statics course notes

Engineering Mechanics: StaticsEngineering Mechanics: Statics

• Mechanics is the branch of the physical sciences d ith th t t f t ti f b diconcerned with the state of rest or motion of bodies

subjected to forces.

Engineering MechanicsEngineering Mechanics

• MechanicsMechanics– Rigid-body mechanics

• Dynamics – the study of objects in motion• Dynamics – the study of objects in motion– F = ma

• Statics - the study of objects in equilibrium– Also F = ma, but a = 0

– Deformable-body mechanics– Fluid mechanics

Newton’s LawsNewton s Laws• 1st Law – If an object is in motion, it will remain in j

motion. If an object is at rest, it will remain at rest, unless a force acts upon it.– When the sum of forces acting on an object is zero, the object g j , j

will remain at rest if initially at rest.• 2nd Law – When the sum of forces acting on an object is

not zero, the sum of the forces is equal to the product of , q pthe mass and its acceleration.– F = ma

• 3rd Law – The forces exerted by two objects on each• 3 Law – The forces exerted by two objects on each other are equal in magnitude and opposite in direction.– For every action, there is an equal and opposite reaction

S l V tScalar vs. Vector

Physical quantity described by a real number

Magnitude and directionby a real number

(magnitude , but no direction)

Ex. mass, speed Ex. weight, velocity

rF

Fr

FF

(magnitude of a vector )Fr (textbook may use bold F)

Vector AdditionVector Addition

Head to tail or Triangle RuleHead-to-tail or Triangle Rule

U + V = V + U

Vector AdditionVector Addition

Parallelogram Rule

Product of Scalar and VectorProduct of Scalar and Vector

VectorVector Subtraction

Unit VectorsUnit Vectors

Vector ComponentsVector Components

Sum of two vectors can be found by simply summing the components.

A man exerts a 60-lb force F to push a crate onto a truck.

(a) Express F in terms of components using the coordinate system shown.

(b) The weight of the crate is 100-lb. Determine the magnitude of the sum of the forces exerted by the man and the crate’s weight.

Th t l l i t d b t bl hi h t d f th t fThe steel column is supported by two cables which extend from the top of the column to the ground and lie in the same plane. The tensions in cables A and B are 13 kips and 10 kips, respectively. What is the magnitude and direction of the resultant force exerted on the top of the column by the p ycables?

50o60o

A BA B

PositionPosition Vector

A position vector can be formed between two physical locations in space. The magnitude of themagnitude of the position vector is the distance between the two points (length).

A surveyor finds that the length of the line OA is 1500 m and the length of the line OB is 2000 m.

(a) Determine the components of the position vector from point A to point B.(b) Determine the components of a unit vector that points from point A toward(b) Determine the components of a unit vector that points from point A toward

point B.

Vectors in 3-DVectors in 3 D

When dealing in three dimensions, a vector can be written in terms of 3 components.

How can we describe a vector in 3‐D (magnitude and direction)?

1) Direction cosines / direction angles2) Double projection3) Find a parallel line

Direction Angles and CosinesDirection Angles and Cosines

xθ or α

yθ or β

Direction Angles and CosinesDirection Angles and Cosines

zθ or γ

There is a relationship between the direction cosines…

1coscoscos 222 =++ γβα

Double ProjectionDouble Projection

y

Ux

z


y

U jU

xUy j

z


y

Ux

Uz k

zUx i

Position VectorsPosition Vectors

Components of a vector Parallel to a LineComponents of a vector Parallel to a Line

• Ex. When want to express a force in vector form p(components), but we only know the magnitude of the force. We can get the orientation of the force from the parallel lineparallel line.

y

Express the force acting on the wall at A in terms of its components.

A3’

T =1.2 kips

xB4’

9’1’

16’

z

y

Find the resultant force (magnitude and direction) given the three-dimensional system below.

60o

60o

1000N

500N

x30o

25o

45o45o

20o

650N150N

700N

z

Dot Product / Scalar ProductDot Product / Scalar Product

• Can be thought of as a measure of the “right-angularity” of two vectors

θcosVUVUvvvv

=⋅Definition,

In terms of components,ˆˆˆ ++= kUjUiUU

v

)ˆˆ()ˆˆ()ˆ̂()ˆˆ(

ˆˆˆ

++++=⋅

++=

++=

ijVUkiVUjiVUiiVUVU

kVjViVV

kUjUiUU

zyx

zyx

vv

v

1 0 0

vv

...)()()()( ++++=⋅ ijVUkiVUjiVUiiVUVU xyzxyxxx

zzyyxx VUVUVUVU ++=⋅vv

So,

Parallel ComponentParallel Component

θcosUU p

vv=

Parallel ComponentParallel Component

θcosˆˆ UeUevv

=⋅

Since the mag. of a unit vector = 1,

θˆ UUvv

And we previously determined,

θcosUUe =⋅

vv

Therefore,

θcosUU p

vv=

( )eUeU

UeU

p

p

ˆˆ

ˆvv

vv

⋅=

⋅=

( )eUeU p

Normal ComponentNormal Component

UUUvvv

+=Knowing the Triangle Rule,

np UUU +

Therefore,

pn UUUvvv

−=

Cross ProductCross Product

• Very useful for determining moments

Notice that result is a vector!!!

Definition,

( )eVUVU ˆsinθvvvv

=×

Notice that result is a vector!!!

What if 0o (Parallel Lines)?

( )eVUVU sinθ=×

Cross ProductCross Product

NOT Commutative,UVVUvvvv

×≠×

( ) ( ) ( )VUVUVU

UVVU

vvvvvv

vvvv×−=×

When multiplying by a scalar,

Distributive:

( ) ( ) ( )

( ) ( ) ( )WUVUWVU

VaUVUaVUa

vvvvvvv×+×=+×

×=×=×

Cross Product: In terms of components

kji)))

Cross Product: In terms of components

zyx

zyx

VVVUUUVU

vv=×

jikji)))))

-yx

yx

zyx

zyx

VVUU

VVVUUUVU

vv=×

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]kVUiVUjVUkVUjVUiVU xyyzzxyxxzzyˆˆˆˆˆˆ ++−++=

Triple ProductTriple Product

kji)))

( ) ( ) zyxzyx

WWWVVVkji

kUjUiUWVUvvv

⋅++=×⋅ ˆˆˆ

UUU

zyx WWW

zyx

zyx

WWWVVVUUU

= Scalar!!!

zyx WWW


( ) VOLUMEWVU =×⋅vvv


( ) VOLUMEWVU =×⋅vvv

W

V

W

U

V

U

V x W (magnitude is area of side!)

Problem 2.108Problem 2.108

Determine the angle θ between the lines AB and AC

a) using the law of cosines

b) using the dot product


The force F = 10i – 4j (N). Determine the cross product rAB X F.

Problem 2.119 ‐ The disk A is at the midpoint of the sloped surface. The string from A to B exerts a 0.2‐lb force F on the disk. If you express F in terms of vector components parallel and normal to the sloped surface, what is the component normal to the surface?

Problem 2.141 – Determine the minimum distance from point P to the plane defined by the three points A B and Cplane defined by the three points A, B, and C.

Forces, Equilibrium, and F.B.D.’so ces, qu b u , a d s

• Terminologygy– Line of Action – the straight line collinear with a force

vector


• Terminologygy– Systems of Forces – a set of forces which can be two

(coplanar) or three-dimensional.

Concurrent forces Parallel forces theirConcurrent forces -their lines of action intersect at a point

Parallel forces – their lines of action do not intersect


• Terminology If the car and driver is th bj t d

gy– External Forces

the object under consideration…

the road exerts an external force on the tires, and

– Internal Forces

the driver’s rear end exerts an internal force on the car seat.


• Terminologygy– Body Forces – act on the volume of the object (e.g.

gravity and magnetic forces)

Tractor beam!

– Surface Forces – act on the surface of an object

Tractor beam!


• Contact Forces– Surfaces

Smooth surfaces only have normal forces.

fNormal force is perpendicular and friction force is parallel to the contact plane.


• Contact Forces– Ropes, Cables, and Pulleys

Tension force is collinear with the cable and acts on box and craneTension force is collinear with the cable and acts on box and crane (equal magnitude and opposite direction).


• Contact Forces– Ropes, Cables, and Pulleys

• Frictionless pulleys only change the direction of the force, not the magnitudethe magnitude.

Frictionless bearings in pulley


• Contact Forces– Springs

Original length of spring, Lo

Spring is stretched to length L

Spring exerts a force on the wall

SWall exerts an equal and opposite force on the spring

Spring constant has units of force per length (lb/ft or N/m)

* Spring force depends on the change in length of the spring.oLLkF −=

Forces, Equilibrium, and F.B.D.’sForces, Equilibrium, and F.B.D. s

• Equilibriumq– Objects within the Newtonian (or inertial) reference

frame are at rest (or constant velocity).

0=∑Fr

Entire train is taken as the Constant velocity

0=∑ xF

0=∑F reference frame.0=∑ yF

Train accelerates

Forces, Equilibrium, and F.B.D.’sForces, Equilibrium, and F.B.D. s

• Free Body Diagramsy g– Used to isolate objects from their surroundings

• Step 1 – Identify the object to isolate• Step 2 – Sketch the object isolated from surroundings• Step 3 – Draw vectors representing external forces

Tension is now anTension is now an internal force!

What is the tension in the rope?

m

Three-Dimensional Force SystemsThree Dimensional Force Systems∑ = 0F

v

• In 2-D, In terms of components,

∑ = 0F

∑∑

=

=

0

0

y

x

F

F

In terms of components• In 3-D,

In terms of components,

∑∑

=

=

0

0x

F

F

∑∑

= 0

0

z

y

F

F

Systems of Forces and MomentsSystems of Forces and Moments

• Moments – rotation at a point due to a force applied at some distance

Sign Convention

Clockwise = negative

Counterclockwise = positive

MP = DF

In terms of components…In terms of components…

FxyyxP FDFDM +=∑

Fy

FxDyIf line of action goes through P…

Dx

FxDy

then MP = 0

Problem 4.24 – The tension in the cable is the same on both sides of the pulley. p yThe sum of the moments about point A due to the 800‐lb force and the forces exerted on the bar by the cable at B and C is zero. What is the tension in the cable?

The weights W1 and W2 are suspended by the cable system shown. The cable g 1 2 p y yBC is horizontal. If the largest moment that can be resisted by the pole DP at point P is 100 lb‐ft, determine W1.

The Moment Vector: Moment About a PointThe Moment Vector: Moment About a Point

…any position vector from the point to the line of action ofpoint to the line of action of the force.

FDMvv

=θiFMvvvFrM

vvv×=

The moment vectorThe magnitude of the

moment vectorIf the position vector is perpendicular

FDM P =θsinFrM Pv=FrM P ×=

The Moment Vector: Moment about a PointThe Moment Vector: Moment about a Point

Moment about point P generated by force F.

Moment vector is drawn according to the rotation and right‐hand rule.

Moment of a Force About a LineMoment of a Force About a Line

The capstan will rotate about the vertical axis (line)

The capstan will not rotate

If h li f i f h fThe line of action of the force must be skew to the line that we want to rotate about.

If the line of action of the force is parallel or intersects the line, then there will be no rotation.

Position vector from any arbitrary point on os o ec o o a y a b a y po oline to any point on Line of Action

FrM P

vvv×=P

The cross product will give us the moment about a point on the line.

but we want the moment about the line…but we want the moment about the line, or the component of the moment vector that is in the direction of the line.

The dot product should give us the component acting in the direction of the linecomponent acting in the direction of the line.

PL MeMv

•= ˆ

but if I want it in vector form

[ ]eMM LL ˆ=v

…but if I want it in vector form…

[ ]LL

Special Cases…Special Cases…

Force is perpendicular to planeLOA of Force

Force is perpendicular to plane containing line

Force is parallel to line

intersects line

FDM L ⋅=

A concrete precast wall section is held by 2 cables, where one cable is not shown (behind wall). The tension in BD = 800 N

Find:

a) Moment about point O of the force exerted by the cable at B.

b) Moment about the x-axis generated by force BD.

c) Angle between cable BD and the vertical side BCc) Angle between cable BD and the vertical side BC.

y

B

3.5 m

4 mA

x3 5 m

O C

z

3.5 m

1 m

D

CouplesCouples

Couples are two forces with equal magnitudes, opposite directions, and

100 N

magnitudes, opposite directions, and different LOA’s

100 N

CouplesCouples

What is the moment generated about point P due to the couple?

100 N

100 N

5 m)5)(100()( +=+ ∑ mNMCCW

7 m

P

...)5)(100()( +=+ ∑ mNMCCW P

mNmNmNMCCW P ⋅−=−=+ ∑ 200)7)(100()5)(100()(

CouplesCouples

100 N

P1 m

1 m

100 N

...)1)(100()( +−=+ ∑ mNMCCW P

mNmNmNMCCW P ⋅−=−−=+ ∑ 200)1)(100()1)(100()( P∑ ))(())(()(

The moment it exerts about any point is the same!

Problem 4.115 ‐ Determine the sum of the moments exerted on the plate by the two couples.

Problem 4.114 – The moment of two couples are shown. What is the sum of the moments about point P?

Same magnitude…negative let’s k th t it i i thus know that it is in the opposite direction

[ ] [ ][ ] [ ]FrFrMvvvvv

−×+×= 21

[ ] [ ] FrrFrFrMvvvvvvvv

×−=−×+×= )( 2121

Remember from vector vvvRemember from vector addition/subtraction 12 rrr =+

The couple moment can be calculated using a position vector from one line‐of‐action to the other.

However, the force vector on which the position vector ends is the one used in the cross‐product.

vvS i lFrM v×= Special case…

DFM DFM =

Since the total force exerted by the In 2 D the couple can becouple is zero, it is often represented

by the moment it exerts.

In 2‐D, the couple can be represented by its magnitude and a

circular arrow.

Examples of CouplesExamples of Couples

The forces are contained in the x‐y planeThe forces are contained in the x y plane.a) Determine the moment of the couple and represent it as shown in Fig.

4.18c.b) What is the sum of the moments of the two forces about the point

(10, ‐40, 20) ft?

Equivalent SystemsEquivalent Systems

The two systems are equivalent if:

1) The sum of forces in System 1 and 2 are hthe same.2) The sum of moments about a point for System 1 and 2 are the same.

FFF

FFvvv

vv

+

=∑∑ 21

DBA FFF =+

( ) ( )vv∑∑( ) ( )

( ) ( ) ( ) FEDDCBBAA

PP

MMFrMFrFr

MMvvvvvvvvv ++×=+×+×

= ∑∑ 21

System 1 is equivalent to a force and y qcouple moment acting at Q

The simplest system that can be equivalent to any system of forces and momentsy y

WRENCH

50 k‐ft

100 k

System 1

0.5 ft

100 k

System 2

If the system of forces is concurrent,

we can represent System 1 as a single force.

If the system of forces is parallel,

we can represent the system as a single force k l iat a known location.

25 lb

Find an equivalent single force and couple moment at point A which can represent the system below.

10 lb

15 lb20 lb

25 lb

5 lb

3’ 3’ 3’ 3’A B

Can the single force and couple moment be replaced by a single force?

Replace the system of forces by a single force passing through O and a couple. Can the reduced system be replaced further by a single force?

F1 200#

Can the reduced system be replaced further by a single force?

y F1 = 200#

F2 300#

3

4(2,8) ft

F2 = 300#

60 deg

4

(6,4) ft

(4,0) ft

60 deg

Ox

F3 = 100#

O

Equivalent Systems: Distributed LoadsEquivalent Systems: Distributed Loads

15 lb20 lb

25 lb

5 lb10 lb

3’ 3’ 3’ 3’A

75 lb

A

8’

A


50 lb 50 lb

3’ 6’ 3’A

100 lb

A

6’

A


Split the 50‐lb forces into four 20‐lb and two 10‐lb forces

20 lb 20 lb 20 lb 20 lb10 lb

and two 10 lb forces

3’ 3’ 3’A

10 lb10 lb

3’

100 lb

A

6’

A


Split the forces such that the distance between them approaches zero.

Total of 100 lb over 12’, so:

between them approaches zero.

12’A

x

ftlb

ftlbxw 33.8

12100)( ==

x

ff“w” is usually associated with distributed loads Function of

distance alongMeasure of intensity of load along a givendistance along

beamof load along a given distance

w(x) = 1.2 k/ft

20’

Replace the distributed load by a single (equivalent) force

The equivalent force can be thought of as the area under the

( ) kipsftft

kipsF 24202.1 ==

24 kips

the area under the distributed load

∫= dxxwF )(10’

∫L

20’

For a uniform distributed load the equivalent loadload, the equivalent load is the area under the loading curve, and it acts at the center. L

L/2

For a triangular distributed load the equivalent load isload, the equivalent load is the area under the loading curve, and it acts 1/3 from the “heavy” end or 2/3 Lfrom the “light” end.

(2/3) L (1/3) L

“plf” means pounds per linear foot (lb/ft)300 plf

plf means pounds per linear foot (lb/ft)

10 f 8 f10 ft 8 ft

lb1500 lb

3.33 ft 14.67 ft

300 plf

150 plf

10 f 10 f10 ft 10 ft

3000 lb

750 lb

3.33 ft 10 ft6.67 ft

3000 lb

750 lb

3.33 ft 10 ft6.67 ft

3750 lb

8.67 ft

500 plf 500 plf

70 plf70 plf

p p200 plf

6 f 6 f 6 f10 f6 ft 6 ft 6 ft10 ft

( ) lbftplfplf 2580)6(7050012 =⎥⎤

⎢⎡ −

( ) lbftplf 600)6(20021

=( ) lbftplfplf 2580)6(70500

22 =⎥⎦⎢⎣

lbftplf 1540)22(70 =

600 lb @ 4ft

2580 lb @ 22ft

1540 lb @ 17ft

( ) lbF 352025801540600↓ ∑

A

( ) lbFy 352025801540600 =++−=↓+ ∑

( ) ( ) ( ) ( )

lbf

MCW A ++−=+ ∑

80540

2225801715404600

lbft ⋅= 80540

ftlbftFx y 882280540

=⋅

== ∑

3520 lb

ftlbM

xA

88.223520

===∑

A

x = 22.88 ft

Objects in EquilibriumObjects in Equilibrium

Forces are no longer concurrent

Wh d li ithWhen we were dealing with concurrent forces, we analyzed equilibrium of a particle.

∑ = 0Fv

Th f dThe forces generated no moment about the point, so we couldn’t use any equations involving moments.

Now that we are analyzing the equilibrium of an object that has

mass and size, we can introduce an equilibrium equation involving

moments.

∑∑

=

=

0

0

M

Fv

v

∑ = 0anywhereM

Generally speaking… 2‐D

∑v ∑

Vector Sum Sum of Components

∑ = 0Fv

∑∑

=

=

0

0

y

x

F

F

∑ = 0anywhereMv

∑ = 0anywhereM

3 equations, so we can only solve for 3 unknowns

Supports and their ReactionsSupports and their ReactionsForces and couples exerted on an object by a support are called reactions

Support

Reactions – represents the i i f drestraining forces and moments

that the support can offer.

Pin SupportPin Support

Pin prevents vertical and horizontal motion (offers an xand y reaction)…

but, pin allows rotation ( t ff i ti(cannot offer any resisting couple moment)

Pin SupportPin Support

Roller SupportRoller Support

Roller can only prevent object from moving in one directionfrom moving in one direction (offers only an x or y reaction)

Pin on wheels!

If A is negative, the roller is lifting off of the ground!

Kind of like a Roller SupportKind of like a Roller SupportThese can offer resistance in both directions along a line of

iaction

A can be positive or negative.

Fixed SupportFixed Support

Object is embedded in the wall.j

A fixed support has the potential to develop all three reactions.reactions.

Free‐Body DiagramsFree Body DiagramsF F

Ax

Ay B

lbB

W = 50 lb

50 lb

Ay

Ax

Free‐Body DiagramsFree Body Diagrams

B30o

60o

2000 lb

20o

2000 lbA

70o


RRx

TT

Frictionless surface

RRy


10 plf

20 ft

(20 ft)10 plf = 200 lb

A

10 plf

10 ft

A

Ax

MAy MA


500 N/m500 N/m

Weight = 20 N/m

A

3 N1 kN‐m

A B

3 N

3 m 3 m 6 m

Free‐Body DiagramsFree Body Diagrams1/2(500 N/m)(6 m)(20 N/m)(12 m)

Bx

3 N1 kN‐m

ByA

3 N

3 m 3 m 4 m 2 m

( )

( ) ( ) ( ) ( ) ( )NAN

MCCW

NN

B

02650016122093121000

0=+ ∑( ) ( ) ( ) ( ) ( )

NA

mmmmmNAmmN mN

mN

08.451

0265002

6122093121000

=

=++−−⋅

Free‐Body DiagramsFree Body Diagrams1/2(500 N/m)(6 m)(20 N/m)(12 m)

Bx

3 N1 kN‐m

ByA

3 N

3 m 3 m 4 m 2 m

( )( ) ( ) BmmNN

F

NN

y

0650011220308.451

0

=+−−+

=↑+ ∑ ( )0

0=

=→+ ∑x

x

BF

( ) ( )NB

BmmNN

y

ymm

92.1285

065002

1220308.451

=

++

If F = 4 kN, what are the reactions at A and B?

Problem 5.18 Draw the free‐body diagram of the structure by isolating it from its supports at A and E. Determine the reactions at Aand E.

Problem 5.55 Suppose that you want to design the safety valve to open when the difference between the pressure p in thevalve to open when the difference between the pressure p in the circular pipe (diameter = 150 mm) and atmospheric pressure is 10 MPa. The spring is compressed 20 mm when the valve is closed. What should the value of the spring constant be?

Problem 5.63 The boom derrick supports a suspended 15‐kip load The booms BC and DE are each 20 ft long The distances areload. The booms BC and DE are each 20 ft long. The distances are a = 15 ft and b = 2 ft, and the angel θ = 30o. Determine the tension in cable AB and the reactions at the pin supports C and D.

120 lb/ft6 ft 5 ft

1 ft

120 lb/ft

BA

self wt = 100 lb/ftself wt. = 100 lb/ft

W1

W2

A) Find W2 for equilibrium if W1 = 400 lb.2 1B) Find the reactions at A and B.

beam. wt. = 70 plf

50o 4

30 ft‐lbA C

50

3

43 ft3 ft4 ft B

Find the reactions.

Bm. Wt. = 70 plf

30 ft‐lbA C

3 ft3 ft4 ft B

How is this different than the previous?

Statically Indeterminate ObjectsStatically Indeterminate Objects

0=∑F

0

0

0

=

=

=

∑∑∑

y

x

M

F

F3 equations, 3 unknowns

0=∑ anywhereMWe can solve…

!NOT ALWAYS!

Statically Ind. – Case 1Statically Ind. Case 1

More unknowns than equations of equilibrium.

How?

More supports than theminimum numberMore supports than the minimum number necessary to maintain equilibrium (redundant supports)

8 unknowns, so (8‐3) = 5 degrees of redundancy (or indeterminacy)indeterminacy)

F = 100 N

BA

1 m1 m

100 N

AxFBD

MA

1 m

BAy

1 m


Improper Supports

…the object will move!

BAF Object takes off

down the road!!!

F0≠∑ xF

Ay By

This occurs anytime all of the reactions are parallelThis occurs anytime all of the reactions are parallel…

BA BAF

3 eq. and 3 unknowns, but we’re still indeterminate!

or concurrent!

BAF

P

BAF

If F causes a moment about point P, then the object rotates!

P

BAF

BA

P

However…

BAF

Properly or improperly supported?Properly or improperly supported?


More unknowns than equations of equilibrium.

How?

More supports than theminimum numberMore supports than the minimum number necessary to maintain equilibrium (redundant supports)

8 unknowns, so (8‐3) = 5 degrees of redundancy (or indeterminacy)indeterminacy)


Improper Supports

…the object will move!

BAF Object takes off

down the road!!!

F0≠∑ xF

Ay By

This occurs anytime all of the reactions are parallelThis occurs anytime all of the reactions are parallel…

BA BAF

3 eq. and 3 unknowns, but we’re still indeterminate!

or concurrent!

BAF

P

BAF

If F causes a moment about point P, then the object rotates!

P

BAF

BA

P

However…

BAF

Properly or improperly supported?Properly or improperly supported?


State whether each of the L‐shaped bars shown is properly or improperly p p p y p p ysupported. If a bar is properly supported, determine the reactions at its supports.

Equilibrium in 3‐DEquilibrium in 3 D

2‐D 3‐D

∑ = 0Fv

∑∑

=

=

0

0

y

x

F

F∑ = 0F

v

∑∑

=

=

0

0

y

x

F

F

∑ = 0Mv

∑

∑

= 0anywhere

y

M∑ = 0zF

∑ = 0Mv

∑∑∑

=

=

0

0

y

x

M

M

∑ = 0zM

6 scalar equilibrium equations!6 scalar equilibrium equations!

Supports and ReactionsSupports and Reactions

Ball‐and‐SocketBall‐and‐Socket

Free rotation, but no translation in any direction.

Ball‐and‐socket joint to connect the housingthe housing of an earth grader to its frame.


RollerRoller

A ball and socket on wheels. Free rotation, free translation in the x‐z plane, but no translation in the y‐direction


HingeHinge

Can rotate freely only about the hinge axis and cannot translate.

Pin usedPin used to support a strut on a tractor.

Why don’t we use only one hinge on ahinge on a door?

The forces Fxexert a coupleexert a couple moment to resist the moment due to the weight.


BearingBearing

Similar reactions to those of the hinge, but in some cases, Some bearings do not restrict

translation in the z‐directiontranslation is allowed in the longitudinal direction.

translation in the z direction.

Journal bearing used toused to support the end of a shaft.

Thrust bearing used to support a drive shaft.


FixedFixed

No translation or rotation.

Problem 5.149Problem 5.149The 80‐lb bar is supported by a ball and socket support at A, the smooth wall it leans against, and the cable BC. The weight of the bar acts at its midpoint.a) Draw the free‐body diagram of the bar.b) Determine the tension in cable BC and the reactions at A.

Two and Three‐force MembersTwo and Three force Members

A two‐force member is subjected to no couple moments and forces f j pare applied to only two points on the member. Weight is neglected.

Notice that the two forces are collinear.

Two‐force members will maintain translational (or force) equilibrium if FA and FB are equal and opposite.

Two‐force members will maintain rotational (or moment) equilibrium if FA and FB are collinear.

Therefore, only the magnitude of the force must be determined, since the direction can be determined by the line AB.

A three‐force member is subjected to only three forces that must be either concurrent or parallel.

LOA’s are parallel and intersect at infinity

0=∑ PMP0=∑ ∞M

TrussesTrusses

A truss is a structure composed of slender members joined together at their end points two force membersend points…two‐force members.

Welded, bolted, or riveted to a gusset plate

Connected with large bolt

A truss can only be loaded at the joints,A truss can only be loaded at the joints, and the members are assumed to be weightless (negligible).

Top chords

Bottom chordsWeb

Top and bottom chords usually carry significantlyTop and bottom chords usually carry significantly higher loads than the web elements.

R f t t i ll l d d thRoof trusses are typically loaded on the top joints whereas bridge trusses are typically loaded on the bottom joints.

Truss Assumptions:1) Loaded only at joints which are assumed to be pinned2) Composed entirely of two‐force members2) Composed entirely of two‐force members3) Neglect weight (weightless)

Warren truss

Tension – force points awayCompression – force points towardCompression – force points toward

Though actual trusses are seldom pinned joints or connections they areThough actual trusses are seldom pinned joints or connections, they are designed such that the couple moments exerted at the joints are small compared to the axial forces!

Truss Analysis: Method of JointsTruss Analysis: Method of Joints

First step is to draw a F.B.D. and solve for the reactions.p


Second, examine the axial forces acting on the joints , g j(equilibrium of a particle)…start with knowns.


Two ways of viewing at the jointTwo ways of viewing at the joint…

T

Cut through member

TAB

TAB

Look at forces on pin

TAC

60oTAC

TAB

60o

500 N 500 N


Thirdly, use the two available equations of equilibrium to y, q qsolve for the two unknowns and verify their correct sense (tension or compression).

0

0

=

=

∑∑

y

x

F

F Cannot use sum of moments since forces are concurrent.∑ y

Member AB is 577N in tensionMember AC in 289N in compression


Lastly, methodically move from joint‐to‐joint, and determine y, y j j ,the axial forces in each member.

Move here next…

…or here.

Determine the force in each member of the truss shown. Indicate whether the members are in tension or compression.

Zero‐Force MembersZero Force Members

Suppose we want to use the method of joints to determine the axial forces in the members…

It would greatly simplify the process if we could identify b hi h d l di (any members which do not support any loading (zero‐

force members).


Therefore members AB and AF are zero force membersTherefore, members AB and AF are zero‐force members under this particular loading.

However, they may not longer be zero‐force members ifHowever, they may not longer be zero force members if the loading changes!


Now there are only 4 joints to consider!Now there are only 4 joints to consider!

H il id tif f b ?How can we easily identify zero‐force members?

Three members, two of which areThree members, two of which are collinear, and no load.

Two noncollinear members and no load.

By inspection, which of the following are zero‐forcefollowing are zero force members?

By inspection, determine all the zero‐force members of the Fink roof truss. Assume all joints are pin connected.

24 kips

36 kips

AA

For the truss shown below, determine all zero‐force members and the axial force in member LOin member LO.

Method of SectionsMethod of Sections

Quicker method of determining the axial force in a particular member.

What if I only care about the axial force in member BC?

Do I still have to go through the method of joints to eventually get to BC?


After determining the reactions at the gsupports, we can make a cut through the truss to find the axial forces in the desired member.

Notice that we have to cut through the member in question (BC).q ( )

After we make the cut, remove one side or the other and include the unknownor the other and include the unknown axial forces.

Our forces are no longer concurrent like they were in the method of joints wethey were in the method of joints...we can use the sum of moments equation!


0=∑ yFWe can find TBCdirectly!

0∑M 0=∑ cMWe can find TBDdirectly!

0=∑ xFNow we can find Tfind TAC

We can only cut through as many members as our equations of equilibrium will allow us to solve.

Th P tt b id t i bj t d t th l di h D t i thThe Pratt bridge truss is subjected to the loading shown. Determine the force in members JI, EI, and DE.

For the truss shown below,a) Identify all zero force members;b) D i h i l f i b BG d AB i h h d fb) Determine the axial force in members BG and AB using the method of

sections; andc) Determine the axial force in member BE using the method of joints.

Frames and MachinesFrames and Machines

Frames Machines

‐ Remain stationary and support loads ‐ Move and apply loads

Frames and machines are two common pin‐connected structures composed of multiforce members.

They are not entirely composed of two‐force members and cannot be analyzed like trusses!

Determine the forces on each of the members of the frame.

Step 1: Analyze the entire structure (a.k.a. find the reactions at the supports).

6 kN

Ax

1 m

BC

Ay 8 kN

1 m

Gx

1 m

E D

Gy

1 m 1 m 1 m

Step 2: Disassemble the frame.Load at joint? It doesn’t matter, but only at one of the joints.

‐13

6 kN

Cx

Ay

T

T

Cy

CyCx

8 kN

T

Two‐force members!

Dx

D

T

Two force members!

5Dx

Dy

Dy

T

Cy

For the frame shown draw the free‐body diagrams of (a) the entireFor the frame shown, draw the free body diagrams of (a) the entire frame including the pulleys and cords, (b) the frame without the pulleys and cords, and (c0 each of the pulleys.

Determine the horizontal and vertical components of force which the pin at C exerts on member ABCD of the frame.

Problem 6.73 – The force F = 10 kN. Determine the forces on member ABC, presenting your answers as shown in Fig. 6.25.

Problem 6.88 – The weight W = 80 lb. Determine the forces on member ABCD.

Problem 6.83 – The mass m = 50 kg. Bar DE is horizontal. Determine the forces on member ABCD, presenting your answers as shown in Fig. 6.25.

Problem 6.133 – Determine the reactions on member ABCat B and C.

Internal Forces and MomentsInternal Forces and Moments

P P

Wh d li i h f bWhen dealing with two‐force members, we can cut through the member to determine the internal force.

P P

The only internal force is an axial force!

Internal Forces and MomentsInternal Forces and Moments

Wh d li i h h f ( )When dealing with three‐force (or more) members, we get more internal forces than just a simple axial force!

M (Internal bending moment)

P

V (Internal shear force)

(Internal axial force)

B

Why can’t we perform a method of sections on a frame?

D E

CB

A C

D

Positive Sign ConventionsPositive Sign Conventions

Positive shear forces make the Positive bending momentsPositive shear forces make the segment rotate clockwise

Positive bending moments rotate the segment upwards…make the segment smile!

Beams Columns

Long, slender structural members which are intended to support loads

Loads are primarily applied parallel to the longitudinal axis and are compressivewhich are intended to support loads

perpendicular to its longitudinal axis.longitudinal axis and are compressive.

Types of Supports

Simply supported

Overhanging

Cantilevered

Fixed

ProppedPropped

3‐span Continuous

Types of Loads

Concentrated

Uniform distributed

Moment/couple/pure moment

Non‐uniform distributed

Combined

10 ft 5 ft

200 lb10 lb/ft

A B

5 ft 5 ft 2.5 ft 2.5 ft

200 lb10 lb/ft

50 lb

A BD C

350 lb

‐50 lb

F 0)( ↑ ∑F 0)( ↑ ∑lbV

F

C

y

200

0)(

=

=↑+ ∑lbV

F

D

y

100

0)(

−=

=↑+ ∑

ftlbMMCCW

C

C

⋅−=

=+ ∑500

0)(

ftlbMMCCW

D

D

⋅−=

=+ ∑375

0)(

200

250

0

50

100

150

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Shear (lb)

‐200

‐150

‐100

‐50 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Distance along beam (ft)g ( )

‐200

0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

‐600

‐400

omen

t (ft‐lb

)

‐1200

‐1000

‐800Mo

Distance along beam (ft)Distance along beam (ft)

25 kN/m

8 kN/m

200 kN‐m

5 m 1 8 m 3 7 m

A BC D

1 5 m5 m 1.8 m 3.7 m1.5 m

Determine the shear force and bending moment at point CDetermine the shear force and bending moment at point C.

Shear and Bending Moment DiagramsShear and Bending Moment Diagrams

‐Plot of the shear force and bending moment along the length of the beam‐Plot of the shear force and bending moment along the length of the beam.

F

2/3 L 1/3 L

F

2/3 F1/3 F2/3 L 1/3 L

2/3 F1/3 F

F

2/3 L 1/3 L2/3 F1/3 F

/ /

x

Between x = 0 and x = 2/3 L

P

M

xFM

MCCW cut

31 0)(

0)(

=−

=+ ∑x

1/3 F

V FxM 31=

Fy 0)( =↑+ ∑

FVFV

31

31 0

=

=+−

F

2/3 L 1/3 L2/3 F1/3 F

/ /

x

Between x = 2/3 L and x = L

F

P

M

x

1/3 F

V

0)()(

0)(21 LFFM

MCCW cut

+

=+ ∑Fy 0)( =↑+ ∑

)(0)()(

32

32

31

xLFMLxFxFM

−=

=−+−

FVFFV

3231 0

−=

=−+−

Relationship Between Loading, Shear, and B di MBending Moment

w

dxwV ∫−=Δ )( wddV

−=

V

dxwV ∫Δ )(

dM

dx

M

dxVM ∫=Δ VdxdM

=

Draw the shear and bending moment diagram.

100 lb

50 lb20 ft20 ft

50 lb

Draw the shear and bending moment diagram.

200 lb10 lb/ft

5 ft 5 ft 2.5 ft 2.5 ft

A B

350 lb

‐50 lb

350 lb

450 ft-lb

300 lb

40 plf

4 ft 3 ft 8 ft

V (lb) 0

M (lb-ft) 0

75 ft-lb 50 lb

10 ft 10 ft

V (lb) 0

M (lb-ft) 0

50 plf

Centroids & Center of Mass/Gravity

Centroid – The geometric center of an object.

2L/3

x

x

y

y


x

y

Lx 32=

Ly 31=

In general…

Centroid only depends on geometry, not density

Coordinates of the centroid, or average position, of an area is…

∫

∫

∫

∫

=

=

A

A

A

A

dA

ydAy

dA

xdAx

2 x

y

y = x2 –x + 1

Determine the x-component of the centroid of the area.

In discrete form, or when dealing with a combination of simple shapes…

∫

∫

∫

∫

=

=

A

A

A

A

dA

ydAy

dA

xdAx

∑∑

∑∑

=

=

ii

iii

ii

iii

A

Ayy

A

Axx

Sum of individual centroids times area

Sum of areas

Centroids of Composite Areas

1

2

x

y

12’

6” 4”

15”

6”

4”

18”

24”

??

==

xy

3”

4”

4”

1”

Find the location of the centroid.


Center of gravity vs. Center of mass

The point at which the resultant weight of a system of particles acts.

Since c.g. depends on the acceleration of gravity, center of mass is a better term to use.

Center of gravity of the solar system???

Center of mass is independent of gravity.

c.g.

∑∑∑∑∑∑

=

=

=

WzW

z

WyW

y

WxW

x

Center of mass

∑∑∑∑∑∑

=

=

=

mzm

z

mym

y

mxm

x

100 lb

300 lb

100 lb

(4,8)ft

(10,6)ft

(5,3)ft Find the center of gravity.

Moments of Inertia

Area Moment of Inertia Mass Moment of Inertia

- used in Dynamics to calculate the rotational motions of objects.

- used to calculate stresses and deflections in beams.

Can be thought of as a measure of the cross-sectional area’s resistance to bending.

F

F

Much easier to bend the board in this orientation…

…than this one.

v.s.

b

h

x

y

3

121

bhI x =

43 95.1)5.2)(5.1(121

inI x ==

We are taking the larger dimension to the third power, therefore we get a larger moment of inertia about the centroidal x-axis.

b

h x

y

…only for a rectangular cross-section…

43 70.0)5.1)(5.2(121

inI x ==

By nature, the cross-section will try to rotate about its centroidal axis.

∫=A

x dAyI 2

∫=A

y dAxI 2

Moment of inertia about x-axis

Moment of inertia about y-axis

AkI xx2=

AkI yy2=

kx (rx ) is called the radius of gyration

∫=A

y dAxI 2

∫

∫=

A

A

dA

xdAx

1st moment of an area

2nd moment of an area

xy 4002 =

100 mm

200 mm

Determine the moment of inertia of the area bounded by the curve and the vertical line.

x

y

x

y

10”

4”

Diameter = 4”

What is the moment of inertia about the x and y-axis for the following shapes?

y’ Diameter = 4”

x’

y

x

What if the shape is removed some distance away from a parallel axis?

C(6,10)in

Parallel Axis Theorem

2'

2'

xyy

yxx

AdII

AdII

+=

+=

d is the distance from parallel axis to the axis of the individual shape.

12’

6” 4”

15”

6”

4”

18”

24”

"72"76.23

==

xy

What is the moment of inertia about the centroidal x-axis of the large beam?

Polar Moment of Inertia

yxO

AO

IIJ

dArJ

+=

= ∫ 2

Resistance to rotation about the origin (or z-axis)

AkJ OO2=OR

where kO is the radius of gyration about the origin.

222yxO kkk +=

Determine the moment of inertia, both Ix and Iy, about the x and y-axis.

Determine the location of the centroid.

Determine the moment of inertia, both Ix and Iy, about the centroidal x and y-axis.

What is the polar moment of inertia?

2” 2-1/8”

Determine the moment of inertia and polar moment of inertia about the centroidal axes.

What is the radius of gyration about the x and y-axis?

Determine the moment of inertia about the centroidal x and y-axis.

Determine the location of the centroid.

3”

1”

4”

4”

1” 1” 2”

2 x φ5/8”

Friction

Book is not in static equilibrium.

Frictional component opposes motion!

Before, we assumed a smooth surface…

…but in reality, the surface has some degree of roughness which contributes to friction. Frictional

component opposes motion!

N f

f

N θ θ = Angle of Friction, or

the Internal Friction Angle

Frictional force only increases as the need arises in order to maintain static equilibrium.

F

f

Nf sµ=

Nf kµ=

No motion

Motion

Impending motion

ss

s Nfµθ

µ=

=tan

kk

k Nfµθ

µ=

=tan

F

f

No Yes

Are the surfaces in motion relative to each other?

Friction force opposes motion

Is slip impending?

F

f

No

Yes

You must determine mag. and direction using eq. of equilibrium.

Friction force opposes impending motion

P 200 lbs

A 200-lb block is initially at rest on the horizontal surface. If µs = 0.35 and µk = 0.2 between the surfaces, find the magnitude of the friction force acting on the block when P is: a) 30 lbs, b) 50 lbs, c) 70lbs, and d) 90 lbs

3 Problem Types

Equilibrium

Using sum of forces and moments for each bar (6 eq. and 6 unknowns), find NA, fA, NC, and fC.

A

B

C

µA = 0.3 µC = 0.5

100-N bars

…then verify that

CC

AA

NfNf

5.03.0

≤≤

3 Problem Types

Impending Motion At All Points

What is the largest force for which the boxes will not slip? The ends of the bar must slip

at the same time.

3 Problem Types

Impending Motion At Some Points

Either block A slips relative to B, or both A and B move together.

The mass of box A is 15 kg, and the mass of box B is 60 kg. The coefficient of static friction between boxes A and B and between box B and the inclined surface is 0.12. What is the largest force for which the boxes will not slip? What is the smallest force for which the boxes will not slip?

The mass of box A is 15 kg, and the mass of box B is 60 kg. The coefficient of static friction between boxes A and B and between box B and the inclined surface is 0.12. What is the largest force for which the boxes will not slip?

What is the smallest force for which the boxes will not slip?

P 200 lb

75 lb

Two blocks resting on each other weigh 200-lb and 75-lb, respectively. If the coefficient of friction between the blocks is 0.5 and between the 75-lb block and the floor is 0.3, determine the largest force P that can be applied without upsetting equilibrium.

F

250 lb

3 ft

5 ft

52 in

Tipping / Overturning

Will the refrigerator tip or slip?

µs = 0.25

As F increases, either the crate will be on the verge of sliding (f = µsN), or …

If the surface is rough (large µs) then N shifts to the corner and the crate tips over.

F W

N

f

F W

N f

x

F

250 lb

2 ft

5 ft

52 in

µs = 0.25

The refrigerator has a weight of 250-lb. If F = 60-lb, determine if it remains in equilibrium.

0.2 m

0.8 m µs = 0.3

30o

The crate has a mass of 20-kg. If P = 80N, determine if it remains in equilibrium.

P

Wedges

A bifacial tool with the faces set at a small acute angle. When pushed forward, the faces exert very large normal forces!

F

Wedges are very efficient when there is no friction…

…however, they may slip out!

The box A has a mass of 80-kg, and the wedge B has a mass of 40-kg. Between all contacting surfaces, µs = 0.15 and µk = 0.12. What force F is required to raise A at a constant rate?

The box A has a mass of 80-kg, and the wedge B has a mass of 40-kg. Between all contacting surfaces, µs = 0.15 and µk = 0.12. What force F is required to raise A at a constant rate?

010sin10cos0)(

211 =+−−

=→ ∑NNf

Fx

010cos10sin8.784

0)(

112 =+−−−

=↑ ∑Nff

Fy

010cos10sin10cos10cos0)(

3311 =++++−

=→ ∑fNNfF

Fx

010cos10cos10sin10sin4.392

0)(

1313 =−++−−

=↑ ∑NNff

Fy

For Block A:

For Block B:

Belt Friction

βµseTT 12 =

The force T2 necessary to cause impending slip of the rope in the direction of T2 is…

Belt Friction βµseTT 12 =

T2 is always the larger of the two forces

β is the angle of contact in radians.

I want the rope to slip in the direction

of T2

100 lb 30o

3 ft 3 ft

6 ft

F

What are the maximum and minimum values of F that may be applied without causing the 6 ft x 3 ft, 100-lb block to slip or tip? The coefficient of static friction between the block and the ramp is 0.5, and the coefficient of static friction between the rope and the fixed drum is 0.3.

P

200 lb

Platform

Find the minimum and maximum values of P which may be applied without causing the 200 lb block to slip. The coefficient of friction between the platform and the block is 0.8; the coefficient of friction between the rope and the drum is 0.2. Assume that the platforms attached to the identical springs (k = 75 lb/in) are only able to move horizontally. Each spring is compressed 1.2 in relative to its original length.

statics course notes

Documents

motion f

force f

sum of forces

i t d b t bl

express f

h t d f th t fthe steel

statics mechanics

study of objects