Topic 2&3 - Stress & Strain

7/30/2019 Topic 2&3 - Stress & Strain

1/62

Mechanics of Solids

Stress and Strain

Dr. Seshu Adluri


2/62

orces - n ro uc on Forces and Moments

External Surface Tractions (Exterior)

Interior (Body Forces, etc)

Internal Equilibrium

External-external

External-internal

Internal-internal

Stress & Strain -Dr. Seshu Adluri


3/62

orces External forces A lied forces, Reactions



4/62

n erna orces

Why do they develop?

Internal equilibrium orce ransm ss on



5/62

orces

Internal forces

and Free body Diagrams



6/62

orces

Normal forces

Momentors on

Bending

Coplanar loads



7/62

na ys s o orces

Identify external forces, geometry, etc.

Draw free-body diagrams of any identified

Apply equations of equilibrium

Solve for unknowns



8/62

ress

Quantifies the intensity of internal forcesin a body



9/62

ress

Normal stress &Shear stress

FA

m



10/62

ress



11/62

ress

UnitsPa kPa MPaMPa GPa si ksi .

(N, lb, kip, mmmm, mm, in.)

Total force and total moment

orce an momen agrams



12/62

orma ress



13/62

n ec s



14/62

n ec s, e c.



15/62

verage ress



16/62

ress e

Stress field isenerall com lex

but can be

a roximated tosimpler models



17/62

Shear Stress



18/62

ear ress



19/62

ear ress



20/62

xamp e



21/62

xamp e



22/62

ress - ev ew

ForcesExternal surface bod forces reactions

Internal

Stress

,General state of stress in a solid



23/62

ra n

Effects of strain



24/62

orma ra n



25/62

orma ra n



26/62

ear ra n



27/62

ra n- e orma on e



28/62

a er a es ng n ens on



29/62

ens e es pec men

ASTM standards areused for testing

ascertain theirproperties

available in thelibrary, online and

w a ora ory



30/62

ater a a ure o es n ens on



31/62

ress- ra n urve - s ee



32/62

as c concep s

Hooks Law

(Robert Hook, 1676) Youngs Modulus

(Thomas Young, 1807)

Elasticity

Linear on near

Plastic deformation

Strain hardening



33/62

ra n energy

.

The energy of the work is transferred to the body. It is stored in the body as strain energy and is released if

the load is elastically removed.

Area under the stress-strain curve gives strain energy

For elastic deformation,

strain energystrain energy = The factor indicates gradual application of the load.


.


34/62

o u us o es ence



35/62

o u us o oug ness



36/62

-for different steels



37/62

- steels



38/62

trengtMaterial Yield MPa Ult.Stress MPa Elon ation %

Aluminum [Al] 20 70 60

Aluminum Alloy 35 - 500 100 - 550 1 - 45

Brass 70 - 550 200 - 620 4 - 60

Brass; Noval 170 - 410 410 - 590 15 - 50

Brass; Red (80% Cu, 20% Zn) 90 - 470 300 - 590 4 - 50

Brick - 7.0 - 70 -

Bronze Re ular 82 - 690 200 - 830 5 - 60

Bronze; Manganese 170 - 450 450 - 620 10 - 35

Concrete (Compression) - 10 - 70 -


Copper [Cu] 55 - 330 230 - 380 10 - 50

Copper Alloy 760 830 4


39/62

trengtMaterial Yield (MPa) Ult.Stress(MPa) Elongation (%)

Glass - 30 - 1000 -

Iron (Cast) 120 - 290 69 - 480 0 - 1

Iron (Wrought) 210 340 35

Magnesium [Mg] 20 - 70 100 - 170 5 - 15

Magnesium Alloy 80 - 280 140 - 340 2 - 20

Monel (67% Ni, 30% Cu) 170 - 1100 450 - 1200 2 - 50

- - -

Nylon; Polyamide - 40 - 70 50

Rubber 1.0 - 7.0 7.0 - 20 100 - 800


Solder; Tin-Lead - 12 - 54 55 - 30

Steel 210 - 1600 340 - 1900 3 - 40


40/62

trengt

.

Stone; Granite (Compression) - 70 - 280 -

- - -.

Stone; Marble (Compression) - 50 - 180 -

Titanium [Ti] - 500 25

Titanium Alloy - 900 - 970 10

Tungsten [W] - 1400 - 4000 0 - 4

Wood; Ash (Bending) 40 - 70 50 - 100 -

Wood; Douglas Fir (Bending) 30 - 50 50 - 80 -

- - -


Wood; Southern

Pine (Bending)40 - 60 50 - 100 -


41/62

material



42/62

oa ng un oa ng - ysteres s



43/62

ress- ra n urve -concre e



44/62

ress- ra n urve -a um n um



45/62

Stress-Strain Curve natural rubber



46/62

ress- ra n urve cas ron



47/62

- (temperature dependent)



48/62

a era s ra n e av our



49/62

a era s ra n o son s a o

lateral

r

=

allongitudinL

=

allongitudin

lateral

=



50/62




51/62


Normally Poissons ratio iscalculated for homogeneous and

lateral

r

.

If material different properties in twolateral

allongitudinL

,as orthotropic material. This willhave two different Poissons ratio

allongitudin=

va ues. For anisotropic materials, the


directions.


52/62

ear o u us



53/62

ear o u us



54/62

ear o u us

( )+=

12

G



55/62

o u

Steel E=200,000 MPa, G=77,000 MPa,n=0.28-0.33

Concrete: E=20,000 30,000 MPa, G=



56/62

o uE (GPa) G (GPa) Poisson's Ratio

Steel E=200,000

MPa, G=77,000 MPa,

u y .

Monel (67% Ni, 30% Cu) 170 66 0.32

Nylon; Polyamide 2.1 - 2.8 - 0.4

n=0.28-0.33

Concrete: E=20,000

Rubber. - -4.0 10-3

. - -1.0 10-3 0.45 - 0.5

Steel 190 - 215 75 - 80 0.27 - 0.33

- - -, a, =

. .

Stone; Limestone (Compression) 20 - 70 - 0.2 - 0.3

Stone; Marble (Compression) 50 - 100 - 0.2 - 0.3

- - -

Wood; Douglas Fir (Bending) 11-13 - -

Wood; Oak (Bending) 11-12 - -


oo ; ou ern ne en ng - - -


57/62

o uE (GPa) G (GPa) Poisson's Ratio

Steel E=200,000

MPa, G=77,000 MPa,

u u .

Aluminum Alloy 70 - 79 26 - 30 0.33

Brass 96 - 110 36 - 41 0.34

n=0.28-0.33

Concrete: E=20,000

r c ompress on - - -

Bronze; Regular 96 - 120 36 - 44 0.34

Bronze; Manganese 100 39 0.34

, a, =Ceramic 300 - 400 - -Concrete 18 - 30 - 0.1 - 0.2

Copper [Cu] 110 - 120 40 - 47 0.33 - 0.36

Copper Alloy 120 47 -

Cork - - 0

Glass 48 - 83 19 - 34 0.2 - 0.27


Iron (Cast) 83 - 170 32 - 69 0.2 - 0.3

Iron (Wrought) 190 75 0.3


58/62

,life



59/62

p n an a tu e

A-frame connector with square

tube (1 ) and simple pin haul4tonne trailer.

Friction coefficient betweenroad and tires: 0.3 (static),

0.14 (rolling)

ng e o e rame a e apex s

Normal stress: y=36ksi, safety factor=2, shear: y=0.6y, S.F.=3

Find thickness of tube, pin diam.

P/2 W/2


P/2 W/2


60/62

p n an a tu e

Total horizontal force in the tube P=W=0.3*4tonne*2.205kip/t=2.646k

Stress in the tube, P/Atube < Y/safety factor. Ne lectin the loss due to inhole area in the tube A = 2b+2h t

Using b=h=1.75, ttube > 2.646*2/(36*(2*1.75+2*1.75)=0.021, Use 1/8

The pin is in double shear

Shear stress in the in P/ 2A < /safet factor Apin> 2.646*3/(2*0.6*36) = 0.18375 sq.in.

Apin=d2/4, dpin > 0.4834. Use 1/2 pin.

P/2 W/2


P/2 W/2


61/62

w a e o a pro em Dead Whale dragged on an ice sheet

20ton whale

Friction coefficient = 0.25 ,

Rope 200 long

Max strain in the rope 1% Nylon rope, E = 1500 MPa

Find rope diam., elongation


Please dont go and kill

whales this is just an example!


62/62

w a e o a pro em= = * * * =. . .

Stress =P/Arope

, strain =/E P/(0.01*E)=2966 sq.mm.

drope>61.4mm, use 2.5 rope (63.5mm)

Actual strain =P/EArope=44.48*1000/(E**d2/4)=0.0094

Elongation =actual*Lro e=0.0094*200*304.8mm/ft=571mmW


WP

Topic 2&3 - Stress & Strain

Documents

Transcript of Topic 2&3 - Stress & Strain