Expression to find Shear Center Video Lecture - Civil Engineering (CE)

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so
okay this is geometric property first
point to notice shear Center is a
geometric property independent of
loading
magnitude
okay so let's say are some arbitrary
cross section okay now in this cross
section that will be both the share
Sigma X Y and Sigma XZ sheis is coming
in this cross section because this is
inclined and is thin-wall that would be
a shear flow like this in this cross
section okay if you recollect in a
previous lecture when he started bending
moment we define the torsional moment T
as integral Sigma XZ y minus Sigma XY Z
times D ax X right okay this was our
definition of torsional moment about a
Torsen origin now I want to take a
torsional moment about the shear centre
at shear Center okay not about some
origin but I want to take the torsional
moment about shear Center the original
coordinate system was y&z like this okay
so now I have to move this by an amount
is it a C and by an amount Y yes see
okay there is a shear Center with
respect to the origin chosen to find
these shear stresses okay then what
happens okay so now what happens when I
incorporate this when I shift the origin
my Y star I had to compute the torsional
moment about T star or let me define a
start coordinate system via
this Y star Z star okay Y star is
related to y as y + y SC y - y SC okay
now similarly Z star is related through
Z - Z - Z SC right okay so now the
torsional moment about T star which is
integral Sigma XZ y star minus Sigma X Y
Z star DAX ax okay it would be integral
Sigma XZ y - y SC - Sigma X Y Z - Z SC
into da X X this I can rewrite it as
integral Sigma X C Y minus Sigma X Y Z
da x - y SC integral Sigma X C D ax ax
ax e ax plus Z SC integral Sigma X Y DX
X Y okay now what happens this is a
initial torsional moment okay - why yes
C times this is nothing but your shear
force V Z there's nothing but the shear
force V Z plus Z plus C times a shear
force V Y and what is a definition of
shear Center the torsional moment of or
a shear Center is zero so this has to be
zero
okay now let's consider a case Byron
if we Z is zero the case that we have
been looking at all along then Z SC is
gone by minus T divided by V Y okay
similarly if V Y is zero that is winning
over the other direction alone which we
will see subsequently ysc would be given
by T divided by V Z okay
here the partial duration of T is
anti-clockwise moment is positive okay
only then from right hand thumb rule
will get the XS the positive x-direction
okay so that's how you are minus T by V
Y okay I see I see okay
so this is how you find this shear
Center okay so basically you find it
torsional moment due to these stresses
now let's ponder a little bit more on
why Sigma XY does not produce any moment
in this case in inter after moment in
this case okay we found that for the
channel section Sigma XY was for channel
section
Sigma X Y was of the form some a naught
plus a 1 Y square R minus a 1 Y square
right and write that as such if you go
back here you will find that if you go
back here all this is some constant
times constant minus some constant times
y square Y is a variable about which
this shear stress varies ok so basically
this how Sigma X Y varies whereas Sigma
X Z varies linearly with the independent
variable which is yes in this case ok so
coming back here I am write this as some
function of minus a 1 times y square
where a 0 and a 1 are independent of
independent of y and z ok the coordinate
positions okay now what am i interested
in I am interested in finding integral
Sigma XY into Z ba x okay now for the
transaction this would be integral a
naught minus a1 Y square dy okay it
varies from minus H by 2 to H way to
okay into Z which varies from which is
DZ which way is from minus bf plus z cg
to is at cg okay
you
this is y this is Z okay Y varies from H
by 2 to H by 2 and s it varies from this
is the CG as a CG or a cross-section it
varies from this is TW the thickness of
the base TW to the centerline of the web
base at z cg it will vary from said cg
minus TW by 2 2 is at cg + TW by 2 okay
so now what happens
what is if I integrate this I will get
it as a 0 into H minus a 1 by 3
into 2 into H cube by 8 right okay that
is this part and the Z part will be
let's see G minus T plus T W by 2 whole
square minus is at CG minus TW by 2
the old squaw by 2 okay so this will be
a naught H minus a 1 H cube by 12 in 2
by 2 ay plus B would be TW a minus TW 2
2 times is a CG ok so that will be the
moment produced due to Sigma X Z into Z
DX
ok now integral Sigma XZ in to Y DX has
to be computed ax Sigma XZ is a linear
function in yes
which will be this I can write it as
some B naught into yes where an S is
measured from yeah that is yes so how is
that related to use it
that will be BF s would be would be
minus bf plus z cg + z ok ok so that
will be B naught into some constant
let's say C naught plus is
that dy DZ dy DZ we're in this Y years
from minus H by 2 plus TF by minus TF by
2 to H by 2 plus T of by 2 and Z is from
minus C naught to minus C naught 2 is
HCG okay so if I do is computation what
I will get this get it as B naught into
TF into c naught into z cg + c naught
plus Z square by 2 which will be is at C
d square minus C naught square by 2 will
be Sigma XZ into Y ok so now I have to
add these to get the net torsional
moment this will be integral Sigma XY
into Z ax minus integral Sigma X Z into
y be ax ok it is superb used to get the
net torsional moment okay that I have to
divide by the resonate torsional moment
so from here I get Z test C as T divided
by B Y ok this expression will include
will be in fact is at cg + e but e is
same as what we had computed before ok
so you run through these calculations to
get through up to this step okay now the
final thing I want to show you is this
integral Sigma XY DX is equal to V Y ok
I tried it for rectangular section and
you can do it for other sections ok for
rectangular section you found that Sigma
XY is V Y by B H cube 12 into H by 2
whole square minus y square divided by 2
okay so now let's integrate this now V Y
has to be integral Sigma XY da x ax
so this will be 12 V Y by 2 bh cube
integrated over
h square by 4 minus y square dy divided
from minus H by 2 to H by 2 the center
thing integrated over DZ from minus B by
2 to B by 2 because I am looking at a
rectangular cross section with this as Y
and this as Z ok so this is a
rectangular cross section of C ger Pyron
resistances H by 2 and this distance is
H by 2 and this width this B by 2 B by 2
each okay so this will be the 6 B Y by B
H cube if I integrate D Z I get B into H
square by 4 into H minus y cube by 3 so
it is 1 by 3 into H cube by 8 into 2
okay so what is this VB cancels HQ H
cubed cancels the 6 into 1 by 4 minus 1
by 12 right okay so this is nothing but
dy by 2 good by 12 into 2 so that will
be equal to V Y ok so basically that's
why integral Sigma XY da is equal to V Y
ok so we'll stop here for today's
lecture thank you
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