Stresses Analysis

11/16/2010

1

Simply Supported Beam witha Centered Load

Relationships:R1 = R2 = F / 2 VAB = R1VBC = -R2

Nomenclature:F = Loading force R = Reaction force V = Shear force M = Moment l = beam length x = location

2xFMAB

=

2)xl(FMBC =

Simply Supported Beam with an Intermediate Load

Relationships:R1 = F . b / lR2 = F . a / l VAB = R1VBC = -R2

lxbFMAB

=

l)xl(aFMBC =

11/16/2010

2

Simply Supported Beam with a Moment Load

Relationships:R1 = -R2 = MB / lV = MB / l

lxMM BAB

=

l)lx(MM BBC =

Simply Supported Beam with an Overhang Load

Relationships:R1 = -F * a / lR2 = F / l * (l + a)VAB = -F * a / lVBC = F

lxaFMAB

=

)alx(FMBC =

11/16/2010

3

Simply Supported Beam with an Uniform Load

Relationships:R1 = R2 = w . l / 2

xw2

lwV =

)xl(2

xwM =

Simply Supported Beam with Twin Loads

Relationships:R1 = R2 = F VAB = FVBC = 0VCD = -FMAB = F . xMBC = F . aMCD = F (l -x)

11/16/2010

4

Cantilever Beam with End Load

Relationships:R1 = V = F M1 = -F . lM = F (x - l)

Cantilever Beam with Intermediate Load

Relationships:R1 = V = F M1 = -F . aMAB = F (x - a)MBC = 0

11/16/2010

5

Cantilever Beam with Uniform Load

Relationships:R1 = w. l M1 = -(w. l 2 ) / 2V = w (l - x)

2)xl(2wM =

Cantilever Beam with Uniform Load

Relationships:R1 = w. l M1 = -(w. l 2 ) / 2V = w (l - x)

2)xl(2wM =

HW

45

P

P=2 angka terakhir (N)w = 5 N/ml= 10 m

A

B

11/16/2010

6

Cantilever Beam with a Moment Load

Relationships:R1 = 0 M1 = MBM = MBV = 0

Fundamental of

Mechanical DesignPart II Mechanics of Material

y

x

z

yzyx

zx

zy

xy

11/16/2010

7

TopicsPart I. Structural Statics1. Introduction2. Force System3. Moment System4. Centroid and Center of Gravity5. Moment of Inertia6. Equilibrium7. Structural Analysis8. Internal ForcesPart II. Mechanics of Material9. Bending Stress10. Transverse Shear Stress11. Torsion12. Stress Analysis13. Deflection14. Statically Indeterminate Beams15. Column

Bending Stress>>

11/16/2010

8

Bending StressStraight member x

xslim0x

=

r

yr

r)yr(lim0x

=

+

=

= rx+= )yr(s

r/cr/y

max

=

y sx

r

c

maxc

y=

maxc

y=

Bending Stress

Force equilibrium

====A

maxmax

AA AydA

cdA

c

ydAdP0

Moment

0ydAso,0c A

max=

===A

2max

A AdAy

cdAydPyM =

A

2dAyIif

IMc

max =

Stress at any intermediate distance y is: IMy

=

maxc

y=

11/16/2010

9

Bending StressExample:Aluminum beam with rectangular cross section (4 x 4 cm) experience force F = 100 N as shown in figure. If length of beam l = 2 m, find the maximum bending stress.

Solution:

MPa69.4m/N1069.4104

121

02,050I

Mc 2684

max ==

==

MB=50 Nm2cm

4cm

D

MD= ?

Bending StressExample:Stress distribution

11/16/2010

10

Physical Properties of Materials

Source: Machine Elements, BJ, Hamrock

Material Density Modulus ofElasticity

(GPa)Yield

Strength(MPa)

Ultimatestrength

(MPa)Ductility,%EL in

2 in

Poisson'sratio

Iron 7870 207 130 260 45 0.29

Low Carbonsteel (AISI 1020)

7860 207 295 395 37 0.30

Stainless steelsFerritic, type 446

7500 200 345 552 20 0.30

Alumunium(>99.5%)

2710 69 17 55 25 0.33

Copper (99.9%) 8940 110 69 220 45 0.35

Bending StressCurved member

=r

d)rR(

=

EdrR

r

o

p

d

r

e

ri

y

k

l

m

n

roR r

CentroidalNeutral axis

==r

d)rR(EEM

11/16/2010

11

Bending StressNeutral axisif Fn = 0

=A

0dA

E, R , , d = constan =

A0dA

r

d)rR(E

=

A0dA

r

)rR(Ed

0dAr

dAREdAA

=

=

Ar

dAAR

Bending StressMoment Mz = 0

==A A

2dA

r

d)rR(E)rR(dAM

=

=

A A

22 dA

r

d)rR(rR

rdA)rR(EdM

dAr

rRrRrRrR

rMA

22

+

=

=

AAAA

2 dArdARdARr

dARrR

rM

11/16/2010

12

Bending StressCont.

=

AAAAdArdARdA

r

dARRrR

rM

)RAAr(rR

rM

=

So,)Rr(Ar)rR(M

=

Bending Stress>>

)Rr(Ar)rR(M

=

If y positive from neutral axis and eRr =

)yR(AeMy

=

===

i

or

rA r

rln

h

rdrb

h.b

rdAAR

0

i

For rectangular cross section:

b

h

For circular cross section:

c 2crrR

22+

=

11/16/2010

13

Bending Stress

The curved beam with centroidal radius 25 cm experiencesmoment M=100 Nm. The beam has rectangular cross section asshown in figure. Determine bending stress for y = 1 cm.

Example:

4cm

6cm

MPa454.110)188.24(12.024

1011008

2

1y =

=

=

cm88.242412.06

2228ln

6R ==

=

Bending Stress

The curved beam with centroidal radius 25 cm experience moment M=100 Nm. The beam has rectangular cross section as shown in figure. Determine bending stress for :1. Top layer2. Centroidal layer3. Bottom layerPredict the stress distribution!

Problem:

4cm

6cm

11/16/2010

14

Bending StressSolution

4cm

6cm

MPa869.310))12.3(88.24(12.024

1012.31008

2

TOP =

=

cm88.242412.06

2228ln

6R ==

=

Bending StressCont.

MPa167.010))12.0(88.24(12.024

1012.01008

2

centroidal =

=

MPa545.410))88.288.24(12.024

1088.21008

2

bottom =

=

11/16/2010

15

Transverse Shear Stress

>>

Transverse Shear StressConcept:

How transverse shear is developed.

11/16/2010

16


Deformation due to Transverse Shear

Cantilevered bar made of highly deformable material and marked with horizontal and vertical grid lines to show deformation due to transverse shear. (a) Undeformed; (b) deformed.


Three-dimensional and profile views of moments and stresses associated with shaded top segment of element that has been sectioned at y about neutral axis. (a) Three-dimensional view; (b) profile view.

11/16/2010

17


ItyA

dxdM

tdxdF

==

yAI

dMdAyI

dMdFarea

==

VdxdM

=

ItyA.V

=

and

so

Transverse Shear StressFormula:

ItyVA

=

V is always the absolute value of the total shear (from the V diagram) on thebeam at the section being investigated (this is usually the point where the shearis greatest).A is always the area on the face of the structural section on either side of theplane at which shear is being investigated. For most beam applications this is thearea above (or below) a horizontal plane through the neutral axis of a beam.y is always the distance from the neutral axis to the centroid of the area A.I is always the moment of inertia of the entire cross sectional area of the beamwith respect to the neutral axis.

t is always the total width of the beam at the plane where the shear is beinginvestigated (this is usually the width at the neutral axis).

ItVQ

= If QyA =or

11/16/2010

18

Transverse Shear StressExample:Aluminum beam with rectangularcross section (4 x 4 cm)experience force F = 100 N asshown in figure. If length of beaml = 2 m, find the maximumtransverse shear stress atcentroidal line.

Solution:

24

84

6

max /10....04.0104

121

)10124)(50(mN

ItVQ

=

==

50 N

Transverse Shear StressProblem:Given:The beam with loading condition shown belowDetermine:The horizontal shearing stress at 2 cm increments from top to bottom of the beam. Plot these stresses - use maximum vertical shear on the beam

3kN 5kN 2kN

80cm 60cm 60cm 80cm

8 cm

8 cm

4 cm

11/16/2010

19

Transverse Shear StressSo

lu

tio

n

Transverse Shear Stress1st increment

2

83m/kN96.563

04.010164121

07.0)02.004.0(5.5It

yVA=

==

2nd increment

2

83m/kN79.966

04.010164121

06.0)04.004.0(5.5It

yVA=

==

4

8

8 7

11/16/2010

20

Transverse Shear Stress3rd increment

2

83m/kN49.1208

04.010164121

05.0)06.004.0(5.5It

yVA=

==

4th increment

2

83m/kN05.1289

04.010164121

04.0)08.004.0(5.5It

yVA=

==

Transverse Shear Stress

Stress distribution

8 cm

8 cm

4 cm

kPa96.563=kPa79.966=

kPa49.1208=kPa05.1289=

11/16/2010

21

Transverse Shear StressMaximum Shear Stress

Torsion>>

11/16/2010

22

TorsionConcept:

TdAcA

max =

TdAc A

2max=

oc

max

dA

TorsionCont.

pmax I

Tc=

TdAc A

2max=

pA

2 IdAif =

pIT

=For general relationship:

= area polar moment of inertia

So,

11/16/2010

23

TorsionPower TransferPower is the rate of doing work orhp = Force x Velocity

=

phT

Where : F = force, Nu = velocity, m/sT = torque, Nm = 2pif = rotational speed, rad/sf = frequency, hertz

hp = F . u

hp = T .

or

TorsionExample:A shaft with diameter 10 mm carries a torque of 30 Nm. Find the maximum shear stress due to torsion!

1044

0

4

0

32 1082.92

)005.0(24

22 ====== pipi

pipi cddAIcc

Ap

MPaITc

p

8.1521082.9005.030

10max =

==

Solution:

11/16/2010

24

TorsionAngle of twist

= dcdxmax

Hookes law:

Gmax

max

=GI

Tcp

max =

then: = dcdxGI

Tc

p

d

maxdx

cA

BSee linear deflection (segment de) :

e

TorsionAngle of twist

d

maxdx

cA

B

GITdxd

p=

==B

A px

xB

A GIdxTd

11/16/2010

25

TorsionExample:A 50 mm-long, circular, hollow shaft made of carbon steel must carry a torque of 5000 Nm at maximum shear stress of 70 MPa. The inside diameter is 0.5 of the outside diameter 72.94 mm. If the shear modulus rigidity (G) for carbon steel is 80 GPa, find the angle of twist !

GITLdx

GIT

GIdxT

p

L

0p

B

A px

x===

rad396 10799.1108010737.105.05000

=

=

4612412444

2 10737.12

10.235.182

10.47.3622

mrrdAI io

Ap

==== pipipipi

o103.0296.5700179.0 ==

Torsion

Problem:1. A shaft is used for electric motor with power 8 kilowatts at

frequency 40 Hz. Find the maximum shear stress if thediameter of shaft is 16 mm!

2. A 50 mm-long, circular, hollow shaft made of carbon steelmust carry a torque of 5000 Nm at maximum shear stress of70 MPa. The inside diameter is 0.5 of the outside diameter.Find the minimum outside diameter !

11/16/2010

26

TorsionSolution (1):

Nm83.31402

8000hT p =pi

=

=

4944

A

2p m10434.62

)008.0(2cdAI =pi=pi==

MPa58.3910434.6

008.083.31ITc

9p

max =

==

TorsionSolution (2):

pi=

pi

pi=pi==

4

o

i4

o4

i4

or

r

3

A

2p r

r12r

2r

2rd2dAI

o

i

5.0c

r

r

r

dd i

o

i

0

i===

366

max

p m1043.711070

5000Tc

I=

=

=

2d

rc oo ==

11/16/2010

27

TorsionSolution (2) cont.:

363o m1051.48r =

364

o

i3

o m1043.71r

r12r

=

pi

So, the minimum outside diameter is 72.94 mm

m03657.0ro =

PR:

Sebuah motor listrik dengan daya 100 kW, putaran 1500 rpm, menggerakan

Sebuah pompa centrifugal dihubungan dengan sebuah poros pejal (solid shaft). Bahan poros adalah stell st 40 dengan modulus elastisitas geser (G) =70 GPA, dan tegangan geser adalah 70 MPA. Panjang poros 40 cm. Tentukan

1. Diameter Poros

2. Sudut puntir poros.

11/16/2010

28

Stress Analysis

.

y

x

z

yzyx

zx

zy

xy

Stress AnalysisStress Element

A B

C

x

y

x

xy

xy

xy

yx

y

x

z

yzyx

zx

zy

xyx

y

yx

xyx

y

Plane Stress

11/16/2010

29

Plane Stress

+

++

= 2sin2cos22 xy

yxyx'x

= 0F 'x+++= cossindAsincosdAsinsindAcoscosdAdA xyxyyx'x

( ) ( ) +++= 2sin2

2cos12

2cos1xyyx'x

++= cossin2sincos xy2y2x'x

Plane Stress Transformation

A B

C

x dA cos

y dA sin

x dA

xy dA cos

xydA

xy dA sin

yx

Plane Stress

yx

xy1

22tan

=

= 0F 'y

Similary, the shear stress in oblique plane can be expressed as:

02cos22sin22d

dxy

yx'x=+

=

Plane of maximum and minimum normal stress:

+

= 2cos2sin2 xy

yx'y'x

So,

11/16/2010

30

Plane Stress

0'y'x =

In this condition,

Principle Normal StressMaximum and minimum normal stress:

( ) xy22

yxyx21min

max'x 22

, +

+==

Plane Strain22

21 221

2,

+

+

=yx

xyyx

=

yx

xy

1tan2

Maximum shear strain,

Principle Normal Strain

22

max 221

+

= yxxy

Orientation of principle axis,

11/16/2010

31

Plane Stress

xy

yx2 2

2tan

=

2'

yx +=

In this condition,

0d

d'y'x

=

xy2

2yx

minmax 2

+

=

Plane of maximum and mimimum shear stress:

Principle Shear Stress

Maximum and minimum shear stress:

Plane Stress

++= 2cos222121

= 2sin2yx

If the principle stress are known, the normal and shear stress at any oblique plane at angle can be determined from following equation:

Stress transformation

11/16/2010

32

Three-Dimensional Stresses

0)2()()(

222

22223

=+

+++++

xyzzxyyzxzxyzxyzyx

zxyzxyzyzxyxzyx

231

3/1

=

221

2/1

=

232

3/2

=

Principle shear stresses

Principle stresses can be determined by finding the three roots to the cubic equation:

Plane Strain

2cos222121 +

+=

2sin)( 21 =

If the principle strains are known, the strains at any oblique plane at angle can be determined from following equation:

11/16/2010

33

Mohrs Circle

xy

x

2

Plane Stress

2

2yx

c

+=

y

y

1

2

1

2

Deflection>>

11/16/2010

34

DeflectionFrom bending stress

r

y=

r

Ey=

m

dxx

dr

ds

dAr

EyP =

= y.ydArEM

r

EIM =EIM

r

1=

= drdsdsd

r

1 =

ds dx and tan = dy/dx

Mdx

ydEI 22

=

m1

x

y

Nomenclature:y = Deflection = Slope M = Moment x = location E = Modulus of elasticity I = Area moment of Inertia

The equation of the elastic curve (moment curvature relation) :Equation of the Elastic Curve

( )

( )

( ) 2100

10

2

21

CxCdxxMdxyEI

CdxxMdxdyEIEI

xMdx

ydEIEI

xx

x

++=

+=

==

Constants C1 & C2 are determined from boundary conditions

11/16/2010

35

Boundary condition:Equation of the Elastic Curve

Three cases for statically determinant beams,

Simply supported beam

0,0 == BA yy

Overhanging beam0,0 == BA yy

Cantilever beam0,0 == AAy

Equation of the Elastic CurveOverhanging beam Reactions at A and C Bending moment diagram

Curvature is zero at points where the bending moment is zero, i.e., at each end and at E.

EIxM )(1

= Beam is concave upwards where the bending

moment is positive and concave downwards where it is negative.

Maximum curvature occurs where the moment magnitude is a maximum.

An equation for the beam shape or elastic curve is required to determine maximum deflection and slope.

11/16/2010

36

A simply supported beam with length l experiences uniform load asshown in figure. Assume that the cross section is constant along thebeam and that the material is the same throughout, implying that thearea moment of inertia I and the modulus of elasticity E are constant.Find the deflection of any x.

Example 1:

q = wo N/m

lx

y

Example

2xw

2LxwM

2oo

=

)xLx2xL(EI24

wy 433o +=

0y,lx0y,0x

==

==

Moment:

2xw

2Lxw

dxydEI

2oo

2

2

=

1

3o

2o C

6xw

4Lxw

dxdyEI +=

21

4o

3o CxC

24xw

12LxwEIy ++=

Boundary condition :1. x = 0, y = 02. x = L, y = 0From boundary condition 1: C2 = 0From boundary condition 2: C1 = - wo L3 /24

ExampleSolution

11/16/2010

37

Example

ft 4ft15kips50psi1029in723 64

===

==

aLPEI

For portion AB of the overhanging beam,(a) derive the equation for the elastic curve,(b) determine the maximum deflection, (c) evaluate ymax.

ExampleSOLUTION:

Develop an expression for M(x) and derive differential equation for elastic curve.- Reactions:

+==LaPR

LPaR BA 1

- From the free-body diagram for section AD,

( )LxxLaPM

11/16/2010

38

Example

PaLCLCLLaPyLx

Cyx

61

610:0,at

0:0,0at

113

2

=+===

===

Integrate differential equation twice and apply boundary conditions to obtain elastic curve.

213

12

6121

CxCxLaPyEI

CxLaP

dxdyEI

++=

+=

xLaP

dxydEI =2

2

Example

=

32

6 Lx

Lx

EIPaLy

PaLxxLaPyEI

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI

61

61

3166

121

3

22

+=

=+=

Substituting,

11/16/2010

39

Example Locate point of zero slope or

point of maximum deflection.

0dxdy

=

=

2

Lx31

EI6PaL

dxdy

=

2m

Lx31

EI6PaL0

L577.03

Lxm ==

Example

=

32

6 Lx

Lx

EIPaLy

Evaluate corresponding maximum deflection.

( )[ ]32max 577.0577.06 = EIPaLyEI

PaLy6

0642.02

max =

( )( )( )( )( )462

maxin723psi10296in180in48kips500642.0

=y

in238.0max =y

11/16/2010

40

Method of superposition

q = wo N/m

L/2x

yF

L/2

Principle of Superposition:Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings

A simply supported beam with length L experiences uniformload q and concentrated load F as shown in figure. Assume thatthe cross section is constant along the beam and that thematerial is the same throughout, implying that the area momentof inertia I and the modulus of elasticity E are constant.Determine the equation of the deflection curve!

Example:


q = wo N/m

L/2x

yF

L/2

11/16/2010

41

If there are multiple loads on beam, you can compute deflections foreach load separately and add the results.Procedure is facilitated by tables of solutions for common types ofloadings and supports.

Solution:


L/2

F

L/2

wo

L/2 L/2

wo

L/2

F

L/2

= +

y(x) = yI (x) yII (x)+

Moment-Area Theorems Geometric properties of the elastic curve

can be used to determine deflection and slope.

=

=

==

D

C

D

C

D

C

x

x

CD

x

x

dxEIM

dxEIMd

EIM

dxyd

dxd

2

2

Consider a beam subjected to arbitrary loading,

First Moment-Area Theorem:

area under (M/EI) diagram between C and D.

=CD

11/16/2010

42

Moment-Area Theorems

Second Moment-Area Theorem:The tangential deviation of C with respect to D is equal to the first moment with respect to a vertical axis through C of the area under the (M/EI) diagram between Cand D.

Tangents to the elastic curve at P and Pintercept a segment of length dt on the vertical through C.

=

==

D

C

x

x

DC dxEIM

xt

dxEIM

xdxdt

1

11

= tangential deviation of C with respect to D

Application of Moment-Area Theorems

Cantilever beam - Select tangent at A as the reference.

ADD

ADD

A

ty

=

=

=

,0with

Simply supported, symmetrically loaded beam - select tangent at C as the reference.

CBB

CBB

C

ty

=

=

=

,0with

11/16/2010

43

Bending Moment Diagrams by Parts

Determination of the change of slope and the tangential deviation is simplified if the effect of each load is evaluated separately.

Construct a separate (M/EI) diagram for each load.

- The change of slope, D/C, is obtained by adding the areas under the diagrams.

- The tangential deviation, tD/C is obtained by adding the first moments of the areas with respect to a vertical axis through D.

Bending moment diagram constructed from individual loads is said to be drawn by parts.

Moment-Area Theorems

For the prismatic beam shown below, determine the slope and deflection at E.

EXAMPLE:

11/16/2010

44

Moment-Area TheoremsSOLUTION:

Determine the reactions at supports.

waRR DB ==

Construct shear, bending moment and (M/EI) diagrams.

( )EI6

waa

EI2wa

31A

EI4Lwa

2L

EI2waA

32

2

22

1

=

=

=

=

Moment-Area Theorems Slope at E:

EI6wa

EI4LwaAA

32

21

CECECE

=+=

=+=

( )a2L3EI12

wa2

E +=

=

+

+=

=

EI16Lwa

EI8wa

EI16Lwa

EI4Lwa

4LA

4a3A

4L

aA

tty

224223

121

CDCEE

( )aL2EI8

way3

E +=

Deflection at E:

11/16/2010

45

Application of Moment-Area Theorems to Beams With Unsymmetrical Loadings

Define reference tangent at support A. Evaluate A by determining the tangential deviation at Bwith respect to A.

Lt AB

A =

The slope at other points is found with respect to reference tangent.

ADAD +=

The deflection at D is found from the tangential deviation at D.

ABADD

AB

tLx

tEFEDy

tLxEF

LHB

x

EF

==

==

Maximum Deflection Maximum deflection occurs at point

K where the tangent is horizontal.

AAK

AKAK

ABA L

t

=

+==

=

0

Point K may be determined by measuring an area under the (M/EI) diagram equal to -A .

Obtain ymax by computing the first moment with respect to the vertical axis through A of the area between Aand K.

Documents

Stresses Analysis