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Mechanics of Solids
Stress and Strain
Dr. Seshu Adluri
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orces - n ro uc on Forces and Moments
External Surface Tractions (Exterior)
Interior (Body Forces, etc)
Internal Equilibrium
External-external
External-internal
Internal-internal
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orces External forces A lied forces, Reactions
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n erna orces
Why do they develop?
Internal equilibrium orce ransm ss on
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orces
Internal forces
and Free body Diagrams
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orces
Normal forces
Momentors on
Bending
Coplanar loads
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na ys s o orces
Identify external forces, geometry, etc.
Draw free-body diagrams of any identified
Apply equations of equilibrium
Solve for unknowns
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ress
Quantifies the intensity of internal forcesin a body
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ress
Normal stress &Shear stress
FA
m
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ress
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ress
UnitsPa kPa MPaMPa GPa si ksi .
(N, lb, kip, mmmm, mm, in.)
Total force and total moment
orce an momen agrams
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orma ress
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n ec s
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n ec s, e c.
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verage ress
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ress e
Stress field isenerall com lex
but can be
a roximated tosimpler models
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Shear Stress
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ear ress
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ear ress
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xamp e
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xamp e
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ress - ev ew
ForcesExternal surface bod forces reactions
Internal
Stress
,General state of stress in a solid
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ra n
Effects of strain
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orma ra n
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orma ra n
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ear ra n
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ra n- e orma on e
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a er a es ng n ens on
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ens e es pec men
ASTM standards areused for testing
ascertain theirproperties
available in thelibrary, online and
w a ora ory
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ater a a ure o es n ens on
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ress- ra n urve - s ee
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as c concep s
Hooks Law
(Robert Hook, 1676) Youngs Modulus
(Thomas Young, 1807)
Elasticity
Linear on near
Plastic deformation
Strain hardening
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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.
Stress & Strain -Dr. Seshu Adluri
.
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o u us o es ence
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o u us o oug ness
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-for different steels
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- steels
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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 -
Stress & Strain -Dr. Seshu Adluri
Copper [Cu] 55 - 330 230 - 380 10 - 50
Copper Alloy 760 830 4
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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
Stress & Strain -Dr. Seshu Adluri
Solder; Tin-Lead - 12 - 54 55 - 30
Steel 210 - 1600 340 - 1900 3 - 40
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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 -
- - -
Stress & Strain -Dr. Seshu Adluri
Wood; Southern
Pine (Bending)40 - 60 50 - 100 -
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material
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oa ng un oa ng - ysteres s
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ress- ra n urve -concre e
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ress- ra n urve -a um n um
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Stress-Strain Curve natural rubber
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ress- ra n urve cas ron
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- (temperature dependent)
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a era s ra n e av our
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a era s ra n o son s a o
lateral
r
=
allongitudinL
=
allongitudin
lateral
=
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a era s ra n o son s a o
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a era s ra n o son s a o
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
Stress & Strain -Dr. Seshu Adluri
directions.
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ear o u us
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ear o u us
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ear o u us
( )+=
12
G
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o u
Steel E=200,000 MPa, G=77,000 MPa,n=0.28-0.33
Concrete: E=20,000 30,000 MPa, G=
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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 - -
Stress & Strain -Dr. Seshu Adluri
oo ; ou ern ne en ng - - -
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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
Stress & Strain -Dr. Seshu Adluri
Iron (Cast) 83 - 170 32 - 69 0.2 - 0.3
Iron (Wrought) 190 75 0.3
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,life
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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
Stress & Strain -Dr. Seshu Adluri
P/2 W/2
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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
Stress & Strain -Dr. Seshu Adluri
P/2 W/2
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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
Stress & Strain -Dr. Seshu Adluri
Please dont go and kill
whales this is just an example!
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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
Stress & Strain -Dr. Seshu Adluri
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