Full transcript
0:01okay guys welcome back to our class
0:04in esai 132 mechanics of deformable
0:07bodies
0:08and how was the exam last week were you
0:11doing good
0:12well i'm gonna check but anyways
0:16we are now on our um
0:19third module or third chapter
0:22in this course and
0:26this chapter is all about
0:30extra loads okay so
0:33can you remember that we have dealt with
0:35extra loads previously
0:36on um i guess on our
0:40first chapter and all those things that
0:44we have studied there
0:45we will um go back to it
0:49and uh dig in more deeply
0:52a topic because there there are still a
0:55lot of things that we should learn
0:58when we are going to deal with extra
1:00loads
1:01okay so this lesson
1:08which is basically your saint venus
1:10principle
1:12and also your elastic deformation of
1:15an actually loaded member okay so
1:19um basically for these two lessons
1:22our main goal is to learn the concept
1:26of the saint venant's principle and also
1:30to solve for
1:31the deformation of an actually loaded
1:33member
1:34so we will learn new formulas in this
1:37topic
1:38and we will first derive them okay
1:41so let's go first on our first topic
1:44which is the saint venus principle
1:46as you can remember on our previous
1:48topic we
1:50have dealt more with materials that
1:52behave elastically
1:54because that is where we usually design
1:57our structural members when the material
1:59behaves elastically
2:01okay plastic
2:04the formation yeah then probably we
2:07cannot
2:08use we cannot maximize the use of that
2:11material
2:11okay and now let's consider dao
2:14a bar and let us um
2:18imagine canon real life situations ah
2:21mahitabha
2:25okay so we have this bar and let's say
2:28this bar is a robbering a bar but
2:32deformation in a bar and
2:37at this specific point
2:41okay i added i applied a force at
2:44a specific point and atom bar and let's
2:47say
2:47that the force that i've applied here is
2:51what you call p so kanina ends atom bar
2:55there is a force of p that is applied
2:58here
2:59and on the other end of the bar naga
3:12so if you are going to imagine
3:15what would be the deformation of this
3:17bar when
3:19added a force of b
3:23saying any deformation is yeah so
3:26there will be the tendency that there is
3:28a localized deformation standing atom
3:46and then assume that these lines are
3:48straight
3:49and voila the appliance for snappy
3:53undeformed panning a bar but now if we
3:56are going to apply
3:59the force of pedi is atombar what will
4:01happen
4:06and what any
4:40okay
4:54foreign
4:56ashai localized my deformation
5:04okay and then paramas mclaren
5:07zoom in
5:37okay what can you observe so
5:41there is
6:04and then once again
6:11so until such a point
6:15let's say during a point knee even out
6:18young
6:33a in an amount of stress there is always
6:36that equivalent amount of strain
6:41let's say this one is
6:44your section a
7:00section
7:57okay certain section b
8:02cannot stress distribution
8:09okay so the body
8:52throughout the whole cross-sectional
8:55area
9:00principle we can disregard kung asaki
9:02apply unloads
9:04material because eventually iniglionia
9:07from
9:08the location where the load is applied
9:11it will eventually even
9:12out okay
9:20but we can just assume distributed
9:24stress evenly distributed and
9:27same applies
9:32another situation
9:49stress distributions
9:53it will create the same
9:56stress distribution at
10:00a certain distance gigan saturn
10:08okay they will create the same effect
10:12because canicella this one and these two
10:15are equal magnitudes
10:31they will always create the same
10:34internal normal stress there is atom
10:38bar okay
10:42we disregard on how the behavior um
10:46shows there is ends
10:53ends so what we are only concerned about
10:56is
10:57the distribution nia
11:12out is also
11:15equal to the distance of largest
11:18dimensions
11:20so say for example
11:37is equal to the largest dimensions
11:40as my largest dimension any is this the
11:42largest dimension
11:45this one no this is the
11:48largest dimension
11:55so therefore this one this d and this
11:59d is said to be equal
12:04a certain point even outside and that is
12:07just basically your st synthetic
12:09principle
12:10it states that the stress and strain
12:13produced at points in a body so kanye
12:16and stress
12:18strain reproduce body sufficiently
12:21removed from the region
12:23of external load application will be the
12:26same as the stress and strain produced
12:28by any other applied external loading so
12:30monaco passable
12:31since this one and this one has the same
12:34resultant
12:35forces then they will also have the same
12:39stress distribution there is atom
12:43cross section at a considerable distance
12:46now
12:46from where the load is applied
12:50finance
13:03but we consider whatever
13:08and then it will just have the same
13:11effect kong givens animal pug apply
13:14unload so simply this one and this one
13:19has the same average
13:22normal stress and that is your st
13:25finance principle let's proceed to our
13:28second topic this one is more
13:30interesting
13:32and your second lesson is about elastic
13:34deformation of an actually loaded member
13:37the goal of this lesson is just to
13:39determine the deformation
13:41of a member that is subjected to extra
13:44loads
13:45we have learned sigouro sauna by using
13:48our knowledge
13:49in on our stress strain diagram
13:54deformation from the concepts
13:57elastic behavior but here we will
14:01create a formula
14:07deformations object so let us
14:11consider this bar
14:29go micro sectional area and then there
14:31is
14:32sectional area okay
14:40distributed
15:11distributed force there is atom bar
15:14exchanging
15:14direction and we all know that this one
15:17will
15:18elongate badong
15:24right it will elongate pad and
15:27this elongation is your delta
15:30and our goal for this um lesson is to
15:34know
15:34your delta so how will you know your
15:36delta
15:37bang delta morning
16:11so let's create a formula out of this
16:13one and
16:14we also should relate our stress and our
16:17strain so atopic solves
16:18atom elongation you should also relate a
16:21relationship with your stress and then
16:24relate to stress modulus of elasticity
16:30elasticity is equal to stress over
16:32strain
16:33by applying our method of sections
16:35before let us take a differential
16:38element
16:39okay now turning rectangle differential
16:42element
16:43in canisius thickness of the
16:46x okay and and atom differential element
16:50from this point
16:51is x okay and if we are going to make a
16:55free body diagram and an atom
16:56differential element
17:02and then we all know that this one will
17:05create a deformation
17:07differential element okay and
17:10this e nut is what you call
17:13your d delta and then since this
17:17is in equilibrium then let's say that
17:20the load application
17:21this one is n as a function of x
17:25and n as a function of x
17:28because
17:40because we have this distributed loan
17:44okay so let's just call it n
17:47as a function of x so therefore what is
17:50the stress
17:50at this differential elementary yeah we
17:54all know that stress
17:55is just equal to normal force
17:59over area and what is our normal force
18:03our normal force is n as a function of x
18:06and what is our area
18:11depending distance from this point
18:15so area also as a function
18:19of x okay
18:22so depending here
18:26depending on
18:31so this is our stress nx over ax
18:35and how about for our strain
18:38strain
18:42is your delta over initial length so
18:46what is your delta here this is your
18:48d delta over your initial length of
18:53dx so this is your
18:56um d delta over dx
19:00strain from this formula that we have
19:04learned on many
19:07lessons ago okay
19:10so how can you relate your stress and
19:12your strain
19:14your stress and your strain is related
19:16on our
19:17modulus of elasticity equation and let's
19:20just
19:21say for now atom modulus of elasticity
19:24will vary also along the x zero
19:27of course material then same resonation
19:31modulus of elasticity but that's just
19:33saying mo depending punch along the
19:35external modulus of elasticity in this
19:37case
19:38okay next generate long-term formula so
19:40what will happen so you learned
19:42that your modulus of elasticity as a
19:44function of
19:45x is equal to what is your stress
19:50nx
19:52over ax
19:56over your strain which is d delta
20:00over d x and what is the goal ganesh
20:12so we have to transfer this delta on the
20:14on the left of our equation and then
20:18manipulate the equation so what will
20:20happen
20:22d delta is equal to
20:26nx dx
20:31over ex
20:53okay so therefore the i
20:56you have this equation in solving your
20:59deformation
21:00your delta is equal to integral of
21:06nx dx
21:09over ex
21:13a x
21:16okay and then you you should evaluate it
21:19at the whole length of the
21:21beam so if this is your length so this
21:24is from
21:25and this length though is from um
21:28canning a beam is let's say five meters
21:30so therefore this is from zero
21:32to five okay so nx dx over
21:36ex ax so this is your general formula
21:40when you are going to solve for the
21:43deformation
21:44of this kind of beam
21:48if you have a distributed loading
21:51along the x and you have a modulus of
21:55elasticity
21:57that will vary along the x which is
22:00usually not possible and
22:04an area which will vary also
22:07along the x but how about sir
22:12if if any atom
22:15kind of beam
22:23familiar kind of beam
22:49the loads are just applied at the ends
22:52of this beam
22:53and then this has the same
22:55cross-sectional area throughout
22:57and then it also has the same modulus of
22:59elasticity throughout
23:01okay of course
23:07of elasticity along its length okay
23:10so therefore this is constant
23:14this is constant and this is constant
23:20therefore
23:23to simplify this one since the kind of a
23:25constant
23:26therefore is
23:29n l over
23:33e e
23:36so this is your formula among
23:40deformation given
23:43not a force nothing like nothing
23:47okay so unsung en
23:50this is the internal
23:53normal force so
24:02morning
24:24an internal normal force
24:28from here to here okay
24:43and this is your l1 this is your l and
24:46this is your elsa among another
24:48okay in your area money
24:52cross sectional areas in your modulus of
24:55elasticity which is constants
26:36uh
26:45p1
26:49p2 p3
26:53then p4 check it out
26:57so let's try to evaluate this one
27:00and then deformation
27:28is equal to the summation of
27:31n l over
27:34i okay because
27:39this is one region
27:46this is one reason
27:53this is one region upon
28:41is the same but there is a change in
28:44exchange unload so another region upon
28:46this yeah
28:49while i change the loading per uncross
28:51sectional area and exchange
28:53so this is another region and then
28:58during a portion of a change of load so
29:02another region and tell me about the
29:05p4 okay and to get the deformation of
29:08this kind of loading
29:10then we all all we have to do is to sum
29:13this
29:13one two three and four regions so once
29:24region one okay so what is that
29:27that is p1 times length na
29:32l1 okay p1 l1
29:36over cross sectional area a1
30:03and then this one is elto
30:07to over uncross sectional area is just
30:10the same
30:11among a1 so a1
30:16times our modulus of velocity which is
30:18the same
30:20then plus canada pod
30:23so during a portion is p one minus pito
30:28putta during a portion and then cutting
30:30a portion delay has the same load okay
30:32the republican play on p3 but
30:34so they represent change so this is
30:36still
30:37p one minus p two
30:40p two okay and
30:44this is your l3
30:49this is your l3 over
30:53any month changing anymore something
30:56like this is your
30:57etuna okay so your areato
31:02times same modulus of elasticity
31:05and then lastly there is a lasting
31:08region well i'm not changing area
31:11but unloading upon an internal load upon
31:14a change so what will happen
31:16so p one minus p two plus
31:19p three now hello dirk p one
31:23unknown p one minus p two unknown
31:26p one minus p two happened
31:29one minus p p2 plus p3
31:33okay times your l4
31:38l4 over
31:42area e2 times your
31:46modulus of elasticity
31:55a certain segment then you should use
31:58the summation of nl
32:00over a okay so let's have an example
32:04okay so i have here a cantilever beam
32:08extra loads apply 70
32:1140 and 80 and the goal of this problem
32:14is so to know that displacement at d
32:17and take note
32:21displacement
32:25negative displacement okay
32:28and also we are asked for the
32:31displacement of point b
32:33relative to c so the back a mag change
32:37man put on length
32:39b to c
32:50so we are looking for your delta
32:54d
32:57how can we solve that so in order for us
33:01to
33:02start we should first know all the
33:05internal forces we should first know all
33:08the
33:08reactions that is happening in this beam
33:12reaction reaction
33:20we don't know this one e
33:23n now the question is how to solve for
33:27your a n
33:28by using summation of forces along the x
33:31assuming among directions a and is going
33:34there positive
33:35so equals to a n minus
33:3980 minus 40
33:42plus 70 equals to zero
33:45so your a ndi is
33:4850. so 50 kilo
33:52newtons and therefore assumption it is
33:55going
33:56to the left okay
34:13okay now what will our what are we going
34:17to do
34:18atonish a graph
34:32a graph so this is your zero
34:38okay
34:41so let's say positive
34:46okay so during a person during a party
35:32minus 80 that is negative
35:3630 kilo newtons okay and one upon
35:39so constantly okay
35:45negative 30 minus 40 is
35:48negative 70 kilo newtons
35:52and then while i load
36:10segmentary
36:37so to get the deformation
36:40summation of n l
36:44over i so
36:50monetary segment monetary second segment
36:53and one atom 30 segment
36:55okay for segmenta normal force nemo
36:58segment one
36:5950 kilo newtons
37:03times length of two meters
37:07over atom cross-sectional area is pi
37:11times zero point zero
37:15twenty-five meters
37:18squared so node
37:37is 200 times 10 to the power of
37:429 newton per
37:45meter squared okay
38:03normal force take note of the sign
38:07the important enzymes atom force which
38:10is negative 30 kilo newtons
38:13multiplied by the length of one meter
38:17over cross-sectional area of pi
38:20times 0.025 squared
38:24times the modulus of elasticity of 200
38:28times 10 to the power of 9.
38:32okay and then last negative gap on so
38:36negative 70 times length of 1.5
38:42over cross-sectional area pi times 0.025
38:46squared times 200
38:49times 10 to the power of 9.
38:54and makal kyota and then
38:58and the answer of your displacements
39:00among d
39:01is just equal to negative
39:0789.1
39:08times 10 to the power of negative
39:113 millimeters
39:18okay so take note an atomic solver is
39:20meters
39:21in a unit so
39:25you convert to millimeters today and as
39:28you can see this is negative
39:35displacement so therefore
39:40okay so again
39:43negative displacement we can say that
39:46this one is compressing
39:48and if positive but on this means a
39:50positive atom displacement
39:52nag
39:55okay we're not done yet because why
39:58we're looking for
40:00the displacement of your of your bee
40:03relative to your sea so
40:06nato pakistan canada
40:11your displacement of your b relative to
40:14c
40:15is equal to
40:20so negative 30
40:23times your length of 1 meter
40:27over your area of pi
40:30times 0.025 squared times
40:34200 times 10 to the power of 9
40:39and that is negative
40:4376.4 times 10 to the power of negative 3
40:48milli meters
41:46so let's have another example let's have
41:48this example another one
41:51so we have this assembly and
41:54can you see uh this one here
41:59is an aluminum chute dao
42:48this is your nc
42:53and what is the
42:56deformation
44:09[Music]
45:00so therefore you can create this
45:02equation
45:04delta c is equal to
45:07delta t plus
45:11delta r
45:14and on positive meaning and positive
45:16mana
45:17we all know that your delta t
45:21or this job will compress
45:27okay so among the formation will go on
45:29this direction
45:44will
46:11okay so let's solve what is your
46:14deformation
46:15among this is 80 kilo newtons
46:19times the length of your job which is
46:22400 millimeters over
46:27areas which is given 400 millimeters
46:31squared times your modulus of elasticity
46:34of your job which is
46:3670 kilonewtons per
46:39millimeter squared this is cancelled
46:42this is cancelled this is cancelled this
46:44is cancelled
46:45so take note of
46:48the millimeter millimeter
46:52and then a billion millimeter okay next
46:54plus
46:5518 upon saturning rod 80 kilo newtons
46:58times
46:59600 over um areas
47:02among rod which is pi
47:06times radius which is 10 and given the
47:09diameter
47:10so 5 squared times modulus of elasticity
47:13of
47:14200 kilonewtons per
47:17millimeter squared okay
47:21so therefore your delta c
47:24is just equal to 4.20
47:28millimeters and that deformation is
47:31going to the
47:33right understood
47:36so therefore
47:41assumption
48:24okay last example we have this problem
48:31and
48:35so we have this load of
48:3990 kilo newtons applied at this point
48:43f any 200
48:46millimeters from here and 400
48:49millimeter shaft from this
48:52beam okay the problem is
48:56we all know the form is yeah
49:00yeah i would say
49:06but since they have different materials
49:09and different cross-sectional area
49:11then probably in any pagani form
49:14okay so i'm cutting a beam
49:17deform among ac obedie
49:22and assume this one is rigid okay
49:25and the question is um displacement
49:38first thing to do is to know
49:42the deformations deformation
49:56d
50:08and by using that idea we can get
50:13the distance of your
50:17delta f so atom goal is to
50:20look for ac and bd
50:24and how can you determine your ac and bd
50:27from your formula n l over a e but the
50:30problem is you still don't know the n
50:32because loads delay
50:35so inside button you need to create an
50:38equation
50:39which is a summation of forces along the
50:42y
50:43going up sequel to
50:46man f ac
50:50plus f b d
50:53minus 90 kilo newtons equals to zero
50:57this has two unknowns so therefore we
50:59need another equation
51:01and let's just say that moment that
51:05there is atom point b
51:06summation of moments at point b
51:09assuming this one is positive equals to
51:13f ac times
51:17600 millimeters so node
51:21minus 90 kilo newtons times
51:24400 millimeters equals to
51:28zero money fec
51:31this is your 90. so therefore
51:34your fac is equal to 60 kilo newtons
51:38and your fbd
51:41is equal to 30 kilo
51:45newtons
51:51so to get the deformation of your
51:55ac
52:0260 times
52:05length of three hundred
52:08over lana this is kilo newton
52:12this is millimeter cross sectional areas
52:15among material ac is made of steel
52:20so 20 millimeters
52:23pi 10 radius
52:27squared modulus of elasticity of your
52:29steel
52:30is 200 gigapascals which is
52:33kilo newton per millimeter okay
52:37so pilaman is here this is 0.286
52:42millimeters
52:44so among the deformation is going down
52:46among ac which is obviously
52:48go down next
52:53delta bd so fbd which is
52:5630 kilo newtons times the formation
53:01original length of 300 millimeters over
53:04pi times
53:07somebody say aluminum 40 oh
53:11yeah so the emitter 20 squared
53:14times modulus of elasticity of aluminum
53:1870. so this one is 0.102
53:23millimeters and this is still going down
53:25of course okay
53:26going
53:29okay so we had these answers but we're
53:32not done yet
53:33because we're looking for delta f
53:38to paddle
53:48[Music]
53:54huh
53:56length and we know that this one is 400
54:00this one is 200.
54:07we know that this one is a triangle and
54:08we can use this slope
54:11slope and we know that this one is
54:15so delta f is equal to
54:20minus delta bd
54:25plus let's call this length as
54:29delta so
54:32fbd plus delta okay
54:53but we know that this one is 0.286
54:58minus 0.102
55:02so this length the i is equal to
55:090.184
55:12okay so that one is 0.184
55:18okay and this length is
55:22600 millimeters and this length here
55:26is 400 millimeters
55:30so what can we say this is similar
55:32triangles
55:47eight 0.184 over six hundred
55:51ba same triangle so equal slope
55:54so rise over and rise over run so
55:56therefore your delta is equal to
55:59zero point one two
56:04milli meters so therefore our final
56:07answer
56:09delta f equals to um bd which is
56:13pilot on bd 0.102
56:18plus 0.12267
56:27so delta f is equal to zero point
56:30two two four six
56:33seven millimeters so
56:37this is your final answer
56:43equal slopes formula okay and then i
56:46wrap this one
56:47which is this one
56:53delta so
57:01so easy breezy
57:04char okay understood so
57:08we have solved three problems and those
57:11three
57:12problems have different situations and
57:15i hope you have learned a lot and as a
57:18summary for this lesson
57:20we have learned the principle of
57:21varegnan's theorem
57:23and second we have learned excel loads
57:26and
57:27we have learned a general formula for
57:29our actual loads
57:30for the deformation of our actual loads
57:32which is nl
57:34over ae and we have solved different
57:37problems
57:38with application of your this
57:41displacement
57:42formula of your excel loads so if you
57:45have
57:46still things that you don't understand
57:48then it's good for you to practice
57:50because there are there are a lot of
57:51problems in the internet so if you have
57:54any questions
57:56comment down your section comment down
57:58your questions and
58:00comment down your questions on the
58:02comment section below
58:03and please like and subscribe to my
58:07channel
58:08and and thank you for listening and see
58:12you again
58:13next week