-------------------------------------------
free product key for office 2016 ...how to activate office 2016 free - Duration: 3:43.
hello guys
as you all can sea that
office 2016 is opened
and the office 2016 is not registered
now i'll show how to register it for free
using simple a trick
mine is already disabled
click on the link from the description
you will get this link
download it into your pc
now extract the containing files using .rar
install the setup
your office will be activated
or
copy the key to you word 2016 activation dialogue box
as you can sea mine is activated
-------------------------------------------
New Kdrama Queen For Seven Days 7日的王妃 (Eng Sub) - Duration: 1:53.
I had a dream
Because love him
Want to be with him
The tears in the world
all came to me
Are you all right?
Lift your head up
Just look up
Oppa
The real one I love appeared
So nervous and distressed to make me crazy
-------------------------------------------
Am I A Good Candidate For Conscious Sedation Dentistry? Greenville Dentist Katie H. Bridges, DMD - Duration: 1:00.
Hi, I'm Dr. Katie Bridges at Wood Creek Dental.
Sedation dentistry is a great option for many types of patients.
Some patients are fearful of the dentist.
Some haven't visited the dentist in years because of a negative experience, a bad gag
reflex, or just simply no time in their schedule.
Sedation dentistry offers you the opportunity to have an extensive amount of dental treatment
done in one appointment, remembering little to none of it.
It's important that you remember it's a stress-free environment with sedation dentistry.
It's a relaxed atmosphere and at Wood Creek Dental, we promise to keep you comfortable.
We would love to talk to you about the world of sedation dentistry and all it offers.
Please call us or visit our website at woodcreekdental.com.
-------------------------------------------
Woman pushes for helmet bill in mom's memory - Duration: 2:07.
HE WAS COLD BUT OTHERWISE OK.
A CLACKAMAS WOMAN IS HONORING
HER MOTHER THIS WOMAN'S DAY BY
TRYING TO SAVE LIVES.
HER MOTHER'S DEATH INSPIRED HER
TO WORK BEYOND A NEW LAW AND
WE'LL GO BEYOND THE HEADLINES
TONIGHT.
IT'S BEEN TWO YEARS SINCE
SHARON BURGE PASSED A WAY.
A LOSS THAT IS STILL RAW FOR HER
DAUGHTER SHAUNA WELLMAN.
I THINK OF HER EVERY OTHER
DAY.
IT WAS JULY 15, WHITE WATER
RAFTING.
IT WAS THROUGH A LOCAL COMPANY,
ONE THAT DID NOT PROVIDE EL
METS.
RIGHT AS THEY HIT THE SECOND
RAPID, THE BOAT TIPPED SIDEWAYS.
BD AND MY MOM FELL INTO THE
RAFT.
SO SHE HIT HER HEAD ON SOMETHING
INSIDE THE RAFT.
THAT'S WHEN SHAUNA NOTICED
SOMETHING WAS NOT RIGHT WITH HER
MOM.
I WENT OVER TO HER AND ASK
HER IF SHE WAS OK AND SHE DIDN'T
KNOW.
SHE COULDN'T TELL ME IF SHE WAS
OKAY.
SHARON WAS LIFTED TO OHSU.
DOCTORS PUT HER AN INCUBATOR.
SHE DIED DAYS LATER.
IT'S BEEN SO DIFFICULT FOR
NOT JUST US, BUT HER MOM IS
STILL HERE.
HER SISTER IS STILL ALIVE.
WE'RE ALL THIS -- YOU KNOW, WE
HEAR IT EVERY DAY.
SO WE DON'T WANT FAMILIES TO GO
THROUGH THIS.
NOW THE FAMILY IS PUSHING FOR
STATE LAWMAKERS TO PASS SENATE
BILL 643 ALSO KNOWN AS THE
HELMET BILL.
IT WOULD REQUIRE SHAUNA CALLS
ATE COMMON SENSE LAW.
THEY JUST HAVE TO HAVE
HELMETS AVAILABLE.
THERE'S NO INQUIREMENT FOR
PEOPLE TO COME IN.
WHILE THE PASSAGE OF THIS BILL
WON'T BRING PER MOM BACK, SHE
HOPES IT COULD SAVE LIVES.
IF YOUR MOTHER'S STILL HERE,
GIVE HER A HUG, TELL HER YOU
LOVE HER BECAUSE YOU NEVER KNOW
WHEN THEY'RE NOT GOING TO BE
THERE ANYMORE.
WE'LL KEEP YOU UPDATED AS
-------------------------------------------
Facebook Live: Top 10 Baby Names of 2016 - Duration: 16:04.
>> GOOD MORNING AND HAPPY
FRIDAY, EVERYONE!
I'M JENNA FROM THE SOCIAL
SECURITY ADMINISTRATION.
THANK YOU SO MUCH FOR JOINING US
THIS MORNING.
>> AND I AM CRYSTAL, ALSO FROM
THE SOCIAL SECURITY
ADMINISTRATION.
WELCOME TO OUR FACEBOOK LIVE!
I'M SO EXCITED TO BE HERE.
AND IN HONOR OF BABY BOYS, AS IF
YOU CAN SEE, I'M WEARING ALL
BLUE, MY NAILS ARE BLUE, I HAVE
A BOY, EVEN THOUGH HE'S NOT A
BABY, I'M REPPIN' FOR THE BOYS.
AND JENNA, SHE KIND OF HAS ON,
AS YOU CAN SEE, A LITTLE PINK
AND A LITTLE BLUE, BECAUSE SHE'S
NOT BIASED, SHE DOES NOT
HAVE KIDS YET.
[ LAUGHTER ]
>> SO TODAY IS A VERY BIG DAY
FOR US HERE AT SOCIAL SECURITY.
TODAY, WE ARE HERE TO BRING YOU
OUR VERY EXCITING AND VERY
SPECIAL DELIVERY OF THE TOP TEN
MOST POPULAR BABY NAMES OF 2016.
WE'LL BE ANNOUNCING SOME OF
THOSE NAMES IN JUST A LITTLE
BIT, BUT BEFORE WE DO THAT, WE
WANTED TO GIVE YOU A LITTLE
BACKGROUND INFORMATION ON HOW WE
GOT STARTED IN THE BABY
NAMES BUSINESS.
>> YES.
AND WE DO THIS EVERY SINGLE YEAR
JUST BEFORE MOTHER'S DAY JUST SO
THAT, YOU KNOW, YOU CAN GET A
CLUE OF WHAT'S GOING ON, ANY
EXPECTING MOMS OUT THERE, THIS
IS FOR YOU.
SO EXPECTING MOMS, DO US A
FAVOR, COMMENT BELOW, DROP SOME
COMMENTS, WHAT ARE YOU NAMING
YOUR BABY?
ARE YOU RELYING ON THIS LIST, DO
YOU HAVE A NAME PICKED OUT?
GIVE US SOME COMMENTS THERE, WE
WOULD LOVE TO HEAR FROM YOU.
>> YEAH.
AND IF YOU ARE AN EXPECTING MOM
FOR THE VERY FIRST TIME,
CONGRATULATIONS, THAT'S VERY
EXCITING.
BUT WE ALSO WANT TO HEAR FROM
OUR EXPECTING DADS, AUNTS AND
UNCLES, GRANDPARENTS, BECAUSE
LET'S FACE IT, WHEN THERE'S A
NEW BABY COMING, IT'S A VERY
EXCITING TIME FOR EVERYONE IN
THE FAMILY.
SO LET US KNOW YOU'RE WATCHING,
WE WOULD LOVE TO HEAR FROM YOU.
AND WE'D ALSO LOVE TO KNOW WHAT
YOU THINK THE MOST POPULAR BABY
NAME IS GOING TO BE FOR THE BOYS
AND FOR THE GIRLS, SO PLACE YOUR
GUESSES BELOW; WE'D LOVE TO HEAR
WHAT YOU THINK.
>> YES.
THIS EXCLUSIVE LIST HAS BEEN A
HIT WITH THE SOCIAL SECURITY
ADMINISTRATION FOR THE PAST
20 YEARS.
>> YEAH, THAT'S RIGHT, CRYSTAL.
SO THE AGENCY BEGAN COLLECTING
BABY NAMES SINCE 1997, LIKE
CRYSTAL SAID, ABOUT 20 YEARS
AGO, WITH NAMES DATING BACK ALL
THE WAY TO THE LATE 1800s.
SO WHAT WOULD HAPPEN WAS PARENTS
WOULD LET US KNOW THE NAME OF
THEIR NEW BABY WHEN THEY WOULD
APPLY FOR A SOCIAL SECURITY
NUMBER WITH US.
SO WE HAVE THE NAME OF EVERY
SINGLE PERSON IN THIS COUNTRY
THAT HAS A SOCIAL
SECURITY NUMBER.
SO THAT MAKES US THE NUMBER ONE
SOURCE FOR THE MOST POPULAR
BABY NAMES.
SO THIS HAS BEEN, YOU KNOW, A
REALLY FUN TRADITION
FOR US HERE.
>> YES.
AND WE ALSO USE THIS TIME OF THE
YEAR TO REMIND YOU TO GET TO
KNOW YOUR SOCIAL SECURITY.
WE WANT TO ALWAYS REMIND YOU
THAT SOCIAL SECURITY IS MORE
THAN JUST RETIREMENT.
WE ARE HERE WITH YOU THROUGH
LIFE'S JOURNEY.
IF I COULD JUST ADD A QUICK
LITTLE STORY, JENNA.
>> OF COURSE.
>> SO WHEN I WAS 15, MY DAD
PASSED AWAY AND I RECEIVED
SURVIVOR BENEFITS.
SOCIAL SECURITY WAS THERE WITH
ME THROUGH THAT LIFE'S JOURNEY;
IT WAS A HORRIBLE TIME FOR US.
BUT WHAT WOULD WE HAVE DONE IF
WE DIDN'T HAVE SOCIAL SECURITY
AFTER LOSING AN INCOME IN
THE FAMILY?
SO WE JUST WANT YOU
TO KNOW THAT.
AND AS IT RELATES TO BABIES, WHY
THIS IS SO IMPORTANT IS BECAUSE
HERE AT SOCIAL SECURITY IS ONE
OF THE FIRST ORDERS OF BUSINESS
FOR A BABY, GETTING THEIR SOCIAL
SECURITY CARD, RIGHT?
>> EXACTLY.
SO THANKS FOR THAT, CRYSTAL.
AND IF YOU ARE TUNING IN FOR THE
FIRST TIME, THANK YOU FOR
JOINING US.
CRYSTAL AND I ARE HERE TO BRING
IN THE TOP TEN MOST POPULAR BABY
NAMES OF 2016, SO STICK AROUND
AND WE'LL LET YOU KNOW THOSE
NAMES IN JUST A BIT.
BUT IN THE MEANTIME, WE WANT TO
GIVE THOSE EXPECTING MOMS AND
DADS A LITTLE BIT OF ADVICE THAT
WE THINK IS IMPORTANT FOR YOU TO
KNOW WHEN GETTING READY FOR YOUR
NEW BABY.
SO NUMBER ONE, MOST IMPORTANT,
IS CHOOSING A NAME FOR
YOUR BABY.
AND THAT'S WHAT WE'RE HERE TODAY
TO HELP YOU DO, IS DECIDE THAT
PERFECT NAME FOR YOU AND
YOUR BABY.
NUMBER TWO, YOU NEED TO GET A
SOCIAL SECURITY NUMBER FOR YOUR
BABY, AND CRYSTAL AND I ARE
GOING TO LET YOU KNOW WHY THAT'S
SO IMPORTANT IN JUST A MINUTE.
AND LASTLY, WE ENCOURAGE YOU TO
CHECK OUT OUR WEBSITE TO LOOK AT
ALL THE PROTECTIONS THAT SOCIAL
SECURITY WILL OFFER FOR YOU AND
YOUR NEW BABY.
>> YEAH, AND IF I COULD ADD
SOMETHING AS AN EXPERIENCED MOM,
I WANT TO LET YOU GUYS KNOW THAT
THE EASIEST AND MOST --
[ NO AUDIO ]
-- GIVING INFORMATION FOR YOUR
BABY'S BIRTH CERTIFICATE, THAT
IS THE BEST TIME FOR YOU TO
APPLY FOR YOUR BABY'S SOCIAL
SECURITY CARD.
RIGHT, JENNA?
>> YEAH, THANK YOU FOR POINTING
THAT OUT, CRYSTAL.
SO LIKE I SAID, THAT IS THE MOST
IMPORTANT THING YOU NEED TO DO
AFTER YOUR BABY IS BORN FOR MORE
THAN ONE REASON.
NUMBER ONE, YOUR BABY NEEDS A
SOCIAL SECURITY NUMBER IN ORDER
TO BE CLAIMED AS A DEPENDENT ON
YOUR INCOME TAX RETURN.
NUMBER TWO, THEY NEED ONE IN
ORDER TO BE ON YOUR HEALTH
INSURANCE PLAN.
AND LASTLY, THEY NEED ONE IN
ORDER TO APPLY FOR BENEFITS FOR
YOUR BABY.
SO YES, YOU NEED TO TAKE CARE OF
THAT AS SOON AS POSSIBLE.
>> YOU KNOW, WE SEE SO MANY
PREGNANT WOMEN WALKING AROUND
AND I OFTEN WONDER TO MYSELF,
HOW MANY NEWBORN BABIES ARE BORN
EVERY DAY?
>> I ACTUALLY KNOW THE ANSWER TO
THAT, CRYSTAL.
>> NO WAY.
>> YEAH, I WAS DOING MY RESEARCH
THIS WEEK GETTING REALLY EXCITED
FOR TODAY'S BIG ANNOUNCEMENT,
BUT THERE ARE APPROXIMATELY
11,000 SOCIAL SECURITY NUMBERS
ISSUED EVERY SINGLE DAY.
SO THAT TURNS OUT TO BE
APPROXIMATELY FOUR MILLION
SOCIAL SECURITY NUMBERS ISSUED
EVERY YEAR.
THAT IS A LOT OF BABIES.
>> THAT'S A LOT OF BABIES.
>> YEAH.
>> IT'S A LOT OF BABIES AND A
LOT OF SOCIAL SECURITY NUMBERS.
AND I CAN IMAGINE THAT NUMBER IS
SO HUGE BECAUSE THERE'S TWINS
BORN AND TRIPLETS AND
QUINTUPLETS, ALL THESE KIND OF
BABIES THAT ARE BORN, RIGHT?
>> SO MANY, YES.
>> SO I JUST WANT TO REMIND
ANYONE WHO IS HAVING TWINS THAT
YOU CANNOT GET A SEQUENTIAL
NUMBER FOR YOUR BABY.
THOSE NUMBERS ARE RANDOMLY
SELECTED, SO YOU CAN'T REALLY
REQUEST THAT THE NUMBER IS IN
ORDER SO THAT CAN HELP YOU
REMEMBER THEM BETTER.
>> YES, ABSOLUTELY.
SO AND I JUST WANTED TO ASK THE
AUDIENCE WHAT YOU THINK THE
THREE STATES HAVE THE HIGHEST
AMOUNT OF BIRTHS A YEAR.
SO LET US KNOW WHAT YOU THINK IN
THE COMMENT BOX BELOW.
THERE'S THREE STATES THAT HAVE
THE HIGHEST NUMBER OF BIRTHS,
AND WE'LL LET YOU KNOW THE
ANSWER TO THAT IN A LITTLE BIT.
BUT LET US KNOW WHAT YOU THINK,
I'M CURIOUS TO KNOW
WHAT YOU THINK.
SO I DON'T KNOW ABOUT YOU,
CRYSTAL, BUT I AM GETTING
SO ANXIOUS.
>> I AM, I'M SO EXCITED.
>> I GET SO EXCITED EVERY
YEAR FOR THIS.
>> LET ME TELL YOU A
FUNNY THING.
>> OKAY.
>> A FUNNY THING THAT THE
AUDIENCE PROBABLY DOESN'T KNOW
IS THIS LIST IS SO EXCLUSIVE
THAT I DIDN'T EVEN KNOW THIS
INFORMATION UNTIL, LIKE, A
COUPLE OF MINUTES AGO.
>> DO YOU THINK YOUR NAME IS
GOING TO BE ON IT?
>> UM, WE'RE GOING TO HAVE TO
WAIT AND SEE.
BUT MY NAME IS REALLY TRICKY.
OKAY.
>> HOW POPULAR IS IT?
>> I'M NOT TOO SURE, I'M NOT TOO
SURE HOW POPULAR IT IS RIGHT
NOW, JUST BECAUSE I HAVE AN
UNFAIR ADVANTAGE.
I SPELL MY NAME LIKE CRYSTAL
WITH A "C," RIGHT?
BUT THERE'S SO MANY WAYS YOU CAN
SPELL CRYSTAL, IT'S LIKE K-H,
C-H, K-R-I, SO MANY WAYS TO
SPELL CRYSTAL THAT I AM REALLY
DISADVANTAGED.
AND THAT IS SO NOT FAIR TO ME.
BUT I DID MY RESEARCH
LIKE YOU DID.
>> YES.
>> AND I DO KNOW THAT MY NAME
WAS MOST POPULAR IN THE YEAR
2000, AND THERE WERE 2,600 BABY
GIRLS NAMED CRYSTAL WITH A "C."
>> THAT'S A DECENT AMOUNT.
>> YEAH.
>> THAT'S A PRETTY GOOD NUMBER.
>> NOT TOO BAD, NOT TOO BAD.
>> YEAH.
SO I LITERALLY JUST LOOKED THIS
UP THE OTHER DAY, I DIDN'T KNOW
HOW POPULAR MY NAME WAS EITHER,
BUT I HAVE SOME GIRLFRIENDS WHO
ARE EXPECTING BABY GIRLS THIS
YEAR, SO I WAS TRYING TO GET
THEM TO CONSIDER NAMING THEM
AFTER ME.
BUT I DON'T KNOW IF THAT'S GOING
TO WORK OUT.
BUT THERE WAS ABOUT 6,000 GIRLS
NAMED JENNA IN 2001
THAT WERE BORN.
SO, THAT'S A LOT!
>> THAT IS A LOT.
>> YEAH.
>> WOW!
OKAY, SO I'M GOING TO ROOT FOR
YOU THAT SOMEONE NAMES THEIR
BABY JENNA, ONE OF YOUR FRIENDS.
>> ANYONE WHO IS THINKING ABOUT
A NAME FOR A GIRL, CONSIDER
IT FOR ME.
>> OKAY.
SO WHILE I'M PUSHING FOR ONE OF
JENNA'S FRIENDS TO NAME THEIR
BABIES AFTER HER, I JUST WANT TO
ASK YOU GUYS A QUESTION.
THIS IS LIKE REALLY IMPORTANT, I
REALLY WANT TO KNOW, WHAT DO
YOU GUYS THINK?
PLEASE DROP IT IN THE COMMENTS,
WHAT DO YOU GUYS THINK IS
IMPORTANT IN NAMING A BABY?
SO DO YOU BELIEVE THAT THE MOST
IMPORTANT --
[ NO AUDIO ]
-- SOMEONE OR DO YOU WANT TO
NAME YOUR BABY SOMETHING
ORIGINAL THAT NO ONE ELSE
HAS THAT NAME?
>> YEAH.
>> YEAH, SO I WONDER WHAT PEOPLE
THINK ABOUT THAT.
>> WELL, I WAS SUPPOSED TO BE A
BOY, SO MY PARENTS THOUGHT THAT
I WAS A BOY DURING MY MOM'S
ENTIRE PREGNANCY, BUT, SO THEY
WERE PLANNING TO NAME ME AFTER
MY FATHER WHO IS A JUNIOR.
SO I WAS GOING TO BE THE THIRD
AND THEY WERE REALLY EXCITED
ABOUT CARRYING ON THE NAME.
BUT WHEN THAT DIDN'T HAPPEN,
THEY HAD TO QUICKLY GO THROUGH A
BABY NAME BOOK, AND THAT'S HOW
THEY PICKED OUT MY NAME.
SO MINE -- MY FIRST NAME ISN'T
NAMED AFTER ANYBODY, BUT MY
MIDDLE NAME IS NAMED AFTER MY
GRANDMA, AND I REALLY ENJOY
HAVING THAT NAME WITH HER AND
THAT CONNECTION WITH HER, BUT I
THINK EITHER WAY IS GREAT AS
LONG AS YOU LIKE THE NAME AND,
YOU KNOW, YOUR CHILD LIKES THE
NAME, OBVIOUSLY.
SO I THINK IT'S GREAT
EITHER WAY.
>> YEAH, I AGREE.
I HAVE A REALLY, REALLY FUNNY
STORY ABOUT THAT.
SO MY SON'S DAD AND I, WE AGREED
ON A NAME EARLY ON IN MY
PREGNANCY, WE AGREED THAT THE
NAME WOULD BE JORDAN SHAQUILLE
IN HONOR OF OUR TWO FAVORITE
BASKETBALL STARS, MICHAEL JORDAN
AND SHAQUILLE O'NEAL.
SO I WAS SO EXCITED, I GOT
EVERYTHING MONOGRAMMED, LITTLE
T-SHIRTS, THE WALL HAD JORDAN ON
IT, THE TRASH CANS HAD JORDAN ON
IT, WENT INTO LABOR, AND JUST
LIKE YOU HAD MENTIONED, LIKE MY
SON'S DAD WAS A THIRD, SO I
REALLY DIDN'T WANT MY SON TO BE
A FOURTH, I'M LIKE HE'S NOT A
KING, BUT I DON'T WANT HIM
TO BE LIKE --
>> WISHFUL THINKING.
>> BUT AS SOON AS I DELIVERED,
HIS DAD PICKED HIM UP AND SAID
HE WANTED TO NAME HIM AFTER HIM.
SO AT THAT POINT I DIDN'T CARE.
>> YEAH.
[ LAUGHTER ]
>> AFTER LABOR, I DIDN'T CARE.
SO NEEDLESS TO SAY, WE HAD TO
CHANGE EVERYTHING THAT WAS
MONOGRAMMED AND REDO THE WHOLE
ENTIRE ROOM OVER AGAIN.
>> YOU'RE JOKING.
>> NO, I'M SERIOUS.
>> OH, MY GOSH.
WELL, I'M SURE NOW YOU CAN'T
IMAGINE HIM BEING NAMED ANYTHING
OTHER THAN WHAT HE IS, RIGHT?
>> RIGHT.
>> YEAH, THAT'S ALWAYS HOW
IT WORKS, SO.
ALL RIGHT.
WELL, THE TIME IS FINALLY HERE,
WE ARE READY FOR THE BIG LAUNCH.
SO ANYONE WHO HAS ANY
LAST-MINUTE GUESSES, PUT THEM
BELOW, BECAUSE WE ARE ABOUT TO
RELEASE THE BABY NAMES.
>> I CANNOT WAIT.
>> I KNOW.
[ NO AUDIO ]
SEE WHAT PEOPLE SAY?
WE'LL LET YOU GUYS -- WE'LL SEE
WHAT YOU GUYS HAVE TO SAY BEFORE
WE RELEASE THE NAMES
REALLY QUICK.
JUST WANT TO GO ON --
>> CHECK YOURS TOO.
>> OH, AND I ALSO WANTED TO LET
YOU KNOW IF ANYONE HAD SOME
GUESSES ON THE THREE STATES THAT
HAVE THE MOST POPULAR -- OR THE
MOST BIRTHS, THE ANSWERS ARE
CALIFORNIA, TEXAS, AND NEW YORK.
SO THERE YOU HAVE IT.
SO ANYONE WHO GOT IT RIGHT, GOOD
FOR YOU, CONGRATULATIONS.
>> ALL RIGHT.
>> BUT LET'S GET STARTED
ON THE NAMES.
I WANT TO -- LET'S GET TO THE
NAMES, THE BIG RELEASE,
LET'S DO IT.
SO OKAY, STARTING WITH NUMBER
10, ETHAN FOR THE BOYS AND
HARPER FOR THE GIRLS MADE THE
LIST FOR NUMBER 10.
AND THEN ELIJAH AND EMILY BEAT
THEM OUT FOR NUMBER 9, ELIJAH
FOR THE BOY, EMILY FOR THE GIRL.
>> AND --
>> A LOT OF "E" NAMES.
>> YEAH, RIGHT, RIGHT.
>> YEAH.
>> AND MICHAEL AND ABIGAIL JUST
MADE THE LIST AT NUMBER 8, WITH
JACOB AND CHARLOTTE BEATING THEM
OUT FOR THE NUMBER 7 POSITION.
>> OH, AND MIA JUST MADE THE
LIST AT NUMBER 6.
I HAVE A NIECE NAMED MIA, THAT'S
REALLY EXCITING.
AND BENJAMIN MADE THE LIST FOR
NUMBER 6 FOR THE BOYS.
>> OKAY, GUYS, THAT'S ALL WE
CAN GIVE YOU.
SO IF WANT TO LEARN THE TOP FIVE
NAMES, YOU HAVE TO JUMP OVER TO
SOCIALSECURITY.GOV TO FIND OUT.
>> YEAH, YOU GUYS GOT TO GO
CHECK IT OUT.
AND WE THANK YOU SO MUCH FOR
JOINING US THIS MORNING AND
STARTING YOUR DAY OFF WITH US.
WE WANT TO LET YOU KNOW THAT YOU
CAN ENGAGE WITH US ON SOCIAL
MEDIA AT ANY TIME,
INCLUDING FACEBOOK.
WE ALSO HAVE A TWITTER ACCOUNT,
A YOUTUBE ACCOUNT, LINKEDIN
ACCOUNT, GOOGLE PLUS ACCOUNT,
AND WE ALSO HAVE A BLOG.
SO WE WANT YOU TO HEAD OVER TO
OUR SOCIAL SECURITY MATTERS BLOG
AND YOU CAN TAKE OUR BABY NAMES
QUIZ, IT'S REALLY FUN, AND SEE
WHAT NAME THAT YOU SHOULD PICK
FOR YOUR BABY.
>> AND YOU CAN ALSO GO TO OUR
YOUTUBE CHANNEL TO CHECK OUT THE
CLOSED CAPTION VERSION OF THIS.
>> ABSOLUTELY.
SO, CRYSTAL, HAPPY MOTHER'S
DAY TO YOU.
>> THANK YOU.
>> EVERYONE, HAPPY MOTHER'S DAY.
WE HOPE EVERYONE HAS A WONDERFUL
AND SAFE MOTHER'S DAY WEEKEND.
>> HAPPY MOTHER'S DAY, EVERYONE!
ENJOY!
I CERTAINLY PLAN TO ENJOY MINE.
>> BYE, GUYS, THANK YOU.
>> BYE.
-------------------------------------------
Strength of Materials II: Strain Transformation, Mohr's Circle for Strain (14 of 19) - Duration: 1:24:31.
Still I have problem with ma. Ma is as important or more important than ra. Ra is a reaction,
ma is a reaction here. You have also here, I don't know what I call there b, so you have
an e also rb and mb, that's four unknown and therefore, you have two extra unknown here.
Not one extra unknown. It was in your homework, previous homework in single directive method
and in superposition method. I have given you this problem which was a fixed support
at both and then there are lots of funny things that some of you are going to do and my guess
is, many of you understand the whole scenario correctly. You make occasional error. You
put-- you lose one or two or three points. But there are some also student which makes
static error which is not supposed to be at this level. So I explain as we go along. So
first of all, you start writing your ei. First of all, this is a differential equation. So
when you write ei-- y over dx squared equal to m as a function of x. Any function can
be given here. However, we have discussed that. You have ra and you have ma at the beginning,
you should start from there. Now, sign, many of you in this class, especially not the other
class, you have difficulty with this sign. Everybody sees the moment of that part is
negative, they put a negative here. That I do not buy it, that is wrong, that is absolutely
wrong. We are not talking about the static moment. We are talking about the internal
moment, the reaction of the beam to the action. Everything that you learn in MA218 for the--
I don't know this is the 10th time I'm repeating there, some of you are paying attention, this
is only for benefit of those that still have a little problem with MA218. Everything that
we did in MA218 when you write m sigma equal to mc over i, that m is not external moment,
it is internal moment. Everybody understand that that you have to go to the body. I don't
care what you have here, you have-- they have r here, m here, load here et cetera. You go
to that section. There is only one n. There is only one shear and there is only one moment.
Everybody understand what I've said? That defines all of your stresses, is that correct
or not? Therefore, we are talking about this m, not that m, is that correct? Therefore
if this goes this way, this must be going that way, yes or no? We go through balancing.
Therefore, this system is a positive system or negative system? Negative, or else, you
can do it like that which I have done it many, many times, the action is negative. The reaction
would be, how many times have you seen me doing that? This has not apparently registered
for some people, at least about 10 or 15 people put this wrong. So it is ra y is positive,
the action is negative. The reaction is the reaction of the beam to that is a positive.
So this become ra x minus zero to the power of 1, 1 because the moment is linear. Is that
correct or not? With the same reasoning, ma is going positive. The reaction to it is negative.
If you draw it that way. Some of you draw it on the other direction. So that gives you
a positive mode, everybody-- but that's OK. If you draw it that way, this should be minus
ma. If you draw it opposite to that, then it should be plus. Everybody understand it,
yes? This system, action and reaction. This is negative, we talked about it many times.
This action and that's reaction. Now, this is this blue one. This is inside the beam.
Is that correct or not? Yes? Anyhow, I think that everybody by now should recognize that,
but I saw that and that was surprising. X minus zero to the power of zero and nothing
else happening until at 1, which is a load constant load. Shear is linear, moment is
a parabola. Then you put again, the action is? Look at the action, if I put the loads
here, the action is positive, the reaction of beam to it is, negative. So minus 12 x
minus 1 to the power-- that's 1b there is not-- to the power of 2 over 2, is that correct
or not? Yes, since this goes all the way down, this is what I don't like. If you miss that
one, you miss probably 6 point or 5 point or 6 point or 7 point, I don't know because
you cannot miss that, I gave you the 10, 12 homework. Six of them has this in it. Is that--
how can you miss that, is that correct or not? Then what? Then this goes all the way
to the end then I have, this is because when I wrote here, it goes all the way to the end,
is that correct or not? Then I have to subtract this much of it which is plus 12x minus y.
Where does it start? X minus 3 to the power of 2 over 2, is that correct or not? That
is your moment equation for entire beam, correct? Then the rest is automatic, I don't care how
you do that because this is very simple, then you said ei, dy with the x or theta equal
to ra x minus zero to the power of 2 over 2 minus mx minus zero. By the way, you can
drop this zero because zero has no effect on your quantity. So minus 12-- but here,
you cannot, but these are not parenthesis. These are-- this is another cardinal mistake
that you or you make in this class. These are not parenthesis, these are? Bracket, bracket
has a special format which we discussed in class with you guys. X minus 3 to the power
of 3 over 6. If you miss that one, then the whole system collapses. So that's why I don't
understand how you did your homework, some of you because this was very much simpler
than your homework. Ra x minus zero to the power of 3 over 6, minus ma x minus zero to
the power of 2 over 2 minus 12x minus 2 to the power of 4 over 24 plus 12x minus 3 to
the power 4 over 24 plus c1x plus c2. Now here come another point which is very interesting,
many of you have seen in the homework that I did on the class on the board. Many times
I put c1 and c2 equal to zero, is that always the case? Of course, not. This is the reason
for it and that reason are these two, is that correct or not? Since I have a moment there,
therefore that means it's not rotating, it's like that. Therefore add x equal to zero,
theta equal to zero or you can say theta a equal to-- because this is for the theta,
for a equal to zero. Therefore, that gives you zero, zero. This is-- some of you calculate
that minus 1 to the power of 3, I warned you about. If you do that, you haven't heard anything
about the singularity method, everybody understand that. So this is negative, therefore negative
must be equal to zero, this is negative zero, therefore c1 become equal to zero. And similarly,
add x equal to zero, y equal to zero but fortunately, many of you knew what you are doing. This
is zero, this is zero. Again, this is negative, therefore zero, this is negative therefore
zero, this is zero c2 equal to zero. And that is-- that's the-- what my objection come to
your static system. You use this and you get c1 or c2 equal to zero, yes or no? What is
this? No, what is that? Is it the fixed support? If you understand what you did at a you should
do it for b as well, yes or no? What kind of class don't know how to do it? Why, probably
they see me doing this they have a recipe in mind, everybody understand what I'm saying
that they follow-- OK, you did that, there was a fixed support, you put c1 equal to zero.
See, I'll out c1 equal to zero, c2 equal to zero, if that's the case and you understood
it, there is no doubt that you should use it for b as well, yes or no? Because at b
also, because that's a fixed support, yes or no? And b also theta equal to zero and
y equal to zero. Some of you y equal to zero, some of you-- you can use it, I told you can
solve it with that one. So far we got rid of c1 and c2. But don't forget that we have
four more or no less than two equation of equilibrium, two equation of equilibrium is
ra plus rb equal to the load or sigma ma or sigma mb equal to zero, yes or no? I need
two more equation. Where are that coming from? From two other boundary condition and two
other boundary condition, this is the key actually equation, those two was in some of
your homework as well. Don't you have an rb here which prevent this from going up and
down, yes or no? Don't you have here an mb there which prevent this from the rotation.
That is that the key to this problem, then these are the one I'm looking for. This is
the one you get. This, you get the point, this you don't. And this here, which many
of you missed, an x equal to 5, k2 equal to? Zero, which gives me the following equation.
You go there and put x equals to 5, so you get this equation. Anyhow, ra5 to the power
of 2 over 2 minus ma times 5 minus 2, 5 minus 1 it become 4 to the power of 3 minus 2, 5
minus 3, it become 2 to the power of 3. And you simplify that equal to zero or you get
the way I have it here. You had a different rate, 12 one half ra or 25 ra if you wish,
minus-- let's do it this way, 25ra minus 10ma, equal to 224. So that's equation one or any
version of that is correct. Then next equation I need actually, I don't need to use any equilibrium
equation. This was simpler than what you've thought. So, an x equal to 5 meter, also y
equal to?
Zero.
Zero. Then I use this equation, this become two equations for?
Two unknown.
Two unknown. So therefore, I go there, so I put this second equation. Let's make it
short. Anyway the second equation, I give you the-- as when you put this one into the
last equation, again, this become 5542 and just simplify that one. That become 125ra
minus 75ma, become equal to 720 or some of you divide that by 6 or whatever, that's OK
too. Usually I try to make it round number because it's easier. Is that correct or not?
Never divide it by 6 or 3 because you get all decimal points. Is that correct or not?
As you see, all I have to do, multiply this by 5. Is that correct or not? And let's subtract
this two together, this becomes 50 and then subtract it, and 75 minus-- it become really
simple. So, if you calculate that, ma become equal to 60 kilometer and ra become equal
to 15.4 kilometer. And then in order to calculate rb and mb, what should I do? I should use
the equation of?
Equilibrium.
Equilibrium. Everybody understand that. Yes or no? The-- And the funny part is about 10
people in the class and I ask you to calculate reaction, you write ra and rb but you didn't
write ma and mb. And I said, please, have some respect for m because m is as important
as ra's. Still, this is the funny part, it's still all of that-- all of the static, still
people do not recognize the moment reaction as part of a deal. And that's why some of
you have a little bit of problem of static there. Is that correct or not? The moments
are there, the moment of the reaction. Anyhow, you write sigma here-- but ra is actually
equal to 15.4 and the total load is 24 so the balance of it goes to rb. Is that correct
or not? Yes? So, anyhow, rb become equal to 8.6 and then you take moment about here, you
have to take ma into account, mb into account. So mb also become equal 11, I suppose, 11
kilometer and meter. So that problem that result that they-- anyhow, still some-- anybody
have not got there Joshua, Cesar Moreno [assumed spelling], Cesar? OK. Nick, Nick Kate [assumed
spelling], yeah, OK. Josh Fa [assumed spelling], Josh? No? Nick? OK. Joshua, Gary? OK. And
Brian Rose [assumed spelling]. They're missing. Anyhow, let's continue with the discussion.
We have lots of work to do because they have to finish this chapter. We are about one class
behind, right, remember? Go back to the-- your notes please. Everybody put notes aside.
Go back into your note. Look at the state of strain that I gave you. Please put this
aside everybody. Go to the previous page in your note because I ran out of time that's
why I have to repeat and I don't want to repeat. Look at epsilon x. Where was epsilon x? When
I started the practice drawing the more circle where a strain. What was epsilon x? Epsilon
x was?
Y.
What was epsilon y?
110.
110? What was gamma?
180.
180. Now, remember, in the stress we draw a square or cubic and we show the sigma. But--
in strain are not like that. This is representation of a strain. This is original. You see, this
is what I said that result of-- look at it. I said the result of what? Strain transformation
of class exercise, that's what we did. Yes or no? Last time, what was the strain I started
with? This one, look at this one, how much is that? Class 34. See, this length which
was one, after a strain become one, 340 micro which is 1.00034. Everyday understand what
I'm saying that. Yes? What is this? What's epsilon y? Hundred ten, if it was one unit
because epsilon is the amount of the formation pair one unit of length. Yes or no? So one
become one plus epsilon. Yes or no? So that is 110. What is this angle in change of this
angle because this was originally 90 degree. Now, is 90 minus gamma which is a positive
scenario I gave it to you last time which is how much? Hundred eighty. Yes or no? This
is original state of those command, show me like that. I have seen people look like that.
They put here something like this. And they put here epsilon x as if this is sigma. This
is not sigma. Everybody understand that this is change in the length. Is that correct or
not? You can guess it looks like that but you cannot put it like that. However, this
I would not ask you to do many times but I want you to realize this is what's going on.
Is that correct or not? Yes? So then, what did we get for? What did we get for the principal
strain? Go back again. What did we work for principal strain? Come on guys. Take a look.
Well how much epsilon a and epsilon b was? Look at this one. Look at how many degree?
Remember how many degree did we have? What's theta p equal to? Nineteen degree, yes or
no? You forgot. Theta p was? Look at this, theta p was? Nineteen degrees. So, we shift
it-- this is what's positive, counterclockwise. So, we went from x, 19. Look at 2 theta p
was 38 degree. It is in your note. So, we went up 19 degree and this is one principal
strain and this is one-- there is no change in angular. So, what was epsilon a? Epsilon,
was a positive, 370. It means it is expansion. Yes or no? Yes. And epsilon y, what was epsilon
b, epsilon maximum, now epsilon minimum. Epsilon minimum was 80. Yes or no? If it is minus,
then you make it smaller. Yes or no? Because minus means compression, plus means tension.
This is the state of stress, principal stress. And principal stress, I'm sorry. Yes or no?
What do you think this is? This is maximum gamma, in plane gamma because I have to add
to this either 40 by 5 degree up or 45 degree from here down. Is that correct or not? Which
we think it's here, 64 degree. Look at it. Epsilon average, look at your-- Epsilon average
was 225for both x and y. Everybody following me? And then gamma was how much? Gamma max
was in plane, was 225. So, this is original state of a strain. This is principal strain.
This is maximum, and what is this? You can read it, exactly. What is that that I didn't
do? But I asked you to do. See, it was a question but I run out of time. What was that?
Thirty degrees.
Rotation of?
Thirty.
Thirty degree here which on the more circle, you have to go 60 degree. You take your x
and y wherever it is, you go 60 degree. We have done that in the past so I don't have
to do it. Is that correct or not? Yes? So, this is what we do. This is the presentation.
I do not usually ask you to do it but it's good for you to understand it because this
is essential for you to see how this works, because some of your homework at least gives
you some idea. If you cannot plot it, that's OK. I do not expect you to do it. However,
this is the format of it. Is that correct or not? Yes? Now, how do we-- Now we got to
the point now. I want to leave it at that and I would-- will go to what we are really
doing. Here is the following. So, let's go back a few pages. Go to the second page of
the hand out. The first page is the outline. We already talk about it. Let's go back here.
You see this is what we are doing in general. We talk about-- this is MA 218. We talk about
this frame and gammas and plane stress which is more circle. Stress transformation which
was chapter 5. Now we are. We are doing exactly the same but this time we are doing with the
strain. Is that correct or not? Now, what is happening is the following. Go to the next
page. Here we go. These are-- Really this is a chart. This is my chart. So you can use
it anyway you want to. It's not-- I haven't seen it in any book. However, you can-- these
are these, call it design process. So put it here in design, put here design process.
Of course, you cannot write it if it is in your computer. I have an extra copy there,
you can take it.
Oh you can write it?
Yeah.
OK, good. OK, this is what we did in MA 218. We-- When we went through this, for all the
chapter, we were able to calculate sigma x, sigma y and gamma xy, yes or no? At the end
of MA 218 we were calculating sigma x, sigma y, gamma xy due to the bending, due to the
torsion, et cetera, et cetera. This is the design process. Then we don't stop there.
What did we do in last chapter in MA 218? We put it through the Mohr's circle to end
up with what? Sigma maximum, sigma minimum, principal stresses and principal maximum,
shear stress, absolute. I didn't put absolute here because that means we have to draw three
circle, yes or no? We went through that as well. Then of course if I need that-- epsilon
was this, I have to [inaudible]. If I need the epsilon, I can use the Hooke's law. Remember
Hooke's law, what's the Hooke's law about? Yeah. Everybody remember Hooke's law, I put
it on the board last time. Epsilon x was equal to what, sigma x over e minus nu sigma y over
e. We don't have sigma z. Is that correct or no? Yes? You want to write it again, go
ahead write it. Sigma y equal to minus nu sigma x over e plus sigma y over e, generally
when we don't have sigma z, this is what will be epsilon x, epsilon y. Of course we have
epsilon c or epsilon z and that is minus nu sigma x over e minus nu sigma y over e. Everybody
knows that. So if you are taking a plate and pulling it and pulling it, the thickness becomes
much smaller because of the two negative, negative. Or if you're pushing it and pushing
it, the thickness become larger. That's the handle, that's the-- if you want to practice
that, this is MA 218 handout, so you can take a look there. But you don't-- I'm sure you
know that or you must know that. So that's-- There, we have three equation. Gamma xy equal
to what, equal to tau xy divided by a modulus of rigidity and also the other two that because
these are the major parts, so I'm putting here. So the Hooke's relation as these-- it
refers here to this one. This is the Hooke's law. You can add sigma z to it if you want
to but if it is three-dimensional there are sigma z over e sigma z that comes from the
third term and the other two gamma but usually isn't there. Now, this is but-- Remember,
there is always that relationship between sigma and epsilon, yes or no? So if I have
all the sigma I can calculate all the epsilon. But if I have all the epsilon, I can calculate
all the sigma. There's no problem there. I'm sitting in my office. I can calculate that.
But the point here in the design process, we go from here to here. Is that correct or
no? Then we make a model of that. Is that correct or no? And we measure the strength
this time. But this time we end up with that? Epsilon x, y and gamma xy through the measurement.
And then we repeat the same process of course everything should match. Is that correct or
no? If the two match together, then our design is valid design or it is good design. Other
words, I have to go modify my design to see what I miss or use a different method or something
is wrong. Is that correct or no? So that design is no longer good because I'm not-- my assumption
is not good, maybe. Or I'm missing something there. Everybody understand that? So the process
are parallel as you see that. The design process is exactly the same. It needed to draw more
circle. It is always through this Hooke's law which is in-- you see I can go back and
forth. And that so I have this, I can find this, I have this then that's no problem there
but this two parallel process to each other, yes or no? Now, the next question is how do
we measure epsilon? So this is what I want to do, this again now new battery. So how
do we come up if I want to start from here, from here I need to have, we call it gauges.
Now, the gauges are shown in this page. So let's go back again, this is a gauge. The
picture of a gauge, this is it. This is the picture of the gauge. Everybody see that?
The gauge is lots of wire here, you see how long this wire is, and everybody and goes
at the end here some of you have taken the lab you know about that, our tag. Everybody
knows from psychic classes that when there are wires there and the length of the wire
changes that resistant changes, yes or no? So this gauges have lots of wirings, it's
very tiny wire there. So they are-- When you put it-- stick it on the top of material and
load it, the length become either longer or shorter. So the wire-- So that measures in
a way measured the strain for you. You see you don't have to know if they have calibrated
ahead of time for you. It connected to the machine, the machine gives you the epsilon.
Is that correct or no? So those are the gauges, our electrical wire you connect that to the
system and you wrote the epsilon. The only problem is this. That if you want to write
it down please, this is the lecture today. Because you have to use this constantly in
all your next setup, homework you have to use that. I don't think I need that anymore
for the time being. So let's put this one off for the time being, see whether we need
it later on. So because we need the board. Now let's say this is the-- your instrument
and then you managed to put one strain gauge here which is in the x direction and manage
one strain gauge here with the y direction. So the a gauge and b gauge give you epsilon
x and epsilon y. So I need it-- To start with, I need epsilon x, I need epsilon y and I need
gamma xy. Is that correct? In order to start my process up, strain gauge process, I need
epsilon x. OK, I have epsilon x. I need epsilon y. I read epsilon y. But how about gamma?
Gamma was as I showed you earlier that was the reason I show those picture to you. Gamma
is the change in the length. Gamma is change in the format of the shape, shape of this.
So I cannot measure that. Is that correct or no? How do we do that? You see, how do
we do that through the process that I gave you? Instead of measuring gamma which is impossible
to measure, we do that. We put another gauge here. Let's say at 45 degree, which we call
it-- First of all, we need three gauges. In order to come up with epsilon x, epsilon y
and gamma xy, obviously we need three gauges not one. Is that correct or no? One gives
me epsilon x, one gives me epsilon-- hopefully this one gives me gamma. Let's see how it
works. Is that correct or not? Yes? So, they come-- usually they come in three and we call
it rosette. So when there are three gauges together, we call it rosette. So please--
we have 45 degree rosette, we have 60 degree rosette which we call it delta. We have some
other kind of rosette or we can now put it on arbitrary location, I can explain that
to you in a minute. Is that correct? But if we buy this and install it all by system,
so I have x, I have y and I have epsilon at 45 degree. But what was the equation you wrote
last time? Remember the general equation that you should-- all of you remember? Epsilon
x prime was equal to what? This is the key equation guys which started-- we started the
lecture a week ago with this one. What's epsilon x cosine square theta, you remember that class?
Epsilon y times sine plus gamma is?
Two gamma and 2.
Not 2. But 2 gamma over 2 but we don't need the 2 and 2. That was for the purpose of matching
it with the stress, is that correct, to use the same idea. But gamma xy sine theta and?
You know, I added 2 and you subtract the 2 to make it-- it looks like a stress plane
strip. That's all I did. Yes, is that correct or no? But this was the equation that came
up from the process. So let's use it for gamma. So here you say epsilon at 45 degree because
that's where you are measuring, yes or no? x prime, your x prime is at 45 degree equal
to epsilon x cosine of how much? Cosine of 45 which is the square root of 2 over 2 squared,
plus epsilon y square root of 2 over 2 square plus gamma xy square root of 2 over 2, square
root of 2 over 2. Those are sine and cosine of 45. As you see, this ends up to be one
half, this ends up to be one half, of course this 2 ends up to be one half. Therefore,
from here I can calculate gamma xy, gamma xy which is here. This 2 and 2 two goes there
become 2 epsilon 45 minus because there are going to the other side, minus epsilon x plus
epsilon y. So through this equation, or through your choice you have two choices, either use
directly this or if you see that, that's the result of this everyone but this is for, only
45 degree. If I have a delta rosette which is 60 degree, I cannot use that for that.
Is that correct or no? Yes? Because the-- it's 45 degree by, right, by anyhow by the
way we discuss it. Here is the most common rosette that you can buy, this is the-- almost
anybody uses that. Is that correct or no? You come up with the epsilon x, epsilon y
and gamma xy. However, you don't have to do that. If by any chance, you cannot do that.
As I said, the next one is the 60 degree rosette which is-- looks something like that. So x,
so it looks like something like that. But then I will explain it to you what happen
there in a minute. So you don't have to worry about it. Those are the detail of it. But
let's say that you don't-- cannot put your gauges in any particular direction. If I put
my first gauge as theta 1 angle, theta 1 angle could be 25 degree, 35, whatever. I put a
gauge at theta 1. I put another gauge at angle of theta 2. This is gauge a. This is gauge
b. And I put another one-- I don't have even to put it that way. I put that one even gauge
c at that direction but remember, this angle state is like a static. So, this is not theta
3. That angle, always-- you always measure it from the positive x. Remember the static
problem that we always-- because that-- remember the sign would change if you use the other
one. So, theta 1, remember, in general. That's why on the graph I also said theta 1, theta
2 and theta 3 or the-- at those chart that I gave you. So, if this is the case, you see
I have still three gauges. But I have-- Did the direction of theta 1, theta 2 and I gave
it to you. Theta 1, let's say 32-- 30 degree, theta 2 may be 45 degree, theta 3 may be 120
degree. Everybody understands very well, I put, it doesn't matter. However, I can write
this equation three times. Everybody understands that. Yes? For theta equal to theta 1, for
theta equal to theta 2 for theta equal to theta 3 and I measure all that which is on
this side. It become three equation for three because here it was one equation, one unknown
because I already had epsilon x and epsilon y. So routinely, this become like that. So,
you write it like that. Epsilon theta 1 which you measure equal to epsilon x cosine square
of theta 1 plus epsilon y sine square of theta 2 plus gamma xy sine theta cosine theta--
there's 1 to cosine theta. So, you put theta what is given. So, cosine become a number
you squared it. So as you see this is one equation for epsilon x, epsilon y and gamma.
Then, you write it again for theta 2. Theta 2 may be 52degree. I don't care, 60 degrees.
So, you write epsilon, theta 2 because you have measured that remember, a, b, c is measured
from your experience. So, you write epsilon x, cosine square of theta 2 plus epsilon y
sine square of theta 2, plus gamma xy, sine theta 2, cosine theta 2. And you write it
one more time for epsilon. Theta 3 as you see it this become three question for three
unknown. Your three unknown is epsilon x, epsilon y, and gamma. But, usually that's
not the case because most of the time you try to put your gauges along the x and y as
you will see in some of your example. I mean, this is silly to put it on. Unless you don't
have a way to do it or you want it to put it at one-- zero degree end up to be at 5
degree so you want to adjust your note so that's theta 1. Is that correct or no? Yes?
You had a question. Somebody had a question, right.
Sine theta-- oh, this one is all 1, this one all 2. So, thank you. This is all theta 1,
theta 1, theta 1, theta1, this is theta 2, theta 2 and then the last one is theta 3,
theta 3. So, three equations. I didn't have even to write it but I wrote it. Then by that,
then we can calculate everything. Now, let's go to a couple of example and see whatever
we can do. You need this. You need three gauges. At any direction, we'll do it. This is the
bottom line. And the last thing you want to connect, you have to connect your knowledge
of this MA 218 with the knowledge of strain. There is nothing new there. Everybody should
be able to do it. What, the only problem that I've seen in the past is people are afraid
of the strain. But they are not afraid as much of the stress but they are both parallel.
Remember that when you have the stress, you can't find the strain. Now, let's to go through
some example one by one. Strain number one is to go to the next page. You will see that
problem. So, you see that problem there. Maybe I should put that one down again, so OK. So
down again. So here, we go the next problem. Let me see. What happened? It's not showing
anymore. But, that's OK. I pulled it out so I cannot show you anymore. Anyhow, you have
it on your handout. Look at the three gauges there. I don't need the board with this one
anymore. So, look at the three gauges there. I says look at problem number 1 or quiz number
6. Look at that, do everybody see that? Yes, that's the one. Everybody should look at this
problem with three gauges like that. This three gauges like that. I do the essential
part of it to show you how this work. Already, you got the message. I'm sure that you know
exactly what to do here but let's say the three gauges are like that. It is given to
you. Is that correct or not? Everybody see that? It says on the surface of a structure
or component. In a space vehicle, the strains are monitoring and we monitoring the strain
informing this in this information to the computer to check it all the time. And we
found out that epsilon a equal to what, 1100 micro. Yes or no? What is epsilon a? Epsilon
a is?
Epsilon a actually is epsilon x equal to 1100 micro. Yes or no? Correct or not? Yes? Are
you looking at your homework? Are you looking at the quiz number 6? Everybody? All right.
So, what's the epsilon y? So, where is yours?
Go get one. There's one on the table. There are a couple or more left on the table. We
taught that. You are not going to understand what I'm talking about by writing this number
down, guys. Zero. If I were you, I would not even understand a part of what we are ever
talking about. You see, this is not the way to handle this course. You have to have--
I gave you many chances to have a copy of this one. Anyhow, epsilon a is epsilon x is
1100. Is it epsilon y is given?
Yes.
Yes. Epsilon y equal to how much?
200.
Epsilon y is equal to 200 micro. We are looking for gamma xy. Yes or no? This is not given,
instead, what is given? Epsilon in the direction of this c, is what you see there? This gauge,
this gauge and this gauge. Is that correct? And this angle being 30 degree, so this is
a, this is b apparently, this is epsilon b, this is the b and this is the c gauges. Is
that correct or not? So, of course I use that for x, that's for I-- so, I said it earlier,
I used that one for gamma. Yes or no? So, therefore, I write here. But I cannot use
the equation of 45 degree rosette because this is not a 45 degree. I have two general
equation. So, epsilon at this, epsilon c which is equal to how much? Epsilon c is given equal
to this capital C. What's given equal to also 200 micro, yes or no? Correct? So, 200 micro
equal to epsilon x. Epsilon x, look at the equation. I erase the equation but I should
have left it there. Epsilon x, epsilon is 1100 times cosine of 150 square. Yes or no?
Yeah.
Now, notice here. Here if you put 30 here, cosine of 30 if it is negative or positive
does not make any different because it's a square so no changes there. So, if you-- is
it up 150 you put 30 no problem there. Everybody understand that? Yes? Plus and minus become
squared it become-- the same thing for epsilon y. So, epsilon y which is 200 times sine of
150 square, so let's write it down. Again, if this is a negative it becomes square. However,
if you make a mistake, there is a big difference in gamma because 130 and 150, sine of 30 and
cosine 30, both are positive. But sine of 150 is positive. Cosine is negative. Everybody
understand what I'm saying that? Therefore if you make a mistake, there you are totally
doing a wrong problem. Is that correct or not? Therefore, definitely will be cosine
of 150 degree and sine of 150 degree, and one of them will be negative. Is that correct
or not? Yes. Correct?
Yes.
So, that's what I'm referring to. I hope everybody understand that. This is 30 degrees, sine
is positive, cosine is positive 150 degree which is this number, sine is positive but
cosine is negative. Everybody understands that. So automatic would come into the-- your
sine. Anyhow, so you must use this angle. So, this is you theta, this is what I'm saying
here. So, please be careful. Anyhow, because I have seen many mistakes. From here you calculate
gamma xy and gamma xy become equal to 1560 micro which is plus. Then it depends what
is being asked. This is the same problem that I did last time. So, I should not do it again.
But briefly, I'm going to have to explain it a few minutes because I have lots of other
problems to discuss with you. So, I'm going to briefly mention that. So, let's say we
are looking for principal strain and maximum shear strain. Is that correct or not? Principal
which are normal strain and maximum shear strain which is the gamma. Is that correct?
I have to plot that Mohr's circle with the same sign convention we did before, so this
become something like that. So, I'm not going to do it but I'm going to give you an answer.
So, epsilon on average, become equal to 650 micro. First of all, you divide this by 2,
so you go like that because you need gamma over 2. Gamma over 2 is 1560 divided by 2,
so that is 780 micro radian. So, that is your gamma over 2 for your problem. Then you plot
it, you plot the circle, et cetera, et cetera. I can show that what happened there at the
end, so this become your circle. So, this is what is going to happen. So, this become
epsilon minimum, and this become epsilon maximum, and this will be epsilon, this will be gamma
over 2 with that arrow, remember that. And all become equal to-- anyway, you use the
graphical method or equation doesn't make anything but this become 900 micro, and epsilon
a or epsilon max become 1550 micro and epsilon b or epsilon minimum, these are principal
boundaries but I call it a and b small because I don't want you to make it with this a and
b because this was gauges 1, gauges b, this is principal. Everybody understand. The 2
point that I'm showing here and there which usually I show it at a and b but that one
become minus 200 micro. Now, what happened there is like that, you go here, you plot
this point. Plotting that point is 1100, you go here, 1100, then you go 780, you have to
go 780 down here, so that will be your x. Then you go 200 here you go 780, so that will
be your y. Then you connect these together, that would be your c, this would be your r.
This is 900 divided by 2, so this is 450, this is 450. And we calculate 450 and 1100,
that gives you the r, et cetera, et cetera. Wish we have done it many time. Is that correct
or not? So, I should not go into that detail. Now, we go that one. So we calculate that
one. Now the last part I want to mention it one more time because-- already. What is the
in plane gamma or in plane maximum shear strain?
R.
It's not the radius. Gamma over 2 is the radius. Is that correct or not? So, please write it
down. Gamma over 2, x, y plane which is this plane which is related to x and y, is equal
to the radius, correct? The radius was?
Nine hundred.
Nine hundred, 900 micro. And then, gamma max in plane x, y, in plane become 1800 micro
which is larger than what we started and here which was 1560. Now, how do we calculate absolute?
Look at it, I'm asking for absolute maximum shear strain. I mentioned that last time.
Answer?
Is it plane strain or plane stress by the way? Don't forget about what I did here. Remember,
I said plane stress is common. Plane strain object must be between?
Two digit. Is this surface that I put a gauge on it, is it between two digit plane or is
it open to the surface? Is that-- So, sigma z must be equal to?
Zero.
Zero. Sigma z must be equal to zero. Epsilon c must not be zero. Is that correct or not?
Because at the surface, look at this, this is the surface of your car. You are putting
a gauge here, your gauge here. The car is being pressed this way, this way is that car.
Is this side getting larger or smaller? Obviously not. This is a plane of stress, is that correct?
We use the idea to plot the Mohr's circle. But our real problem always are plane stress
unless you put your object between two rigid body which I told you last time to all your
cookies. Is that correct or not? The soft material between the two harder material or
relatively rigid body. Is that correct or not? We don't have that often unless you put
a plastic material inside a concrete hole and everybody-- they try to push it, which
doesn't have any room for expansion. Everybody understands in the z [inaudible] and that's
sort of all you get the plate, soft material, rubber, you might put it between two piece
of metal, is that relative to that metal, it can be considered rigid. So, therefore,
there is no epsilon. You can't push it this way, push it this way. It wants to go up.
You don't let it go up, epsilon is zero but sigma z is not zero, but that's very not common.
Is that correct? If I've seen the surface of the shuttle there it must be plane of stress.
Is that correct or not? Plane of stress mean what? Means that sigma z equals to zero but
epsilon z is not equal to zero, is that correct or not? Yes. What was epsilon z? I gave it
to you last time. Epsilon z was equal to? I calculated it for you. Do you remember that?
Yes?
Negative epsilon?
That's right. I use the-- I set and use the equations all of everything I already discussed
that. But I gave you also equation that epsilon c equal to epsilon z equal to minus nu 1 minus
nu 2 times epsilon, it-- I can't put a, b or x, y or what because they are all constant.
The sum is constant. Is that correct or not? Yes? So what will be epsilon c, can you guess
it?
No. The sign of it, can you guess it? What is epsilon max? Is it plus or minus?
Minus.
So it means lots of tension in this direction. What's the epsilon minimum, it is?
Minus.
Extension or compression?
Compression.
Compression, means you have an object that being stretched this way by sigma x, is that
correct, a lot? But in this y, it is compressed but this is much little with-- compared because
you have this one. This side is going to get the smaller or larger?
Larger.
See, if this was not there, and you are stretching it, this side becomes smaller. Is that correct
or not? The effect of this is less than effect of that. Is that correct or not? So, in other
words, epsilon z or epsilon c equal to minus nu sigma x over e minus nu sigma y over e.
Is that correct or not? Sigma y, x is tension so you get huge negative. Sigma y is in? Compression.
What's the smaller quantity? Of course, minus time. Minus become? Plus, but does not overcome
that. Is that correct? So, I expect epsilon-- when I see this I expect my epsilon c to be?
Negative.
Negative number, is that understood? Everybody understand about that? However, if this was
small tension but this was not compression, this side would go into become larger. Is
that correct or not? The z take this become larger because I'm putting more pressure on
it than more tension. Is that correct or not? Therefore, in that case, epsilon c would that
been become? So, you have some idea ahead of time what you expect but let's see whether
that happened here. So, epsilon c equal to epsilon z equal to minis 0.29, did I give
you points enough there?
Or point three, did I give you 0.03? Thirty-five both sides. Point 35 for this problem, 1 minus
0.35, 1550 minus-- what did I put here? I think this is 250. What did I put? You know,
this become maximum, did I gave you maximum or minimum? No, not yet?
Yeah.
This is 1550. This is 250, I have it here. I don't know. So that's 250, yes. So minus
250, so you put minus 250 here. So you do it together, it becomes minus 700 micro. So
what I'm saying that, if you are looking for absolute maximum, gamma will not be in the
x, y plane which is this circle, will be in the x and z frame because your-- this become
minus 700, and then your second circle become like that and your third circle become like
that. We have done-- so you need the diameter of the larger circle, yes or no? Correct?
It's the same thing we did for stress. So-- But the point is epsilon c is not equal. Any
time, you're bumping off your car, is epsilon c equal to zero? You stretch it this way,
you stretch it, and it's going to change a little bit. Of course, you don't see it by
eye. Everybody see-- because these are all micro. Two hundred micro, what is that? Point
00002, everybody, four zero, 2 that's not something -- Yes?
What?
We can't, no, you can't. Obviously, the whole idea is the Poisson ratio. The relationship
between epsilon z and epsilon x is also Poisson ratio. If you -- but if you stretch something
in one direction, that was one of those early question. That was not a good question, that's
what I'm saying there. Look what you are asking me. What is Poisson ratio?
Well, because one of the home works that did give you mu.
Oh, I gave you g and e, yes?
Yeah.
Then you have-- you use the formula to calculate that. Is that-- what we can't-- mu-- the problem
is not solved. Because nu is the whole key. If I stretch something in one direction, nu
part-- nu Poisson of it goes to the whole Hooke's law that I repeat it. I give it to
you as the handout. This all depends on the value of Poisson ratio. However, if in a problem
they were not given which I'm going to do within a minute. And so there is a relationship
between g and e and--
Nu.
-- nu. Is that what you are referring to? Yes. So, your question is different. If the
problem is not-- you said can I solve this problem with nu? No you cannot solve it without
nu. All of it. If g is given and e is given, there is a relationship like that. I was going
to do it through another example. That's the relationship between g and e and u. So you
can get it from that equation. Because nu again is not measurable, g and e, you can
measure it in the lab, g by the torsion, e by the tension. So you measure-- that's not
what you have taken. You know what I'm talking about, yes or no? Correct? Before the lab--
great.
For this problem, why did he said epsilon c equal to epsilon z?
Because that's what-- no changes that. Epsilon c and z are not-- this is only rotation of
xy.
Epsilon c given here is 200 micro.
OK. This is reference to xyz axis. This is the gauge abc. That's a capital C. Those are
abc, that's why I've changed the capital to-- those have nothing to do with this.
OK.
Yes? Those are principle. The three principle. You see, this is xyz become abc, small abc.
It has nothing to do with this one. This is the measurement. That's why I've changed--
actually exactly what I did, I call it here epsilon a or something. Yeah. Epsilon max,
epsilon minimum, more than. This is x reference to x, y, z axis there. Alright. Yeah. You
know that, you know it already. Is that the correct one? Let's do some more example now.
Let's go to problem number six in your handout. There are many on the handout. This is now
you have to be very careful about this type of problem. So, exactly what you have to do
depends on your knowledge of, again, back MA218. So go to handout. There are six problem
there on the handout. And problem number 6. So, I wish I could get that one, look I took
it out. So it isn't sure. So, let's see whether this come back again. So, if you have current
back again, yes you can do it. OK. So just while I'm taking it out. So-- OK. It's week
seven-- alright let's go back again. Look at this problem. This problem we did that
one. This problem we did already. That's the one with a nu, just talked about it. And these
are your homework. We do that. Let's go through this problem. Problem number six. We kind
of have to do that. This is what, on and-- ok. It will come in a bit. You have it in
your handout. You know I'll put this for the benefit of camera. What you all have in guys
in your handout. So, go to problem number six in you handout. And take a look at problem--
handout problem number six. Take a look, read the question and can you answer that quiz.
That was your quiz, today can you answer that question. If yes then you can go home all.
Alright. Come on guys. The only thing I can recommend here to you is this guide. Although
we are talking all about the strain and the strain relationship, but you should not look
at it this way. You should go back to MA218, look at it as state of a stress. So everything
depends. So, everything and we talked about it in this lecture, it depends on the state
of stress first. Yes or no. Look-- I started the lecture by saying that we may or may not
have a sigma x, yes or no? Correct. Then we may or may not have a sigma y. And then we
may or may not have a tau xy-- I put it all positive, yes or no? In all your homework
problem, MA218, 219, you end up with this is which is a plane stress. Is that correct
or not? Sequence z of course equals to zero. Yes or no? Yes?
Yeah.
All right. What is this? Of course it's a plane of stress. Is that correct or no? Now
here is the question. We have put a gauge there, gauge a and b. The gauge reading of
gauge a is equal to it's given, 100 micro. Yes or no? So, let's put it there. The gauge,
this is the system, this is a force p, a torque t and here at that point, I put here a gauge
like that and a gauge like that. And that direction is how much? That this is x, this
is y and that direction is, how many degree? Forty degree.
Forty five.
Oh 45, whatever is there. I don't see if I'll be-- Is that correct or not? Yes? OK. The
question is, if epsilon, this is again the reference what you just asked India and like
y a couple of minutes ago. This is epsilon a and epsilon b. Epsilon a is measured after
the loading. Before the loading, of course, there is no stress, no strain. After the loading,
epsilon is a is 100 micro. Epsilon b is minus 55 micro new, Poisson ratio is a given equal
to 0.29 for this problem, 0.29. It is all there. Modulus of elasticity is 30 times 10
to the power of 6 psi or 30,000 ksi, that's the same thing. It's there. And the question
is, what is p and what is t? Come on guys. Now, did I give you the location of the gauge?
Notice this is r gauge, two gauges actually, c and I put it there. Is this important where
I put my gauge? No. Don't say no. You have to have an answer for this. You see, although
you said no, they knew the answer or you just said no because everybody says no? Which is
true? Which is the true case?
No because--
No why?
--you didn't put any of this.
Why?
You didn't put any lengths?
Why? OK. So, can I put it anywhere?
So in this case--
That's what I'm asking you.
Yes.
So, why? OK. You say yes.
I guess I don't know.
So, that's the question. You see, by saying that no or yes, what's your reasoning? And
that's exactly what I'm giving you a quiz for. The quiz I ask for all the reason. If
you don't know reason, why you are putting theta equals to zero here? Then we need to
cut or change a to b, don't put that b, then I realized that you did not 100% are there
that I want you to be. Is that correct or not? That's exactly my point. Yes.
That's not a good question. I mean, the answer is correct in a way but that's not the way
I'm looking at this one, it goes back to basic. When you have lecture 1 MA 218, when you have
a p here in this cross section sigma is everywhere. Yes or no? So, does it make any difference
whether it's on top or on the bottom or on this side? No. Is that correct or not? That
is p. However for the torque, when we have the torque here, what the shear stress was
outside? Wasn't that outside equal to constant of tc over j? So, it makes a difference?
No.
No. However, if this was a beam, that's what some of your homework is. And I put here a
load. Does it make a difference where I put my gauge?
OK.
Of course it makes a big difference now. See, this stance that I'm looking for, so you have
to go to your background and check that here. Normal stress is uniformly distributed in
the cross section, including on the surface. Yes or no? It's going tension, all tension.
Yes or no? Here or here or here or here or here is all the same, except I don't want
to put my gauges near the support because I'm afraid of stress concentration. Remember
the stress concentration ideas, so I'm trying to stay away from that. Is that correct? And
what is the route on the tension every sigma in every cross section? This is all making
the story short, both at the tau and the sigma is all similar in all the cross-section. That's
the answer that I was looking for. However, if it become a beam, not even the distance
x become important. This way, y also is important. Yes or no? Mx changes the value of the m.
Yes or no? Oh this is MA 218, the further I go the more moment do I have. So, I have
two know where I'm putting it. Yes or no? This space sigma equal to what? Sigma equal
to m, y over? You know, all the neutral access the effect is zero as you go up is tension
or comprehension of one side is the tension, one side is compression. Is that correct or
not? Vice versa, so, the definition is where you put it here is totally dependent on all
the state of stress at that point. Is that correct or not? Which will take us to some
of your homework? I don't know whether you look at those homework or not, I'm going explain
it before I did here. It does not make any difference, everybody agreed. All the answer
I heard is correct because it's uniformly distributed. Yes or no? Therefore, the location
does not matter. Let's finish the problem, then I get-- go back to that one. Now, first
thing first, how does my state of stress look like that? Go back. This is the whole idea.
This is interesting. If I give you the p and I give you the t and I give you the dimension,
by the way the diameter of this rod is 1 and a half inch, which is written there, everybody
can calculate the state of stress because this is very elementary for you. Are there
a normal stress, the p over a and tc over j. Yes or no? So therefore, in every point
here or there or their including point c, you have a normal stress. Sigma equal to p
over a because it is in tension. Yes or no? Correct? So sigma x equal to p over a and
then tau, which the way it looks like because it's going this way. So, the tau will be like
that. We have gone through that. This is very basic. The tau, which is tau xy equal to tc
over j. I have to give it minus because this is going downward. Is that correct or not?
Yes? Correct?
Yes.
This would be the state of a stress. Correct? OK. Now, is this going to create epsilon x?
Now, the key as you saw in that graph is the Hooke's law. What's the Hooke's law? And see,
I have to use more board here. The Hooke's law is the following. This is in general.
Here is particular to that problem. So I'm going to erase that, and then epsilon x equal
to what guys? Sigma x over e minus nu what?
Sigma z
There is no sigma z. Is that correct or no? Do we have a sigma y here?
No.
No. So, therefore this become equals. This cannot be anymore elementary than that. Everybody
understand what I'm saying that the relationship between a stress and a strain is as I've showed
you in that graph earlier, I can go back again and look at it everywhere, which was everywhere.
I go back before looking at that one more time. Look at it. Hooke's law, Hooke's law,
Hooke's law. Anytime I have Hooke's law, I can go from the stress to strain. You have
seen it many times in there or vice versa. Is that correct or not? That is the idea I'm
asking you to use in this set of homework which is very simple. Now, if you look at
it carefully, it becomes very simple. However, you have to remind yourself. Anytime I have
sigma, I can put it in the Hooke's law to come to the epsilon or I have epsilon, I put
it to the Hooke's law to get to-- that's as simple as that. They are parallel to each
other. Is that correct or no? Nevertheless, we go there so therefore-- now sigma x also
equal to in this problem only p over a. I'm looking for p. So, if I give-- I have sigma,
I can solve it. But sigma equal to e times epsilon because there is no sigma y. There
is no sigma z. Is that correct or not? Therefore, sigma x, equal to epsilon x times modulus
of elasticity, epsilon x was given equal 200 times 10 to the power of minus 6, multiplied
by 30 times 10 to the power of 6 psi. So that this and this drops out. So it becomes 3000
psi. So, the sigma x is 3000 psi. There-- It doesn't take a genius. This is nature of
MA218, to calculate the value of?
P.
P. Is that correct or not? So, therefore, you calculate the p equals to sigma times
the area and sigma is 3000 psi and the area is pi times radius squared which is 0.75 squared.
You can calculate that one because the diameter was given. This is a circular shaft and you
calculate the p to be equal to 55,299 or 5300 pound. So, for 5300 pound loading here, epsilon
a will be reading that 100 micro. Is that correct or not? Yes? You could have done by--
I could put the load here. You could calculate epsilon in reverse. Yes or no? Now, the next
question is what is the value of torque? OK. What's the torque? Look at these guys. Torque
is related to tau is equal to tc over j. Yes or no? But what is tau? What do I need there?
What is the relationship between stress and the strain? Shear strain? This is Shear stress
and shear strain, gamma. Gamma was equal to, I wrote it a few minutes ago. Gamma xy was
equal to what? Tau over modulus of rigidity. Tau xy divide it by-- the only question is,
that's why probably you were asking. Here, I have given you new and modulus of elasticity
but I did not give you the modulus of rigidity. I was going to bring it here. You asked earlier.
Is that correct or not? Therefore g equal to what? First, let's calculate the g based
on these two, g become equal to e. Again, divide that to one plus nu so I give you that
11.6 times 10 to the power of 6 psi. So if I have the g, if I have the gamma, I have
the g like the other one. I can calculate the tau. If I have the tau, I can calculate
the t, by noting reverse order. Is that-- Everything is in reverse order. It's nothing
new. You all know the answer. Is that correct or not? Yes? But in-- instead of going forward,
we are going sort of backward. Everybody understand that? We are having the testing and we want
to see what kind of loading caused this kind of gauges, is that kind. Even if it'd break,
we want some time to do that, what kind of forces this kind of damage to the structure,
which is by itself is another art. Is that correct or not? Yes? All right, anyhow. So
what should we do now? OK. Now, what should I do? You see here, I had epsilon x. So what
do I need? Look at the problem. I need the gamma. But gamma is not measurable. Instead
of that, what do I have? I have that epsilon at? At 45 degree, didn't I ask you to what
to do to get that? Yes or no? It's again, it's reverse. So you can use the equation
that I gave you. Epsilon of 45 degree equal 2, you can use that equation. Or, I'm sorry,
gamma xy equal to 2 epsilon or 45 degree minus epsilon x plus epsilon y. I just gave you
a while ago. Yes or no? Yes? Correct?
Yes.
All right. So let's use it that one. Become two times, what's epsilon 45 degree? Epsilon
45 degree is?
It-- We read it because it's needed there instead. It is?
Negative 55.
Negative 55. So it is minus 55 minus epsilon x. Epsilon x is? Epsilon a is?
One hundred.
One hundred plus what? Epsilon y, what's epsilon y?
A.
A? A. You see, that's the mistake, common mistake. Look, how many of you immediately,
you said epsilon y is zero because I didn't measure it? You all assumed epsilon y is zero?
Yes or no?
Right and you know again as that-- that's what I was referring earlier. That is the
Poisson ratio, yes or no? If that is an epsilon x, nu Poisson goes to epsilon? I don't need
to do measurement because I only have two unknown, if I had another unknown there, I
needed three measurement. Everybody understand what I'm saying there? However, yes, write
it down. I called it the 4 with the epsilon y equal 2 minus nu sigma x over e, plus sigma
y over e, minus the other one, the other one is zero, sigma y you all just said those.
So it actually become very standard minus nu tau epsilon x, which is really the definition
of the Poisson ratio. Whatever you are producing, what direction, nu Poisson? What's that other
one? Is that correct or not? Yes? So therefore, if epsilon x was 100, then epsilon y become
equal 2, minus 0.29 times 100. So it is minus 295. So do not jump to the conclusion. This
is always true. It is not measured but it is there. I can calculate that. Is that correct
or not? I did not measure because I could calculate. One measurement was enough. Actually,
if you measure it, it should be showing negative 29. Is that correct or not? If all the instrument
is correct. Therefore, you put here minus 29, not zero and you calculate gamma xy and
gamma xy become equal to gamma xy become equal to somewhere here. Minus 180 times 10 to the
power of minus 6 or you can write it in micro, micro radian. We don't have to put it, so
minus 8, 181. As soon as you have the gamma, it is needless to say, you immediately can
calculate tau xy. So you calculate tau xy. Tau xy, equal to g times gamma, g was 11.6,
times 10 to the power 6, multiplied by minus 181, times 10 to the power of minus 6. But
notice again, if you apply this 6 business that I always ask you to use, tau become minus
2100 psi. And it become minus, notice t is positive but tau is negative. And that's the
answer that came by itself, so because the measurement was like that. Yes?
What is the unit for the gamma right there?
Gamma, it hasn't-- doesn't have a unit because it's a micro radian. But in radian-- but this
is unitless, it's just inch over an inch millimeter about this radian. Yes?
Oh, yes.
All right. The same thing with epsilon, epsilon doesn't have any-- its just inch over inch,
millimeter over millimeter. It's the ratio that we need. Anyhow, so therefore no don't
worry. I have the most important problem yet there. Yet there, still we have that. Did
you see how this problem is solved, OK? Then we take-- that takes us to the next problem
which is this problem. So you have homework like that in the book or homework like that
again. I've mention that for your beam. If there is a beam here, there is a load here
where you put your gauges is very important. Is that correct or not? So you have to go
to a state of a stress. Find what we did earlier. This is you have homework like that. You missed
it, some of you. You had homework like that. I thought in this class, I gave a problem
like that. It was a p here. Is that correct or not? Yes? Yes? Now, you're getting similar
problem. That part of it is not a question. That part of it, you already know. You have
a homework like this. I'm going to mention to you because I don't have yet. There is
here, a force p and there is here a force q. Is that correct or not? And then you've
measure something here, somewhere here, you measure your strain and then you want to calculate
the value of p and q which is exactly this problem, a little bit different. Because that
one now, it is bending involved as you see. OK? Now, look what happened here. If you cannot
do that problem, certainly you cannot this problem. That's why I gave you earlier those
homework in order that you for at least through the quiz to learn it. Is that correct or not?
Now, what you have to do here, you have to draw the free-body diagram either the upper
part or the lower part. So I'm aggressive because we are out of time. I'm just saying
that this one is one is p. Yes or no? This one is q. This is your n. This is your shear.
Is that correct? You have a shear force in this cross section. You have normal forces.
There is cross section. Also, you have depending on this lpl. You also have a-- moment, so
you put your gauge here or here, or here, or here. That's much a big difference, depends
where you are. Is that correct? But you have to be able to come up with your state of stress.
That defines your strain. But if you have done it in the past, it should not be a problem.
But if you haven't done it in the past or you couldn't do it in the past, you have a
problem with the static which many of you did. Remember in this class especially-- yes?
You could not even bring those forces down here which is very simple as you know by now,
everybody knows that. However, I don't know which homework they gave you, they-- whether
they put it here or here. I just-- if you are here, that's sigma equal to q over area,
minus q over area, minus this is pl, everybody under-- minus m, y over i, et cetera. So what
I'm saying that, you may have a sigma y in the form of compression. If you are on this
site, compression to compression, if you are in this site, compression and tension. So
it-- let's go to this site, so you have definitely compression minus q over a, minus my over
I because we are not at the end, we are not here. We are somewhere near the centroid of
the section, depend on the center of that. You also have a shear stress like that. That
shear stress this time is not tc over j. It is vq over it, et cetera, et cetera. But if
you quite frankly look at it, this is exactly what you see there. Everybody understand that,
this is one stress and one shear. One normal stress and one shear stress. That's also is
one normal stress and shear. What is the x? It is y. And then the rest is the same. The
static of the problem changed, the strain part of it change. Let's get-- Give me five
more minutes because I did this for the other class. I need to mention it because this is
very interesting problem, sorry. I have to go somewhere too but I take five minutes of
your time because this is interesting problem. Go to problem number 3. So go to problem number--
sorry that I have to-- it is here. I put it here on the board too. I hope it is still
there. So go to problem number 3, give me another five minutes. If you want to leave,
you have something to do. But I suggest you stay because this is very interesting problem.
It's simple problem but at the same time, it's very interesting problem. You've got
there a bunch of problem coming like that. Because this is a pressurized tank, yes or
no, correct? Let's go quickly there. First of all, if it is a pressurized tank, what
is sigma 1? Remember the two sigma, we had sigma for sphere, sigma 1 was equal to sigma
2. One was longitudinal, the other one was over their circumference of that but this
is a sphere, this not a cylinder. In this one, x and y doesn't matter whether this is
x and y or this is x and y. They put it anywhere you want to. Is that correct or not? Because
this is spherical shear, it doesn't matter. But this was equal to pressure inside or inside
divided by 2t, yes or no? Correct? Now knowing that, which is from the beginning of the class,
that's the pressurized tank and we put here and didn't put a gauge, we put a line here
20 millimeter long, originally was 20 millimeter. After we load it, that increased by how much?
That increased by 0.012 millimeter increase. That's plus. So that's what we did. And the
outside diameter is-- the inside diameter, outside, the r inside is equal to 1000 millimeter.
The thickness of the tank is 10 millimeter. So the ratio is more than 10, the ratio is
100, so I can use the thin-walled theory which I have and that's what I'm using there. Is
that correct, but the ratio should have been larger than 10, the ratio is 100. Everybody
see that? Yes. So that works. You don't have to worry about this. So all of this is changing.
Now, the question is what's the value of the pressure inside? So I'll put this gauge, there
is a pressure inside. I don't know what. This gauge tell me what's the pressure inside.
And don't you want to know every time you are in there, let's say, we're living in tolerance
near the refinery, what is the pressure inside the tank all the time, what is related, yes.
So because the pressure can go up and down. What you see is through the strain gauges.
That's what I said. This problem is interesting. You don't have to put the strain gauge inside.
You can put it outside, yes or no?
Yes.
Yes? The question is if I have a reading of what it said that we have what, there is something
there. It says what? So let's read there, oh, that's it. This become like that. Is that
correct or no? Again, let's go quickly. So epsilon must be equal to delta over l. That's
the definition of epsilon, yes or no, right? The delta is-- How much this time? The delta
is 0.012 millimeter. Original length was 20 millimeter. So you divide that. That becomes
600 times 10 to the power of 6, or 600 micro plus. Why? Because it's increasing. Notice
of course everybody knew the tank is going to expand. Yes or no? The location, not important,
as I said because x and y doesn't matter here. Yes or no? I can put it anywhere. So this
is what we did. So here is your x. So this is the gauge epsilon x. This is a and that's
epsilon a, which is epsilon x. Is that correct or not? Yes? Now, what's the state of stress?
Again, I said everything depends on the-- now please, when you go to this homework,
start from this theta. Although we are talking about the strain but you should think of a
stress because that makes it simple a lot for you. What's the stress like on top of
this tank? Sigma 1 and?
Sigma 2
And both of them are?
Equal.
Equal. So that's the state of stress. This is the key to the problem, guys. Sigma 1,
going this way or it doesn't matter because this is a sphere. And sigma 2 and both of
them are equal to each other. Yes or no? Correct? Therefore, what's the epsilon x equal to?
Epsilon x equal to what? Sigma x over e? Yes, minus nu sigma y over e. Yes or no? But sigma
x equal to sigma, sigma 2 equal to sigma, the same sigma, yes or no, right? And that
is-- Let's put it that way. Let's go to that equation like what we get. That is equal to
p times ri, ri is-- well how much, 1000 divided by 2 times 10 millimeter because the thickness
is 10 millimeter. So it becomes 50p. So all I'm saying that sigma 1 equals to sigma 2,
this 50p, this is 50p, so all you have to do, e is given, nu is given, nu is given,
nu is not given or is given. I don't know which of us the nu, nu was given.
Point 3.
Yeah 0.3. So this was given. Sorry to keep you here, 0.3 is given. So you put it here
everything. So epsilon x is equal to epsilon x, we measure it, 600 times 10 to the power
of minus 6, equals to 50p, times y minus nu divided by e. Because sigma x is equal to
sigma y. That's what I did. Nu is given, e is given, you calculate the p. Is that correct
or not? Yes? The p become equal to, please write it down. The p become 3.43 megapascal.
After you calculate the p equal to [inaudible], you can calculate sigma 1 equals to sigma
2 equals to 50 times 3.43 megapascal, become 171 megapascal. So that's the stress of that.
It's all simple. Very simple. What is interesting that I kept you here, let's draw the circle
for the stress and for--
Strain.
Strain. So you see something that you haven't seen in the past. OK. This is for a stress
guide. For stress, this is sigma versus tau, with the arrow. Is this sigma x or is this
principal? Is this principal or not?
It is not a principal.
These are principal, because there is no shear or stress. Yes or no? So this one is how much?
171. Yes or no? The other one is how much? 171. Have you seen a circle with zero radius?
That is your first circle, because sigma 1 is equal to sigma 2. So your inclined shear
is zero. Is that correct because your circle is zero. Now what's out-of-plane shear? The
out-of-plane shear, what is sigma z? Is there a sigma z anything going this way?
No.
No. So sigma z equals to-- this is in other words sigma max, sigma minimum both of them
are 171. This sphere and that one, both principal, yes or no? Not if you have seen in that before.
But sigma z equal to zero because there is no sigma like that. So this is your tau max
x and z or y and z-- it doesn't matter, both of them applies, equal one half of sigma equal
to one half of 171. Now, let's go to a strain. Epsilon versus gamma-- oh wait, I want you
to see both of them. Now epsilon, what was epsilon equal to? The epsilon x was equal
to 600, yes or no? Yes, 600 micro. What is epsilon y? Does that make any difference?
It's the same thing. Epsilon y also is 600 micro, so both of them here, 600 and 600.
What is epsilon z? Is epsilon c zero?
No.
No. That equation I gave you. Do you remember that? Epsilon c equal-- because epsilon c,
this is the thickness of check-- tank changes, yes or no? Actually there was one homework
I asked you to calculate the changes, the thickness of that. And that one, epsilon c
equal to minus nu, 1 minus nu, epsilon a plus epsilon b. And since both of them are positive,
that one should be negative and you do that and you draw that last circle and that is
your maximum gamma. So that will be it. So--
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Message for your night 15 May - Duration: 0:54.
For more infomation >> Message for your night 15 May - Duration: 0:54. -------------------------------------------
Helps for Hosts: Be flexible! (Don Schnicker) - Duration: 0:59.
Hi, my name is Don Schnicker,
and I've been serving as a small group host at Saddleback Church.
I'd like to share some insights in the one area that
I think it's important as a small group host, and that is...
Make sure you maintain flexibility in your small group.
We want to obviously accomplish all the five purposes
within the group constraints,
but in any given time meeting to meeting
it's important that we focus on what's important that time.
Each small group kind of develops its own personality based on their members
and based on what's going on in their lives in any given time.
So we have to focus on if we need to be working on ministry,
or we need to be working on bible study.
We need to work on what's important for the group,
and make sure we keep everybody focused on individual things that time.
Key, Be flexible!
-------------------------------------------
WI woman catches monster fish, and then goes into labor on boat - Duration: 1:45.
For more infomation >> WI woman catches monster fish, and then goes into labor on boat - Duration: 1:45. -------------------------------------------
Video: Potential record warmth for the week - Duration: 3:09.
GOOD ADVICE ANYWAY.
ED: IF YOU LOOK BEHIND US,
THERE'S AN IMPLICATION THAT WE
ARE GOING TO GET WARMER.
YOU CAN SEE THE WARMER
TEMPERATURES TO THE WEST AND
THEIR CLEARLY HEADED HERE.
LOOK AT THE RAINFALL SINCE THE
RAIN BEGAN SATURDAY AND SATURDAY
NIGHT, IT'S BEEN RAINING WELL
INTO TODAY.
OVER 2.5 INCHES IN FAIRHAVEN,
TWO INCHES IN GARDNER.
MUCH OF THE REST OF THE REGION
FALLING BETWEEN ONE AND TWO
INCHES.
THAT HAS HELPED WITH THE POLLEN
COUNT BUT YOU CAN CLEARLY SEE
WHAT HAPPENS THE NEXT COUPLE OF
DAYS.
TREE POLLEN, MERTZ, -- MAPLE AND
BIRCH AND POPULAR.
WE LOST THE RAIN BUT NOT THE
CLOUDS YET.
WE STILL HAVE THAT STORM
SPINNING AROUND TO THE EAST BUT
AT LEAST THE DRIER AIR IS COMING
IN AT LOW LEVELS OF THE
ATMOSPHERE.
NOW WE ARE PRETTY MUCH LEFT WITH
THE CLOUDS.
THAT WILL GO DURING TONIGHT AS
WELL.
SO A WHOLE DIFFERENT WORLD
TOMORROW WHEN YOU HAD SUNSHINE
IN MUCH MILDER TEMPERATURES AND
DRY AIR.
OVERNIGHT MORE NIGHT, SOME HIGH,
THIN, CIRRUS CLOUDS, THROUGH AND
IN THE REALLY WARM IT WILL BE
HERE WEDNESDAY AND THURSDAY.
WE ARE TALKING 80'S TO AROUND 90
DEGREES.
HARD TO BELIEVE AFTER WHAT WE
JUST EXPERIENCED.
THIS IS SHOWING YOU SOME OF THE
STORM REPORTS, MOST OF THIS IS
TREE DAMAGE AND LIMBS COMING
DOWN.
WE STILL HAVE A WIND ADVISORY IN
EFFECT THAT INCLUDES THE
HARTFORD AND SPRINGFIELD AREA
UNTIL 7:00 TONIGHT.
THAT SEEMS TO BE WHERE THE
STRONGEST WINDS HAVE BEEN.
A LITTLE BRIGHTENING UP THE SKY
AND TEMPERATURES JUMPED UP A
LITTLE MORE.
THAT ALLOWED THE STORM WINDS TO
WORK DOWN TO THE GROUND.
AS WE GO THROUGH TONIGHT, THE
WINDS WILL EASE UP SOME.
THERE IS BOSTON, STILL KIND OF
OVERCAST BUT THAT REPRESENTS AN
IMPROVEMENT BECAUSE IT IS NOT
RAINING.
WE'VE SEEN THE MILDER AIR AND
ALBANY IS ONLY 70 DEGREES RIGHT
NOW.
ABOUT 10 DEGREES MILDER THAN
YESTERDAY AT THIS TIME.
TONIGHT THE LOWS WILL BE IN THE
40'S AND LOW 50'S AS THE SKIES
CLEAR IN THE WARMER AIR TO THE
WEST, 80 IN CINCINNATI AND 88 IN
ST. LOUIS.
THIS TIME, THIS IS THE ERA THAT
IS HEADED OUR WAY.
SO A BIG RECOVERY TOMORROW WITH
THE SUNSHINE AND A FEW
FAIRWEATHER CLOUDS WELL INTO THE
70'S FOR HIGHS.
CHECK IT OUT OVER THE NEXT SEVEN
DAYS.
YOU CAN SEE THE WARMING TREND
MIGHT BE A LITTLE CONSERVATIVE
ON WEDNESDAY.
I THINK WE WILL SEE MANY SPOTS
WELL INTO THE 80'S TO PERHAPS
CLOSE TO 90 ON THURSDAY.
THE COLD FRONT COMES THROUGH AND
IT DOES TURN COOLER AS WE HEAD
TOWARD THE WEEKEND BUT THAT IS
CERTAINLY A LOT DIFFERENT THAN
THE SEVEN DAY FORECAST I WAS
SHOWING YOU A WEEK AGO AT THIS
TIME.
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