PROBLEM 3. A POWER OF ONE THOUSAND KILOWATTS IS TO BE TRANSMITTED AT 800 RPM BY MEANS OF A STEEL SHAFT WITHOUT EXCEEDING A TORSIONAL STRESS OF 80 MPa OR TWISTING THROUGH MORE THAN 1° IN A LENGTH OF 20 DIAMETERS. DETERMINE THE FOLLOWING: A) DIAMETER OF SOLID SHAFT REQUIRED B) IF Di = 0.5 Do, WHAT HOLLOW SHAFT DIAMETERS ARE REQUIRED? C) WHICH IS LIGHTER FROM A) AND B), AND BY WHAT PERCENTAGE? CONSIDER EQUAL LENGTH AND SAME MATERIAL. C) SOLID SHAFT B) HOLLOW SHAFT PROBLEM 4. A BELTED TRANSMISSION SYSTEM IS LOADED AS SHOWN. THE INPUT SHAFT TAKES ITS POWER FROM A 150 kW MOTOR THAT RUNS AT 100 RPM. THE OUTPUT SHFT DRIVES A CERTAIN MACHINERY EQUIPMENT. DETERMINE THE FOLLOWING: A) TORQUE AVAILABLE AT EACH OF SHAFTS A, B, AND C. B) TORSIONAL STRESS INDUCED ON THE 60 mm DIAMETER OUTPUT SHAFT. C) LENGTH OF BELT REQUIRED TO CONNECT EACH PAIR OF PULLEYS. D) MAXIMUM BENDING MOMENT ON SHAFT B, CONSIDERING A BELT COEFFICIENT OF FRICTION OF 0.25.

Transcript

00:01 So for problem number three, a, it's solution, the diameter of solid shaft, we have to solve for that first.

00:23 So this is the torque t.

00:44 So we can use 2 pi n t divided by 60 to solve for p.

01:01 So then the torque is equal to 60p divided into 2 pi n.

01:19 So this becomes 60 times 10 to the 6 power watts divided into 2 pi times 800 rpm.

01:49 So this will then equal approximately 11 ,936 .6 .6 newton meters of torque.

02:12 All right, so now we go on in problem three, part a.

02:18 We can use the torsional stress.

02:46 So in this case, our torque with respect to the torsional stress, t is just a 16t divided into pi diameter cubed.

03:09 So then d is equal to cube root of 16t, divided into pi times taw so this becomes the cube root of 16 times 11 ,936 .6 newton meters divided into pi times 80 times 10 to the six power pascales so we yield approximately 0 .125 meters and this is equal to 125 millimeters.

05:09 So now we move on to the hollow shaft diameters.

05:15 This will be d and d .i.

05:23 So d will be the outer diameter, and d .i is the inner diameter.

05:39 So the outer diameter, 0 .5d and thus the inner diameter.

05:46 Point 5d so using torsional stress can take taw make this equal to 16 t d and divided into pi d to the fourth power minus di to the fourth power then times a d times 10 to the power pascal's.

06:34 So this is calculated as 16 times 11 ,936 .6 newton meters, then times the outer diameter divided into pi times d to the fourth as in the outer diameter to the fourth minus 0 .5d this quantity to the fourth power.

07:26 So if we solve for d, as in the outer diameter, if you work through this expression, it will get 0 .134 meters or 134 millimeters, and then the d -i is equal to 0 .5d, so that is worked out to 67 millimeters.

08:05 So for c of problem 3, we have the weight comparison.

08:32 So we can start with the area of solid shaft, call this a solid.

08:41 This equates to pi over 4 times d squared.

08:55 It's equal to pi over 4 times 1 .25 meters squared.

09:06 So this is approximately equal to 0 .012.

09:16 1227 square meters.

09:24 So next we have the area of hollow shaft called us a hollow.

09:45 This is pi over 4 times the outer diameter squared minus the inner diameter squared.

10:07 So this is equal to pi over 4 times 0 .134 meters squared.

10:24 Minus 0 .067 meters quantity squared.

10:39 So you can put these two in brackets here.

10:47 So this is approximately equal to 0 .01 .06 square meters.

11:02 So the hollow shaft is lighter.

11:06 Our weight difference.

11:07 So the percent difference is the a solid minus the a hollow.

11:35 Divided into a solid, then times 100%.

11:55 So this is 0 .01227 square meters minus 0 .016 square meters, then divided into 0 .0127 square meters, then multiply times 100%.

12:53 So if you work this out, you will get 13 .6 % as the difference.

13:04 So that is 8, b, and c for problem 3.

13:08 And now we can move on to problem 4.

13:13 So this solution set will now be raised...

Question

Please give Ace some feedback