FE Statics Review

Sanford Meek Department of Mechanical Engineering

Kenn 226 (801)581-8562 meek@mech.utah.edu

Statics

forces = 0!

moments = 0!

2  

Vectors

Scalars have magnitude only Vectors have magnitude and direction

3  

Vectors

Vector addition and subtraction !A+!B =!R

!R!!B =!A

A

-B

R

A

B

R

B

A

4  

Vectors

Resultants

sin!A

=sin"B

=sin#C C

A

B

Law of sines

Law of cosines

C2 = A2 +B2 ! 2ABcos!

α

β

γ

5  

Vectors

Cartesian coordinates

The only system that is generally used in statics

x, y, z axes i, j, k unit vectors

6  

Vectors

2 2 2x y zA mag A A A= = + +A

!

ˆ ˆ ˆ ˆyx zAA A

A A A A! "! " ! "= = + +# $# $ # $

% & % &% &

AA i j k!

ˆ ˆ ˆx y zA A A= + +A i j k

!

ij

k A

7  

Direction cosines

ˆ ˆ ˆ ˆyx zAA A

A A A A! "! " ! "= = + +# $# $ # $

% & % &% &

AA i j k!

The components of the UNIT vector,

are the cosines of the angles made by the vector with the coordinate axes.

cos

cos

cos

x

y

z

AAAAAA

!

"

#

=

=

=

2 2 2cos cos cos 1! " #+ + =8  

Position Vectors

!rA = xAi + yA j+ zAk!rB = xB i + yB j+ zBk

!rAB = xB ! x( ) i + yB ! yA( ) j+ zB ! zA( ) k

Final point (head of arrow) – initial point (tail of arrow)

9  

Force Vectors Simply the force magnitude times the unit position vector

ˆ ˆ ABAB

AB

F F Fr

= = =rF F r!!

ˆ( 5')ˆ( 3'cos20 )ˆ(3'sin 20 6')

AB = !

+ ! °

+ ° +

r i

j

k

!

10  

Dot Product Use: find the component of a vector along a certain direction – component of force along a particular axis

cos

x x y y z z

AB

A B A B A B

!=

= + +

A BA B

! !i! !i

A!

!

11  

Dot Product Use: find the component of a vector along a certain direction – component of force along a particular axis The result is a scalar.

cos

x x y y z z

AB

A B A B A B

!=

= + +

A BA B

! !i! !i

A!

!

12  

Dot Product

13  

What part of A is aligned with the line? What part is perpendicular?

cos ˆa aA !=A u!

ˆ au!A

!=!A"!Aa

ˆcos aA ! = A u!i

Projected component (scalar):

=perpendicular part

Moment (Torque)

14  

Only  the    perpendicular    part  of  force  gives  torque  

Cross Product

15  

IEEE  logo  

Remember Right Hand Rule

Cross Product

16  

In terms of i, j, k

i x j = k j x k = i k x i = j

j x i = -k k x j = -i i x k = -j

i x i = 0 j x j = 0 k x k = 0

i ⇒j ⇒k ⇒i positive

i ⇐ j ⇐ k ⇐ i negative

Moment of a Force

17  

posiBon  vector  from  O  to  any  point  on  the  line  of  acBon  of  F  

Primary use of the cross product in statics

Moment About an Axis

18  

Triple scalar product

In this application, axis “a” is the y-axis, making unit axis vector ua simply base vector j.

Resultant Moments

19  

Resultant Moments

20  

Force Couples

21  

answers.com  

Equal and opposite forces acting at a distance

No net force

Force Couples

22  

These  all  have  the  same  net  moment!  

Q:  Why  is  this  a  useful  observaBon?  A:  It  speeds  up  computaBons.  Look  for  couples!  

Force Couples

23  

d  r!

M Fd= !

=

rM F! !!vector connecting any two

points on lines of actionperpendicular distance between fo

The force at th

r

e

ces

tip of

d

=

=

=

r

F r!

!

!

Equivalent Force-Moment Systems

24  

PRINCIPLE OF TRANSMISSIBILITY

P  

F  

P   F  

d  

M=Fd   F  

For  both  of  these,  the  equivalent  load  at  P  is  the  same…  

You can slide any force along its line of action without affecting the equivalent load!

Equivalent Force-Moment Systems

25  

An equivalent shifted force always exists for a set of coplanar forces. The net force for coplanar forces is itself in the plane. The moment for coplanar forces is perpendicular to the plane.

For these, you can always find a location to place the net force such that the moment about “O” is unchanged.

( )R OM

dF

=

Distributed Loads

26  

Demand  the  moment  to  be  the  same:  

Equilibrium

27  

forces = 0!

moments = 0!

No acceleration – at rest or constant velocity

Equilibrium

28  

Free Body Diagram

Define coordinate system Write all known and unknown forces and moments in terms of the coordinate system

Sum all forces to zero Sum all moments to zero

Support Reactions

29  

zero horizontal resistance. has vertical resistance. zero moment resistance.

has horizontal resistance. has vertical resistance. zero moment resistance.

has horizontal resistance. has vertical resistance. has moment resistance.

Two-Force Members

30  

A  two-­‐force  member  is  a  body  or  part  that  is  subjected  to  exactly  two  forces,  and  no  moments.    The  equilibrium  equaBons  then  imply  that  those  two  forces  must  be  equal  in  magnitude,  opposite  in  direcBon,  and  be  collinear  (i.e.,  act  along  the  same  line  of  ac4on.)  

Trusses

31  

Usually, one support is a roller to allow expansion and contraction from temperature changes

•  A structure composed of two-force members – tension or compression

•  Loads are applied only at the joints •  Joints are frictionless pins •  Member weight is insignificant

Trusses

32  

Method of Joints – FBD at each joint (pin joints)

Method  of  joints:  apply  force  balance  at  pins.    Since  force  balance  has  only  two  equaBons  (ΣFx=0  and  ΣFx=0),  pick  joints  that  have  only  two  unknown  forces!  

Force balance gives: C  T  

T  

As  part  of  your  checks,  first  guess  tension  “T”  or  compression  “C.”  

Trusses Method of Joints – FBD at each joint (pin joints)

Joint  (pin)  FBD  has  the  applied  loads  and  link  loads.    Link  (spar  or  element)  FBD  has  only  axial  load!      Links  must  be  two-­‐force  members.  (thus  equal  and  opposite  forces)  

33  

Trusses Method of sections – make a ‘cut’ in the structure

34  

Trusses Method of sections: Make cuts with no more than 3 unknowns.

Apply 3 statics equations (ΣFx=0, ΣFy=0, ΣM=0) to get the 3 unknowns!

35  

Trusses Steps for solving truss problems

•  Find the reactions by looking at the truss as a single rigid body

•  Use the method of joints if the unknown forces to be solved are near or at the joint with a known (applied) force.

•  Use the method of sections if the unknown forces to be solved are not at or near the known (applied) forces.

36  

Frames Not necessarily two-force members – loads can be applied anywhere on themembers

37  

Frames Internal forces

They  don’t  appear  in  this  FBD:  

38  

Internal Forces

39  

Make a virtual cut in a member and analyze the internal forces

Internal Forces

40  

Sign conventions

Internal Forces

41  

Load, shear, and moment relationships

•  Shear is the integral of the load

•  Moment is the integral of the shear

•  Beam slope is the integral of the moment

•  Beam deflection is the integral of the slope

42  

Friction

configuration matters!

43  

Friction

F P=( )F Nµ!

N W=

To be in equilibrium, friction is however large it needs to be treated as a pin support at an unknown location

moment  balance:  

44  

Friction

Frictional limit: At some point, P will reach a peak  value,  Fs,  large  enough  to  start  dragging  the  block.    

s sF Nµ=

Impeding motion

45  

Friction

F = µsN

F = µkN

46  

Friction

sF Nµ!( only at impending motion)sF Nµ=

Methodology:  assume  that  fricBon  is  a  pin  support  located  at  an  unknown  posiBon.  Find  these  three  unknowns  (horizontal  and  verBcal  pin  reacBons,  as  well  as  pin  locaBon)  by  enforcing  three  equaBons  of  equilibrium.  Validate  “no  slip”  assumpBon  by  confirming  that  F  <  µsN.  

47  

Friction To  slip,  both  ends  must  slip,  which  implies  that  BOTH  ends  will  be  at  impending  slip.  Seing  FA=  µANA  and  FB=  µBNB,  the  FBD  has  THREE  unknowns:  NA,  NB,  and  angle  θ.  Use  the  THREE  equilibrium  equaBons  to  solve  for  them!    Answer:  

48  

Wedges

49  

Screws Screws are wedges around a cylinder

How many threads?

Lead = Number of threads x Pitch

Pitch = 1/(number of threads per inch)

50  

Screws

Upward Motion Self locking

51  

Screws Downward Motion

Non-­‐self-­‐locking  

FricBon  angle  is  smaller  than  the  thread  (wedge)  angle  

The  moment  must  resist  the  moBon  

Self-­‐locking  

FricBon  angle  is  bigger  than  the  thread  (wedge)  angle  

The  moment  must  start  the  moBon  

52  

Belts

Tension  is  different  at  each  end  because  of  fricBon!  (Contrast  with  a  pully)    

53  

Belts

dT dT

µ !" =

54  

Belts Tension increases in the direction of impending motion

2 1( )2

1

T eT

µ ! !"= 2!1!

If  fricBonless  (µ=0),  then  T2=  T1    (like  pulley).    With  fricBon,  however,  T2>  T1  (consistent  with  limit µ→∞).  

Usually written as: where β is the angle of contact

Moment of Inertia

about  an  axis  polar  moment  of  inerBa  

Moment of Inertia Parallel axis theorem

Use this if you know Moment of Inertia about the primed axes passing thru the centroid, but you seek it with respect to some different set of parallel axes.

!

d2 = dx2 + dy

2

Moment of Inertia Radius of Gyration

The equivalent moment of inertia of a body that is a point mass at a distance (the radius of gyration) from the axis.

!

Ix = Akx2

kx =IxA

!

Iy = Aky2

ky =IyA!

Io = Ako2

ko =IoA

Moment of Inertia

Moment of inertia for Composite Areas:

Add them together (subtract for holes).

Make sure that all the moments of inertia for all parts are found about the same axis.

Use tables and parallel axis theorem.

Centroids - Lines

Centroids - Areas

Centroids – Composite Bodies