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Statics

Centroid

Centroid (Geometric Center): the arithmetic average position (center) of particles in an object; the intersection of

all the hyperplanes that divide the shape into two parts of equal moment

V=

body

dV= i

Vi (volume)

r =

body

rdV/OdV

V(centroid; rigid body)

r =

i

miri

i

mi(centroid; system of rigid bodies or particles)

x=

body

x dV

V=

i

mixi

i

mi(xcoordinate of the centroid)

y=

body

y dV

V=

i

miyi

i

mi(ycoordinate of the centroid)

z= bodyz dV

V=

i

mizi

i

mi(zcoordinate of the centroid)

to find the centroid of complex shapes, add or subtract sections from easy to compute spaces

IF an area or line possesses two axes of symmetry, THEN its centroid must be located at the intersection of

the two axes

IF a figure possesses two axes of symmetry at a right angle to each other, THEN the point of intersectionof these axes is a center of symmetry

IF a volume possesses three axes of symmetry, THEN its centroid must be located at the intersection of thethree axes

A=rL (theorem of Pappus-Guldinus; surface areas)

V=rA (theorem of Pappus-Guldinus; volumes)

L=arc length of generating curve A=area of generating surface =angle of rotation in radians r=distance from centroid of generation object to axis of revolution 0

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Center of Mass

Center of Mass (CM)(r): the mass-weighted average center of a system of particles

m=

body

dm= i

mi (mass)

r =

body

rdm/Odm

m(center of mass; rigid body)

dm=dV IF the density of an object (2D or 3D) is constant, THEN the center of mass is at the location of the

centroid

whenever a body has an axis of symmetry, the center of mass always lies on that axis the center of mass doesNOThave to be within the body

to find the center of mass of complex shapes, add or subtract sections from easy to compute spaces

r =

i

miri

i

mi(center of mass; system of rigid bodies or particles)

x=

body

x d m

m=

i

mixi

i

mi(xcoordinate of the center of mass)

y=

body

y dm

m =

i

miyi

i

mi(ycoordinate of the center of mass)

z=

body

z d m

m=

i

mizi

i

mi(zcoordinate of the center of mass)

Center of Gravity

Center of Gravity (CG): the gravitational force weighted average center of a system of particles; the point through

which the gravity on an object acts

in a uniform gravitational field, the center of gravity is the same location as the center of mass

IF g has the same value at all points on a body, THEN its center of gravity is identical to its center ofmass

the earths gravitational field is NOT exactly uniform on any object, but is usually close

W=

body

dW= i

wi (weight)

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rcg=

body

rdw/Od w

W(center of gravity)

dw =g d m=gdV when a body acted on by gravity is supported or suspended at a single point, the center of gravity is

always at, directly above, or below the point of suspension

a body supported at several points must have its center of gravity somewhere within the area boundedby the supports or else it will tip over symmetries are very useful in finding the center of gravity to find the center of gravity of complex shapes, add or subtract sections from easy to compute spaces

r =

i

wiri

i

wi(center of gravity; system of rigid bodies or particles)

x=

body

x dW

W =

i

wixi

i

wi (xcoordinate of the center of gravity)

y=

body

y dW

W=

i

wiyi

i

wi(ycoordinate of the center of gravity)

z=

body

z dW

W=

i

wizi

i

wi(zcoordinate of the center of gravity)

= r mg = rw (the total gravtitational torque)

Locate the Center of Gravity of Any Object

1. suspend the object from a single point and make a vertical line from this point

2. suspend the object again from a different point and make another vertical line from this point

3. the center of gravity is at the intersection of the two points

First Moment of Area

Static Moment (First Moment of Area): contributes to shear stress; the moment from solid mechanics

Moment (First Moment of Mass) (Moment of Force): ; the moment from classical mechanics, distance times

force

Qyz=

body

x dV (first moment of area;yz plane)

Qxz=

body

y dV (first moment of area;xz plane)

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Qyx=

body

z dV (first moment of area;xy plane)

Qx=

body

y dV=yA (first moment of area;x-axis;2D)

Qy=

body

x dV=xA (first moment of area;y-axis;2D)

IF a volume possesses a plane of symmetry, THEN the first moment with respect to that plane is zero AND

the centroid of the volume is located in the plane of symmetry

to find the first moment of area of complex shapes, add or subtract sections from easy to compute spaces

Area Moment of Inertia

Axis of Rotation(ri): an axis perpendicular to the plane of motion and and always a constant distance from everyparticle in a system during a rotation if there is no linear, translational motion

Area Moment of Inertia (Second Moment of Area)(I): a measure of how resistant a shape is to bendingPolar Moment of Inertia (Polar Moment of Inertia of Area)(J): a measure of how resistant a shape is to torsion

Area Radius of Gyration(r): a measure of how resistant a shape is to buckling

IO=

body

r2dV/OdV (area moment of inertia)

Ix=

body

y2 +z2

dV (area moment of inertia;x-axis)

Iy=

bodyx2 +z2 dV (area moment of inertia;y-axis)

Iz=

body

x2 +y2

dV (area moment of inertia;z-axis)

JO=

body

r2 dV (area moment of inertia; polar)

JO=Ix+Iy (polar moment of inertia relation) the SI unit of area moment of inertia is them4 the moment of inertia depends on the choice of axis

to find the area moment of inertia of complex shapes, add or subtract sections from easy to compute

spaces

ASSUMES all area moments of inertia use the same axis you can NOTassume that all of the mass is concentrated at the center of mass and multiply by the

square of the distance from the center of mass to the axis; this is WRONG

Ixy=

xy dA (product area moment of inertia;x andy axes)

when one or both of the x and y axes are axes of symmetry for the areaA, the product of inertia Ixy iszero

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rL=

IL

A(area radius of gyration)

rx=

Ix

A(area radius of gyration;x-axis)

ry=

Iy

A(area radius of gyration; y-axis)

rz=

Iz

A(area radius of gyration;z-axis)

rO=

JO

A(area radius of gyration; polar)

r2O=r2x+ r2y (polar area radius of gyration relation)

Mass Moment of Inertia

Mass Moment of Inertia (Moment of Inertia) (Second Moment) (I): the tendency of a body to maintain itsangular velocity, even if it is zero; the rotational analogue to mass

Product of InertiaIi j

: the moment of inertia about the j-axis when a given object is rotating about the i-axis

Mass Radius of Gyration (Radius of Gyration) (k): the distance from the axis of rotation axis where all of themass of a body could be concentrated in a ring to produce the same mass moment of inertia as the body

IO=

body

r2dm/Odm (mass moment of inertia)

Ix=

body y2 +z2 dm (mass moment of inertia;x-axis)

Iy=

body

x2 +z2

dm (mass moment of inertia;y-axis)

Iz=

body

x2 +y2

dm (mass moment of inertia;z-axis)

I= i

mir2i (mass moment of inertia; system of rigid bodies or particles)

the SI unit of mass moment of inertia is thekg m2

the moment of inertia depends on the choice of axis

you can NOTassume that all of the mass is concentrated at the center of mass and multiply by thesquare of the distance from the center of mass to the axis; this is WRONG

kL=

IL

m(mass radius of gyration)

kx=

Ix

m(mass radius of gyration;x-axis)

ky=

Iy

A(mass radius of gyration;y-axis)

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kz=

Iz

A(mass radius of gyration;z-axis)

kO=

JO

A(mass radius of gyration; polar)

r2O=r2x+ r2y (polar mass radius of gyration relation)

IOL=Ix2

x+Iy2

y+Iz2

z 2Ixyxy 2Iyzyz 2Ixzxz (product mass moment of inertia)

IOL=Ix2x+Iy2y+Iz2z (product mass moment of inertia; principle axes of inertia)

Ixy=

xy dm (product mass moment of inertia;x andy axes)

Iyz=

yz dm (product mass moment of inertia;y andzaxes)

Ixz=

xz dm (product mass moment of inertia;x andz axes)

[I] =

Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz

(inertia tensor)

I =

Ixx 0 0

0 Iyy 0

0 0 Izz

(principal inertia tensor)

Moment of Inertia Theorems

Iz=Ix+Iy (perpendicular axis theorem)

IO=I+Ar2 (parallel axis theorem; area moment of inertia)

Ixy=Ixy+xyA (parallel axis theorem; product area moment of inertia)

IO=IG+ mr2G/O (parallel axis theorem; mass moment of inertia)

Ixx=Ixx+ m

r22+ r23

Iyy=Iyy

+ mr23+ r21

Izz=Izz+ m

r21+ r22

Ixy=Ixy+ mr1r2

Iyz=Iyz+ mr2r3

Ixy=Ixy+ mr3r1

(parallel axis theorem; inertia tensor components)

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Stretch Rule: the moment of inertia of a rigid object is unchanged when the object is stretched parallel to the axis

of rotation, (without changing the distribution of mass except in the direction parallel to the axis)

Surface of Revolution: a surface that can be generated by rotating a plane curve about a fixed axis

Body of Revolution: a body which can be generated by rotating a plane area about a fixed axis

Imbalance Eccentricity: the distance between the bodys center of rotation and its center of mass; a measure of

the degree imbalance in a rotating object

a significant imbalance eccentricity leads to vibrations

Fluid Statics

p=gz (pressure on body submerged in fluid)

ASSUMES density is constant

w (z) = pA (z) (weight on body submerged in fluid)

F =

dF =

A

pn dA (total force due to pressure on a surface)

MO=

A

dMO=

A

rO (pn) dA (total moment due to pressure on a surface)

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Archimedess Principle: when a body is completely or partially immersed in a fluid, the fluid exerts an upward

force on the body equal to the weight of the fluid displaced by the body

Buoyant Force: the upward force exerted by a fluid on a body completely or partially immersed in the fluid

Fz=gV (upward force from Archimedess principle)

the line of action of the buoyant force passes through the center of gravity of the displaced fluid, NOTthe body

a floating object displaces its weight a submerged object displaces its volume

Pascals Law: pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and

the walls of the containing vessel

P= F1

A1=

F2

A2AND F2=

A2

A1F1 (Pascals law)

in a hydraulic lift, the output force can be greater than the input force, but the smaller force must go alonger distance due to the conservation of energy

Internal Forces

Loads on Solid Bodies

Force Torque

Normal to Plane of Surface tension/compression bending

In Plane of Surface shear torsion

Concentrated load problems

1. use force and moment balance on a FBD of the entire beam to determine any missing loads or reaction

forces if possible

2. make two cuts between each of the loads and reaction forces

only one cut between loads is necessary for the shear force

to find the bending moment values at the shortest and longest lever arm, two cuts are needed

3. use force balance to determine the shear force

the shear force is whatever additional force is needed to achieve static equilibrium for the given FBD

4. use moment balance to determine the bending moment

the bending moment is whatever additional moment is needed to achieve static equilibrium for the given

FBD

5. connect the values on the shear force and bending moment diagrams with straight lines to complete the

graphs

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Distributed load problems

1. use force and moment balance on a FBD of the entire beam to determine any missing loads or reaction

forces if possible

2. use force balance to determine the shear force at easy points like supports

3. use moment balance to determine the bending moment at easy points like supports

4. use the derivative/integral relationships between applied load and shear force to determine the change in

shear force between the known values from step 2

5. use the derivative/integral relationships between shear force and bending moment to determine the change

in bending moment between the known values from step 3

a distributed load on a beam can be replaced by a concentrated load

the magnitude of the concentrated load is equal to the area under the load curve line of action of the concentrated load passes through the centroid of that area

FR=

dF =

A

w dA (total force due to distributed load on a surface)

x=

Axw d A

Aw dA

(xcoordinate of where resultant force is applied)

y=

Axw d A

Aw dA

(ycoordinate of where resultant force is applied)

this formula can be used to replace distributed loads with an equivalent concentrated load

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Stress

Normal Stress

Stress: the force per unit area, or intensity of the forces distributed over a given section

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the SI unit of stress is the PascalPa

the US customary units of stress are:

the pound per square inchpsi kilopound per square inchkip

the unit pound is ambiguous:

pound-force: lbf4.448222N pound-mass: lbm0.45359237kg

Normal Stress(): stress due to loads normal to the objects surface

Axial Loading: forces directed along the longest axis of a rod or similar long, thin member

Concentric Loading: the line of action of the concentrated loadsP andP passes through the centroid of thesection considered

a uniform distribution of stress is only possible in concentric loading

= limA0

F

A

stress at a point

=P

A

normal stress;

axial loading

exactly correct for a load perfectly evenly distributed over the end of the bar generally only correct in the middle of the bar

=P

Acos2

normal stress;oblique load

Shear Stress

Shear Stress(): stress due to loads parallel to an objects surface

ave=P

A

average shear stress over section

this usually is NOT useful because the shear stress varies significantly over most types of sections

=P

Asin cos

shear stress;

oblique load

Single Shear: shear with one plane

Double Shear: shear with two planes

Splice Plate: a plate laid parallel to primary members at a joint to provide reinforcement; lead to double shear

ave=P

A=

F

A single shear 11

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ave=P

A=

F/2

A=

F

2A

double shear

Bearing Stress (b): the shearing stress on the member that a fastener passes through, rather than on thefastener itself

Bearing Surface: the surface of contact between a fastener and the members it is joining

b=P

A=

P

td bearing stress

A=projected area of hole t= thickness d=hole diameter A

Stress Vector

Stress Vector (Traction) (t): the total force per unit area in the limit at a given pointP for a plane normal to avectorn; the total stress at a point

traction depends on both its location, point P and its orientation, the plane normal ton

free surfaces have zero traction

the stress normal to the free surface r= 0 the stress tangent to the free surface does not necessarily equal zero

equal and opposite reaction forces result in equal and opposite tractions

t (P, n) = limS

0

F

S lim

A

0

F

A=

dF

dA traction vector A=a given plane area S=a given surface = (t(P, n) n)n =|t|cos ()n

normal stress from traction

= t(P, n)

shear stress from traction

the normal stress vector and the shear stress vector are components of the traction vector

Surface Force: a force that acts on an internal or external surface of a body

Body Force: a force that acts through the volume of a body

body forces are typically described as force per unit volume

tx

ty

tz

=

xx xy xz

xy yy yz

xz yz zz

nx

ny

nz

traction vector;

Cauchy formula

t =Tn

traction vector;

Cauchy formula;

matrix equation

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Stress Tensor

Second Order Tensor

Ti j

: a mathematical object that obeys the transformation rule T=Tt

material elements are drawn as cubes for visualization, but are really the intersection of three mutually

orthogonal planes intersecting at a point

all stresses are shown in the positive direction on a stress element

T=

T11 T12 T13

T21 T22 T23

T31 T32 T33

=

11 12 13

21 22 23

31 32 33

=

xx yx zx

xy yy zy

xz yz zz

=

x yx zx

xy y zy

xz yz z

Cauchy stress ten

S=

x m yx zxxy y m zyxz yz z m

deviatoric stress tensor

the index convention for the stress tensor is: the first index is the facethe component acts on the second index is the directionthat the component acts in rows correspond to faces columns correspond to force directions

the sign conventions for the stress tensor are: positive shear stress is stress in the positive direction of the second index positive normal stress is positive in tension (outward)

T=Tt stress tensor transformations i j= ei ej

component of transformation matrix

ei, ej, ek= unit vectors in old coordinate system ei, ej, ek=unit vectors in new coordinate system

Principal Stresses

det (T I) =0

principal stresses eigenvalue equation

3 I12 +I2 I3=0

principal stresses polynomial equation

I1=x+ y+ z

I2=xy+ xz+ yz 2xy 2xz 2yzI3=det (T)

stress invariants;

from general stresses

I1=1+ 2+ 3

I2=12+ 23+ 31

I3=123

stress invariants;

from principal stresses

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the invariants are all independent of the coordinate system

1=2 cos

13

+ 13I1

2=2 cos

13

+ 120

+ 13I1

3=2 cos

13

+ 240

+ 13I1

principal stresses

a= 13I21I2

b= 1

3I1I2 I3 2

27I31

c=

127

3

=arccos b

2c

=

13

a

constants;

principal stresses formulae

must be in degrees

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Static Equilibrium

xxx

+yx

y+

zxz

+ Fx=0

xy

x+

yy

y+

zy

z+ Fy=0

xzx

+yz

y+

zzz

+ Fz=0

equilibrium equations;

linearly elastic body;

rectangular coordinates

rrr

+1

r

r

+rz

z+

1

r(rr) + Fr= 0

rr

+1

r

+z

z

2

rr+ F=0

rzr

+1r

z

+zzz

+1r

rz+ Fz=0



cylindrical coordinates

rrr

+1

r

r

+ 1

rsin

r

+

1

r

2rr + r cot

+ Fr=0

rr

+1

r

+ 1

rsin

+

1

r

cot + 3r

+ F=0

r

r+

1

r

+

1

rsin

+

1

r2 cot + 3r+ Fz=0



spherical coordinates

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F= body force surface forces do NOT enter the equilibrium equations ASSUMES the body is continuous ASSUMES the body has NO cracks

i j= 0 i= j

1 i= j Kronecker delta function

Stress Transformations

Free Surface: a surface of an object or face of a unit cube that isnt touching anything (except air) and has NO

forces acting on it

Plane Stress: when there are two faces of a cubic element that are free surfaces

when working with plane stresses, we define the z-axis to be perpendicular to the two free surfaces

this makesz=xz=yz=0Principal Stresses: the normal stresses that occur when the object is rotated so that there are NO shear stresses

Principal Planes of Stresses: the planes in three dimensions containing the principal stresses

no shearing stress is exerted on the principal planes

the planes of maximum shearing stress are at45 to the principal planes

Octahedral Planes: the planes whose intersections with the principal coordinate planes make45angles with theprincipal coordinate axes

x=x+ y

2 +

x y2

cos (2) + xy sin (2)

y=x+ y

2 x y

2 cos (2) xy sin (2)

xy=x y

2 sin (2) + xy cos (2)

plane stress transformation

x+ y=x+ y plane stress transformation

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Mohrs Circle for Stress

(x ave)2 + 2xy= R2

equation of Mohrs circle;

plane stress

ave= x+ y2

center of Mohrs circle;

plane stress

R=

x y2

2+ 2xy

radius of Mohrs circle;

plane stress

forplanestress, in3D, two of Mohrs circles must have a principal stress that is zero for Mohrs circle,positivenormal stresses aretensionand negativeare compression for Mohrs circle, positive shear stresses create clockwise rotation and negative create counter-

clockwise rotation

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Principal Stresses

max, min=x+ y

2

x y

2

2+ 2xy

principal stress;

plane stress

max=ave+R

maximum principal plane stress

min=ave R minimum principal plane stress tan (2p) = 2xy

x y

angle of rotation to principal planes of stress

the principal planes are p and p+ 90

max=0.5 |max min|

maximum shear stress;

general3Dstress

max=R=

x y2

2+ 2xy

maximum in-plane shear stress

tan (2s) =x y2xy

angle of rotation to planes ofmaximum in-plane shear stress

Strain

General Assumptions

ASSUME the material is:

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linear elastic homogenous isotropic NO plastic deformation NO creep

NO fracture

NO fatigue NO corrosion NO embrittlement NO chemical or environmental effects NO relativistic or quantum mechanical effects the continuum hypothesis holds

Normal StrainRelative Displacement

B/A

: a displacement where both rods move, but by unequal amounts

Strain(): the deformation per unit length

strain is a dimensionless quantity

it is customary to keep the units rather than canceling them out microstrain, written as=xxxx =xxxx 106 is a common way of writing small strains

True Strain: actual strain accounting for a decrease in cross sectional area during loading

Engineering Strain: the strain computed by using the original area rather than the changed area under loading

Stress-Strain Diagram: a curve characteristic of the properties of a given material; independent of the dimensions

of the specimen used

determine the yield point by making a0.2%offset so =0.002and make a line parallel to the elastic part ofthe curve and find the intersection

=

L0

P

AEdx

displacement;

variable cross section

= i

PiLiAiEi

displacement;

different cross sections

=L L0= PLAE

displacement

B/A=B A=

PL

AE

relative displacement

i j=uixj

=uj

xi

normal strain definition

larger displacements may have smaller strains

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positive displacements may have negative strains zero displacement may have a non-zero strain different displacements may have the same strain

=

L0

engineering strain

t= L

L0

dLL

=ln LL0

true strain

= limx0

x=

d

dx

normal strain at pointQ

Shear Strain

Average Shear Strain (xy): the change in angle between the x and y axes on the deformed body from theundeformed body

shear strain is unitless

shear strain must be measured in radians

xy=

2

average shear strain;

experimental form

xy=xy

2 =0.5

v

x+

u

y

shear strain;

tensor form

G= E2 (1 +)

= 3k(1 2)2 (1 +)

shear modulus

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Strain Tensor

Eulerian Coordinate System (Current Coordinate System): a coordinate system defined on a deformed body

Lagrangian Coordinate System (Reference Coordinate System): a coordinate system defined on an unde

formed body

x=ux+ 0.5u2x+ v2x+ w2xy=uy+ 0.5

u2y+ v

2y+ w

2y

z=wz+ 0.5

u2z+ v2

z+ w2

z

xy=uy+ vx+ uxuy+ vxvy+ wxwy

xz=uz+ wx+ uxuz+ vxvz+ wxwz

zy=vz+ wy+ uyuz+ vyvz+ wywz

strain-displacement relations;

large displacements

subscripts foru,v, andw are partial derivatives

xx=u

xxy=yx=

u

y+

v

x

yy=v

yxz=zx=

u

z+

w

x

zz=w

zyz=zy=

w

y+

v

z


small displacements;

Cartesian coordinates

rr=ur

rr=r=

ur

ur

+1

r

ur

=1

r

u

+ur

rrz=zr=

urz

+uz

r

zz=uz

z

z=z=1

r

uz

+u

z



cylindrical coordinates

rr=ur

rr=r=

1

2

ur

ur

+1

r

ur

=1

r

u

+ur

r==

1

2r

1

rsin

u

+u

u cot

= 1

rsin

u

+ ursin + u cos

r=r=

1

2

1

rsin

ur

+u

ru

r

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r=u

r

=1

r

v

+

u

r

r=1

r

u

+

v

r

v

r



polar coordinates;

plane stress

=

11 12 13

21 22 23

31 32 33

=

xx 0.5xy 0.5xz

0.5yx yy 0.5yz

0.5zx 0.5zy zz

Cauchy strain tensor

the Cauchy strain tensor follows the same sign conventions as the Cauchy stress tensor the Cauchy strain tensor is symmetric the strain values listed on a differential solid element arei j NOTi j

this means they need to be divided by two before being put into the strain tensor

i j= 0.5

uixj

+uj

xi

strain tensor components;

compact form

=

u1x1

0.5

u1x2

+u2x1

0.5

u1x3

+u3x1

0.5

u1x2

+u2x1

u2

x20.5

u3x2

+u2x3

0.5

u1x3

+u3x1

0.5

u3x2

+u2x3

u3

x3

Cauchy strain tensor;

explicit form

=

x m 0.5yx 0.5zx0.5xy y m 0.5zy0.5xz 0.5yz z m

deviatoric strain tensor

the deviatoric strain tensor is symmetric

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Strain Compatibility Equations

22xy

xy=

2xxy2

+2yy

x2

22yz

yz=

2yy

z2 +

2zzy2

22zxzx

=2xx

z2 +

2zzx2

2xxyz

=

x

yz

x+

zxy

+xy

z

2yy

zx=

y

yz

xzx

y+

xy

z

2zzxy

=

z

yz

x+

zxy

xyz

strain compatibility equations;3D

i j,km+ km,i j ik, jm jm,ik=0

strain compatibility equations;

3D; index notation

2f(x, y)

xy=

2f(x, y)

yx

2Dplane strain

2xy2 +

2y

x2 =

2xy

xy

strain compatibility equation;2D

ASSUMES linear strain-displacement relations

Principal Strains

=t

strain tensor transformations

i j= e

i

ej component of transformation matrix

ei, ej, ek=unit vectors in old coordinate system ei, ej, ek= unit vectors in new coordinate sy

det ( I) =0

principal strains eigenvalue equation

3 I12 +I2 I3=0

principal strains polynomial equation

I1=x+ y+ z

I2=xy+ xz+ yz 2xy 2xz 2yzI3=det ()

strain invariants;

from general strains

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I1=1+ 2+ 3

I2=12+ 23+ 31

I3=123

strain invariants;

from principal strains

the invariants are all independent of the coordinate system

1=2 cos

13

+ 1

3I1

2=2 cos13

+ 120+ 13

I1

3=2 cos

13

+ 240

+ 13I1

principal strains

a= 13I21I2

b= 13I1I2 I3 227I31

c=

127

3

=arccos b

2c

=

13

a

constants;

principal strains formulae

must be in degrees

Strain Transformations

Plane Strain: when all of the deformations of an object occur in parallel planes; when two faces of any unit cube

have no strains

plane stress and plane strain never occur simultaneously UNLESS the Poisson ratio=0

x=x+ y

2 +

x y2

cos (2) +xy

2 sin (2)

y=x+ y

2 x y

2 cos (2) xy

2 sin (2)

xy= (x y) sin (2) + xy cos (2)

plane strain transformation

x+ y=x+ y

plane strain transformation

z=zx=zy=0

general plane strain

1=x cos2

(1) + y sin2

(1) + xy sin (1) cos (1)

2=x cos2 (2) + y sin

2 (2) + xy sin (2) cos (2)

3=x cos2 (3) + y sin

2 (3) + xy sin (3) cos (3)

general strain rosette

xy=2OB (x+ y)

shear strain;

strain rosette;

45angles

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R=

x y2

2+xy

2

2 radius of Mohrs circle;plane strain

forplanestrain, in3D, two of Mohrs circles must have a principal strain that is zero

Principal Strains

max, min= x+ y2

x y2

2+

xy2

2 principal strain;

plane strain

max=ave+R

maximum principal plane strain

min=ave R

minimum principal plane strain

tan (2p) = xyx y

angle of rotation to planes of

maximum in-plane shear strain

the principal planes are p and p+ 90

max=|max min|

maximum shear strain;

general3Dstrain

max=2R=

(x y)2 + 2xy

maximum shear strain;

in plane

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Bulk Stress and Strain

h=x+ y+ z

3

hydrostatic stress

V= VfV0

volume change

Vf =x (1 + x) + y (1 + y) + z (1 + z)=xyz (1 + x+ y+ z+ xy+ ez+ yz+ xyz)

final volume after deformation

e=x+ y+ z

bulk strain AKA dialation definition

bulk strain is independent of the coordinate system bulk strain is approximatelythe volume change per unit volume ONLY for small deformations

e=1 2

E(x+ y+ z) bulk strain

e=pk

=3 (1 2)

Ep

bulk strain;

hydrostatic pressure

k=B= E3 (1 2)=

EG

3 (3G E)

bulk modulus

x=x+ e

y=y+ e

z=z+ e

Lam equations

=2G

= E

(1 +) (1 2)

Lam constants

Poissons Ratio

Saint-Venants Principle: the strains that can be produced in a body by the application, to a small part of its

surface, of a system of forces statically equivalent to zero force and zero couple, are of negligible magnitude at

distances which are large compared with the linear dimensions of the part

= lateral strainaxial strain

=yx

=zx

Poissons ratio

Poissons ratio ASSUMES that the material is homogeneous and isotropic 1

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=0.5for constant volume materials; typically water, organic tissues, rubbers, etc. Poissons ratio is unitless

x=xE

axial strain

y=z=xE

lateral strain

E2G

=1 +

elastic deformation relation

Metamaterial: an artificial material engineered to have properties that may not be found in nature; metamaterials

usually gain their properties from structure rather than composition, using small inhomogeneities to create effective

macroscopic behavior

Auxetics: materials that have a negative Poissons ratio

auxetic materials typically have high energy absorption and high fracture resistance

Thermal Stress and Strain

=1

L

L

T

P

coefficient of linear thermal expansion

constant characteristic material property strain per unit temperature change units areC1

P=AET

force to produce thermal stress

thermal strain may be avoided by applying forces to create thermal stress only applies for a homogeneous rod of uniform cross section

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T=P

A=ET

thermal stress;

ignores non-thermal stress

T=LT

thermal expansion/contraction

T=T=TL

thermal strain

IF the material can expand freely, THEN there is NO thermal stress

xx= 1

E(x y z) + T

yy= 1

E(y x z) + T

zz= 1

E(z x y) + T

xy=yx= 1

Gxy

xz=zx= 1

Gxy

yz=zy= 1

Gxy

generalized Hookes law;

3Dthermal strain;

isotropic material;

ASSUMES uniform thermal expansion

Constitutive Laws

Generalized Hookes Law - Compliance Form

Constitutive Equations: a set of equations relating the stress and strain in a material

the constitutive matrixSis symmetric

Anisotropic: properties that are different in every direction

Orthotropic: properties that are the same throughout two or three mutually orthogonal planes of rotational sym-metry, but different between the planes

Transverse Isotropic: properties that are the same in one plane and different normal to this plane

Isotropic: properties that are the same in all directions

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xx

yy

zz

yz

zx

xy

=

S11 S21 S31 S41 S51 S61

S12 S22 S32 S42 S52 S62

S13 S23 S33 S43 S53 S63

S14 S24 S34 S44 S54 S64

S15 S25 S35 S45 S55 S65

S16 S26 S36 S46 S56 S66

xx

yy

zz

yz

zx

xy

compliance matrix;

linear anisotropic,

homogeneous material

fully ansiotropic materials have21 independent elastic constants the general ansiotropic compliance matrix is symmetric

xx

yy

zz

yz

zx

xy

=

1

Exyx

Eyzx

Ez0 0 0

xyEx

1

Eyzy

Ez0 0 0

xzEx

yzEy

1

Ez0 0 0

0 0 0 1

Gyz0 0

0 0 0 0 1

Gzx0

0 0 0 0 0 1

Gxy

xx

yy

zz

yz

zx

xy

compliance matrix;

linear orthotropic,


orthotropic materials have9 independent elastic constants the orthotropic compliance matrix is symmetric

xx

yy

zz

yz

zx

xy

=

1

Epp

Epzp

Ez0 0 0

pEp

1

Epzp

Ez0 0 0

pzEp

pzEp

1

Ez0 0 0

0 0 0 1

Gzp0 0

0 0 0 0 1

Gzp0

0 0 0 0 02 + 2p

Ep

xx

yy

zz

yz

zx

xy

compliance matrix;

transverse isotropic,


transverse isotropic materials have5 independent elastic constants the transverse isotropic compliance matrix is symmetric

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xx

yy

zz

yz

zx

xy

=

1

E

E

E0 0 0

E

1

E

E0 0 0

E

E

1

E0 0 0

0 0 0 1

G 0 0

0 0 0 0 1

G0

0 0 0 0 0 1

G

xx

yy

zz

yz

zx

xy

compliance matrix;

linear isotropic,


isotropic materials have2 independent elastic constants the isotropic compliance matrix is symmetric

xx

yy

xy

=

1

E

E 0

E

1

E0

0 0 2 + 2

E

xx

yy

xy

compliance matrix;

linear isotropic,

homogeneous material;

plane stress;

rectangular coordinates

rr

r

=

1

E

E0

E

1

E0

0 0 2 + 2

E

rr

r

compliance matrix;

linear isotropic,


plane stress;

polar coordinates

xx

yy

xy

=

1

E

E0

E

1

E0

0 0 2 + 2

E

xx

yy

xy

compliance matrix;

linear isotropic,


plane strain

Generalized Hookes Law - Stiffness Form

=E

Hookes law;

normal stress and strain

xy=Gxy

yz=Gyz

zx=Gzx

Hookes law;

shear stress and strain

ASSUMES

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xx

yy

zz

yz

zx

xy

=

C11 C21 C31 C41 C51 C61

C12 C22 C32 C42 C52 C62

C13 C23 C33 C43 C53 C63

C14 C24 C34 C44 C54 C64

C15 C25 C35 C45 C55 C65

C16 C26 C36 C46 C56 C66

xx

yy

zz

yz

zx

xy

compliance matrix;

linear anisotropic,


the constitutive matrix Cis symmetric theSfor compliance and Cfor stiffness is in fact backwards, yet correct fully ansiotropic materials have21 independent elastic constants the general ansiotropic stiffness matrix is symmetric

xx

yy

zz

yz

zx

xy

=

1 yzzyEyEz

yx+zxyzEyEz

zx+yxzyEyEz

0 0 0

xy+xzzyEzEx

1 zxxzEzEx

zy+zxxyEzEx

0 0 0

xz+xyyzExEy

yz+xzyxExEy

1 xyyxExEy

0 0 0

0 0 0 Gyz 0 0

0 0 0 0 Gzx 0

0 0 0 0 0 Gxy

xx

yy

zz

yz

zx

xy

stiffness mat

linear orthotro

homogeneous m

= 1 xyyx yzzy zxxz 2xyyzzxExEyEz

orthotropic materials have9 independent elastic constants the orthotropic stiffness matrix is symmetric

xx

yy

zz

yz

zx

xy

=

1 pzzpEpEz

p+zppzEpEz

zp+pzpEpEz

0 0 0

p+pzzpEzEp

1 zppzEzEp

zp+zppEzEp

0 0 0

pz+ppzE2p

pz (1 +p)E2p

1 2

p

E2p 0 0 0

0 0 0 Gzp 0 0

0 0 0 0 Gzp 0

0 0 0 0 0Ep

2 + 2p

xx

yy

zz

yz

zx

xy

stiffnes

transverse

homogeneo

=(1 +p) (1 p 2pzzp )E2pEz

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transverse isotropic materials have5 independent elastic constants the transverse isotropic stiffness matrix is symmetric

xx

yy

zz

yz

zx

xy

= E

(1 +) (1 2)

1 0 0 0 1

0 0 0

1 0 0 00 0 0

1 22

0 0

0 0 0 0 1 2

2 0

0 0 0 0 0 1 2

2

xx

yy

zz

yz

zx

xy

stiffnes

linear

homogene

isotropic materials have2 independent elastic constants

the isotropic stiffness matrix is symmetric

xx

yy

xy

=

E

1 2E

1 2 0E

1 2E

1 2 0

0 0 E(1 )

1 2

xx

yy

xy

stiffness matrix;

linear isotropic,


plane stress

xx

yy

xy

=

E

1

2

E

1

2

0

E1 2 E1 2 0

0 0 E

2 + 2

xx

yy

xy

stiffness matrix;

linear isotropic,homogeneous material;

plane strain

Metals

F= Fa+ Fr= krm

+ B

rn

total force;

Fa= krm

attractive force;

Fr= Brn

=e

r

repulsive force;

U= k

(m 1) rm1 B

(n 1) rn1

potential energy;

dU

dr= k

rm+

B

rn=0

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= 1

A

dU

dr

E=ao

A

2k

a3o

e

ao

Ceramics

P=(WD)/(W S)/ =

WDW S

volume fraction of pores

Glass

Glass Transition Temperature(Tg): the temperature at which the viscosity of the glass is equal to1012 Pa s

(T) =0e(Q/RT)

glass viscosity;

Arrhenius-type relation

ln

0

=

Q

R

1

T

glass viscosity;

Arrhenius-type relation

ln =A + B

T T0

glass viscosity;

Vogle-Fulcher-Tamman

Concrete

() = E0

1 +

E0EC

2

c

+

c

2

E0=33w1.5

c psi

oct= 1 e

G0 oct

=0

1.0 + 1.3763 5.36

c

2+ 8.586

c

3

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Fibrous Composites

x

y

xy

=

1

Exyx

Ey0

xyEx

1

Ey0

0 0 1

Gxy

x

y

xy

general orthotropic, homogeneous material;

plane stress

ASSUMES plane stress

this is reasonable for thin plates and sheets

z=yz=xz=0

fiberous composite;

plane stress ASSUMPTION

yxEy

=xy

Ex

general orthotropic, homogeneous material;

plane stress

A=Af+Am

cross-sectional area;

fiberous composite

A=total cross-sectional area Af=fiber cross-sectional area Am=matrix cross-sectional are

Vf=Af

A

volume fraction;

fiber content;

composite

Vm=Am

A

volume fraction;

matrix content;

composite

x=f+ m

total strain;

fiberous composite;

parallel to fibers

y=f=m

total stress;fiberous composite;perpendicular to fibers

Ex= VfEf+ VmEm

Youngs modulus of elasticity;

fiberous composite;

parallel to fibers

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Ey= EfEm

VfEm+VmEf

Youngs modulus of elasticity;

fiberous composite;

prependicular to fibers

xy= Vff+ Vmm

major Poisson ratio;

fiberous composite;

plane stress

Gxy= GfGm

VfGm+VmGf

shear modulus;

fiberous composite;

plane stress

Polymers

U= 0.5U0 (1 cos (3)) potential energy;simple rotation60about spine axis

(t) =

t0

G (t s)d (s)ds

ds =G (0) (t) +

t0

(t s)G (s)ds

ds

=e+ v=

E+

c

strain;

Maxwell model

=

c

E

c strain;

Kelvin-Voigt model

J=00

cos ()

dynamic elastic modulus

J=00

sin ()

dynamic loss modulus

Rubbers

natural rubber is too soft for many applications and is sticky at room temperature

sulfur is added to rubber to increase its strength

sulfur creates cross-linking between the polymer molecules

white lead, magnesium oxide, and litharge, among other materials, can be added to accelerate the vulcan-

ization process

rubber is black because of the addition of carbon black to it

carbon black filler increases the strength of rubber

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Stretch(): the ratio of a deformed length to a material allowing large deformations such as an elastomer likerubber or biological tissues

=1 +

uniaxial stretch vs uniaxial strain

t(1) = d

d1=

2c1e

k(21+2

11 3)

2

1 21

+2

c2

21 +213++ c3

31 + 2

(1, 2, 3) =N

p=1

p

p (p)

strain energy density;

Incompressible Biological Tissues

0= f(, ) + ceF(a,)

f(, ) =1211+ 2

222+ 3

221+ 241122

F(a, ) =a1211+ a2

222+ a3

221+ 2a41122

0=cea1

211

0=ce

F(a,)

w () = C

e

0.5(21)2 1

Wood

LR

EL=

RLER

ANDLTEL

=T LET

ANDRTER

=T RET

condition for orthotropic model

Orthotropic Parameters for Selected Woods

Wood LREL

RLER

LTEL

T LET

RTER

T RET

Douglas Fir 0.179 0.152 0.159 0.186 3.034 3.378

Oak 0.427 0.421 0.648 0.614 2.137 2.068

Quaking Aspen 0.386 0.513 0.288 0.581 10.433 12.800

Beech 0.228 0.228 0.255 0.269 2.206 2.275

Birch 0.207 0.214 0.179 0.200 4.206 4.825

=E

An

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4 () =4

x4+ 2

4

x2y2+

4

y4=0

biharmonic equation;2D

p= (x, y, z) =

(x a)2 + (y b)2 + (z c)2

general solution to biharmonic equation

a,b, andc are complex numbers

Particular Solutions to the Biharmonic Equation Include: any polynomial of order three or less any polynomial with coefficients chosen to satisfy the biharmonic IF is a solution to the Laplace equation: x;y;z; x2 +y2 +z2

(x, y) =G

2h

4

27h3 hx2 +y2+x3 3xy2 Prandtl stress function

Shafts and Torsion

Torsion

Transmission Shaft: a shaft that transmits power from one point to another

when acircularshaft is subjected to torsion, every cross section remains plane and undistorted

the formulas here are ONLY valid for circular cross section shafts

the torsion formulas cannot be used where the loading couples are applied at or near abrupt changes in the

diameter of the shaft

the torsion formulas can only be used within the elastic range of a given material

J=

2 dA

polar moment of inertia

J=0.5r42 r41

polar moment of inertia;

circular tube

r1=inner radius r2=outer radius

T=

(dA) =

dF= maxJr

torque on shaft

T=torque applied to shaft =distance from center axis of shaft

=T

J=

rmax

shear stress;

torsion

max= TrJ

max shear stress;

torsion

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min= r1r2

max

min shear stress;

torsion

the minimum shear stress in a solid circular shaft is zero

=

L0

T

JGdx

angle of twist;

elastic range;

variable shaft

= i

TiLi

JiGi

angle of twist;

elastic range;

composite shaft

= T LJG

angle of twist;

elastic range;

homogeneous shaft

E/B=E B= T L

JG

relative angle of twist

=

L=

rmax

shear strain; torsion;

is in radians

max= rL

= Tr

JG

max shear strain;

torsion

Shaft Design

J

c=

M2 + T2

max

all=

M2y+M

2z+ T

2

max

all

shaft design requirement;

failure by shear stress

a typical shaft analysis requires: 1 torsion diagram 1axial loading diagram 2shear force diagrams

2bending moment diagrams

Torsion In Noncircular Shafts

max= T

c1ab2

max shear strain;

solid rectangular cross section shaft

= T L

c2ab3G

angle of twist;

solid rectangular cross section shaft

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a=wider cross section length b=narrower cross section length c1, c2=coefficients found in a table based on the ratio b

a

T=2qa

torque;

thin-walled hollow noncircular shaft

a=area bounded by the center line of the wall cross section

q=t= constant

shear flow;


ASSUME wall thicknesstis small compared to other dimensions

= T

2ta

shear stress;


= T L

4a2G ds

t angle of twist;


Beams and Bending

Beams

Beam: a solid object with one dimension (the length) significantly longer than the other two dimensions (the cross

section); a long, thin member with transverse loading; usually oriented horizontally

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S-Beam (American Standard Beams): an I-beam with narrow flanges

W-Beam (Wide Flange Beams): an I-beam with wide flanges

Prismatic Beam: a beam with a uniform cross section

Nonprismatic Beam: a beam with a varying cross section

nonprismatic beams allow for a better optimized design

Transverse Loading: a load applied at a right angle to the surface of the length of a beam

transverse loading only creates bending and shear in a beam

oblique loading creates bending, shear, and axial forces on a beam

Beam Cross Section Transformations

Transformed Section: the new equivalent cross section of a beam assuming homogeneous material with constan

modulus of elasticity

n=E2

E1

scale factor for transformed beam sections

To Transform a Beam Section:

1. keep the section of material corresponding toE1 constant

2. multiple the cross section width of the material corresponding toE2 byn

x=nx0 transformed cross section width

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Beam Parameters

I=

y2 dA

area moment of inertia

I= bh3

12

area moment of inertia;

solid rectangle

I= 4

r42 r41

area moment of inertia;annulus circle

relative to the neutral axis find neutral axis by finding average height of materials; ([amount] [height])

[total amount]

to find the moment of inertia of more complex shapes, find the moment of inertia of a large, simpleregion and then subtract the necessary parts

S=

I

c

elastic section modulus

S=|M|all

elastic section modulus;

beam of constant strength

Pure Bending

Pure Bending: a bending moment is applied to a beam while the shear force is zero and no torsional or axial loads

are applied

the moment of the couple is the same about ANY axis perpendicular to its plane, AND is zero about ANY

axis contained in that plane

any cross section perpendicular to the axis of the member remains plane

the plane of the cross section passes through C

at any point of a slender member in pure bending, there is a state ofuniaxial stress

concaveside (upperportion) of a beam is in compression

convexside (lowerportion) of a beam is in tension

Neutral Surface: a surface parallel to the upper and lower faces of the member, where x andx are zero

Neutral Axis: the line of the intersection the neutral surface and a transverse cross section of the beam

longitudinal normal strainx varies linearly with the distance y from the neutral surface

in the elastic range, the normal stress varies linearly with the distance from the neutral surface

the first moment of the cross section about its neutral axis must be zero

as long as the stresses remain in the elastic range, the neutral axis passes through the centroid of the section

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x=MyI

=yc

max

flexural stress

max= McI

=M

S

maximum absolute value of stress

m=|M|max

S=

|M|max cI

largest normal stress

Smin=|M

|min

allow

minimum allowable section modulus

x= y

=y

cmax

flexural strain

max= cp

max absolute value of strain

Curvature(): the reciprocal of the radius of curvature

Anticlastic Curvature: the curvature of a traverse cross section with radius of curvature of a bending beamdue to Poisson ratio expansion and contraction

=1

=

M(x)

EI=

maxc

curvature

[Anticlastic Curvature] = 1

=

anticlastic curvature

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y=y

z=

y

normal strain;

bending

General Bending

shear forces create shear stresses

there are NO shear forces in pure bending bending moments (couples) create normal stresses

x=P

AMzy

Iz+

Myz

Iy

normal stress;

general eccentric loading

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x=MzyIz

+Myz

Iy

normal stress;

unsymmetric bending

x= PA

MyI

normal stress;

eccentric loading

tan =

Iz

Iy tan angle between neutral axis andz-axis;

unsymmetric bending

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R= A

dA

r

distance from the neutral axis to the center of curvature;curved beam

x= MyAe (R y)=

M(rR)Aer

normal stress;

curved beam

Beam Shear

Shear Flow(q): the horizontal shear per unit length

Shear Center: the point O of a section where the line of action of of the applied load P intersects the axis o

symmetry of the end section; the point where an applied load can create bending without twisting

H=V Q

Ix

shear force

q=H

x=

V Q

I

shear flow

V=vertical shear force in transverse section t= beam thickness perpendicular to the shear

Q=first moment of section about the neutral axis I= area moment of inertia of cross-section abo

Q=Ay

Qfor typical beam

shear always flows towards the neutral surface and the neutral axis of the beam the total shear flow must equal the shear force in both magnitude and direction shear flowq times the distancexbetween fasteners along the length of a beam gives the shear force

on each fastener

ave=H

A=

V Q

It

average horizontal shear stress

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max= 3V2A

maximum horizontal shear stress;

narrow solid rectangular beam

max= VAweb

maximum horizontal shear stress;

I-beam

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Load, Shear, and Bending Moment Diagrams

Concentrated Loads Only

make a free body diagram with cuts on both sides of every load and reaction

shear is constant between loads/reactions

bending moment should be the same on either side of a given load/reaction

bending moment varies linearly between loads/reactions

by convention, positive shear pushes left side of the beam up and the right side down

by convention, positive bending moment makes the beam bend down in the middle

shear is force needed for Fy=0 of FBD of loads and reactions

if the force needed for balance Fy=0 is down, shear ispositive, ifup,negative

bending moment is moment needed for Mcut= 0 of FBD of loads and reactions

if the bending moment needed for balance Mcut= 0 is down, the bending moment is positive, ifupnegative

General Load

a bending moment or couple creates a jump in the bending moment diagram equal to its magnitude

a pure bending moment or couple does not effect the shear diagram

Singularity Functions

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x an dx =

x an+1 n01

n + 1x an+1 n0

singularity function integration

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Beam Load, Shear, Bending, and Deflection

dM

dx= V

slope of moment curve

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dV

dx=w

slope of shear curve

M(x) =

V dx

bending moment

V(x) =

w dx

shear force

u (x) = (x) = 1EI

M dx

beam deflection slope

u (x) =y (x) = 1

EI

udx

beam deflection

First Moment-Area Theorem: the [area under the(M/EI) diagram between two points] is equal to [the anglebetween the tangents to the elastic curve drawn at these points]

Second Moment-Area Theorem: the [tangential deviation tC/Dof Cwith respect toD] is equal to [the first moment

with respect to a vertical axis through Cof the area under the(M/EI)diagram between Cand D]

Euler-Bernoulli Beam Deflection

EI

flexural rigidity

d4y

dx4=w (x)

EI

elastic curve;

prismatic beam

EIy (x) =

dx dx dx w (x) dx +

1

6

C1x3 +

1

2

C2x2 +C3x +C4 elastic curve solution

it is often useful to use this version of the Euler-Bernoulli beam equation for analyzing distributed loads

d2y

dx2=

M(x)

EI=

1

elastic curve equation

EIy = x

0

dx

x0

M(x) dx +C1x +C2

elastic curve solution

it is often useful to use this version of the Euler-Bernoulli beam equation for analyzing pointloads

Boundary Conditions:

yA=0 MA=0 yB=0 MB=0

simply supported beam

yA=0 yB=0

overhanging beam

yA=0 A=0 VB=0 MB=0

cantilever beam

hinged beams should be treated as two separate beams with a connection that only transmits axial loads

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Beam Bending Boundary Conditions

Condition y (x) (x) M(x) V(x)

Free End 0 0

Free End with Applied Force

Free End with Applied Moment

Hinge 0

Roller Support 0

Fixed Support 0

Clamped Support 0 0

Curved Beams

Beams on Elastic Foundations

EI4v

x4+ k(x) v (x) =0

EI4v

x4+ kv (x) =0

= 4

k

4EI

has units of inverse length

q (x) =k(x) v (x)

v (x) =ex (c1 cos (x) + c2 sin (x)) + ex (c3 cos (x) + c4 sin (x))

general solution

v (x) = e

x

2EI3[P cos (x) + M(cos (x) sin (x))]

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Beam Design Procedure

ASSUMEallow is the same in tension and compression

ASSUME cross section is symmetric about the neutral axis

ASSUME bending mode of failure

ASSUME the prices all materials are the same, only the amount of material used matters

1. findallow for the chosen material from design specifications and material properties or allow= UF.S.

2. draw shear and bending moment diagrams to find|M|max

3. findSmin fromSmin=|M|min

allow

4.

(a) for a timber beam, pickb andh of the cross section to satisfy any constraints and1

6

bh2 =S

Smin

(b) for a rolled steel beam, pick the lightest beam available per unit length to satisfy any constraints and

SSmin the most important design consideration is usually the location magnitude of the largest bending moment on

the beam

short beams, especially those made of timber, may fail in shear under a transverse load

DMD+ LMLMU

LRFD;

bending beam

MU= SU

ultimate bending moment strength

Columns, Buckling, and Stability

Columns, Buckling, and Stability

Stability: the ability of a structure to support a given load without experiencing a sudden change in its configuration

Stable: returning to the original equilibrium position

Unstable: diverging further from the original equilibrium position

Buckling: deformation due to compressive stress that causes large changes in alignment by folding or collapsing

the structure

Column: a vertical prismatic member under axial loading

Critical Load (Pcr): the maximum load under which a perturbed column will return to its original equilibriumposition

Effective Length(Le): the length of a pin-ended column having the same critical load as the given column

Load Eccentricity(e): the distance between the line of action of the loadP and the axis of the column

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ASSUME columns are straight, homogenous prisms

L=length Le=effective length r=radius of gyration e=eccentricity of the load

L

r=Slenderness Ratio

Le

r=Effective Slenderness Ratio

d2y

dx

2+p2y=0

buckling differential equation;

centric load

y=A sin (px) +B cos (px)

buckling equation general solution;

centric loading

p2 = PEI

load parameter

Pcr=2EI

L2e

critical load;

Eulers formula;

centric loading

Pcr= 2EI

L2

critical load;

Eulers formula;

pinned ends

cr= 2E

(Le/r)2

Eulers critical stress

cr= 2E

(L/r)2 Eulers critcial stress;

pinned ends

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d2y

dx2+p2y=p2e

buckling differential equation;

eccentric load

y=A sin (px) +B cos (px) e

buckling equation general solution;

eccentric loading

ymax=e

sec

PEI

L2

1 =esec2

PPcr

1 maximum allowable deflection;eccentric loading

max=P

A

1 +

ec

r2sec

P

EI

L

2

=

P

A

1 +

ec

r2sec

2

P

Pcr

maximum stress;

eccentric loading

P

A=

max

1 + ecr2

sec

12

P

EILe

r

max1 + ec

r2

secant formula;

eccentric loading

ASSUMESI=Ar2

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Column Design - Centric Loads

real column design is a very empirical process with various standards organizations detailing accepted meth

ods for calculating a safe, appropriate design

cr=1 k1Ler

straight line critical stress

cr=2 k2

Le

r

2 parabola critical stress

cr= 3

1 + k3

Ler

2 Gorden-Rankine critical stress formula

Column Design - Structural Steel

L/r4.71

EY

cr= 0.877e

critical stress;

structural steel;

centric loading

e= 2E

(L/r)2

parametere

all = cr1.67

AISC required safety factor

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Column Design - Aluminum

L/rx all =

C3

(Lr)2

empirical aluminum column buckling stress;

centric loading

C1=140MPa C2=0.874MPa C3=354000MPa 6061-T6aluminum alloy;

SI units

C1=20.3ksi C2=0.127ksi C3=51400ksi

6061-T6aluminum alloy;

english units

C1=213MPa C2=1.577MPa C3=382000MPa


SI units

C1=30.9ksi C2=0.229ksi C3=55400ksi


english units

Column Design - Wood

all =CCP

maximum allowable stress;

solid column;

single piece of wood

CP= 1 + (CE/C)2c

1 + (CE/C)

2c

2CE/C

c

column stability factor

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CE= 0.822E(L/d)2

parameterCE

c=0.80

parameterc;

sawn lumber

c=0.90

parameterc;

glued laminate wood

valid for any length of column - short, intermediate, or long

ASSUMES a rectangular cross section with sidesb anddandd< b

Column Design - Eccentric Loads

P

A+

Mc

Iall

allowable-stress method

sometimes results in an overly conservative design

P/A

(all )centric+

|Mx|zmax/Ix(all )bending

+|Mz|xmax/Iz(all )bending

1

interaction method

P/A(all )centric

+ Mc/I

(all )bending1

interaction method;

in plane of symmetry

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Initially Curved Columns

Nonlinear Elastic Columns

Rayleigh-Ritz Method

Exact Elliptic Integral Buckling

Perturbation Methods

Compound Buckling

Plates and Shells

Thin-Walled Pressure Vessels

Pressure Vessel: a closed container designed to hold gases or liquids at a pressure substantially different from

the ambient pressure

Thin-Walled Pressure Vessel: a pressure vessel where the thickness is substantially less than the surface area

typically, the diameter is at least 10 times (sometimes cited as 20 times) greater than the wall thickness

Hoop Stress: stresses along the circular cross section of a cylinder

Longitudinal Stress: stresses running along the length of a cylinder

when a pressure vessel is subjected to external pressure, the formulas are still valid

the stresses will be negative since the wall is now in compression instead of tension

p=gage pressure r=inner radius t= wall thickness

1= pr

t

hoop stress;

thin-walled pressure vessel

2= pr

2t

longitudinal stress;

thin-walled pressure vessel

1 and 2 are the principal stresses

max=2= pr

2t

maximum in-plane shearing stress;

cylindrical thin-walled pressure vessel

1=2= pr

2t

stresses;

spherical thin-walled pressure vessel

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max=0.51= pr

2t

maximum in-plane shearing stress;

spherical thin-walled pressure vessel

Leak Before Break Design

cc t

condition to avoid brittle fracture

cc=a= K2Ic

2t

critical crack length

ASSUMES a small initial surface flaw this criterion is conservative for cracks originating IN the material instead of at the surface IFc>a THEN this criterion is insufficient to ensure leak before break

2c=crack surface length a=crack depth t= pressure vessel wall thickness t= maximum stress in pressure vessel wall cc=a F= 1 S=t KIc can be found for a given material in tables

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Thick-Walled Cylinders

r=a2pi b2pe

b2 a2

(pi pe) a2b2b2 a2

1

r2

=a2pi b2pe

b2 a2 +

(pi pe) a2b2b2 a2

1

r2

z=r=rz=z=0

stress; thick-walled cylinder;

subjected to internal and external pressures

r(r) + (r) =2 a2pi b2pe

b2 a2

sum of stresses; as a function of radius;

thick-walled cylinder;


r= 1

E(r)

= 1E

( r)

z=E

2

a2pi b2peb2 a2

r=rz=z=0

strain; thick-walled cylinder;


u (r) =

1

E a2pi b2pe

b2

a2

r+

1 +

E (pi pe) a2b2

b2

a2

1

r

radial displacement; thick-walle

subjected to internal and externa

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pi=inner pressure pe=external pressure a=inner radius b=outer radius pressures pi and pe are positive when compressive r+ = [constant] angular displacementv () =0 due to symmetry

Interference Fit (Shrink Fit) (Press Fit): joining a larger and smaller thick-walled cylinder together without fasten-

ers by making the inner cylinder bigger than than the inside of the larger cylinder and raising the temperature o

the outer cylinder during assembly and then cooling to room temperature

ai=inner radius of inner cylinder bi=outer radius of inner cylinder

ao=inner radius of outer cylinder bo=outer radius of outer cylinder

c=inner-outer cylinder interface radius

ui (bi) = pbiEi

b2i a2i (1 i) b2i + (1 +i) a2i interface displacement of inner cylinder;

interference fit cylinders

u0 (ao) = pao

Eo (b2o a2o)

(1 o) a2o+ (1 +o) b2o interface displacement of outer cylinder;


=ui (bi) uo (ao)

deformed radius difference;

intereference fit

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p=

c

1

Eo

b2o+ c

2

b2o c2+o

+

1

E

c2 + a2ic2 a2i

i1

interface pressure approximation;


ASSUMESbiao=c a negative pressure is compressive

r(r) =

1

Er(r)

E ( (r) + z (r)) + T radial strain;


u (r) =rT

radial displacement;

interference fit cylinders;

unconstrained thermal expansion;

used to find the temperature difference needed to manufacture an interference fit

Internally Pressurized Spherical Membranes

Plates

x (x, y, z) = E

1 2(x+y)

y (x, y, z) = E

1 2(y+x)

xy (x, y, z) =Gxy

x (x, y, z) =z 2w

x2

y (x, y, z) =z 2w

y2

xy (x, y, z) =2z2w

xy

r(r, , z) =zE1 2

2w

r2 +

r

w

r+

r22w

2

(r, , z) =zE1 2

2w

r2 +

1

r

w

r+

1

r22w

2

r (r, , z) =2zG

1

r2w

1

r

2w

r

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r(r, , z) =z 2w

r2

(r, , z) =z

1

r

w

r+

1

r22w

2

r (r, , z) =2z 1r2

w

1

r

2w

r

Shells

Contact, Holes, and Stress Concentrations

Contact Stress

z (x, y) = 3P

2ab

1 x

2

a2y

2

b2

normal stress in ellipitical contact area;

Hertzian contact stress

z= 3P2ab

a=ca 3

3P

1 21

/E1+

1 22

/E2

2 (11+ 12+ 21+ 22)

b=cb3

3P1

21/E1+1

22/E2

2 (11+ 12+ 21+ 22)

ki=1 2i

Ei

a=2

P (k1+ k2)R1R2

L (R1+R2)

half width of contact region;

parallel cylinders in contact;

loadP

z= 1

P (R1+R2)

R1R2L (k1+ k2)

maximum normal stress;


loadP

z= 1

P

RL (k1+ k2)

maximum normal stress;

plate and cylinder in contact;

loadP

ASSUMES NO tangential loads

ASSUMES NO plastic deformation

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x=z

2 + (1 2) ba + b

y=z

2 + (1 2) aa + b

principal stresses;


loadP; center of contact area

a=ellipse semi-axis inx-direction b=ellipse semi-axis iny-direction

ASSUMES NO tangential loads ASSUMES NO plastic deformation

Holes in Plates

r= 0.50

1 a

2

r2

+

1 + 3

a4

r44 a

2

r2

cos (2)

=0.50

1 +

a2

r21 + 3 a

4

r4

cos (2)

r=0.50

1 3 a4

r4+ 2

a2

r2

sin (2)

stress; hole in 2D infinite rectangular pla

uniaxial tension;finite plate width

r(r) =3 +

8 2

a2 + b2 +

a2b2

r2 r2

(r) =3 +

8 2a2 + b2 +

a2b2

r2 1 + 3

3 + r2

Stress Concentrations

Stress Trajectories: curves that define the direction of the largest tensile stress or compressive stress at each

point of the curve

K=maxave

stress-concentration factor

max=KT c

J maximum shear stress;

torsion at shaft fillet

m=KMc

I

maximum stress;

critical cross section

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Problem Solving Methods

Statically Indeterminate Loadings

Statically Indeterminate: a problem where the reactions and or the internal forces cannot be solved for using only

the laws of statics

a structure is statically indeterminate whenever it is held by more supports than are required to maintain its

equilibrium

these methods can be applied to any type of loadings - axial, torsion, etc.

Basic Geometry Method

1. use free body diagrams to get an equation relating the loads

2. use geometry to get an equation relating the displacements

(a) IF the displacements are equal, THEN set them equal and substitute the formulae for displacements

(b) IF the total displacement is zero, THEN set the sum of the displacements equal to zero and substitute

the formulae for displacements

3. solve the system of two equation from steps one and two to find the loads

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Superposition Method

1. ignore all but one of the reaction forces

(a) ALWAYS use all of the load forces

2. solve the problem for the reaction

3. repeat the problem for each of the reaction forces

(a) the last reaction force can be found using static force balance

4. add (superimpose) each of the solutions to get the actual solution

Combined Loadings

1. sketch diagrams of the object

orthographic drawings are usually better than isometric for complex problems

include dimensions

include external forces

do NOT include internal forces

2. draw free body diagrams of the object in each view

3. determine any unknown external forces such as reaction forces

write force balance in each direction

write moment balance in each direction

4. draw shear force and bending moment/torsion diagrams for the length of the object

include diagrams for all3 force directions and3 moment directions

5. superimpose the graphs on each other

6. calculate the needed geometric properties A,Q,J,Ixx,Iyy,Izz

7. find the cross section with the maximum shear stress

(a) ASSUME the object with fail in shear

(b) ASSUME the forces from are negligible

8. Find the infinitesimal area of the failure cross section with the greatest shear stress

(a) use the fact that the torsion shear stress is greatest on the outer surface

(b) use the fact that the bending normal stress is greatest at the ends away from the neutral surface

(c) assume that all forces originate from the centroid

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Energy Methods

U=

x10

P dx

strain energy

U= 0.5P1x1

strain energy;

elastic deformation

u=

10

xdx

strain energy density

u= 21

2E

elastic strain energy density

u=dU

dV

strain energy density;

infinitesimal element

U=

2x2E

dV

elastic strain energy;

uniaxial normal stress

U=

L0

P2

2AEdx

strain energy;

uniaxial centric normal stress

U= P2L

2AE

strain energy;

uniaxial centric normal stress;

uniform cross-section

U=

L0

M2

2EIdx

strain energy;

pure bending

Iis about the neutral axis

u=

xy0

xydxy

strain-energy density;

plane shear stress

U= 2xy

2G dV elastic strain energy;

plane shear stress

U=

L0

T2

2GJdx

strain energy;

torsion; circular shaft

U= T2L

2GJ

strain energy;

torsion; circular shaft;

uniform cross section

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u= 1

2E

21+

22+

23 2 (12+ 23+ 31)

strain energy density;general state of stress

uv= 1 26E

(1+ 2+ 3)2

volume change strain energy density;

general state of stress

ud= 1

12G

(1 2)2

+ (2 3)2

+ (3 1)2 distortion strain energy density;

general state of stress

Castiglianos Theorem

xj=U

Pj

Castiglianos theorem

j= U

Mj

slope of beam;

point of application of coupleMj

j=U

Tj

angle of twist;

section of shaft where torque Tj is applied

xj=U

Pj=

L0

M

EI

M

Pjdx

beam deflection;

point of application of load Pj

xj=U

Pj=

n

i=1

FiLi

AiE

FiPj

truss deflection;

point of application of loadPj

Virtual Work

Rayleigh-Ritz Method

The Finite Element Method

Basic Steps in the Finite Element Method

Preprocessing Phase

1. create and discretize the solution domain into finite elements; subdivide the problem into nodes and elements

2. ASSUME a shape function to represent the physical behavior of an element; a continuous function is as

sumed to represent the approximate behavior (solution) of an element

3. develop equations for an element

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4. assemble the elements to present the entire problem. Construct the global stiffness matrix

5. apply boundary conditions, initial conditions, and loading

Solution Phase

1. solve a set of linear or nonlinear algebraic equations simultaneously to obtain nodal results, such as dis

placement values at different nodes or temperature values at different nodes in a heat transfer problem

Postprocessing Phase

1. obtain other important information such as values of principal stresses, heat fluxes, etc.

FEM Requirements

the displacement field within an element must be continuous

when nodal displacements are given values corresponding to constant strain, the displacement field must

produce the constant strain state throughout the element

this is so that, as elements are taken smaller (in the h-version) and the actual strain becomes nearlyconstant within them, the finite element method will reproduce this state of constant strain

in this manner, the finite element solution converges to the exact solution as the finite element mesh ismade finer

the element must accurately represent rigid body motion

when the nodal displacements correspond to rigid body motion, the element must produce zero strainand zero nodal forces

compatibility must exist between elements

the assumed displacement field must not make the elements separate or overlap this condition is frequently violated

invariance: the element should have no preferred directions

Basic Equations

[k] = [ke]

global vs element stiffness matrix

[Q] = [Qe]

global vs element force matrix

[d] = [d]e

global vs element displacement matrix

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[] = []e

global vs element stress matrix

[f] =N[d]

displacement field from nodal displacements; FEM

m=rq

number of nodal coordinates

r=number of coordinates per node q=number of nodes

Q=K[d]e

global force matrix; FEM

Q=global force matrix K=global stiffness matrix

[Q]e=ke [d]e

nodal force matrix; FEM

this is analogous to the basic Hookes law F= kx

[]e=B [d]e element strain; FEM []e=D []e

element stress; FEM

[Q]e=nodal force matrix ke=element stiffness matrix N= shape function matrix [d]e=nodal displacement column vector B=special type of derivative of N D=constituitive relation matrix, such as Hookes law stiffness matrix

Affine Element: an element with linear approximating polynomials

Isoparametric Element: an element with approximating polynomials of an order greater than one

affine elements always have straight edges

isoparametric elements edges may be or become curved

cond (A) =A A1

condition number;

squaren nmatrix

Basic 1D FEM Equations

L

dN1dx

dN2dx

TdN1dx

dN2dx

AE dx

u1u2

=

L

N1 N2

TA dx +

N1 N2

TF|L0

k=

L

dN1

dx

dN2

dx

TdN1

dx

dN2

dx

AE dx

stiffness matrix;

1D FEM

L

N1 N2

TA d x=

L

N1A dx

L

N2A dx

body forces;

1D FEM

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N1 N2

TF|L0=

FL

FR

surface forces;

1D FEM

N=3 mnodal coordinate matrix B=

D=

Basic 2D FEM Equations

W=

A

xx

x+

xy

y+ bx

ux+

xy

x+

yy

y+ by

uy

dxdy=0

virtual work;

2D triangular elemen

u= [N] [u]

nodal displacements from virtual displacements

u (x, y) =

uiNi (x, y) ujNj(x, y) ukNk(x, y)

v (x, y) =

viNi (x, y) vjNj(x, y) vkNk(x, y)

2D element nodal displacements

u

v

=

Ni Nj Nk 0 0 0

0 0 0 Ni Nj Nk

ui

uj

uk

vi

vj

vk

virtual nodal displacements;

2D triangular element;

no midpoint nodes

u

v

=

Ni Nj Nk 0 0 0

0 0 0 Ni Nj Nk

ui

uj

uk

vi

vj

vk

nodal displacements;


no midpoint nodes

B=

x0

0

y1

2

y

1

2

x

N

2D

D=

1 0 0

0 1 0

0 0 2

[R (E, )]

constituitive relation matrix;


no midpoint nodes

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ANSYS Workflow

1. input material properties

2. set up the geometry

either third party CAD software or the built in program can be used

3. set up the mesh

add refinements to important areas and areas of interest

4. set up the physics

body forces

surface forces

boundary conditions

5. set up the desired results

displacements

strains

stresses

failure criterion

6. solve for the desired results

review results

reality check results against analytical approximations

see if mesh is adequate

see if constraints and boundary conditions are adequate

use the probe tool to check the actual stress values because the color scales may be deceiving

check numerical methods with a similar problem that can be solved analytically

analytical solutions are likely to ignore stress concentrations

7. optimize design parameters

ANSYS Design Optimization

1. complete standard design workflow

2. under the appropriate sections such as Geometry or Results, check boxes for the parameters to be con-

sidered

(a) check the parameters to be changed in the optimization such as various dimensions

(b) check the parameters to be optimized or to serve as constraints such as volume or maximum stress

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3. create aGoal Driven Optimizationmodule

4. under Design of Experiments, set the upper and lower bounds on the parameters

5. clickPreview, followed byUpdate

these steps are very computationally intensive

6. go to theResponse Surfacesection to create graphs

7. under the Optimization section, set which parameters need to serve as constraints or be optimized and

how

8. click Update Optimization

9. save the best design candidate using Insert as Design Point under Optimization followed by Duplic

Design PointunderParameter Set

10. under Parameter Set, set the best design candidate as the design used with Copy Inputs to Current

by right clicking the desired design point

11. clickUpdate All Design Points

12. verify the accuracy of the results

ANSYS Strain Gauge Simulation

Failure Criterion

Yield Criterion

Yield Criterion (Ductile Failure Criterion):

for uniaxial stress, the maximum normal stress before yielding is the yield stressx

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Maximum Shear Stress Criterion (Trescas Hexagon): a structure wont fail as long as the maximum value o

the shear stressmax in the structure is less than the shear stress at yielding for a tensile test

s=s

2 =MAX

|1 2|2

,|2 3|

2 ,

|3 1|2

maximum shear stress criterion;

principal stresses

s=MAX(|1

2

|,|

2

3|,|

3

1|) maximum shear stress criterion;

principal stresses

max=|a b|

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the maximum distortion energy criterion is somewhat more accurate than the maximum shear stresscriterion for predicting yield in torsion

H(X Y)2 + F(Y Z)2 + G (Z X)2 + 2N2XY+ 2L2Y Z+ 2M2ZX=1

orthotropic yield criter

X-Y-Z axes are aligned with the planes of material symmetry H+ G= 1

2oXH+ F=

1

2oYF+ G=

1

2oZ

2N= 12oXY

2L= 1

2oYZ2M=

1

2oZX

Fracture Criterion

Fracture Criterion (Brittle Failure Criterion):

flaws in a material tend to weaken the material in tension, while not appreciably affecting its resistance to

compressive failure

Maximum Normal Stress Criterion (Coulombs Criterion):

N= MAX (|1| ,|2| ,|3|)

maximum normal stress criterion

X= uN

MAE 3272 Notes

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