Hello, and welcome back, since
we have estimated.
the moment of inertia about the X.
in the last lecture.
we will be able to estimate a the radius of gyration .
By saying that,
Kx=sqrt of Ix /A .
our
A=b*h for the rectangle, where b is a breadth, h
is height, the moment of inertia about the cg is
b*h^3/12.
So Kx =
We open one bracket.
We are writing (b*h^3/12)
all divided by b*h, which is the area inside the bracket
raised to the power of 1/2.
b will go with b, h will be go with h.
So we are left with h^2/12
Raised to the power of 1/2.
Which k x = h/
2 *SQRT of 3.
Next step is how to estimate.
The moment of inertia.
About.
y
y c g.
The moment of inertia,
for the rectangular section.
at y axis passing by the cg.
By the first method, which is parallel axes theory.
We are going to introduce our strip again, we have
the external two axes x and y.
And we are.
We have X G and Y G, passing by the cg.
Which is apart by b/2.
and
h/2
Above the corner.
for the area of the strip, the width is dx and the height= h.
The moment of inertia about the y,
which is the external.
vertical
axis.
=the integration.
From x=0 , the strip as if it started from x=0.
It will end at x =b.
But this time, we are going to multiply the area of the d A*x^2.
and since the d A = d x *h.
inside  integration sign, we are going to write the d x*h*x^2.
h, being constant,
then it will come out
We will integrate from x=0  to x=b.
of x^2*d x, which will  give us (x^3/3)*h
But we are going to say , x start point
x=0 and end point =b
Iy=h*1/3.
*(b^3-0)
At the end, it we give us ,h*b^3/3
If we are going
If you want to estimate
The moment of inertia about XG.
Then by using,
the parallel axes theory, we're going to deduct
the previous value.
We will deduct the from it,  the multiplication of the area (A*
*x bar^2)
x bar^2, where x bar =b/2.
IyG=h*b^3/3.
(-) the area, (b*h.
*(b/2)^2.
At the end, we have (h*b^3/12), since
We have estimated that IyG.
we can get,
the radius of gyration.
but first, as we have done
in the previous Ix , we're going to use the second method.
Second method is directly estimate the moment of inertia about
yG
our strip, this time,
will be apart by x, horizontal distance,
the same strip width=dx, and the height
but our x value,for starting  the integration from (-b/2) to (+b/2).
then Iy G
at the cg
center of gravity
= the integration from (-b/2)
to (+b/2)
of (dA) which is
=d x*h
multiplied by
x
raised
to the power of 2
x^2
h being constant, it will be out.
then, we have the integration ...
(x^2*dx), which will give us x^3/3
inside bracket
and our starting point from x=-b/2, to x=+b/2
then Iyg=h*
h/3 will come out
for the upper bound is
x^3=(b/2)^3
which will give us b^3*
1/8 *1/3 =1/24
if we get 3 out, then we have new bracket(b^3/8-
and inside another small bracket of
(-b^3/8)
(-)*(-)=(+)
So
We have
h/3*(2b^3)*
1/8
which is
h*b^3/12, the same result ,
which we have obtained from
the parallel axes theory
We move to the radius of gyration.
in the y axis.
kgy
=sqrt
of IyG / A
Iyg=b*h^3/12
and A=b*h as numerator
all under the sqrt
So this
h will go with h , one b goes with one b,
we are left with b^2
/12 under the square root.
Then we have b/2
multiplied by
sqrt of 3
Thanks a lot, our next,
topic
will be
the product for
the rectangular section, thanks a lot and
Goodbye.
