Structural Member Properties
Moment of Inertia (I) is a mathematical property
of a cross section (measured in inches4) that gives
important information about how that crosssectional area is distributed about a centroidal axis.
Stiffness of an object related to its shape
In general, a higher Moment of Inertia produces a
greater resistance to deformation.
©iStockphoto.com
©iStockphoto.com
Moment of Inertia Principles
Why did beam B have greater deformation than
beam A?
Difference in Moment of Inertia due to the
orientation of the beam
Calculating Moment of Inertia - Rectangles
Calculating Moment of Inertia
Calculate beam A Moment of Inertia
=
 1 .5 in .   5 .5 in . 
3
12
 1 .5 in .   1 6 6 .3 7 5 in . 
3
=
=
12
249.5625 in.
12
= 2 0 .8 in .
4
4
Moment of Inertia – Composite
Shapes
Why are composite
shapes used in
structural design?
Beam Deflection
Measurement of deformation
– Importance of stiffness
– Change in vertical position
– Scalar value
– Deflection formulas
–
Beam Structure Examples
What Causes Deflection?
Snow
Live Load
Walls, Floors,
Materials, Structure
Dead Load
Roof Materials,
Structure
Dead Load
Occupants, Movable
Fixtures, Furniture
Live Load
Loading
Snow
Live Load
Walls, Floors,
Materials, Structure
Dead Load
Roof Materials,
Structure
Dead Load
Occupants, Movable
Fixtures, Furniture
Live Load
Types of Loads
Factors that Affect Bending
Material Property
– Physical Property
– Supports
–
Physical Property - Geometry
Beam Supports
Beam Deflections
Spring Board
Deflection
Bridge Deflection
Calculating Deflection on a
Spring Diving Board
Pine Diving Board Dimensions:
Base (B) = 12 in.
Height (H) = 2 in.
P
250 lb
72 in.
 Max ?
Known:
Pine (E) = 1.76 x 106 psi
Applied Load (P)= 250 lb
Deflection of Cantilever Beam
with Concentrated Load
 max = P x L3
3xExI
P
L
 max
Where:  max is the maximum deflection
P is the applied load
L is the length
E is the elastic modulus
I is the cross section moment of inertia
Moment of Inertia (MOI)
Moment of Inertia (I) is a mathematical
property of a cross section (measured in
inches4) that is concerned with a surface
area and how that area is distributed about
a centroidal axis.
Calculating Moment of Inertia (I)
I = (12 in.)(2 in.)3
12
I = (12 in.)(8 in.3)
12
I = 96 in.4
12
I = 8 in.4
Cantilever Beam Load Example
Known:
Pine (E) = 1.76 x 106 psi
Applied Load (P) = 250 lb
72 in.
 max =
P x L3
3xExI
 max =
(250 lb) (72 in.)3
(3) (1.76 x 106 psi) (8 in.4)
 max =
(250 lb) (373248 in.3)
(3) (1.76 x 106 psi) (8 in.4)
P
250 lb
 Max
Cantilever Beam Load Example




max = (9.3312 x 107 lb)(in.3)
(5.28 x 106 psi)(8 in.4)
max = (9.3312 x 107 lb)(in.3)
(4.224 x 107 psi)(in.4)
max = (9.3312 x 107)
(4.224 x 107 in.)
max = 2.21 inches
Calculating Deflection on a
Pine Beam in a Structure
Beam Dimensions:
Base (B) = 4 in.
Height (H) = 6 in.
Length (L) = 96 in.
L
P
 max
Known:
Pine (E) = 1.76x106 psi
Applied Load (P)= 200 lb
Deflection of Simply Supported
Beam with Concentrated Load
L
 max =
L3
Px
48 x E x I
P
 max
Note that the simply supported beam is pinned at one end.
A roller support is provided at the other end.
Where:  max is the maximum deflection
P is the applied load
L is the length
E is the elastic modulus
I is the cross section moment of inertia
Calculating Moment of Inertia (I)
I = (4 in.)(6 in.)3
12
I = (4 in.)(216 in.3)
12
I = 864 in.4
12
I = 72 in.4
Simply Supported Beam Example
Known:
Pine (E) = 1.76x106 psi
Applied Load (P) = 200 lb
96 in.
P
 max
 max =
P x L3
48 x E x I
 max =
(200 lb)(96 in.)3
(48)(1.76x106 psi)(72 in.4)
 max =
(200 lb)(884736 in.3)
(48)(1.76x106 psi)(72 in.4)
Simply Supported Beam Example




max = (1.769472 x 108 lb)(in.3)
(8.448 x 107 psi)(72 in.4)
max = (1.769472 x 108 lb)(in.3)
(6.08256 x 109 psi)(in.4)
max = (1.769472 x 108)
(6.08256 x 109 in.)
max = 0.029 inches