Uploaded by izzythehippogriff

Lecture 10 Notes

advertisement
ENGR-2530
Strength of Materials
Monday and Thursday
Lecture 10
Lecture Outline
•
•
•
•
•
Centroid.
Moments of Inertia.
Pure Bending.
Bending Deformations.
Bending Stress
2
CONCEPT OF CENTROID
The centroid, C, is a point defining the
geometric center of an object.
The centroid coincides with the center of
mass or the center of gravity only if the
material of the body is homogenous (density
or specific weight is constant throughout the
body).
If an object has an axis of symmetry, then
the centroid of object lies on that axis.
In some cases, the centroid may not be
located on the object.
First Moment of an Area
First moments of the area A with respect to x and y axes:
Centroid
The centroid of the area A is defined as the point C of coordinates
and which satisfy:
Centroids of common shapes
Determination of Centroid of a
Composite Area
Example
Determine the location of centroid:
Second Moment or Moment of Inertia
Moments of Inertia of the area A with respect to x and y axes:
Moments of Inertia of Common Shapes
Moment of Inertia of a Composite
Area
Parallel Axis Theorem:
Distance from x to x’
Example
Determine the moment of inertia Ix with respect to the
centroid x (horizontal) axis that passes through
centroid:
Moment of Inertia
For area 1 with respect to its centroid axis (x’):
For area 1 with respect to general centroid axis
(x):
Similarly for area 2:
Moment of Inertia
Example
Calculate Moment of Inertia : Vertical and
Horizontal Directions with respect to centroid
In-Class Assignment
Locate the centroid and Calculate the moment
of inertia around the horizontal (x) Axis
Summary
For Centroids:
For Moments of Inertia:
Polar Moment of Inertia
y
x
J = 12 π r 4
Pure Bending
Pure Bending:
Prismatic members
subjected to equal
and opposite
couples acting in
the same
longitudinal plane
Symmetric Member in Pure Bending
Equilibrium 
Internal forces in any cross section are
equivalent to a couple.
Bending moment = Moment of couple
Symmetric Member in Pure Bending
Fx = ∫σ x dA = 0
M y = ∫ zσ x dA = 0
M z = ∫ − yσ x dA = M
The only nonzero stress
component is σx
Bending Deformations
Beam with plane of symmetry in pure bending:
member remains symmetric
bends uniformly to form a circular arc
cross-sectional plane passes through arc
center and remains planar
length of top decreases and length of bottom
increases
a neutral surface (length does not change)
must exist: parallel to upper and lower
surfaces
stresses and strains:
- negative (compressive) above neutral
plane
- positive (tension) below it
34
Download