The aluminum rod bc g 26 gpa is bonded to the brass rod ab g...

Name: the aluminum rod bc g 26 gpa is bonded to the brass rod ab g 39 gpa knowing that each rod is solid and the rods ab and bc have diameters of 14 mm and 12 mm and lengths of 200 mm and 300 mm r 42609
Uploaded: 2022-01-24T17:26:04-08:00
Duration: 10 min 52 s
Channel: Priyanka Khokher
Description: the aluminum rod bc g 26 gpa is bonded to the brass rod ab g 39 gpa knowing that each rod is solid and the rods ab and bc have diameters of 14 mm and 12 mm and lengths of 200 mm and 300 mm r 42609

The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB (G = 39 GPa). Knowing that each rod is solid and the rods AB and BC have diameters of 14 mm and 12 mm and lengths of 200 mm and 300 mm respectively, determine the angle of twist at B and C. 40 Nm 40 Nm A B C Brass Aluminum

Transcript

00:01 Hello, in this question we are given a road a, b, c, d, okay? this is initially a b.

00:15 Then after this, the next portion is b c and then it is cd.

00:28 This is the complete road and this portion cd is hollow from the inside.

00:39 Okay, this is a, this is b.

00:42 This point over here is c and this is at last d okay now we are given that a torque at a is acting is 800 this is 800 newton meter i guess yeah newton meter and at b it is 1600 newton okay so when we do draw free body diagram a reaction force here would be what plus 1 ,800 that would be 2400 newton.

01:28 Okay, so when we draw free body diagram, the torque inside this ab would be 800 newton meter.

01:37 And in bc and cd, both the torque would be 2440, 2 ,400 newton meter.

01:55 This can be explained by doing the free body diagram.

01:59 Let me draw it okay so if this is a b a torque of 800 norton meter is acting okay so to balance this a torque over here must be acting that would be equal to 800 newton meter it is not kilo -neuton meter it is newton meter let me correct it so now for b to d if this is 800 acting over the hair then this would be acting over here and an external torque is also acting over here that would be 1600 newton meter that must be balanced by 1600 newton meter okay and we need to calculate the rotation at a rotation at a can be written as root what is the tors and formula t over i b is equal to tau over y is equal to g theta over l okay so i will break it into three portions and theta can be written over here as t l by g j okay j all the portions can be calculated as we are given the diameters okay the diameter for the first portion is 36 m m so j of aluminum is geo -hillemium is what? pi, let me put my screen on the laptop on that charge for a two second.

04:08 Yeah, now, let us continue.

04:09 J for aluminium would be pi into d.

04:14 D is 36 to the power 4 divided by.

04:21 Similarly, j for aluminum and then brass.

04:29 J for brass can be written as diameter of the brass is 60 m m.

04:33 J for aluminum that is j for brass sorry that too for solid portion only not brass solid portion that would be going to pi into 60 to the power 4 over 30 and j for brass portion but hollow portion it would be what pi into 60 to the power 4 minus 40 to the 4 divided by 32 okay length is also given to us length for ab is 400 meter length is 400 for bc is 375 and for cd is 250 okay now let me write g values also g is for aluminium it is 27 gpa and for brass, it is 39 gp.

06:01 Okay, so total theta at a can be written as theta between d to c plus theta between c to b plus theta between b to a...

Question

The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB (G = 39 GPa). Knowing that each rod is solid and the rods AB and BC have diameters of 14 mm and 12 mm and lengths of 200 mm and 300 mm respectively, determine the angle of twist at B and C. 40 Nm 40 Nm A B C Brass Aluminum

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