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Tle shafi consists of a titanium alloy tube AB and solid stainless steel shaft BC Tube AB has a length of L, = 40 in , an ourside diameter of d, = 1.75 i, and a w...

Question

Tle shafi consists of a titanium alloy tube AB and solid stainless steel shaft BC Tube AB has a length of L, = 40 in , an ourside diameter of d, = 1.75 i, and a wall thickness oft = 0.125 in. Shaft BC has a length of Lz = 50 in and a diameter of d2 = [.25 in. Use G 6,500 ksi for tle tube AB, and G L.OOO ksi for the shaft BC The torque Tc 290 |b fi acts at pulley € in the direction shownDetermine thc maxumum shear stress in thc shaft BC What Is the shcar stress in the shaft BC at rdial dista

Tle shafi consists of a titanium alloy tube AB and solid stainless steel shaft BC Tube AB has a length of L, = 40 in , an ourside diameter of d, = 1.75 i, and a wall thickness oft = 0.125 in. Shaft BC has a length of Lz = 50 in and a diameter of d2 = [.25 in. Use G 6,500 ksi for tle tube AB, and G L.OOO ksi for the shaft BC The torque Tc 290 |b fi acts at pulley € in the direction shown Determine thc maxumum shear stress in thc shaft BC What Is the shcar stress in the shaft BC at rdial distance of Inch on any cross section? What thc maxunum shear strain in the shaft BC? Determinc the Torque Te required at pulley B so that the rotation angle (of twist) of pulley € With respect t0 A IS zero What Is the nonal stress in the shaft BC?



Answers

The solid pulley in Figure $P 9.22$ consists of a two-part disk, which initially rotates counterclockwise. Two ropes pull on the pulley as shown. The inner part has a radius of $1.5 a,$ and the outer part has a radius of 2.0a. (a) Construct a force diagram for the pulley with the origin of the coordinate system at the center of the pulley. (b) Determine the torque produced by each force (including the sign) and the resultant torque exerted on the pulley. (c) Based on the results of part (b), decide on the signs of the rotational velocity and the rotational acceleration.

This question we have given Demeter of self. Hey, baby equals two. 20. And Demeter or soft City course two. 25. And diameter of soft E f is equals to 40. Mm. And allowable shearing stress. How is equal to 60 Mega pastor. Okay, so, no. As the peripheral peripheral motions of the gears are equal so we can write between soft A B and C D s t c d divided by r c equals two. T divided by r b. Okay, so from here, T c d will be equals two RC by our be manipulated by d. Okay, so, by putting the value of r, c and R B R c and R B R the radius of gear C and B so talking city will be equals two 75 divided by 30. These values are directly return from the given diagram, so this will be equals to 2.5 time So deep. Okay. Similarly, for e f and C d. We can write e f divided by uh huh and equals two d c d. Divided by are the okay. So from here store in years, it questions R f divided by are the multiplied by TC. So by putting the value of TCD we get d e f. It calls to R F from the diagram is 19 and RD from. The diagram is 30 and TCD is 2.53 from 2.5 tea from this equation. Okay, so from here, port in EF consult to be seven point five. Okay, so, no, the magnitude of torque in soft baby will be equals two. The baby 22 t equals two. Bye bye. Who r a B square? Well played by allowable shearing stress top. So by putting values, he will be equals two by multiplied by. And you This is cute. Okay, multiplied by allowable stress is 60 divide by two. So from here we get capital t equals two 94 points. 248 Hello, Newton. But Kelly Newton Mm. Okay. So now the magnitude of tort in the South City will be equal to the silly calls to buy by two. Our city, you. But Blair by yeah, So now putting values, we get t c d equals to buy. Played by 12.5 multiplied by 60. Divide by two. So from here D C D is 25 times of t equals two by multiple away, 12.5 q multiplied by 60 developed by two. So from here we get equals two. 73 0.6 Newton. And then Okay, so now similarly, the magnitude of torque in this soft EF he's d e f equals two bye bye to, uh, yeah, holy skew, but played by Tom. So now the ear is close to 7.5 t close to by multiplying by r e f is 20 Q multiplied by 60 divided by two. So from here, capital T comes out to be 100.5. Hello, Newton. Mm. So the smallest value of T obtained from the three cases will be the largest thought that can be applied. So t maximum is equals to the smallest thought, and we have smaller store equals to 73.6. Clue. Newton Emma. This can be converted and can be written as 73.6 Newton meter. This will be the maximum thought that can be applied. Okay, So the answer for this problem he's

So here in this problem were given this circular pipe on the magnitude off p A Were given us 10 killer Newton on we We were given us 20 killer Newton. No, the diameter D one. As you can see in this figure, de even were given us 50. Mm. Similarly, d two were given us 60. On deep three were given us 55. On defoe were given us 65. Mm. No. We have to calculate the exile normal stress in the each segment. So let this segment B one and let this segment Beato now be calculating the stresses in each segment. So for that will be first calculating the cross sectional area. So a one will be called to by by four day, two square minus d one square. So the two square minus d one square. So we're going to pluck the values supply by four. Data were given us 60. Mm. So 60 square minus D one. We have 15 men, so 50 square. So we'll get the area. Of course. Sexual area as 8. 63 0.93 Square. Now. Next will be calculating the core sectional area of second segment that is a two will be equals two by before D four square, minus d three square. So d four square minus D three square. Now put the values so by before D four, we have 65 square 65 square. Similar. This is 55 square. Andi will get the CRA sectional area A to US 9. 42 0.47 Emam Square. Now we can easily calculate Distresses two Sigma one will be equals toe PB divide by a one. So PB we have 20 million London. So 20 times we're converting the units cologne eaten into Newton for that will be multiplied with 1000. There's much off Newton divide by a one. We have 8. 63 0.93 square. So we're going to solve this Onda. We'll get the value as 23 0.15 Newton. Poor mm square on in Newton Per Mm. Square means make up our school. So this is distress in the segment one? No. Next. It will be calculating the stress in the segment too. So Sigma Tau will be equals two b b minus b A. Because if you see the forces at any construction, it will be P B minus B A. So 20 minus 10 means 10 killer Newton on again will be converting killer Newton Inter Newton so multiplied with 1000 There's much off Newton divide by a one a 12 story. So it on a do we have nine, actually. The mistake here it should be 9 42. So 9. 42.47 square. Now I'm going to block the values. So 9. 42 point 47 square Onda we get 10 point six or one newton mm square. So, Newton, um, square means make up our school. So this is a stress in a 2nd 2nd segment. So this answer on next year will be solving the second part. Now, in the be part, were given that the elongation in the upper pipe Delta one is 3.29 On the elongation and the bottom part, Delta Tau were given us 1.25 And we have to calculate the normal exile normal stress in each segment. So strain and the section one would be quills toe elongation over the original length. So here, effective elongation would be Delta one minus Delta Tau. So Delta one, minus Delta Tau over L one. Now we can build the values. Delta One, we have 3.29 Delta two is 1.25 On l one. We have 300 of them, so we'll get the strain. US 6.8 times 10 to the power minus tree. So this is the strain in the first segment. Next, it will be calculating the strain and the second segment. So if silent to in this case, elongation would be Delta Tau only. So Delta Tau over eight. So Delta Tau, we have 1.25 men. Divide by L two. We have 400 men, so we'll get the strain as 3.12 times 10 to the power minus three. So this is the strain in the second segment. So I hope you have understood the problem. Thank you.


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