DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Midterm Review
EGR 140 ENGINEERING MECHANICS -- STATICS
Instructor: Dr. Yiheng Wang COPYRIGHT 2012 DANVILLE COMMUNITY COLLEGE
DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Chapter 2: Force Vectors
EGR 140 ENGINEERING MECHANICS -- STATICS
Instructor: Dr. Yiheng Wang COPYRIGHT 2012 DANVILLE COMMUNITY COLLEGE
Vector Algebra
3 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Cartesian Vector Representation:
z
y
x
j i
k
kjiA zyx AAA
A
Ayj
Axi
Azk
Vector Algebra
4 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Magnitude of a Cartesian Vector:
z
y
x
j i
k 22
yx AAA
A
Ayj
Axi
Azk
A
A'
22
zAAA
222
zyx AAAA
Vector Algebra
5 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Direction of a Cartesian Vector: Coordinate direction angles a, b and g.
z
y
x
A
Ayj
Axi
Azk
A'
a b
g
Vector Algebra
6 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Unit Vector:
z
y
x
A
Ayj
Axi
Azk
A'
a b
g
uA
kjiA
uA
A
A
A
A
A
A
zyxA
kjiu gba coscoscos A
1coscoscos 22 gba 2The magnitude of uA is 1,
kji
kji
uA
zyx
A
AAA
AAA
A
gba coscoscos
Addition of Cartesian Vectors
7 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
y
z
x
O
F1
F2 F4
F3
kjikjikjiBA
zzyyxx
zyxzyx
BABABA
BBBAAA
kjiFF zyxR FFF
Position Vector
8 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
z
y
x
r
yj
xi
zk P(x,y,z)
kjir zyx
Position Vector
9 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
z
y
x
r
B(xB,yB,zB) kjirrr
ABABAB
ABAB
zzyyxx ---
-
A(xA,yA,zA) rA
rB
kjirrr
BABABA
BABA
zzyyxx ---
-
22
ABABAB
ABABAB
zzyyxx
zzyyxxF
rFF
---
k-j-i-
ruF
2
Force Vector Directed Along a Line
10 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
B
A
F
r
u
Dot Product
11 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
q
A
B
zzyyxx BABABAAB qcosBA
1800cos 1
AB
BA
The magnitude of the projection of vector A along an axis aa (the direction of which is specified by unit vector ua) is:
aaA uA
DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Chapter 3: Equilibrium of a Particle
EGR 140 ENGINEERING MECHANICS -- STATICS
Instructor: Dr. Yiheng Wang COPYRIGHT 2012 DANVILLE COMMUNITY COLLEGE
Conditions for particle equilibrium
13 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
0FF R
The 0 here is a vector zero.
0
0
y
x
F
F
0
0
0
z
y
x
F
F
F
2-D problems: 3-D problems:
Springs and Cable-Pulley Systems
14 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Springs
Cable-Pulley Systems
ksF
Assume the cable is continuous and there is no friction, the cable is subjected to a constant tension, T, throughout its length, regardless what q is.
Positive s for pulling and negative s for pushing.
Free-Body-Diagram (FBD)
Step 1: Identify the problem
Step 2: Isolate the object
Step 3: Identify all the external forces acting on the object
Step 4: Determine the direction and magnitude of each force (in a chosen coordinate system)
15
What to include
The object of interest
All the external forces
Coordinate system
What to exclude
All external contacts and constrains
Forces the object exerts on other objects
Internal forces
16
Determine the force in each cable to support the 40-lb crate.
Example 3.7 (P 106 in the textbook):
1st step: Free-body diagram Connection at point A is isolated
from the environment
Appropriate coordinate system is defined
Applied force (W=40 lb) is demonstrated.
All unknown forces are demonstrated (FB, FC and FD).
2nd step: Express each force as magnitude
multiplied by the respective unit position vector.
kji
kjiruF 848.0424.0318.0
843
843
222
ABABB
BB
AB
BB FFr
FF
kji
kjiruF 848.0424.0318.0
843
843
222
ACACC
CC
AC
CC FFr
FF
iF DFD
)lb(40kW
3rd step: Use particle equilibrium condition.
0FF R
040848.0848.0
0424.0424.0
0318.0318.0
CBz
CBy
DCBx
FFF
FFF
FFFF
)lb(15
)lb(6.23
D
CB
F
FF
DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Chapter 4: Force System Resultants
EGR 140 ENGINEERING MECHANICS -- STATICS
Instructor: Dr. Yiheng Wang COPYRIGHT 2012 DANVILLE COMMUNITY COLLEGE
What is moment?
22 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Moment is a vector property that describes the rotational effect (or rotational tendency) about a point produced by a force.
Also known as torque or moment of force.
Components of moment
23 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Moment center: the point about which the rotational effect is produced.
Moment arm: the perpendicular distance from the moment center to the line of action of the force.
Magnitude: the quantified strength of the rotational effect.
Sense and Direction: defined by the moment axis, which is perpendicular to the plane that contains the force and the moment arm. The moment will cause a counterclockwise rotational effect about this axis.
FdMO
Moment center
Moment arm
Determine the sense and direction of moment: right-hand rule
24 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
M
F
r
FrM
O
Curl your right-hand fingers from the moment center to the direction of force, then your thumb will represent the moment axis. The rotation tendency will be counterclockwise about the moment axis.
How to calculate moment?
25 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Scalar formulation:
Vector formulation:
Principle of transmissibility:
qsin rFFdMO
FrM O
FrFrFrM 321OThis suggests that moment can be calculated using a position vector, r, from
moment center, O, to any arbitrary point on the line of action of the force, F.
Quick review of vector cross product
26 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
kji
kji
BA
xyyxxzzxyzzy
zyx
zyx
BABABABABABA
BBB
AAA
This is how you remember it:
Principles of moments
Moment is a vector, therefore it has all the properties of a vector.
27 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
21 FFrFrM O
332211 FrFrFr
FrM
OR
Moment of a force about a specified axis
28 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
zyx
zyx
aaa
aa
FFF
rrr
uuu
Mzyx
Fru
The result is a scalar (magnitude of the projected moment), and the projected moment vector is:
aaa M uM
Similar to what was introduced in Section 2.9, on how to calculate the magnitude of projected force along a specified axis.
Couple moment
29 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
A couple is defined as two parallel forces that have the same magnitude, but opposite directions, and are separated by a perpendicular distance d.
The moment produced by a couple is called a couple moment:
Scalar formulation:
Vector formulation:
FdM
FrM
F
F
r
The position vector, r, can be any arbitrary vector from the line of action of F to the line of action of F. (Again, principle of transmissibility.)
Couple moment
A couple moment is a free vector, i.e., it can act at any point since M depends only upon the position vector r directed between the forces.
In other words, the magnitude and direction of couple moment M will not change with different reference point of evaluation.
30 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
General rules of equivalent system replacement
31 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
MMM
FF
OOR
R
Resultant force from Section 2.6
Resultant moment All the moments produced by forces
All the original couple moments
Reduction of a simple distributed loading
32 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
A
A
L
L
RLO
R
ALR
dA
xdA
dxxw
dxxxwx
FxdxxxwM
AdAdxxwF
bxpxw
Pressure function
Reduction of a simple distributed loading
33 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
l
l
2
lx
x3
2lx
wlFR
lwFR max2
1
DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Chapter 5: Equilibrium of a Rigid Body
EGR 140 ENGINEERING MECHANICS -- STATICS
Instructor: Dr. Yiheng Wang COPYRIGHT 2012 DANVILLE COMMUNITY COLLEGE
Conditions for rigid body equilibrium
35 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
0MMM
0FF
OOR
R
The 0 here is a vector zero.
The resultant moment is zero with respect to any arbitrary point O.
0
0
0
O
y
x
M
F
F
0
0
0
0
0
0
z
y
x
z
y
x
M
M
M
F
F
F
2-D problems: 3-D problems:
Note: you may choose alternative sets of 3 independent equilibrium equations. (P 214 of textbook)
2-D Support reactions
36 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
2-D Support reactions (cont.)
37 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
2-D Support reactions (cont.)
38 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Note: for 2-D support reactions, only these three on this slide have more than one unknown.
3-D Support reactions
39 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
3-D Support reactions (cont.)
40 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
3-D Support reactions (cont.)
41 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
3-D Support reactions (cont.)
42 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Procedure to solve rigid-body equilibrium problems
1. Draw a free-body diagram. 2. Demonstrate all the applied forces and couple
moments. 3. Identify all the support reactions.
a. If a support prevents translational effect in a direction exerts a force in that direction.
b. If a support prevents rotational effect about an axis exerts a couple moment about that axis.
4. Set up the scalar equilibrium equations (overall 3 equations for 2-D problems and 6 for 3-D problems), and solve for unknowns. You can definitely also solve the problems in vector format using the 2 vector formula equilibrium equations if you prefer so.
43 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
Two-Force Members
44 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
A two-force member has forces applied at only two points on the member.
For equilibrium, the two forces must have the same magnitude, act in opposite directions, and have the same line of action.
The line of action of the two forces must pass through the two points where the forces act on.
Three-Force Members
45 DCC EGR 140 ENGINEERING MECHANICS -- STATICS by Dr. Yiheng Wang
A three-force member has forces applied at only three points on the member.
For equilibrium, the three forces must be either concurrent, or all parallel to each other.