ME 108 Statics - Gaziantep Üniversitesierklig/me108/0_lecture1.pdfUnits used in Engineering Statics – SI Units Dimension Unit SI Symbol Length (l) meter m Mass (m) kilogram kg Time

ME 108 Statics

Ahmet Erkliğ [email protected]

Office No: 315 4th floor of Mechanical Engineering Departmen

What is Engineering Mechanics? Engineering Mechanics is a discipline that studies the response of

solids, structures, fluids and devices made of these materials to

mechanical, thermal, and other types of load events. It emphasizes a

basic science approach that draws heavily on physics, mathematics

and computer science.

Mechanics is crucial to addressing modern problems in:

• aerospace engineering • mechanical engineering

• civil engineering • nuclear engineering

• materials science • geological engineering

• agricultural engineering • biomedical engineering

• electrical engineering (electronics packaging)

• chemical engineering (polymeric material modeling)

Particles vs Objects

• Particles:

–Point mass

–No geometry

–Rotation is not important

• Objects

–Contain mass

–Have geometry

–Rotation is important

The basic principle we will use is Newton’s 2nd law

of motion

We will spend considerable time becoming

proficient in applying this law to engineering

problems.

What is Statics?

Statics is the study of particle and rigid body structure

equilibrium. Static means unchanging position in time.

amF

The Basic Concept

If an object is at rest (or moving with a constant velocity), we can say quite a lot about the forces acting on it – and within it!

Importance of Statics

What You Will Learn in Statics?

• How to determine the resultant of multiple forces and distributed forces.

• How to determine the moment of forces about points and axes.

• How to draw “free body diagrams” and identify all of the forces acting on a body.

• How to use “equilibrium equations” to determine unknown or unspecified forces and moments acting on (and within) bodies, including structures and machines.

• How to identify and quantify friction forces. • How to determine centroids, centers of mass/gravity, and

moments of inertia for shapes and objects.

And what will you do with these competencies???

• Use information about forces and moments to analyze and design parts, assemblies, mechanisms, and structures that are appropriate to their intended (and sometimes unintended) functions.

• Use information about centroids, centers of mass/gravity, and moments of inertia in support of the above.

Why is Statics Important?

• ME 108 is the fundamental building block for all of engineering courses.

• Good habits developed in Statics will help you in subsequent courses.

• Performance in Statics is factored into your “Math/Science GPA” for admission to a department

ME 208 dynamics ME 223, 307-308

advanced dynamics

robotics

satellite mechanics

vibrations

advanced mech of

materials

stress analysis

structural engineering

machine design

finite element analysis

others

others

ME 108 Statics ME 223

mech of materials 1

Tips for Success in Statics (i.e., what is expected!)

• Do all assigned reading – before class.

• Solve extra problems

• Don’t spend more than 20 minutes/problem. Get help!

• Attend lecture regularly.

• Ask questions!!!

Courtesy During Lectures

• Arrive for lecture on time.

• No sleeping, reading newspapers, conversations, etc. during lecture. I expect your attention to be focused on statics for 50 minutes.

• Please wait until class is dismissed before leaving.

Text Book

Text Book: Engineering Mechanics Vol-1;Statics

By J. L. Meriam and L. G. Kraige

Auxiliary Book: Vector Mechanics for Engineers; Statics

By Ferdinand P. Beer and E. R. Johnston

Engineering Mechanics – Statics

By R.C. Hibbeler

Many books related with the Statics in the Library

Grading

2 midterm 2 x 20%

Homeworks+Quiz 20%

final 40%

100%

Exams

1st midterm 26-07-2011

2nd midterm 09-08-2011

final ??-08-2011

Course Overview

• Basic principles

• Force systems

• Equilibrium

• Structural analysis and machines

• Centroids, distributed load systems

• Beams

• Area and mass moments of inertia

Fundamentals, units, calculations

• Units

– Length – need to know position and geometry of objects

– Time – need to determine succession of events

– Mass – related to amount of stuff in a body, found using gravitational attraction

– Weight – force due to gravity acting on a mass, W=mg, where g=9.8m/s2

Fundamentals, units, calculations

• Basic Quantities

– Force – push or pull on a body, can be direct (contact) or indirect (no contact)

–Moment – turning effect caused by a force applied at some distance away from the axis of rotation

Fundamentals, units, calculations

Engineering Concepts

– Idealizations – all real problems are simplified

to some degree

– Particle – mass acting is if it were concentrated

at a singe point

– Rigid Body – particle collection in a shape that

doesn’t change with applied force

– Concentrated Force – force acting as if it were

at a single point

1st law: A particle remains at rest, or continues to move in a

straight line with uniform velocity, if there is no unbalanced

force acting on it.

2nd law: The acceleration of a particle is proportional to the

resultant force acting on the particle, and is in the direction of

this force.

3rd law: The forces of action and reaction between interacting

bodies are equal in magnitude, opposite in direction, and co-

linear.

Newton’s laws of motion

case staticfor 0

amF

amF

Newton’s law of gravitational attraction

Consider two particles of mass m1 and m2. Newton stated:

F G

m1m2

r 2

F = force of attraction between the two particles.

G = universal gravitational constant, experimentally found

to be = 66.73 (10-12) m3/(kg-s2).

r = distance between particles.

weight of a particle

Consider a particle with mass m1=m, and let the earth’s mass be

denoted by m2. Letting W=F where we call W the weight, we

can write

W=mg where g = acceleration due to gravity = Gm2/r2

Observation: We commonly think of g as a constant. However, in

our application of F=Gm1m2/r2, the earth is not a particle, it has

nonuniform density, and r depends on position. Nonetheless,

commonly accepted values for g are:

g = 32.2 ft/s2 = 386.4 in/s2

= 9.81 m/s2

Units used in Engineering Statics – SI Units Dimension Unit SI Symbol Length (l) meter m Mass (m) kilogram kg Time (t) second s Derived Unit Basic Units Symbol Area (A) m2 - Volume (V) m3 - Density (ρ) kg/m3 - Velocity (v) m/s - Acceleration (a) m/s2 - Force (F) kgm/s2 N Energy (E, U, K, etc.) kgm2/s2 J

Be consistent when using units!

Units System length mass force time

English foot, ft slug==lb-sec 2/ft pound, lb second, s

metric (SI) meter, m kilogram, kg Newton==kg-m/s2, N second, s

Remark: the relationship among these units is governed by

f=ma. By selecting 3 fundamental units, a 4th “derived” unit can

be obtained. The derived units are:

English system:

metric (SI):

1 slug f

a

1 lb

1 ft/s2

or...1 lb (1 slug)(1 ft/s2 )

1 Newton ma 1 kg 1 m/s2

Suggestion: Avoid

slugs and simply

work with

fundamental units.

Unit Conversions

If p=expression, to change the units of p, multiply the

r.h.s. by dimensionless factors of unity until the desired units

are obtained.

e.g., Convert the velocity v = 10 m/s to English units.

Since 1 ft 0.3048 m, 1 ft

0.3048 m

0.3048 m

0.3048 m1

v 10 m

s1 10

m

s

1 ft

0.3048 m1

32.808 ft/s

Unit Conversions Length Mass

1 km = 0.62137 mi. 1 kg = 2.2 lbs

1 mile = 5,280 ft = 1.6093 km 1 lb = 453.59 g = 16 oz.

1 meter = 1.0936 yds 1 metric ton = 1000kg

1 inch = 2.54 cm (exact) 1 slug = 32 lb = 14.594 kg

Volume Pressure

1 mL = 1 cm3 1 atm = 760 torr (mm Hg)

1 L = 1 dm3 = 1.0567 qts. 1 atm = 14.70 lb/in2

1 gal = 4 qts = 8 pts = 3.7854 L 1 bar = 100,000 Pa

1 in3 = 16.4 cm3 1atm = 101325 Pa

1 qt = 0.946 L

Force Energy

1N = 0.2248lbf 1 J = 1 kg-m2/s2

1N = 2.248x10-4 kip 1 cal. = 4.184 J

1 kip = 1000lbf

1N = 7.233 poundal (lbm ft/s2)

Useful to reduce numbers to manageable size,

especially SI units.

factor prefix symbol

109

giga

G

106 mega M

103 kilo k

10-3 milli m

10-6 micro

10-9 nano n

Prefixes

Numerical calculations Dimensional Homogeneity:

In addition to being numerically correct, an equation must

also be dimensionally correct. Always carry units along with

your calculations.

Accuracy:

You cannot create accuracy with your calculations.

e.g., 12.34 * 12.3 = 151.782

In lengthy calculations, retain one or two extra significant

digits*. Your final results should be reported with the same

accuracy as your data (in this case ... 152.)

* Occasional exceptions to this.

example 1

• convert 175 lb/ft3 to SI units:

• convert 6 ft/hr to SI units:

• convert 1.13 kN-m to English units:

A:

a) 27.5 kN/m3

b) 0.508 mm/s

c) 833 ft-lb

example 2

Two steel spheres with 100 mm diameter are placed

beside one another at the earth’s surface. Compute the

gravitational force W (weight) and the force of mutual

attraction F in Newtons.

The unit weight of steel is

= 0.284 lb/in3

A: W = 40.3 N, F = 1.13 (10-7) N

Problem Solving Strategy 1. Interpret: Read carefully and determine what is

given and what is to be found/ delivered. Ask, if not clear. If necessary, make assumptions and indicate them.

2. Plan: Think about major steps (or a road map) that you will take to solve a given problem. Think of alternative/creative solutions and choose the best one.

3. Execute: Carry out your steps. Use appropriate diagrams and equations. Estimate your answers. Avoid simple calculation mistakes. Reflect on / revise your work.

FORCE SYSTEMS - Force

Definitions:

A scalar is a quantity that is completely characterized by a

single number. Mass, volume, and length are examples of

scalars.

A vector is a quantity that has both magnitude (or size) and

direction. Velocity, force (e.g., weight) and position are

examples of vectors.

Common notation for vector quantities:

F or F

Vector Notation To report a vector, its magnitude and direction need to be

stated. We will discuss a few convenient methods for doing

this including:

• Simple statement of magnitude and direction relative to

some arbitrary reference direction (effective for 2-D).

e.g.,

• Statement in terms of rectangular Cartesian components

(effective in both 2-D and 3-D ... more on this shortly)

Vector Nomenclature:

30°

Line of

Action

Head

or Tip

Tail

Basic Vector Operations

There is more than one way of adding vectors:

• parallelogram addition

• head-to-tail: note that this can be done in any order

Addition:

Multiplication: • multiplication of a vector by a scalar changes its

magnitude but NOT its direction.

Subtraction:

• equivalent to multiplying the vector to be subtracted

by -1 and adding it to the vector from which it is

being subtracted.

Summary:

A

B

R = A+B

R = A+B

R = B+A

A

A

A

B

B

B

Scalar multiplication: s(A + B) = sA + sB, s is a scalar

Vector addition is commutative: A + B = B + A

Resolution of a vector into components

Imagine two directions, a and b in 2-dimensional space.

Provided a and b are not parallel, any vector can be

resolved into components in the a and b directions.

a

b

R

a

b

R A

B

R = A + B

Vectors A and B are often called the projections of R in

the a and b directions, respectively.

more ...

more on resolution of vectors ...

When the a and b directions are

orthogonal (i.e., a and b intersect

at a right angle) it is usually

straightforward to determine the

vector components.

When the a and b directions

are non-orthogonal, it is usually

more work to determine the

vector components. The law of

sines and law of cosines may be

useful.

a

b

R A

B

a

b

R A

B

more ...

law of sines

law of cosines

a

c

C

b

A B

A

sin a

B

sin b

C

sin c

C = A

2 B

2 2ABcosc

Rectangular Components:

A force F is said to have been resolved into two rectangular

components if its components are directed along the

coordinate axes. Introducing the unit vectors i and j along

the x and y axes,

x

y

q

Fx = Fx i

Fy = Fy j

F

i

j

F = Fx i + Fy j

Fx = F cos q Fy = F sin q

tan q = Fy

Fx

F = Fx + Fy 2 2

When three or more coplanar forces act on a particle, the

rectangular components of their resultant R can be obtained

by adding algebraically the corresponding components of the

given forces.

Rx = SRx Ry = SRy

The magnitude and direction of R can be determined from

tan q = Ry

Rx R = Rx + Ry

2 2

Addition of Several Vectors

• Step 1: resolve each force into its components

• Step 2: add all the x components together and add all the y components together. These two totals become the resultant vector.

• Step 3: find the magnitude and angle of the resultant vector.

Illustration

Magnitude and Direction

Example 1

Combine the two forces P and T, which act on the

fixed structure at B, into a single equivalent force R.

P = 500 N

T = 200 N

3m

T

P

A

B

C D

5m

75o

R

P = 500

T = 200q

q

4.48

75cos53

75sin5tan

AD

BD

Law of cosines:

NR

R

cabbac

5.396

)4.48cos()500)(200(2500200

)cos(2

222

222

Law of sines:

4.48sin

5.396

sin

200

q

2.22q

NR 5.396

22.2

Example 2

A barge is pulled by 2 tugboats. The resultant of the forces

exerted by the tugboats is a 5000 N force directed along

the center axis of the barge. Determine tension in each rope

if α =45 degrees

Barge

A

B

30

α

105sin

5000

30sin45sin

21 TT

1

2

3660

2590

T N

T N

Example 3

• Given: Three concurrent forces acting on a bracket.

• Find: The magnitude and angle of the resultant force.

Solution

Solution

Summing up all the i and j components respectively, we get,

Example 4

A ring supports three forces. What is the resultant

force applied to the ring?

more ... A: R = 998 N @ 134.9°

• Hand Solution

This may be simplified to:

more ...

Add force vectors to get the resultant.

Because vector addition is commutative, the order of addition is

not important.

Example 5

25 F = 110 N

Wind

A sail boat tacks into the wind such that the force

perpendicular to the sail is 110 N. Resolve this force

into two components, one parallel to the keel and one

perpendicular to the keel.

A: F= 46.5 N

F = 99.7 N

• Hand Solution