This document discusses moment of inertia, which is a property of an object's shape that represents its resistance to changes in motion. It defines moment of inertia as the second moment of a force or area. The document explains how to calculate the moment of inertia of different beam sections using thin strips and the parallel axis theorem. It also discusses finding the neutral axis, which is the axis where compressive and tensile forces balance. Sample problems are provided to demonstrate calculating moment of inertia for different laminate shapes.

1. STRUCTURAL DESIGN I
STRUCTURAL MECHANICS
MOMENT OF INERTIA
LECTURE 10

2. MOMENT OF INERTIA
• Inertia is the resistance of any physical object to a change in
its state of motion or rest.
• It is represented numerically by an object's mass.
• Moment of Inertia is also known as the second moment of a
force or an area.
• (the first moment is the product of force and the arm)

3. The figure shows beam sections divided up into a very large number of thin
strips parallel to the neutral axis. If we multiply each strip area by the square
of its distance from the NA and sum up all the quantities obtained, we will
obtain the value of INA for the given beam section.
INA = (a1 x y1²) + (a2 x y2²) + (a3 x y3²) +……
It will be observed that the value of the moment of inertia has nothing to do
with the material of the beam but is only a property of its shape.
.

4. The value of MoI of certain sections are:

5. PARALLEL AXIS THEOREM
This theorem states that “ The Moment of Inertia of a lamina about any axis in
the plane of the lamina equals the sum of the following:
i) Moment of Inertia about a centroidal axis parallel to the axis about which
moment of inertia is to be calculated
ii) Product of the area of the laminate and the square of the distance
between the above centroidal axis and the axis about which the moment
of inertia is to be calculated.”

6. b
dX X’
b
d
X X’’
I about XX’ = bd³/12
I about XX’’ = IXX’ + Ay²
= bd³/12 + (bd×(d/2)²)

7. POSITION OF NEUTRAL AXIS
We found that the strip load was given by the expression E x ay
R
As long as y is measured downward all the strip loads will represent tension.
If we put y negative i.e. measured the distance from NA , the load would be
compression.
∑E x ay or E x ∑ay (since E and R are constants) will therefore represent a
R R
summation of a large number of positive and negative quantities. But as the total
compressive force is equal to the total tensile force as C and T forma couple
E ∑ay = 0.
R
∑ay = 0 means that the axis , from which y is measured , passes through the
center of gravity of the section.
The neutral axis of a beam section therefore passes through its center of gravity.

8. The figure shows
the position of
neutral axis of
some sections.

9. X’X
d
c
b
a
Problem: To find the Moment of Inertia of the lamina shown above about XX’axis .
By parallel axis theorem:
IXX” = (IA + (ab × (b/2)²) + (IB + (dc × (c/2)²) where IA = ab³/12 and IB = dc³/12
A
B
Find IXX’ if a = 3, b = 8, c = 2 and d = 8
While the MoI or I of standard sections are known, I of differently shaped
sections can be computed by the same theory

10. An I section lamina has top flange 8cm x 2cm, Bottom flange 12 cm x 4 cm and web
3cm x 16 cm. Calculate the centroidal moment of inertia of the lamina.
To solve this question first the CG has to be located.
Then the I of each part around the CG has to be added
Let CG lie at a distance y from bottom of bottom flange

12. A
C
B
Finding CG
(8×2×21)+(16×3×12)+(12×4×2)=y×(16+48+48)
Y = (336+576+96)÷112=1008÷112=9 cm
Ixx = ((8×2×2×2/12)+(8×2×(21-9)²)) + ((3×16×16×16/12) + (3×16 ×(12-9)²))
+((12×4×4×4/12)+(12×4×(9-2)²))
=(5.33+2304)+(1024+432)+(64+2352)=2309.33+1456+2416=6181.33cm⁴
XX