MD III LoadStress

8/12/2019 MD III LoadStress

1/29

1

Load and StressAnalysis

Section III


2/29

2

Introduction about stresses

Shearing force and bending moment

diagrams Bending, Transverse, & Torsional stresses

Compound stresses and Mohrs circle

Stress concentration

Stresses in pressurized cylinders, rotatingrings, curved beams, & contact

Talking Points


3/29

3

Assume downward force as negative and upwardforce as positive; and counterclockwise moment as

positive and clockwise as negative. Loads may act on multiple planes.

Introduction about stressesBody Diagram-i. Static Equilibrium and Free

0F 0M0T


4/29

4

The load is applied along

the axis of the bar(perpendicular to thecross-sectional area) and itis uniformly distributedacross the cross-sectionalarea of the bar.

The normal stress can betensile (+) or compressive(-) depending on thedirection of the appliedload P.

The stress unit in N/m2orPa or multiple of this unit,

i.e. MPa, GPa.


Cont.ii. Direct Normal Stress & Strain

A

P

E

Assuming elasticity

oLL

AP

E

E

A

LPL o

oL

L

Hookes Law


5/29

5

Sometimes, a body is subjectedto a number of forces acting onits outer edges as well as atsome other sections, along thelength of the body. In such case,the forces are split up, and theireffects are considered onindividual sections. The resultingdeformation of the body is equalto the algebraic sum of thedeformation of the individual

sections. Such a principle offinding out the resultantdeformation is called theprinciple of superposition.


Cont.

n

E1

o

A

LPL

Principle of Superposition:

L1 L2 L3

d1 d2

d3

L3

L2

L1

d3

d2

d1

P1

P2

P3

P4


6/29

6


Cont.Example on Principle of Superposition:

A brass bar, having cross sectional area of 10 cm2is subjected to axial forces asshown in the figure. Find the total elongation of the bar (L). Take E= 80 GPa.

L= -150 m


7/29

7

For engineering materials, n= 0.25 to 0.33.

For a rounded bar, the lateral strain is equal to the reduction

in the bar diameter divided by the original diameter.


Cont.s Ratioiii. Poisson

StrainAxialStrainLateralRatiosPoisson' n

x

z

x

y

n or

s Law:From Hooke E

xx

E

xzy

n

For 1D stress system ( )1D stress

system

0 zy

For 2D stress system ( )0,0 zy

yxx

En

1 xyy

En

1and


8/29

8


Cont.s Ratio:Example on Poisson

A 500 mm long, 16 mm diameter rod made of a homogenous, isotropic material isobserved to increase in length by 300 mm, and to decrease in diameter by 2.4 mmwhen subjected to an axial 12 kN load. Determine the modulus of elasticity andPoissons ratio of the material.

E = 99.5 GPan= 0.25


9/29

9


Cont.iii. Direct Shear Stress & Strain

Assuming elasticity The load, here,

is applied in adirectionparallel to thecross-sectionalarea of the bar.

AQ

G

StrainShear

Gis known asmodulus of rigidity

Single & Double Shear

The rivet is subjected

to single shearThe rivet is subjectedto double shear A

Q

2

n12G

E

Relation betweenn, andG,E

Q

Q


10/29

10

Shearing Force (S.F.) and BendingMoment (B.M.) Diagrams

Simply supportedbeam

Cantilever beam

Sign Convention

Relationship between shear forceand bending moment

dx

dMQ

QdxM Or

diagramforceshearunder theareaTheM


11/29

11

Shearing Force (S.F.) and BendingMoment (B.M.) Diagrams - Examples

i. Concentrated Load Only:


12/29

12


ii. Distributed Load Only:


13/29

13


iii. Combination of Concentrated and Distributed Load:


14/29

14


iv. If Couple or Moment is Applied:


15/29

15

Bending, Transverse, &

Torsional stresses

I

yM

i. Bending Stress

where, Mis the applied bending moment (B.M.) at a transverse

section, Iis the second moment of area of the beam cross-section

about the neutral axis (N.A.), i.e. , is the stressat distance yfrom the N.A. of the beam cross-section.

dAyI 2


16/29

16

ii. Transverse Stress

Bending, Transverse, & Torsional

stresses

Cont.

Ib

yAQ

where Qis the applied vertical shear force at that section;A

is the area of cross-section abovey, i.e. the area between y

and the outside of the section, which may be above or below

the neutral axis (N.A.); y is the distance of the centroid of

areaAfrom the N.A.; I is the second moment of area of the

complete cross-section; and b is the breadth of the section atposition y.

or dAyIb

Q

d

b

R


17/29

17

iii. Torsional Stress

Bending, Transverse, & TorsionalstressesCont.

J

T

where Tis the applied external torque; is the radial direction

from the shaft center; Jis the polar second moment of area of

shaft cross-section; ris the shaft radius; and is the shear

stress at radius .

J

rT

max

4

2

1rJ 44

2

1io rrJ

Solidsection

Hollow shaft

when torsion is presentNote:

Ductile materials tends to break in a plane perpendicularto its longitudinal axis; while brittle material breaks alongsurfaces perpendicular to direction where tension ismaximum; i.e. along surfaces forming 45oangle withlongitudinal axis.

Ductile material Brittle material


18/29

18

Compound stresses and Mohrs

circle

Machine Design involves among other

considerations, the proper sizing of a machinemember to safely withstand the maximumstress which is induced within the memberwhen it is subjected separately or to anycombination of bending, torsion, axial, ortransverse load.


19/29

19

Compound stresses andMohrs circleCont.

Maximum & Minimum Normal Stresses

2

2

(min)

2

2

(max)

22

22

xy

yxyx

n

xy

yxyx

n

Stress State 3D GeneralStress State

2D StressState

D Case:2For

Where:

xis a stress at a critical point in tension or compression normal to the

cross section under consideration, and may be either bending or axialload, or a combination of the two.

yis a stress at the same critical point and in direction normal to the x

stress.

xy is the shear stress at the same critical point acting in the plane normalto the Y axis (which is the XZ plane) and in a plane normal to the X axis(which is the YZ plane). This shear stress may be due to a torsionalmoment, transverse load, or a combination of the two.

n(max)and n(min)are called principal stresses and occurs on planes thatare at 90to each other, called principle planes also planes of zero shear.

Note: x, y, zall+ve, xy,yx, zy, yz, xz, zxall+ve.Due to static balance, xy=

yx, zy= yz, and xz= zx.

Counterclockwise (CCW)

Clockwise (CW)


20/29

20


yxxy

22tan

maxat the critical point being investigated is equal to half of the greatest difference ofany of the three principal stresses. For the case of two-dimensional loading on a particle

causing a two-dimensional stresses; The planes of maximum shear are inclined at 45with the principal planes.

2

1

2minmax

2

2

max nnxy

yx

)maxMaximum Shear Stresses (

The planes of maximum shear are inclined at 45with the principal planes.

The angle between the principal plane and the X-Y plane is defined by:


21/29

21

Compound stresses and MohrscircleCont.

s CircleMohr It is a graphical method to find the maximum and minimum normal stresses

and maximum shear stress of any member.

From the diagram:x= OA, xy= AB, y=OC, and yx= CD. The line BED

is the diameter of Mohr's circle with center at E on the

axis. Point B represents the stress coordinates x,xyon

the X faces and point D the stress coordinates y,yxon

the Y faces. Thus EB corresponds to the X-axis and ED to

the Y-axis. The maximum principal normal stress max

occurs at F, and the minimum principal normal stressminat G. The two extreme-value shear stresses one

clockwise and one counterclockwise, occurs at H and I,

respectively. We can construct this diagram with

compass and scale and find the required information

with the aid of scales. A semi-graphical approach is

easier and quicker and offer fewer opportunities for

error.2-D


22/29

22


Principal Element

True views on the various faces of the principal element

MaxMin

maxis equal to half of the greatestdifference of any of the three principalstresses. In the case of the below figure:

3-D

2

13113ma x

where, 3223211221,

21


23/29

23

Example: A machine member 50 mm diameter by 250 mm long is supported at one end as a

cantilever. In this example note that y= 0 at the critical point.

Compound stresses and MohrscircleExamples.

: Axial load only:1Case : Bending only:2Case

In this case all points in the member are subjected to

the same stress.

MPa83.32MPa,65.7

0

MPa65.71096.11015AP

m1096.110504A

(max)(max)(max)

33

2323

nn

xy

x

(Shear)MPa60.302

on)(CompressiMPa1.61,0

MPa1.61

:BpointAt

(Shear)MPa60.302

0(Tension),MPa1.61

MPa1.61641050

102510250103

:ApointAt

(max)(max)

(min)(max)

(max)(max)

(min)(max)

43

333

n

nn

x

n

nn

x

I

yM

I

yM


24/29

24


: Torsion only:3Case :: Bending & Axial Load4Case

In this case the critical point occur along the outer

surface of the member.

(Shear)MPa7.2625.53

on)(CompressiMPa5.53,0

on)(CompressiMPa5.531.6165.7

:BpointAt

(Shear)MPa4.3428.68

0(Tension),MPa8.68

(Tension)MPa8.681.6165.7

:ApointAt

(max)

(min)(max)

(max)

(min)(max)

nn

x

nn

x

I

yM

A

P

I

yM

A

P

(Shear)MPa7.40

on)(CompressiMPa7.40

(Tension)MPa7.40

MPa7.40321050

1025101

0

(max)

(min)

(max)

43

33

n

n

xy

x

JrT


25/29

25


: Bending & Torsion:5Case :: Torsion & Axial Load6Case

(Shear)MPa9.502on)(CompressiMPa4.81

(Tension),MPa3.20

MPa7.40

MPa1.61

:BpointAt

(Shear)MPa9.502

on)(CompressiMPa3.207.402

1.61

2

1.61

(Tension),MPa4.817.40

2

1.61

2

1.61

MPa7.40

MPa1.61

:ApointAt

(min)(max)(max)

(min)

(max)

(min)(max)(max)

2

2

(min)

2

2

(max)

nn

n

n

xy

x

nn

n

n

xy

x

J

rT

I

yM

(Shear)MPa9.402

on)(CompressiMPa1.377.402

65.7

2

65.7


65.7

2

65.7

MPa7.40

MPa65.7AP

(min)(max)(max)

2

2

(min)

2

2

(max)

nn

n

n

xy

x

J

rT


26/29

26


: Bending, Axial Load, and Torsion:7Case

(Shear)MPa7.482on)(CompressiMPa5.75

(Tension),MPa9.21

MPa7.40

MPa3.531.6165.7

:BpointAt(Shear)MPa3.532

on)(CompressiMPa197.402

8.68

2

8.68


8.68

2

8.68

MPa7.40

MPa8.681.6165.7

:ApointAt

(min)(max)(max)

(min)

(max)

(min)(max)(max)

2

2

(min)

2

2

(max)

nn

n

n

xy

x

nn

n

n

xy

x

I

yM

A

P

JrT

I

yM

A

P


27/29

27

s circle:Example on Mohr The stress element shown in figure has x= 80 MPa

and xy, = 50 MPa (CW). Find the principal stressesand directions.


Locate x= 80 MPa along the axis. Then from x,

locate xy= 50 MPa in the (CW) direction of the axis to

establish point A. Corresponding to y = 0, locate yx= 50

MPa in the (CCW) direction along the axis to obtain point

D. The line AD forms the diameter of the required circle

which can now be drawn. The intersection of the circlewith the axis defines max and minas shown.

3.514050tan2

:isCW toaxis-Xthefrom2angleThe

MPa246440MPa,1046440

MPa644050

1-

max

(min)(max)

22

(max)


28/29

28

Stress Concentration

Occurs when there is sudden changes in cross-sections of membersunder consideration. Such as holes, grooves, notches of variouskinds.

The regions of these sudden changes are called areas of stressconcentration.

Stress-concentration factor (Kt or Kts)

The analysis of geometric shapes to determine stress-concentration

factors is a difficult problem, and not many solutions can be found.

o

ts

o

t KK

maxmax Theoretically


29/29

29

Stresses in pressurized

cylinders, rotatingrings, curved beams,

& contact

Documents

MD III LoadStress