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Strength of Materials I: Review Principles of Statics, Internal Resultant Loads (1 of 20)

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[Music]

but before we go to chapter one I would

like you to refresh your memory about

some of the principle of a static which

is very important for you

those are equilibrium so everybody is

familiar with the equilibrium of course

we have 2d problem and we have 3d

problem so 2d problem generally the

forces have X component and y component

so you have three equations I'm not

going to put it down here for

equilibrium so anytime you draw a

Freebody diagram there is it forces and

there could be moment but the moment is

one-dimensional usually if you are

working on the XY plane so the moment

will be about the z axis or the K

direction yes sir so those are the three

equations in 3d you are going to have

six equations and those six equation

again is Sigma FX equal to zero Sigma FY

equal to zero Sigma FZ equal to zero

which are the component along X Y Z also

three moment which are the moment about

x axis y axis and z ax you should be

able to do that through this exercise

one of this problem was 3d and you are

supposed to do that so and in this class

we never use cross-product you should be

able to use your right hand rule by

taking moment about x axis y axis and z

axis actually if you look in the book

they never use our bodies are the

simpler problem but you should be able

to do that I hope that you have

practiced that in your static class so

there are four so I'm going to be

putting it down Sigma M about x axis

equal to 0 Sigma M about the y axis

practice that one and seek for M about

the z axis which in general we can put

it in a vector for

but this is the breakdown of it in a

scalar equation then we need the

centroid and of course in the centroid

we have X bar which is equal to

summation of X I a I divided by

summation of a I and we have Y bar which

is summation of Y I AI divided by

summation of us of course everybody

knows that the denominator are the areas

yes or though what are the numerator any

idea product X and ay no it's not a

product of X and it has a meaning very

important meaning and it's going to

appear several place in in a strength of

material I'm going to for time being

because I don't know how many of you

remember that which is very important

this is qy we call it and this is QX it

has a meaning it has a name it has some

special specification and you through

the example I'm going to show you one

more time because in aesthetic I've put

a lot of emphasis on this Q which is not

as exactly we say the product of X and

ay it is has a lot of meaning on it and

it's going to be used in strength of

metal material alone okay I'm glad you

put it in the red that means you are

going to pay attention to that yes or no okay

okay

and of course we are going to have

moment of moment of inertia of course

moment of inertia

everybody knows the moment of inertia of

a rectangular shape with respect to the

axis that passes through each Center is

equal to what 1/12 of base times height

cubic that's right and then we have for

the circle as well which is PI over 2

these are IX so let's call it I X equal

to that IX I'm sorry it is PI over 4

radius ^ 4

of course the unit is inch ^ 4 see this

is very important to recognize this

process a is the area yes or no it has

the unit of length ^ - yes

q we call it moment first moment of area

and that the unit of length to the power

of 3 why because it's a product of X and

area everybody understand and the moment

of inertia which sometimes is called

second moment will be linked to the

power of four because the Y square da or

X square da as you recall from static

yes or no correct I hope you remember

that yeah okay there is now this is not

sufficient to know the moment of inertia

of special shape like this you need to

know parallel axis theorem so that is

very important parallel axis theory so

therefore you have here you say i X

equal to I X prime X prime passes

through the centroid of the object plus

the area times distance squared don't

worry about it I'm going to do an

example and do all of this in one

example that you see how is being

applied okay so those are the subject

that I'm sure you already know from your

aesthetic especially if you just pass

the static yes hopefully with a or b i

had one student who got c by the way i

want you know the last past quarter he

asked me to lower - C - because he said

he wants to take it again so I want you

to know this happened I actually I have

one also in 219 he did the same thing

I have to a student requesting me to

reduce the C - C - in order to that's

that important would be for you in

future classes static because you are

going to use it a lot you will see it

here anyhow let's do our first example

about this subject before I go into the

real strength of material so go let's go

back to this problem there are a couple

of problem in your handout

then I decided to do this for the other

class so I'm going to do the same for

you guys and therefore draw the picture

is not in your handouts but this picture

is simple so we can do it here so here

let's say this is the cross-section on

of an object that you are designing and

let's say that it is 75 millimeter by 50

attached to a plate here and like this

these are not in a scale so therefore I

just put here 200 millimeter is the

width this is the thickness of the plate

is 25 millimeter the height here is 150

millimeter these are all in millimeter

if I am not putting some of them just

just to save time and here the thickness

of this plate also is 25 minute

this is the object made of these three

pieces together this is the cross

section of a beam or a column in future

in this class that's what we are going

to talk always about the cross section

of the beam or column is that understood

or a structure this is the cross and now

what we want to do or what is suppose

you are supposed to do in later chapter

which is chapter 4 or 5 always is this

first you are required if you want to do

this strength of material first it is

required to find the centroid of this

object so of course as you see it is not

symmetrical with respect to X but it is

symmetrical with respect to Y so you

have to make a choice here you can put

your x axis here or here on top or

bottom the most customary one is to put

your x axis here your voices again you

have a choice but it is silly to put

your Y axis here the y axis should be

right at the middle because it is

symmetrical so that's our system now so

this is the y axis then we want to

determine the location of the centroid

which I am going to show it with a dot

somewhere here we don't know here is 200

times 25 here is 75 maybe it's a little

bit up there so there

is the location of the centroid of this

object right at the middle because it's

symmetrical so therefore since I chose

this as X so therefore this distance

from centroid to this x-axis we are

going to call it Y bar then we have to

go to this equation we don't care about

X bar because X bar is equal to 0 for

this y-axis yes or no it's symmetrical

so X bar is you now how do we calculate

Y bar now here was was my question

earlier guys what is this

numerator you said it is product of a

times I it's more than that that's it okay

okay

first smaller very good all right now

first of all we are going to divide that

into three area area one area two and

area three so this is the composite

section method and then denominator is

very simple denominator is the area yes

or no so there are four let's put the

area as from top so 75 times 50 plus 150

times 25 plus the last one a 3 which is

200 times 25

so denominator is very simply adding all

the areas no numerator we call it moment

first moment and we call it Q why why a

QX I'm sorry because look it's in rivers

so QX because you are taking moment

about the x-axis the idea is this if

this was flat like that first of all we

are assuming this plate display display

they are all made of the same material

with the same thickness if this doesn't

work in future we will have that in

chapter 4 we get a beam part of it is

wood part of it is still this system

doesn't work so immediately it comes to

your mind or it has to be the same

material so why not change the wood into

still or a still into wood so if you

understand this you understand that

problem immediately so based on that the

same material area would be representing

the weight yes or no yes if

this has a weight where would you put

the weight there you would put the

weight of this object of light system

where you would put it where you would

put it here yes

so that numerator is the first moment of

that area with respect to x-axis and

that's why we call it QX is that current

so we want to write it down here this is

first moment of the area of whatever

area you have about the x-axis of this

similarly the other one is the first

moment about the y-axis so boys knowing

that then the Y become very because in

the in practice some people use okay 75

times 50 represent the weight yes or no

correct if they are the same thickness

is saying who cares about the thickness

and the density because they are all

having the same yes so if you have a

plate like this and it's to a square

foot and one is square foot the weight

of 2 is square is twice as much as one

because they are made of the same

thickness the same material correct or

not yes

so if the weight is there that's the

weight yes or no where do you put the

weight here yes so of course this weight

was usually downward but you are

assuming it's going like that like it

this is flat and the moment of that

about x axis what this time do I need

now look the distance will be there if

they wait here and you take a moment

about here that distance is from here to

here yes or no prey so the distance is

25 150 and 25 so that is 200 yes or no

so there are 4 times 200 as if you are

taking the moment of this weight about

the x axis assuming that was flat this

is how you should have learned this

stuff in the static as well because this

is really what they cheese so then we go

to the second one when we go to the

second one don't forget about this sea

board this is something that you are we

are trying to

this area now the second area this is

the second area let's do it this way now

what's the mass of that that again the

mass represent the area represent the

mass so that is for the weight so it is

equal 250 times 25 where is the location

of that mass it is at the center and

what distance do I need so you can

calculate the distance immediately that

this time you need 75 plus 25 times

hundred so having this in mind life

become much simpler you never make a

mistake by this assuming you are

calculating the moment about certain

axis is that understood yes and that

last one is simple the last one is how

much 200 times 200 times 25 and then

multiplied by what the mass is here and

I need this distance is that correct

that this nurse is 12 and is that how

you learn your aesthetic class yes no

you just fill out the for the table yes

bad bad idea very bad idea everybody

under that is the difference in how you

learn this material so I want you to do

the same thing here in a strength of

material always I'm going to talk about

the concept more than the formula I give

you all the formula all the tools in

today's age with the internet with the

computer they are available you don't

need to remember any of the formula or

any of the equation that's given in the

book yes only only you have to

understand what is good for and how to

use it the most important part this is

the tool will be given to you you know

how to use it that's what I'm talking about

about

all everybody who's using centroid you

should take this at the moment everybody

about certain axes everybody abut but

that determining your all the A's are

all the distances is that understood yes

notice there is a big difference between

the two anyhow this ends up I'm not

going to put all the number then I did

it for you so end up to be equal to 95

millimeter so that now we solve this

problem the century

here is somewhere here a little bit

lower so the centroid here now is at

ninety five millimeter that means this

distance from here to here is 70

millimeter and this distance this is not

in a scale I'm sorry that's eighty

millimeter and so so far because the

total was hundred so question number one

was answer the centroid everybody with a

little bit of detail of that question

number two is their moment of inertia

now before I go to the moment of inertia

I want to ask you a question now that I

did that we are going to change this X

this is X is going in static they put an

X here they asked you to calculate

moment of inertia with respect to that x

axis in a strength that's not what we

are going to do it's the strength when

we go to chapter four you will see any

time we do a bending design the cross

section of a beam you need to calculate

the centroid of that cross section then

you need to take write it down the

moment of inertia with respect to the

axis that passes through their centroid

not any arbitrary axis it has a meaning

we will give you the name that's the

procedure so every problem in design

always start with this to first design a

beam design or column even shaft will go

through the same order you always have

to find the centroid for we did that

then we remove this X as this X does not exist

exist

we put our x axis here which passes now

through their center actually it has a

name later on you'll see in chapter or

we call it neutral axis there is a

reason for it that they will talk about

it in chapter four or five that has a

name that goes through the centroid yes

or no now you have to calculate the

second question this is the second

equation you want to find out the moment

of inertia with respect to the axis that

passes through their century so new axis

yes or no correct so let's calculate if

I see my X fact axis here was there and

that was the Q value if I put my xx

here tell me this is the area above an

area below of this x-axis are equal or

this is six by two this is six by two do

you think the centroid is their question

I want to know how how you learned your

centroid at this testing my testing to

know how you learned the stuff this is

the difference in way of learning is the

centroid they're probably not yes or no

right now the area above is six by two

the area below that C is six by two the

two area are equal is that the case

no so what should be equal in order Y

bar C when I put your X here Y bar is

equal to zero yes or no if Y bar is zero

that quantity must be equal to zero for

this x axis not that exact says yes in

order this to be the XL means that two

of the above and few of the below must

be equal and opposite to each other or

total Q must be equal to 0 this is Rises

I can see that you did not learn your

static in that format let's say this is

a seesaw let's say this is 20 inches and

this is 20 inches and you put here an

area which represent the weight exactly

like that this two area are the same yes

or no let's assume these two area are

the same correct this is the centroid

yes however if I want to move that to

this area to this at the centroid and

this become 100 and this become 300 is

this the centroid but look the area on

the left and area on the light are equal

so what is not equal Y this is not

centroid because of the moment everybody

see that this moment if W W times 300 is

much larger that W times 100 in order

this to be centroid I need to put

another depth 3w there is that correct

or now is that centroid 3 w times 100

equal to 300 time w yes so what is equal

there the moment the cube is that

understand that's the cue you are taking

moment of that with respect you look 3 w

time 100 equal to voided you can do it

this way or you can do it vertically and

you're--you they use this 2,000 years

ago as a scale I mentioned that in my

aesthetic class remember don't make a

scale that used in the Roman time they

put one weight on one side and the lots

of weight on the other side because they

are balancing their movement yes that's

what the cue I'm talking about this is a

century but look the area is this is

three times area than that the area

should you put it X through the centroid

the moment of the lower area and the

moment of the first moment the cue of

lower area and the cue of above area

must be equal and opposite this is very

important for future therefore I can't

calculate that let's calculate that now

for this problem I can see that some of

you are looking a little bit surprised

because the other class will response

was a little bit better I must tell you

that because I want always to put you in

the competition with the other class in

order to get a better result everybody

understand her yeah you are aside from

you being in competition with your

classmate you are going to be in

competition with the other class always

I check that too so remember that I have

this area above oops it doesn't work so

no good so I can go red so notice the

red area you have this kind of problem

in a static which is the area above the

centroid and you have the blue area

which is the area below the centroid I

said this to area if you check it I

don't have time to do you will see that

the areas are not the same so what is

the same write it down one more time the

cue of a bob and if you're below must be

equal on sign or opposite like the

seesaw I show you there and therefore

let's check that one so if I go with

their cue of the above now notice the X

is not there X is here this distance is

80 this distance is 50 and I'm

calculating Q of this two area is that

understood yes so I can add that

together this is the Q of above so is

equal to fifty times seventy five times

what now you give me that second number

the third number that's the area times

where is the mass is here

what distance do I need to take the

moment about the new x-axis yes or no so

what distance is that 80 plus 25

half of 25 the bus is here so this is 25

plus 80 is that correct you're taking

movement about the x-axis rail but

correct we call it you X so there are 4

x 105 so that is it now for this little

red area so that is a key by 25 time

what distance come on guys give me that

distance the mass is here so what this

time do I need I need 40 so just one

second so if you multiply this together

if my number is correct so if you

multiply that together you will get this

number four four hundred seventy three

thousand four hundred seventy three

thousand seven hundred fifty was the

unit see that's what again another

subject that you want to understand here

area come on guys listen to me area is

length to the power of two doesn't

matter millimeter to the power of two

inch to the power of two what is the Q

you just said it mathematically is area

times distance which we call it u so

what's that then the unit of that length

to the policy automatic if you think

about it this is the logic about it

that's what I want you to think the

logic of that tells you since I'm

multiplying area about the distance

which like a weight that I'm doing that

was not weight this area because area

represent the weight because it's made

of the same material and the same

thickness remember that yes that is the

idea behind it

therefore this become length to the

power of three so automatically become

millimeter to the power you don't have

even to think about it because you know

that you shouldn't know that this is the

fact we are going to use that often in

me2 18 and then of course Q of the block

you had a question now before I go very

okay now if you go one further there was

the unit of moment of inertia because

moment of inertia is the second moment

you have to multiply it by another

distant therefore you become to the

power of four so it goes from a to Q

remember that a Q I is that that in that

rest we use sometimes Q in a strength of

material area everybody knows what the

area is so is Q is the first moment I is

the second moment or moment of inertia

which I'm going to do next so Q up there

below area of the in this case blue area

now I have this Q and that U with

respect to this X I'll now you put the

number yourself 200 times 25 is the area

so you don't need to use sign you know

that this moment like the C so what that

moment if this moment goes like that

this moment goes like this everybody

understand it yes or no correct okay

that's the idea of the right hand rule

okay 200 times 25 times what what

distance this is the center what

distance do I need 70 plus 12 and half

so that is 82 and half trick so 82 and

half plus this little area yes or no

that little area is 70 times 25 times

what times 35 I'm glad okay so when you

do that calculation at home you will see

that for this end up to four hundred

seventy three thousand seven hundred

fifty millimeter that means our nice 95

is correct because the Q of a Bob and Q

of the below with respect to the neutral

axis must be equal and opposite yes or

no don't forget that for when we get to

later chapter is that understood

can I erase that down yes now the next

is moment of inertia which of course

requires parallel axes to you

so again I let's calculate IX I'm going

to bypass iy because iy we don't need it

that so I X equal to but I just put the

number there just to show you one more time

time

this works but then the number is not

essential here so IX is equal to

actually if you look at it this one this

is one of the quizzes by there I forgot

to ask if this is one of the quizzes I

have as a student to do in the past if

you go to the quiz page you will see

that picture there I'm sorry I didn't

mention that go to the page and go to

page I don't know go go go further than

last pages last couple of pages go

through no I just no no no no there to

page 14 you see that picture there I

should have mentioned that is that

everybody see that picture exactly like that

that

yes yeah that's part of a question not

the whole quail you have 15 minute to do

all of that plus the first part of it

but you can do it when you do your

homework of course yes after you do the

homework or practice this then you can

do all of that easily everybody

understood it now IX what is IX I need IX

area 1 I X of area 3 and IX of area 2

let's do the first one now how do I do

that first I had knee to the centroid

that's the centroid and this is X Prime

is that correct or the X prime passes

through the centroid of a want to worry

about it I just wanted to know that this

type of question was going to be asked

from you in first quiz yes so there was

I act concentrate on this one I don't

want you to miss that

what is IX prime first you see that you

need IX prime which is passes through

the centroid which is 112 I already

given it to you

112 of base time height cubic yes or no

so it is 1/12 of base what is the base

base is this area base is 75 you are

quiet or at times but I'll need input 50

to the power of 3 but that's not

sufficient that's the moment about X

prime and I'm taking moment about the X

so the therefore requires parallel axis

theory of the distance between the two

axis is d is that correct or not

plus the area area which is 7

five times 50 times distance squared

distances fear we already calculate that

that's 80 plus 80 plus 25 that is 105 to

the power of two don't forget that so it

becomes millimeters to the power of four

everybody that is the moment of inertia

of area one let's do moment of area of

the third one so this is for the second

one this is for the third one the third

one is easy so I'm doing that first is

that correct what's the moment of

inertia of this with respect to that

x-axis with the same routine again I

have to put a X prime axis here yes or

no correct

you guys remember your parallax story

yes or no if you don't do that you could

not pass this class parallel axis story

is essential part is like calculating

area anytime you are designing something

later on in this this book moment of

inertia replaces the area everybody

understand as simple as you calculate

the area of a circle or a square or

rectangular you should be able to

calculate moment of inertia this week is

the best week to refresh your memory or

do whatever learning to do if you want

to do a static without learning this is

no good

notice what I put here three subject

equilibrium centroid at the moment have

these are the most important part that

is going to be repeated again and again

and again so don't be surprised I can

see some surprises on your eye

by the way throughout the course I look

into your eyes I want you to look into

my I know exactly exact how my lecture

registered to you

I can see it immediately so don't hide

anything from me so don't put your head

down I want to look into your eyes this

is the technique I have to work before

so I hopefully work with you guys

because if I don't see that it is not

there is some problem there I may

explain it one more time everybody

understand what I'm saying this so

therefore now the third part I need the

moment of inertia with respect to this X

prime which is again 1/2

well of base time height cubic yes or no

so it is 1/12 of base which is 200 times

25 because that is routine plus again

there is a distance between X Prime and

X because we are taking moment of

inertia with this this is the question I

asked you to do yes or no

so therefore the distance again is 70

plus 12 and so 82 and a half so area

I need the area 200 times 20 time

distance square which is 82 and a half

that's right

to the power of 2 so you can calculate

that now remaining the middle part so

remember that you have to do exactly the

same thing here too so if I consider it

being at one piece still I have to write

112 base base is 25 times height is 153

however does this X passes through the

centroid of that piece no it passes

through the centroid of the whole object

but the centroid of this piece only this

space is where this is 150 where is the

centroid of that 1/100 so it is 75 so I

have a 5 millimeter distance that's my

distance between the two axes yes or no

everybody see what I'm talking about yes

so there you see this is the centroid of

a 1 this is centroid of a 2 you need the

centroid of a 3 which is at 75 and 75

everybody understand so there is a

distance of 5 millimeters so you have to

use that plus the area so plus 25 times

150 I'm sorry that's the area times 150

but that has to be multiplied only by 5

to the power of 2 so that would be your

I X I Y is simple I'm not going to do it

or don't waste your time so that is the

centroid this is the review of the cue

centroid moment of inertia yes a little

five yes all right

so this is I put it here a little bit

bigger is that correct or not where is

our total centroid not first of all this

is 150 where is the centroid of this

page shape

this here so here should be 75 here

should be but where is our x-axis that

I'm taking moment about is it there look

the distances are given it is 70 and 80

so where is the x axis the x axis is

here see that so the distance this

become your X Prime and this is your X

that the distance is 5 the set s that

blue dot is the centroid of the entire

picture is not the centroid of that

seventh centroid of 150 by 25 is that 75

everybody knows that yes okay all right

all right good

everybody clear about that ok now let's

go to the first example now what did

discussion that we have here which is

very important you guys know this is the

first time you are going through this

rate of material this was already a t so

I'm going to erase that

now the first subject that I've I want

to discuss with you guys which is not in

the book but it is very important

actually it is in Chapter seven of your

static book is external load versus

internal or call it actions not note

because that gives you a different idea

there now there is a big difference

between a static and a strength of

material and you have to understand that

that's exactly what's sitting there

external load versus external example

let me give you an example here that you

will see what happened there when we do

this kind of problem let's say we have

here a beam here everybody should draw a

legend let's say this beam is 8 foot long

long

it is a

in here roller there and I say the

weight of this beam is 100 pound per

foot how do you find a reaction this is

such a simple abc's of aesthetic

everybody knows that so what's the

weight of the beam now I just want to

show you some very important factor here

the weight of the beam is 800 yes or no

100 counter foot it's 8 feet it is 800

where do you put it at the middle of the

beam because it's the same weight is

that there's no problem there don't be

this don't be surprised some of you are

surprised maybe you are afraid to say

that that is something you have done it

many many times even in high school yes

or no yes so you put the 800-pound there

so don't worry about it I'm not talking

about this yet because that is the total

weight W is equal to 100 pound per foot

multiplied by 8 feet feet and feet drops

out so it's 800 pound you put it at the

centroid and then you calculate the

reaction now you are removing this you

calculate the reaction here and here it

is symmetrical so this become 400-pound

this is 504 and this is 400 because it's

symmetrical yes or no this you have done

it many many times yes or no that's no

problem there the problem is this in a

strength of material we do the same

thing when we want to calculate the

reaction which was fine now let's say

somebody wants to because this object

first of all in a static object usually

will rigidbody if you go to chapter 4 or

5 don't remember is that if you chapter

4 it was the equilibrium of rigid but

right now you are not only more longer

rigidbody if I have a beam like that you

put it your hand you put a load at the

middle like that put it load at the

middle this object is going to bend if

this was a rigid body it who not being

two bending is that correct and that

bending is the issue we want to address

in future chapter 4 and chapter 4 now

when you look at this the bending are

different from this point to this point

to that point somewhere is going to go

down more than the other so we have to

go to go inside the body

and see what is happening in this body

this body this is what we are going to

do in the next few classes hold it with

you it's going to be like that pull it

this is purely tension we are going to

do week one or two then we can put it

like a column and push it down which

would be purely don't look at it that's

a buckling that's Amy 2:19 we talked

about that okay that it if this was

purely compression it would be

compressed down yes or no chapter one

and two this is the preview of coming

attractions guys so free will be about

tension and compression or two Force

member depending what's happening inside

the body which we call it a stress

strain etcetera so all of that coming in

future then in Chapter three instead of

force we are going to apply a moment

look at my tongue a moment like that

that moment like that moment like that

which is the moment about x-axis look

what happened to this object this object

is going to be twisted we call it

torsion or twist is that the last

chapter three in chapter four which was

just showing it the object is going look

at the moment the change this is this

moment now this moment causes this to

bend that's chapter four and five

chapter six is something else and

Chapter seven we put it all together so

we just have fun to get that let's put

it this way anyhow so what I'm saying

that you have to go inside the body

which you have done it in a static once

remember in a static when you went to

the trusses you were determining the

internal tension or compression remember

the internal forces you were cutting it

and finding the internet that's exactly

what we are doing here however look what

problem I have here let's say that this

is four feet and four feet and I want to

go here three feet here and make a cut

here and determine what is happening

inside the body of the material because

that determines the strength of material

to what which chapter to go now if I

draw it like that this is three feet and

here I put four hundred pound there is

no here forces here there is a force

here 400 is that correct

is that correct

what about the weight of the beam you

see that that for her 800 years with

that is because you were looking at

entire system this you cannot do you

have to go back to the original system

the original system is this guy's a

uniform node of no big very careful a

uniform load of intensity of how much

100 pound per foot because this beam and

the other beam are going to deflect

differently everybody understand that

now if I cut it at 3 feet how much dough

do I have there 300 everybody understand

what I'm saying that so that balance so

that changes the scenario so now don't

write it that all I want you to

understand that that there is a 300 here

then we are going to find out the

internal forces at this point which

would be either horizontal force or

vertical force or a moment exactly so

instead of now know that you understand

remember always do it this way that's

the you your problem number 9 and 11 is

all about that for aesthetic purposes

you can put one single load abandon you

want to cut the beam you have to go to

original back to original no the load

could be uniform it could be in triangle

Road slope road etc it's even a parabola

that you have to do integration

everybody understand what I'm saying

that so if you are looking at the weight

of the wing of airplane it's not the

constant it must be some some parabola

yes or no so then you have to use the

integration you learn it all about that

now let's go to an example in the book

in the handout but I've way to put it

there I don't want you to look at the

solution we are subject still is

internal action so this is the the

structure there is a beam here a b c is

a pin connection at k and also it is a

pin connection and right here this is C

this is the

and this is paint connection to the wall

these two are at the same level and this

distance is six feet that is four feet

and four feet exactly what I said there

and the load here is a uniform load so

we are considering the weight of the top

beam we are not considering the weight

of this one for neglecting that and the

weight is given 50 pound per foot this

is the question and we want to calculate

the internal actions at point B while

you guys are standing there outside for

looking from the window what's the

problem there let's let me check those

you want to come in you're welcome

why are you standing there watching for

next class okay all right okay so how do

we do that guy we want to calculate the

internal force of the what should i do

first you know that from a static you

have to find the reaction at all the

support yes first yes all right so of

course you have to draw a free body

diagram of the member let's say ABC

correct this is the free body diagram

come on guys these are basically static

that we are not talking about here at

Point a it is a pin so what should I put

that point eight guys a x and a y now

hot water would the load there remember

the load is a uniformly distributed load

of 50 pound per foot now if I want only

reaction I can replace it with one

single load what's that single though

that single load is 50 times eight so it

is 400 so I can put here at 400 right at

the middle 400 pound here now we come to

this point that point seems to be a

the pain to CX and CY Harvick looking at

member CD member CD is a two-force

member so the force should be in the

direction of CD so I should not put

their two forces I should put one force

in the direction of CD which the slope

is as you see there the slope is four

run three rice because eight and six

which is the same is that career corner

yes now do you think this item is under

tension or compression this is Ferris

less than of a strength of material is

it under tension is going to be crush or

it's going to be expanded what do you

think putting a load on top crush so it

should be in compression or if I put it

in tension mode the answer will come

negative remember but since I do it

absolutely it is in compression I'm

going to put it in compression mode

knowing the answer comes positive is

that understood

so which side is for compression upward

or downward your God past me2 814 not in

my class come on guys going toward the

joint is compression going away from the

joint is in tension yes or no all you

know it don't be afraid express it I

want to participate everybody understand

it yes if you know it that's okay if you

don't that's a good word to two doesn't

matter we can the reason I'm saying that

because I want everybody be on board you

either knew it or no right when you give

me the wrong answer that's good too

because if I correct you never for you

never forget it you understand what the

process worked that's what I'm saying

that don't be afraid you are stood and

you are in class you need to make a

mistake but you have to commit yourself

to something everybody on this either

fifty-fifty chance that you are either

correct or incorrect however that's the

learning process so if you are learning

there but these are something of course

you should know it from the past yes so

let's call it f CD is that correct so

that's the F remember CD right the rest

now is very simple now I have only three

unknown so I can solve it yes or no by

taking moment about point a of course

you see the rest this is the free body

diagram or

essential part of your analysis in every

structure so Freebody diagram you cannot

bypass that so Sigma M at or about a

equal to 0 would give you nothing there

so you give you this is for feet and

that's for feet so there is 400 times 4

and that is negative because it's going

clockwise then here it has two component

horizontal component doesn't have any

mode we got that that load goes through

this point I didn't draw it correctly

but goes through that point

so let's opinions there let's say that

so therefore since it goes through that

so the horizontal component does that

have a load we're up in the moment and

the vertical component has a moment but

vertical component is 3/5 yes or no so

therefore 3/5 of the force FCB

time distance of 8 and is this way so it

goes that way therefore it is positive

become equal to 0 this is the force

that's the distance you calculate fcd

become equal to 333 point 3 usually we

go three-digit however here I went to

four-digit to calculate fcd as soon as

you calculate fcd you can calculate a x

and a y which is not a big deal here

everybody can do that so let me give you

the answer so ax become equal to minus

two hundred sixty six point seven pound

and a y actually become two hundred huh

so this is the extent of a static which

everybody can do if you can draw of

course the correct free body diagram

correct now next the question that we

are embracing to do what happened if I

cut it at point B now look point B is

here first of all you have a big problem

is 400 is right on top of point B but is

that a correct assumption put in the 400

then I just already mentioned that

that's that that's invalid everybody

know that so if I want to cut it I have

to go back to that so we go back here

and make a cut here everybody understand

so you

have some load on the left some road on

that some part of it depends where you

are here you are at the middle so there

are four next is free body diagram of a

B so we are going to put it in the

middle here so here it is the free body

diagram of a now what force do I have at

a the force do I have at a we already

been determined is a negative two

hundred sixty six point seven so this

rod has a two hundred points they're

going this way yes or no prick let's go

a step by step before I put anything

else that requires here to do what this

is point B let's forget about anything

else because I haven't put the rest of

it because of this force I should have

an internal force going to the rod so

this beam or this rod is on there

tension or compression tension of two

hundred sixty six point seven which will

be the subject of next to lecture

everybody the rest of those problem are

like that so here I put here a force

here now in the static you call this

vxvy and MBEs change that that's going

to change now is not DX it has a name

this force is normal to the

cross-section yes or no therefore we

call it normal force so please don't go

forth give you normal stress normals

great give normal strain anytime you

have a section you have a column all the

load coming on the section everybody

understand that is force pushing the

column down everybody this force it's

not BX anymore if we call it and be

normal why not about one more time

normal to cross section to the cross

section of the beam everything is about

the cross section in me2 18 write it

down in your note every material that we

are talking about is about the cross

section of an object and is of course

laying come into the picture don't get

me wrong the length determined

everything but when you design something

this column has certain cross section

why this is this big because there is

lots of load on it if it was less load

the cross section become

smaller everybody understand what I'm

saying the leg of this chair you are

sitting there is there about 3/4 of an

inch diameter everybody understand if

you put a giant here you have to make it

four inch diameter everybody understand

what I'm say the weight determining the

the usefulness of the ideas nevertheless

this is normal force do not please do

not make it B X B Y B Z this is normal

force and this force would be what this

force going up on the other side will be

going down so it would be shear forces

everybody the one that in the section we

are going to share cursor me he

generally say it like that so these are

they have name of course you said there

is a moment other words it will not be

in equilibrium this is MV 2 now the

first two classes we are going to go

through the normal forces and shear

forces actually next subject which I had

10 more minutes I will talk about normal

forces and shear forces and stresses due

to that now hold on one sec well upward

or downward

what VB yes okay alright and then MB so

now chapter 1 chapter 1 and 2 is about

this - if your video chapter 3 we don't

have it here because you don't have a

torsion here chapter 4 and 5 is about

their bending and I said that and at the end

end

now let's put the load thing they lured

here was how much now this is only 4

feet so the load is like that is I just

put it in fashion because I don't want

that so how much load is that now the

load is 4 feet times what times 50 so it

is only 200 because the half of the beam

yes so where do I put that 200 at the

center like that this is 200 pound and

the distance is 2 feet and 2 feet

because that's the center of the

rectangle this is the loading system

that Freebody diagram does not show it

everybody understand that

don't make that mistake this doesn't

understand ok

now everything is simple so what's the

value of M b and b will be two hundred

sixty six point six seven pound

it is in the tension mode so if it was

like this it would be in the compression

mode once the shear force oh I'm sorry

we forgot something here there was an a

why here see you through this discussion

I forgot to put there so let's remove

this and do it correctly so there was a

to hunt this was a boy as well is that

correct or not yes

but notice if you look at your handout

they put this one downward not upwards

so be careful here you have a choice to

put it there is a reason for it I cannot

explain it out to you actually I will

explain to you later on for time being

you are going only by static either put

it upward or downward the answer come

according to where you put it is that

correct but if you want to understand

that we have to wait until we talk about

this the shear and what happened to the

sign of that everybody understand sign

of that normal force is very simple if

it is tension we call it plus you have

seen it in the study if it is

compression we call it - not a

statically plus about it you just

because it's pull or push so that we

talk about that later on today we don't

have done let's calculate so MB of

course is two hundred sixty six point

seven pound and it is in tension if I

want to calculate shear force I have to

write Sigma FY equal to zero yes or no

correct sick boy oh boy equal this was

Sigma FX what I did not use it Sigma FY

equal to plus 200 - 200 plus shear at B

of course shear at me at that time

become equal to zero because the balance

is zero there is that correct order yes

and then the moment how do I calculate

the moment the last this is for chapter

four and five how do we calculate moment

notice this blue ones or internal

remember the subject was external versus

internal in MA - eighteen all the

chapter is all about this blue one and

what happens to the object here due to

this blue one is that understood which

in a fact I am going to tell you is that

the real load that real note has to be

in the different format we talk about

the stresses etc etc but statics tell me

I should have something

that is that career corner yes and then

if I want to calculate the end where

should I take the moment about take the

moment about point he usually this is

the difference from a static now this

time because I want to avoid all of that

you should always take the moment about

point B not the point

again this is different from a static

notice here I took the moment about a

because I wanted the reaction that was a

static this is the strength now I take

off you always take the moment because

you don't want these two to appear in

your equation everybody so take the

moment about point B where you cutting

it so Sigma M where should I put it here

so I can erase that now Sigma M become

Sigma M at B equal to zero so you have

two hundred times for going negative

this one doesn't have any movement plus

two hundred times two this way positive

Plus don't forget there is M be sitting

there yes or no plus M be correct equal

to zero now I want you to understand

that this is totally different from that

some people in this class even in me2 19

they make a mistake between the two MB

is the amount of moment sitting at that section

section

yes sir Sigma M be equal to zero means

taking moment about point B and summing

it up everybody understand I have made

that very clear aesthetic unfortunately

some of you eventually make make a mistake

mistake

that means this moment this moment this

moment this moment if they have any and

this together must be equal to zero

everybody clear about that yes so as you

see it ends up like that many people

write this and say equal to Mb or some

dating and B equal to zero because they

said they write it like that they think

this is M be some people write it like

that totally right when we put Sigma M

at B this is a sentence summation of the

moment about point B equal to zero

everybody understand it gave that

pattern correct all right so

did you see what I did here so MV become

equal to Mb become 400 positive could

discover the pound foot that means that

direction is correct there a whole beam

is going to bend like this not like that

yes which is this how much time do we have

have

I don't know what happened two minutes

how much it's two minutes over twenty

thirty two okay okay next thing the

first thing we are going to do next next

class which is now looking at fcd

everybody under and then this rod would

be only on their tension or compression

in these cases comforter then I talked

about the stresses strain etcetera

etcetera about members which fuel is the

strength of material all right okay I'll

let you go this time

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