Unit 2 R
Review
Stat
tics
Statics Prin
The laws of motion de
interaction of forces
–Newton’s First Law o
An object in a state of rest or u
be so unless acted upon by ano
–Newton’s Second Law
Force = Mass x Acceleration
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nciples
escribe the
acting on a body
of Motion (law of inertia):
uniform motion will continue to
other force.
w of Motion:
Statics Prin
Newton’s Third Law of
For every action force
and opposite reaction
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nciples
Motion:
e, there is an equal
n force.
Equilibri
Static Equilibrium:
A condition where ther
forces acting upon a pa
the body remains at res
constant velocity
SUM OF ALL FORCES
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ium
re are no net external
article or rigid body and
st or continues at a
EQUALS ZERO
Structural Mem
• Centroid: center of gravity
is in state of equilibrium i
centroid
• Moment of Inertia: Stiffne
its shape. a higher Mome
greater resistance to defo
• Modulus of Elasticity
Ratio of stress to strain. Inh
mber Properties
y or center of mass. Object
if balanced along its
ess of an object related to
ent of Inertia produces a
ormation.
herent to the material.
Right Triang
SOHCAHTOA
Be able to use Right triangle properties or Pyth
gle Review
Sin q = O/H
Cos q = A/H
Tan q = O/A
hagorean’s Theorem to solve for a hypotenuse
Vectors: have mag
y - axis
100 lbs
30
x - axis
The vector has a magnitude of 1
lbs, a direction of 30 degrees CC
from the positive x axis. Its sens
up and to the right.
gnitude, direction and sense
Fx = F * cos q
Fx = 100lbs * cos 30
Fx = 87 lbs
100 Fy = F * sin q
CW Fy = 100lbs * sin 30
se is Fy = 50 lbs
Forces in Tensio
and Compressio
A force is a pus
object on anoth
A tensile force expan
lengthens the object
on.
A comp
compre
the obj
on
on
sh or pull exerted by one
her.
nds or
t it is acting
pressive force
esses or shortens
ject it is acting on.
A moment of a force is a measu
body to rotate about a point or
It is the same as torque.
A moment (M) is calculated usin
Moment = Force * Distan
M = F* D
ure of its tendency to cause a
axis.
ng the formula:
Always use the
perpendicular
nce distance
between the force
and the point!
Typically it is assumed:
•A moment with a tendency
(CCW) is considered to be
•A moment with a tendency
considered to be a negative
y to rotate counter clockwise
a positive moment.
y to rotate clockwise (CW) is
e moment.
FBDs are used to illustrate and
given body.
Roller:
Pin
Connection:
Fixed
Support:
calculate forces acting upon a
Fy
Fy Fx
Mo Fx
Fy
Draw a FBD of th
AB
C E
D
Pin-Connected Pratt Through
Truss Bridge
he pin at point A:
TAB
TAC TAD TAE
Free Body Diagram of pin A
(If you consider the third dimension, then
there is an additional force acting on point A
into the paper: The force of the beam that
connects the front of the bridge to the back of
the bridge.)
Steps for finding
1. Draw a FBD of the ent
2. FX = 0
3. FY = 0
4. M = 0
5. Use the above equatio
forces (substitute bac
6. Redraw the FBD with
Reaction Forces
tire system
You may need to sum
moments about more
than 1 point
ons to solve for reaction
ck into 2 or 3)
reaction forces
Step 1: FBD of syst
A
Ay
tem Each block is 1’ by 1’
B
CC
Cy
Step 2: Sum Forces
A
s in X direction = to zero
50 lb + Cx = 0
B
Cx = -50 lbs
Therefore, Cx = 50 lb
pointing left, not
right
CC
Cy
Step 3: Sum Forces
A
s in Y direction = to zero
-100 lb + Ay + Cy = 0
B I can not solve this
further
CC
Cy
Step 4: Sum Mome
Each block is 1’ by 1’
A
ents = to zero
Sum mom. about C = 0
B -Ay*6’ + -50lb*5’ +
100lb * 4’ = 0
-6 Ay +150 = 0
Ay = 25 lb
CC
Cy
Step 5: Use other equa
A
ations to find unknowns
-100 lb + Ay + Cy = 0
B -100 + 25 lb + Cy = 0
Cy = 75 lb
CC
Cy
Step 6: Redraw FBD
A
B
C 50 lb
75 lb
Truss R
Review
Steps for findin
1. Solve for Reaction force
a. Draw a FBD of the entir
b. FX = 0; FY = 0; M
c. Use the above equation
2. FBD of each joint (use
3. FX = 0; FY = 0 at eac
4. Solve for forces
5. Draw final FBD
ng Truss Forces
es
re system
M=0
ns to solve for reaction forces
vector properties)
ch joint
Truss Example
A
Ay
B
CC
Cy
Truss FBD with solved
A
Reaction Forces
B
C 50 lb
75 lb
Steps for findin
1. Solve for Reaction force
a. Draw a FBD of the entir
b. FX = 0; FY = 0; M
c. Use the above equation
2. FBD of each joint (use
3. FX = 0; FY = 0 at eac
4. Solve for forces
5. Draw final FBD
ng Truss Forces
es
re system
M=0
ns to solve for reaction forces
vector properties)
ch joint