Dasar Perancangan Mekanikal Bab 1

1Jos Istiyanto, ST. MT.

Departemen Teknik MesinFakultas Teknik Universitas Indonesia

DTM FTUI

Mechanical DesignPart I. Fundamental

Mechanical DesignPart I. Fundamental

y

xz

yzyx

zxzy

xy

2Departemen Teknik MesinFakultas Teknik Universitas Indonesia

DTM FTUI

Introduction>>


DTM FTUI Definition

What is Design?Design is the transformation of concepts and ideas into useful machinery.A machine is a combination of mechanism and other component that tranforms, transmits, or uses energy, load, or motion for specific purpose. (hamrock)Fundamental decisions regarding loading, kinematics and the choice of materials must be made during the design of a machine.


DTM FTUI Definition

What is Mechanics?Mechanics is the branch of physical science that deals with the response of bodies to the action of forces.

Three areas of mechanics:1> The mechanics of rigid bodies

- statics (equilibrium of bodies)- dynamics (accelerated motion of bodies)

2> The mechanics of deformable bodies (Mechanics of Material)

3> The mechanics of fluids


DTM FTUI The Mechanics of Rigid Bodies

What is a rigid-body?A rigid-body is a solid which does not deform when force are applied (idealisation).

Statics is concern with bodies that are acted on balanced force and hence are at rest or have uniform motions. equilibrium bodies

Dynamics is concern with accelerated motion of bodies


DTM FTUI Mechanics of deformable bodies

Is the branch of mechanics that deals with internal force distribution and the deformations developed in actual engineering structure and machine components when they are subjected to systems of force.

Mechanics of MaterialsStrength of Material


DTM FTUI Fluid Mechanics

Fluid mechanics is the branch of mechanics that deals with liquids and gases at rest or in motion.

Fluid - Compressible- Incompressible


DTM FTUI Example of Application

..


DTM FTUI

Force System>>


DTM FTUI Force System

Forces and Their Characteristics

The characteristics of a force are as follows:

Its magnitudeThe amount or size of the force

Its direction (orientation)

Orientation of the line segment used to represent the forceIts point of application

The point of contact between the two bodies. A straight line extending through the point of application in the direction of the force is called its line of action.


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Principle of Transmissibility

The external effect of a force on a rigid body is the same for all points of application of the force along its line of action.

PUSH PULL

Rigid Ring : Deformable Ring :


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Classification of Forces

With respect to the area over which they act :

2. Concentrated force

1. Distributed force


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Force system

a. Concurrent forcesThe action lines of all forces intersect at common point

b. Coplanar forcesAll forces lie in the same plane

F1F2

F3

Any number of forces treated as a group constitute a force system.

F1F2F3


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Force systemc. Parallel forcesThe action lines of the forces are parallel, but the orientation of the forces do not have to be same.

d. CollinearThe forces of a system have a common line of action.

F

F


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Force system

Graphical method

The parallelogram law

F2R

F1O

F2

F1

F2

F1O

RR

The Triangle law


DTM FTUI

Force system

Basic of analytical method

The cosine lawc2 = a2 + b2 - 2 ab cos

The sine law

a

b

c

== sinc

sinb

sina


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Force system

Analytical method

Direction of Resultant

F1

F2

R

Magnitude of Resultant:

++= cosFF2FFR 212221 RsinFsin 2 =

= RsinFsin 21


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Force systemExample :

F1=800 N

F2=500 N

450

300

Determine the magnitude of the resultant R and the angle between the horisontal axis and the line of action of Resultant

5.1206R45cos500.800.2500800R

cosFF2FFR022

212

22

1

=++=

++=

0

o121

04.17

5.120645sin500sin

RsinFsin

=

=

=

= + 17.04o = 47.04oF1=800 N

F2=500 N

450

300

R

10


DTM FTUI

Moment System>>


DTM FTUI Moment System

Moment and their characteristics

The moment of a force about a point or axis is a measure of the tendency of the force to rotate a body about that point or axis.

A moment has both magnitude and a direction vector quantityThe magnitude of moment M as defined as the product of the magnitude of a force and the perpendicular distance d from the line of action of the force to the axis.

MO = MO= FdPoint O = the moment center.

11


DTM FTUI Principle of Moment : Varignon's Theorem

The moment M of the resultant R of a system of force with respect to any axis or point is equal to the vector sum of the moment of the individual forces of the system with respect to the same axis or point.

R = F1 + F2 + + FnMO = RdR = F1d1 + F2d2 + + Fndn


DTM FTUI Application of Varignon's theorem

MR = R.d = R(h cos )MA = A.a = A(h cos )MB = B.b = B(h cos )

R cos = A cos + B cos

MR = MA + MB

h

AR

B cos A cos

B

b

d

a

Ax

y

O

12


DTM FTUI Vector Representation of a moment

MO = r x FMO = r x F = rFsin e(0 180o)

rd F

d Fr1

23 1

r2

r3

r1sin 1 = r2sin 2 = r3sin 3 = d


DTM FTUI Position vector

rA = xA i + yA j + zA krB = xB i + yB j + zB krA = rB + rA/B

rA/B = rA - rB

rA/BrA

z

yx

F

rB

13


DTM FTUI Moment of a force about a point

If force F and r are expressed in Cartesian vector form: F = Fx i + Fy j + Fz kr = rx i + ry j + rz k

So the moment Mo about the origin of coordinate O : MO = r x F

= (rx i + ry j + rz k) x (Fx i + Fy j + Fz k)= (ry Fz - rz Fy )i + (rz Fx - rx Fz )j + (rx Fy - ry Fx ) k = Mx i + My j + Mz k



>>

yr

z

yx

F

MO

x

z

O

The moment Mo about the origin of coordinate O can also be expressed in determinant form as :

MO = r x F = What is about two-dimensional case ?zFyFxFzryrxr

kji

222 zMyMxM ++=oM

14



Alternatively, Mo can be written as MO = MO e

Where e = cos x i + cos y j + cos z k

oxxcos M

M=oy

ycos MM=

ozzcos M

M=


DTM FTUI Moment of a force about a line (axis)

MOJ= M = (MO . en) en= [(r x F) . en] en= MOJ en

zFyFxFzryrxr

kji

J

r

z

yx

F

MO

O

en

MOJ= MO . en = (r x F) . en = . en

15


DTM FTUI Moment of a force about a line (axis)

Or alternatively as :MOJ= MO . en = (r x F) . en =

zFyFxFzryrxrnzenyenxe

J

r

z

yx

F

MO

O

en


DTM FTUI COUPLES

A couple is a system force whose resultant force is zero but whose resultant moment about a point is not zero

MA = F1d = F2dMA = rA/B x F1

= F1d en

The characteristics of couples1. The magnitude2. The sense (direction of rotation) 3. The orientation (axis about which rotation is induced)

en

rA

z

yx

F1F2

n

d

16


DTM FTUI COUPLES

Several transformation of a couple can be made without changing any of external effect of the couple on the body

dF

F

dF F

2d0.5F 0.5F

dF F

M = Fd


DTM FTUI COUPLES

Resolution of a force into a force and a couple

F

O

p

rd

F

FF O

p

FO

p

M = Fd

17


DTM FTUI

Moment systemExample :

A parcel is lifted by a fork lift truck. Determine the moment of the weight with respect to the point A at the two border positions!


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Centroid & Center of Gravity>>

18


DTM FTUI

Center of Gravity

The center of gravity G is a point which locates the resultant weight of a system of particles.

The weights of the particles is considered to be a parallel force system. The system of weights can be replaced by a single weight acting tat the Center of Gravity.


DTM FTUI

=

=n

1iiR WWTotal Weight

nn332211RR Wx~Wx~Wx~Wx~Wx K+++=x location:

nn332211RR Wy~Wy~Wy~Wy~Wy K+++=y location:

nn332211RR Wz~Wz~Wz~Wz~Wz K+++=z location:

19


DTM FTUI

=

=

=

=

=

= === n1i

i

i

n

1ii

n

1ii

i

n

1ii

n

1ii

i

n

1ii

W

Wz~

zW

Wy~

yW

Wx~

x

particleitheofweightWparticleitheofscoordinatez~,y~,x~

gravityofcentertheofscoordinatez,y,x

thi

thiii


DTM FTUI

=

=

=

=

=

= === n1i

i

i

n

1ii

n

1ii

i

n

1ii

n

1ii

i

n

1ii

m

mz~

zm

my~

ym

mx~

x

particleitheofmassWparticleitheofscoordinatez~,y~,x~

massofcentertheofscoordinatez,y,x

thi

thiii

Center of Mass

20


DTM FTUI

Centroid for a Body

Consider a body to be a system of an infinite number of particles


DTM FTUI

=

=

=

=

=

= ===1i

i

i1i

i

1ii

i1i

i

1ii

i1i

i

W

Wz~

zW

Wy~

yW

Wx~

x

particleitheofweightWparticleitheofscoordinatez~,y~,x~

gravityofcentertheofscoordinatez,y,x

thi

thiii

21


DTM FTUI

===

dW

dWz~z

dW

dWy~y

dW

dWx~x

( )VdWd

bodytheofweightspecific

=

volume unit per The weight

=

=

=V

V

V

V

V

V

dV

dVz~

zdV

dVy~

ydV

dVx~

x


DTM FTUI

CENTROID

The centroid C is a point which defines the geometric center of an object. Its location can be determined by formulas similar to those used for center of gravity or center of mass.

22


DTM FTUI

===V

V

V

V

V

V

dV

dVz~

zdV

dVy~

ydV

dVx~

x

Centroid of a Volume

For a homogeneous body, the center of gravity Gcoincides with the centroid C of the volume V of the body; the coordinates of C are defined by the relations


DTM FTUI

Centroid of an Area

===A

A

A

A

A

A

dA

dAz~

zdA

dAy~

ydA

dAx~

x

23


DTM FTUI

Centroid of an Area

==A

A

A

A

dA

dAy~

ydA

dAx~

x

For two dimensional body,

These integrals are referred to as the first moments of area A with respect to the y and x axes, and are denoted by Qy and Qx , respectively


DTM FTUI

===L

L

L

L

L

L

dL

dLz~

zdL

dLy~

ydL

dLx~

x

Centroid of a Line

24


DTM FTUI

Composite Bodies

If a body is made up of several simpler bodies then a special technique can be used.Procedure Divide body into several subparts. If the body has a hole or cutout, treat that as negative

area. Centroid will lie on line of symmetry. Create Table and calculate centroid.

y

x

20 mm 30 mm

36 mm

24 mm


DTM FTUI

AreaBody xc yc xc A yc A

25


DTM FTUI

theorems of Pappus-Guldinus

CL

x

2y

y

The theorems of Pappus-Guldinusrelate the determination of the area of a surface of revolution or the volume of a body of revolution to the determination of the centroid of the generating curve or area. The area A of the surface generated by rotating a curve of length L about a fixed axis is

A = 2yLwhere y represents the distance from the centroid C of the curve to the fixed axis.


DTM FTUI

theorems of Pappus-Guldinus

x2y

yA

C

The volume V of the body generated by rotating an area Aabout a fixed axis is

V = 2yAwhere y represents the distance from the centroid C of the area to the fixed axis.

26


DTM FTUI

Distributed load

x

w

O B

xdx

L

w

dW

x

w

O B

x

L

W

W = A

P

C

The concept of centroid of an area can also be used to solveproblems other than those dealing with the weight of flat plates.For example, to determine the reactions at the supports of abeam, we replace a distributed load w by a concentrated load W equal in magnitude to the area A under the load curve and passing through the centroid C of that area.


DTM FTUI

x

y

z

P(x,y,z)

xel

yel

zel

z xel = xyel = yzel = z

12

dV = z dx dy

dydx

When a volume is bounded by analyticalsurfaces, the coordinatesof its centroid can bedetermined by integration.To avoid the computationof triple integrals, we canuse elements of volumein the shape of thinfilaments (as shown).

27


DTM FTUI

Denoting by xel , yel , and zel the coordinates of the centroid ofthe element dV, we write

xV = xel dV yV = yel dV zV = zel dVIf the volume possesses two planes of symmetry, its centroid C is located on their line of intersection.


DTM FTUI

dx

xz

y xelxel = x

dV = r 2 dx

If the volume possesses two planes of symmetry, its centroid Cis located on their line of intersection. Choosing the x axis to lie along that line and dividing the volume into thin slabsparallel to the xz plane, the centroid C can be determined from

xV = xel dVFor a body of revolution, these slabs are circular.

28


DTM FTUI

1 cm

1 cm

2 cm 3 cm

3 cm

Locate Centroid of the Composite Area

Problem


DTM FTUI

1 cm

1 cm

2 cm 3 cm

3 cm

123

29


DTM FTUI

Segment A (cm2) x y xA yA1 4.5 1 1 4.5 4.52 6 -1 1.5 -6 93 1 -2.5 0.5 -2.5 0.5

A = 11.5 xA = -4 xA = 14

cm22.15.11

14AAy~

y

cm348.05.114

AAx~

x

===

===


DTM FTUI

1 cm

1 cm 2 cm 3 cm

3 cm

12

3

30


DTM FTUI

Segment A (cm2) x y xA yA1 4.5 1 1 4.5 4.52 9 -1.5 1.5 -13.5 13.53 -2.5 -2.5 2 5 -4

A = 11.5 xA = -4 xA = 14

cm22.15.11

14AAy~

y

cm348.05.114

AAx~

x

===

===


DTM FTUI

y

x

20 mm 30 mm

Problem

36 mm

24 mm

Locate the centroid of the planearea shown.

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DTM FTUI

Moment of inertia>>


DTM FTUI Moments of InertiaDefinition of Moments of Inertia for Areas: Used in formulas for Mechanics of Materials, Fluid Mechanics, Structural Mechanics.

Consider area A in x-y plane.By definition:

These expressions are referred to as Moments of Inertia about the x and y axes. They may also be called second-moments of inertia.

dI = y dA and dI = x dAWe integrate to obtainI and I for area A.

I = y dA and I = x dA

X2

Y2

X Y

X2

Y2 X

X

Y

Yr

dA

A

32


DTM FTUI

Definition of Moments of Inertia for Areas (Contd)

The second moment about the origin (or z-axis) is defined as the polar moment of inertia.

dJ = r dA ; J = I + I since r = x + yI , I and J > 0 , Units are ft in , m and mm

O2

O X Y2 2 2

X Y O4 4 4 4, .


DTM FTUI Parallel-Axis Theorem for Area

If MOI is known about the x-axis or y-axis through centroid (C) of area, it is possible to determine MOI about a parallel x-axis or y-axis using the parallel axis theorem.

dI = (y + y) dA , I = (y ) dA + 2y y dA + (y dA

But y dA = 0, so that:

I = I A y ; I = I A x ; and J = J A ( x + y ) .

X2

X2

X X'2

Y Y'2

O 02 2

+ + +

) 2

Y

X

X

Y

Y

X

r

C dA

33


DTM FTUI

Radius of Gyration

Radius of Gyration has units of length.Often used in Column design.

k =IA

, k =IA

, k =IAX

XY

YZ

Z


DTM FTUI Moments of Inertia for Area by Integration

If area defined by mathematical function, integration gives MOI

If differential area has differential size in two directions, double integration is used.

If differential element has differential size in only one direction, single integration is used.

X

X

Y

Y

dX

34


DTM FTUI Moments of Inertia for Area by Integration

Procedure for AnalysisIf single integration is used to obtain MOI, specify dA with

finite length and differential width.Case I - Length of element is parallel to axis, so all parts

of element lie at at same distance x from axis-y. Case II - Length of element oriented perpendicular to axis,

so moment arm varies along element. Determine dI of element.

X

X

Y

Y

dX


DTM FTUI

Moments of Inertia for Composite Areas

Composite area consists of group of connected simple shapes.

If MOI of parts about common axis can be determined, then MOI of the composite is algebraic sum of parts.

I = I , I = I , I = IX Xi

Yi

Yi

Zi

Zi

i

35


DTM FTUI Moments of Inertia for Composite Areas

(Contd)Procedure for AnalysisComposite Area Moment of Inertia about reference axis.1. Composite Parts. Divide area into composite parts.

Indicate perpendicular distance from centroid of parts to reference axis.

2. Apply Parallel Axis Theorem. Determine MOI of each part about centroidal axis parallel to reference axis. Use parallel axis theorem to calculate MOI of parts about reference axis.

3. Sum MOI of parts. Calculate MOI of component by summing MOI of parts. If any part is a hole, subtract the MOI of hole in making summation.


DTM FTUI Composite Area Moment of Inertia

EXAMPLEGiven: Composite Area about x-axis as shown.Find: Moment of Inertia of area about x-axis.

100mm

75mm

75mm

25mm

X X

36


DTM FTUI Composite Area Moment of Inertia

Solution

100mm

75mm

75mm

25mm

Approach Ay

Part mm mm I

ctCircle

a bd r

X

: )

) ( ) ( )

Re .

)( ) ( ) ( )

I = (I

A(mm y(mm) Ay I Ay

15,000 75 84.38E6 28.12E6 112.50E6- 1,963 75 - 11.04E6 - 0.31E6 - 11.35E6

I = 101.15E6(mm I (b) I

X X

2 2X

2

a

b

X2

X X

+

+

= =

2

4 4

3 412 4

X X

Dasar Perancangan Mekanikal Bab 1

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