Problem
North South University
Assignment Two
CEE 212: Solid Mechanics: Spring 2016
Calculate the reactions
Show the calculations
for summation of forces
to obtain Shear Force
Show the calculations
of summation of Shear
forces to obtain bending
moment
Com

Problem
North South University
Assignment One
Calculate the reactions
Consider a three force
member and take a
section anywhere across it
Calculate the Axial Force,
the Shear force and the
Bending moment at that
section
CEE 212: Solid Mechanics: Spring 20

Introductory Lecture
Axially loaded members
Axis of the bar
P
P
L
P
P
L+
Beams
Axis of the beam
P
P
L
Beams must resist loads applied laterally or transversely, i.e. perpendicular
to their axes.
Axis of the beam
Supports
Roller support
Movement along only

Chapter 9
Stress Transformation
y
xy
x
x
xy
State of stress in an element
y
Is this the maximum/failure stress condition?
Y
y
Y
x
xy
xy
x
xy
x
xy
xy
x
y
x
X
y
X
x
C
xy
x
xy
y
xy dA
C
x dA
x dAcos dAcos
xy
A
B
xy dAsin
Area BBCC dA
Area

Stress
Types of stresses
Axial stress
Shear Stress
Torsion stress
Bending stress
Examples of Axially loaded members
Tension member
Compression member
Tensile Stress
b
P
Stress =
Force/area=P/A
Note: Area is taken as area of contact between the internal pl

Chapter 6
Bending
y
y
a
b
x
c
z
d
a' b'
c'
d'
Basic assumption: Plane sections normal to the beam axis remains plane
after bending.
Length of a fiber at the axis remains the same, whereas ab is contracted
and cd is elongated.
O
y
d
M
a'
g
e
c'
M
b'
h
f
y

Chapter 7
Transverse Shear
P
P
P
a
a
P
P
SFD
Pa
P
+M+dM
+V
M + dM
M
+M
+V
dx
+M+dM
+V
M dM y
My
I
I
x
+M
+V
dx
dx
h
t
dx
a
FA
e
b
d
f
h
c
g
FA
area
abcd
FB
area
abcd
FB FA
My
M
dA
I
I
FB
ydA
area
abcd
M dM y
M dM
dA
I
I
ydA
area
abcd
How the equili

Beam Statics
What structural elements are called beams ?
Axis of the beam
P
P
L
Beams must resist loads applied laterally or transversely, i.e. perpendicular
to their axes.
Axis of the beam
Diagrammatic conventions
Beam is a one-dimensional structural ele

y, q
q (x) load per
unit length
+M+M
+V
x
a
+M
x
x
L
F
y
+V+V
0, ve V qx V V 0
V
q
x
x
M a 0, clockwise ve, M Vx M M qx 0
2
M
qx
V
x
2
Applying the limit, x 0,
dV
q
dx
dM
V
dx
dV d dM d 2 M
q
dx dx dx dx2
SF diagram by
Integration/summation
q x
V