Unit 2 Review - cusd80.com

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

©iStockphoto.com

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

©iStockphoto.com

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

©iStockphoto.com

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