egr 140 summer 2012 midterm review.pdf

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


Chapter 2: Force Vectors



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


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


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


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


y

z

x

O

F1

F2 F4

F3

kjikjikjiBA

zzyyxx

zyxzyx

BABABA

BBBAAA

kjiFF zyxR FFF

Position Vector


z

y

x

r

yj

xi

zk P(x,y,z)

kjir zyx

Position Vector


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


B

A

F

r

u

Dot Product


q

A

B

zzyyxx BABABAAB qcosBA

1800cos 1

qq

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


Chapter 3: Equilibrium of a Particle



Conditions for particle equilibrium


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


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


Chapter 4: Force System Resultants



What is moment?


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


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


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?


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


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.


21 FFrFrM O

332211 FrFrFr

FrM

OR

Moment of a force about a specified axis


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


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.


General rules of equivalent system replacement


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


A

A

L

L

RLO

R

ALR

dA

xdA

dxxw

dxxxwx

FxdxxxwM

AdAdxxwF

bxpxw

Pressure function

Reduction of a simple distributed loading


l

l

2

lx

x3

2lx

wlFR

lwFR max2

1


Chapter 5: Equilibrium of a Rigid Body



Conditions for rigid body equilibrium


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


2-D Support reactions (cont.)




Note: for 2-D support reactions, only these three on this slide have more than one unknown.

3-D Support reactions


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.


Two-Force Members


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


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.

egr 140 summer 2012 midterm review.pdf

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