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Chapter 4.1 - Mechanics of Deformable Bodies (ESci132) — Transcript

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0:00hello guys good morning and uh welcome
0:03back
0:03to our class here in esai 132
0:06your mechanics of deformable bodies and
0:11yay your christmas break is fast
0:13approaching
0:14and what are your plans for christmas
0:18or do you have any plans or are you
0:21excited of taking a break from all of
0:25these stresses that we have right now
0:27so it's up to you
0:31yeah let's finish our topic first before
0:34we'll
0:35go to our christmas break sure
0:38so we are now done on our um
0:41third module our third module that
0:45talks about extra loads and we already
0:48knew how to solve for the forces
0:51the internal forces in an actually
0:53loaded member and we also knew
0:56statically actually loaded members and
0:59also thermal stresses
1:01okay and on our next module
1:04on our fourth module we will now be
1:07talking about
1:08torsion so as you can remember torsion
1:11is
1:12a moment that is exerted
1:15along the longitudinal axis of isha
1:19perhaps so specifically for this lesson
1:22we will talk about
1:24the torsional deformation of a circular
1:26shaft
1:27and later on we will generate an
1:29equation
1:30um for us to be able to solve for the
1:33deformation
1:34of a material due to torsion
1:37and we will learn torsion
1:40formula okay our goal for this morning
1:44is that first is we need to understand
1:47how the deformation of a material due to
1:50torsion works
1:52and number two is that you should be
1:55able to generate a formula that will
1:59um show the shear stress
2:03of a torque loaded member and
2:06number three is that you should be able
2:08to solve for
2:10these kinds of problems especially
2:13for these shear stresses for
2:16shafts that are subjected to torque
2:20okay again again and again
2:23torque is a moment that tends to twist a
2:26member about
2:27its longitudinal axis okay
2:30so this one is
2:35more applicable
2:38to our machines since our machines have
2:42this what you call
2:43shafts and in order for the mechanical
2:46parts of that machine to work
2:48then these shafts should be able to
2:50rotate
2:51because that is the means of
2:53transmitting power from your
2:55engine to your machine in order for it
2:57to
2:58work so therefore designing your shaft
3:01is very important is very critical in
3:04making a machine so i think you are very
3:08um familiar on how a material will
3:11deform
3:12when it is subjected to torque we all
3:15knew
3:16that when a material so let's say we
3:19have a circular
3:20rubber material okay this one is made of
3:24rubber
3:26and we are putting guide lines in this
3:30rubber
3:39okay so if this end
3:42at the rear part of this quan
3:46of this shaft is fixed on a wall so can
3:49you
3:49fix this as a wall fixed
3:56wall knee and then
4:00this part right here
4:03applying a torque
4:08what will happen to our material
4:11so again the rear end of this
4:15shaft is fixed on a wall and the
4:18front end on this shaft is
4:21subjected to a torque so what will
4:24happen
4:25to your material when it deforms
4:29well obviously it will
4:32look like this so this material will
4:35deform
4:36like this okay when you apply a torque
4:40going what kind of direction is this one
4:44counterclockwise direction so it will
4:46look like this
4:47so as you can observe this um
4:51lines right here still remains the same
4:55it didn't change
4:58in position or whatever but as you can
5:02observe
5:02this longitudinal lines they instead of
5:06having
5:07it horizontally
5:11and if we are going to look at it here
5:13on this quan
5:15this section here
5:18and a change in angle okay
5:22and also take note the shape of the
5:25cross section didn't change
5:27morning atonguan circles right here and
5:31angemo longitudinal lines and twist
5:34and your radial lines which are this one
5:38and this one and they remain straight
5:40but
5:41instead of having it on an initial
5:43position of the ray
5:45knee toyo
5:48into that direction so therefore
5:52there is an angle of twist
5:57and that is what we need later on
6:00topic okay
6:04observe so atoni observe
6:07phenomenon okay so let's observe again
6:10this shaft
6:13and again the other end the rear end is
6:17fixed on a wall and the
6:18front end is subjected to a torsion to a
6:21torque
6:22okay so what will happen
6:26so observe this line right here so this
6:29is your radial line i'm cutting a more
6:31india line
6:32it extends from the center of this
6:34circle into the outermost part of the
6:36circle
6:37and then atom extend from the front to
6:40the back
6:40so money yeah okay and what will happen
6:43so this is your initial
6:45opposition's emo radio lines
6:49but what will happen if we are going to
6:52um twist this one
6:59so on
7:02since this one is fixed on the wall it
7:04will still remain at this
7:06position but on the front end of this
7:10one
7:14more
7:17and then
7:24so what happened so there
7:27is a change in
7:31angle right there so this is the initial
7:35position of the
7:36of your radial line before your torque
7:39is added
7:40but later on after you have applied a
7:43torque in this end then it rotated
7:47at a certain angle and that angle is
7:51what you call your fee so this is your
7:55fee
7:56and
8:09then you will observe that your fee
8:12is smaller compared saturnfisa
8:16okay and
8:48let's study this portion right here
8:53on your plate and this one is located
8:57x away from the rear end
9:00okay
9:03so atona is a plate okay
9:06so again this is your circular element
9:11unknown element element
9:16in canada element is from a distance of
9:19x from the rear okay
9:22and again differential elements among
9:27material always and thickness anatomy is
9:30the x okay so
9:35plate again and then on the thickness
9:37atom plate is the x
9:39and this is the center of
9:42our plate
9:46and given the regios now from the center
9:50to the outermost
9:51part is let's call c
9:54okay and we are taking an
9:57element a small element from this
10:02um
10:15but later after the deformation what
10:17will happen to your element
10:19it will rotate you can say only
10:28as what is illustrated is
10:42this is what happens to your element
10:44after the deformation
10:46i'm gonna hit the side view
10:50okay which is initially
10:56initial
11:09and then there are two angles right here
11:12that you should observe so as you can
11:14recall due to
11:16stress lesson but
11:20initially when your material is a
11:22rectangle
11:23and it is subjected to a shear stress
11:26perhaps by unknown direction saying
11:28shear stress then it will deform like
11:30this
11:36so asaragani and atom um shear strain
11:39ray
11:40atom shear
11:45so this at the same time this is also
11:48your
11:48shear strain and
11:52since this one is located on the
11:53outermost part of this cross section
11:56then this is your maximum shear strength
12:01then probably an atom minimum shear
12:03strain is located at the
12:05um center of this um
12:09cross-section because as you can observe
12:12uncanny atom deformation due to your
12:15shear stress
12:16will depend along the regional distance
12:19from the center of this material
13:05[Music]
13:29probably zero because it is located at
13:32the center
13:53until it reaches to the outermost part
13:56which is your maximum shear strain
13:59material okay understood
14:03so let's um say
14:07instead of having a rectangle
14:11another rectangle there is
14:27should read on that reference that i
14:30gave you okay so
14:37okay okay
14:41let's say it is from a distance
14:44of row okay
14:47so on distance
14:51row on the other hand and distance
14:55outermost part and atom
15:06a certain distance from your center
15:10row okay and your shear strain
15:14the pitze mongolo is your gamma which is
15:17money
15:18shear strain application
15:35angle of twist okay along this
15:40element this plate element okay
15:49explaining
16:07foreign front view neha
16:10front view
16:15okay so morning elements
16:21element at point b
16:26okay but after the deformation
16:29i'm going to hit a bow knee change
16:32angle and cunning angle
16:47rectangular element
17:00top view
17:16so morning
17:27okay so this is your differential
17:30element before the deformation
17:34differentiate elemental
17:38differential elements
17:44the differential element so anya
17:47thickness any morning dx okay
17:51which is kanye okay but after the
17:54deformation
17:55and about rectangular element
18:09and this angle right here is now your
18:12gamma
18:13okay and then the
18:16location from here to here is your row
18:20okay and your angle right here is your
18:24fee okay
18:54how can your divi and your gamma be
18:57related
18:58again your gamma is your shear strain
19:02and your defeat is your angle of twist
19:05monasia
19:07so how can they be related
19:28[Music]
19:47in order for us to equate this the
19:50relationship between your
19:52shear strain and your
19:55angle of twist so how can they be
19:58related
19:59since equal manchester length let's say
20:03an arc length is
20:06r arc length
20:10arc length
20:14your arc length is equal to unsamara
20:17regions
20:18times the regios
20:35times your
20:38angle which is d
20:42okay so rho times your
20:45your angle in radians and also at the
20:48same time
20:49it is equal to the arena
20:52angle which is b
20:56dx
20:59times so which
21:15arc length
21:19okay and for you to be able to get your
21:21r
21:22which is this one it's a front view
21:25row times your dv okay
21:36dx which is this one dx
21:40times your gamma okay
21:50[Music]
21:56and ultimately you will have this
21:58equation gamma
22:00is equal to rho
22:03d v over d
22:07x
22:11okay and if we're going to manipulate
22:14this equation
22:14we can have gamma over rho
22:18is equal to um
22:23d v over dx
22:28in canon gamma overall
22:33and since your dv and your dx
22:36is constant throughout the ban
22:51or any parts of this element
22:55so therefore you can also change this
22:58one instead of having gamma over
23:00rolan pinaka outermost you can equate it
23:04into
23:05gamma max over c is equal to
23:09gamma over rho okay so
23:13still these two are the same and
23:15therefore
23:17we can also arrive on an equation
23:21by substituting this one since this one
23:23is equal to this one
23:27then you will have this equation
23:31gamma is equal to rho
23:35times gamma max
23:38over c okay so
23:43this one is your equation that will
23:46relate
23:53element from the center to the outermost
23:55part
23:56and for your shear strain
23:59okay so no
24:17generate long-term equation out from
24:19this concept we have right here
24:22and we arrived in this equation and this
24:24equation will be applied
24:26later on as we are going to generate a
24:29formula
24:30and that is what you call your torsion
24:32formula
24:33okay so i'm cutting a tongue torsion
24:35formula it will relate
24:37among um shear stress
24:40in your torque applied
24:44okay so when an external torque is
24:47applied to a shaft
24:49then it will also generate a
24:51corresponding internal torque within the
24:53shaft
24:54okay and we will develop an equation in
24:57this segment right here
24:59for us to relate this internal torque
25:02and
25:02the shear stress distribution of your
25:06shaft okay so again let's imagine that
25:09our material
25:11is behaving linearly elastic
25:14so therefore when you say linearly
25:16elastic
25:17stress there will also be a
25:20corresponding
25:21increase of your strain and their
25:24relationship is linear
25:26so therefore so as you can remember on
25:28our equation
25:30you have your gamma is equal to
25:33rho over c times your gamma max
25:38and again if we are dealing with
25:41linearly elastic materials then we can
25:44just
25:45simply change this um
25:48shear strain here into your shear stress
25:51so
25:52we can also apply this one
26:05so for example we have again that
26:08cross-section
26:10front view and then ma
26:17initial positions in radial line before
26:20the deformation
26:22but after you have you have applied a
26:24torque then it rotated
26:26right there okay and you have this
26:29differential element right here
26:32okay and your element
26:36will obviously be at its maximum again
26:40maximum strength is the outermost part
26:44absolutely
26:51so therefore this is your gamma max
26:54again
26:55and since this one is linearly elastic
26:58then we can say that this is also your
27:00tau max
27:21so therefore we can just use these two
27:24equations
27:25interchangeably depending on
27:30okay and so let's call this one as your
27:34differential area your d a and
27:38the ba as you can remember formulas
27:41shear strain i'll shear stress
27:45your shear stress is equal to your shear
27:48force over your
27:50area d v over d
27:54a okay so therefore your dv is just
27:57equal to
27:58tau d a okay so this is your
28:02differential force
28:04and formula differential force but
28:07if you're looking for the torque
28:10a torque is the same moment yeah
28:35so therefore your dt your differential
28:38torque
28:38is just equal to your dv
28:42times your
28:53and since you learned that your tau is
28:55just equal to this one
28:58then you can
29:01expand it further as rho
29:05times rho over c tau
29:09max d a
29:13and simplifying it further
29:16you will have rho squared over c
29:20tau max d a
29:23and take note these
29:27two are constant
29:30constant
29:34distance from the center to the
29:35outermost part
29:38constantly your
29:41shear stress at the maximum
29:45of your cross section so constantly
29:48so we can um move that out and then
29:51atoni integrate so therefore i'm
29:54hittable is
29:55t is equal to tau max
30:00over c integral of
30:03rho squared d a okay
30:08if you can observe this one this is
30:11familiar
30:13statics of rigid bodies
30:19statics sauna
30:31this one is your moment of inertia this
30:34is your moment of inertia of formula
30:36and this one is a special case because
30:39this one is a polar moment of inertia
30:42it is from your cross-sectional circle
30:55whatever okay so therefore we can
30:58simplify this one
30:59this one is your j and if you know
31:03the polar moment of inertia of a
31:05circular shaft then
31:07this will you can simplify that one and
31:10therefore
31:10we can simplify this equation further to
31:13get your
31:14shear stress maximum immune
31:22over your polar moment of inertia
31:26so this is your
31:30torsion
31:58so this is your general formula for your
32:00torsion
32:01your t is your torque your c
32:04is the outermost distance from the
32:07center which is basically cigoro your
32:09radius
32:10and your j is your polar moment of
32:13inertia okay can you remember what is
32:15the formula of our polar moment of
32:17inertia
32:18your polar moment of inertia for a
32:19circular shaft is just equal to
32:22pi over 2 c to the
32:26fourth okay and this one is for your
32:29solid
32:32cross section
32:35and for your liquid oh no for your
32:38hollow cross section
32:40you have this one
32:44pi over two
32:47outers um to the fourth power
32:51minus inner to the
32:54fourth power
32:58for your hello
33:04okay
33:07formula now we can have our example
33:12let's have an example i think
33:17gloomy atom weather so that gloomy
33:18pudding
33:21feeling okay so um in this
33:24first example we have to compare two um
33:27materials
33:28um two shafts the first one is your
33:32solid shaft and the second one is your
33:35halo tube shaft
33:36okay
33:48perhaps okay so you have this 100
33:52millimeter solid shaft
33:54and you also have this hollow shaft
33:57which is 100 millimeter for the
34:00radius okay so
34:03again we are looking for the maximum
34:05torque that can be applied to this one
34:07before it
34:08behaves plastically so how are we going
34:10to solve this one
34:11can you hear me because it's raining one
34:14day
34:15so anyways first we have to determine
34:19the formula we need our formula
34:22tau max
34:25is equal to t c over
34:29j we need this one because we are
34:32already given
34:33that the maximum allowable shear stress
34:36for the two materials is 75 megapascals
34:40so that value we are going to use in
34:43this
34:45um tau max right here okay because this
34:48is the maximum shear stress
34:51material okay
34:55your c is obviously your um
34:58the outermost rages of your material
35:02and your j is the polar moment of
35:03inertia we already have this one
35:06we already have this one this is what
35:08we're looking for
35:09and we don't know our j but r j
35:13is solvable so let us first solve for
35:16the
35:16polar moment of inertias okay
35:40times 100 millimeters
35:45but okay let's make it meters
35:50so 0.1 meters
35:53to the fourth power and that is equal to
35:590.1571
36:01times 10 to the negative 3
36:04meters to the fourth power
36:08this is the moment of inertia of your
36:12solid shaft for your hollow shaft
36:18pi over 2 minus
36:22c outer which is 0.1 meters
36:26to the 4 minus
36:300.075 meters to the
36:33fourth and that is equal to
36:380.1074
36:40times ten to the power of negative three
36:43meters to the fourth
36:47and now we can now solve for the maximum
36:50shear stress
36:51that your solid and your hello cha
36:54can handle okay so
36:58for solid
37:02shaft
37:07okay so for your torque semo solid is
37:09equal to
37:12j tau max
37:17solid over c
37:20which is money
37:23zero point one five seven one times ten
37:26to the power of negative three
37:29meters to the fourth and then
37:33your maximum shear stress allowable
37:37shear stress that is 75 megapascals
37:42newton per meter so that is 75
37:46times 10 to the power of 6 newton
37:50per meter right
37:54divided by your c the same will see
37:570.1 meter so therefore
38:03your maximum torque that can be applied
38:07among um solid shaft
38:11is 118 kilo newton
38:18meter okay
38:24so you can apply up to 118 kilo newton
38:29meter in a torque
38:32before it will reach to its
38:35guild stress okay take note
38:41next how about
38:44for our hello
38:47all right for
38:50hello
38:57th is equal to j
39:01times tau max
39:04hello over c this is equal to
39:09your j say among hello is
39:120.1
39:16times 10 to the power of negative 3
39:20meters to the fourth times your j
39:24which is oh no your tau max which is
39:2875 times 10 to the power of 6
39:32a newton per meter
39:36squared squared
39:40over your sea obviously
40:040.1 meters so therefore
40:07your maximum torque is 80.5
40:14kilometers meter okay
40:18so these are your answers so as you can
40:21see
40:23your solid um shaft can handle greater
40:27torque compared to among
40:29hello shaft which is very obvious okay
40:34ariana versa solid shaft comparison
40:38okay and also it is indicated a certain
40:42problem
40:43we are asked for the stress right here
40:47the stress right here so this
40:50is your ts
40:59and then we are also asked for the
41:04[Music]
41:18and we are going to use the torque t
41:21with this value 80.5 okay
41:24so but dylan is a max because this is
41:27not the maximum na
41:29tao um
41:36is equal to 80.5 kilo newton
41:41meter times
41:45and kenny instead of simulation
41:53distance from here to there which is
41:560.75 meter
41:58oh no 0.075
42:03meters over your polar moment of inertia
42:06of 0.1074
42:10times 10 to the power of 3.
42:14and your stress at c
42:17at this point is 56.2
42:23mega pascals and
42:26these are your answers for
42:29number one
42:32okay
42:40another we'll have three examples
42:48okay so for our second example we have
42:51this
42:52um shaft that is supported by two
42:56bearings
42:56and kenny shaft
43:00applied array okay and we have to
43:03determine the shear stress
43:05at points a and b of this material so
43:09can it point a and b
43:17we have to determine the shear stresses
43:21at those
43:22points so
43:26so first we have to determine the torque
43:30at section a and how can we determine
43:34the torque at section a a
43:36well we have to make a diagram
43:50and then third network
43:54this is 42.5
43:58keep inch
44:01this is 30 and this is
44:0512.5 okay
44:08this one rotates
44:11counterclockwise this one rotates
44:14clockwise and this one rotates clockwise
44:17so to show you graph
44:20counter clockwise
44:34apply
44:4912.5 counter clock clockwise
44:520 and this is your
44:55section a so therefore
44:59your torque at a is equal to
45:0412.5
45:06kip inch
45:09okay and not long okay so now
45:12also the formula t
45:16tau max is equal to t c over
45:20j so we already have our torque
45:23we knew our c or we knew our row which
45:26is given there
45:29tau is equal to
45:53is located on the outermost part then
45:56this is your sea
45:57okay
46:11so they are the same equation
46:17and then outermost okay
46:22let's proceed so we still don't know our
46:25j
46:26so we have to look for our j polar
46:28moment of inertia
46:29is equal to pi over 2 times
46:33c to the 4 and then
46:37this one is equal to pi over 2
46:40times 0.75
46:43inch which is equal to 0.497
46:49inches to the fourth power
46:52okay so now we can solve for stress at a
46:56okay a is the outermost part so that is
47:01your t max
47:02tau max is equal to great 12.5
47:07kilopounds inch
47:10times your c which is 0.75
47:14inches over your j which is
47:210.497 inches to the
47:23fourth so
47:26is keep kilo pounds per inch squared
47:29which is
47:30the unit of stress and that is
47:33your 18.9
47:38kilopounds per square inch
47:41so that is your tau max or your
47:45ta okay
47:49and now for your b tau b
47:54is equal to twelve point five
47:58times canadian c instead of c
48:02rho nancy distance elements
48:070.15 inch
48:11over zero point four hundred
48:14ninety seven inch
48:17to the fourth so therefore your tau b is
48:20equal to
48:233.77 k
48:25s i
48:30okay soyon
48:38i hope you understood this concept on
48:40how to apply this
48:41equation because this is very important
48:44as we proceed to the next topics okay
48:50for our last example we have this pipe
48:53now
48:54[Applause]
48:56torque okay an end a is supported
49:00against the wall
49:02and um wrench i'm getting wrenched
49:05appliance
49:09okay so we have to determine the shear
49:12stress
49:13at sea
49:16okay at the inner and outer walls of
49:19the central portion of the pipe which is
49:21at point c
49:24sound again i did any couple oh sorry
49:32so this is your wrench when i center
49:36maybe applying a 18 newton's theory then
49:38i apply the ring
49:41and
50:01torque is just equal to the summation of
50:04moments
50:05summation of moments at point b going
50:08clockwise is equal to 80
50:13times 200 plus
50:1680 times 300 so therefore
50:20your torque is just equal to 40
50:24newton per meter okay so that is your
50:27torque
50:29and basically this is the internal
50:31torque
50:32of the hello troop shaft
50:35okay it is given
50:40the dinner ages is 40 millimeters and
50:42the outer edges is 50 millimeters
50:48okay and what we're looking for is
50:51the tau which is equal to t rho
50:54over j or tc over j
50:57tau max okay and
51:01a sub polar moment of inertia which is
51:04pi over two outer radius which is
51:100.05
51:12fourth minus 0.04
51:16fourth
51:20and this is equal to 5.796
51:26times 10 to the power of negative six
51:29meters to the fourth okay now proceed
51:33you can solve for the outer
51:37shear stress which is kane and the inner
51:39shear stress which is kane
51:41so one of the outer shear stress which
51:43is
51:4440 newton meter
51:48times your c which is 0.05
51:53meter over 5.796 times 10 to the power
51:58of negative 6
51:59meters so meters meters so newton per
52:03meter
52:04to the fourth squared and the answer is
52:08zero point 345
52:12mega pascals
52:16and that is your shear stress on the
52:18outer
52:20part of the tube shaft and for the inner
52:22part of the shoe
52:23shaft that is 40
52:27newton meter multiplied by 0.04
52:31meters divided by 5.796
52:36times 10 to the power of negative 6
52:38meters to the 4
52:39and your inner shear stress is equal to
52:440.276
52:46mega pascals so these are your answers
52:52task discussion solution okay
52:56so this is your torsion formula
53:00and by the help of your torsion formula
53:03then you have related the shear stress
53:08and your torque the applied torque to
53:10your shear stress and that is very
53:12important for you to
53:13specialize you're going to solve
53:14problems like this
53:16so as a summary for this lesson and you
53:19already
53:20knew what are torque loaded members
53:23and say relationships yeah shear stress
53:27cyan shear strain and whatever choo choo
53:29we also have generated the torsion
53:31formula
53:32which is your tau max is equal to tc
53:35over j or your tau is equal to 0 over j
53:38and then lastly you have able to solve
53:41problems
53:42in relation to your torque and your um
53:45sheer stresses so molasses
53:50because that is so very long and
53:54let's continue the next topic on the
53:56next week
53:57and that will be our last topic
54:00for this year delhi school year but for
54:04the year 2020
54:06okay so stay tuned and
54:09see you again on the next thing bye

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