mae 3272 notes
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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
equilibrium equations;
linearly elastic body;
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
equilibrium equations;
linearly elastic body;
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
strain-displacement relations;
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
strain-displacement relations;
small displacements;
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
strain-displacement relations;
small displacements;
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,
homogeneous material
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,
homogeneous material
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,
homogeneous material
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,
homogeneous material;
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,
homogeneous material;
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,
homogeneous material
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,
homogeneous material;
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;
thin-walled hollow noncircular shaft
ASSUME wall thicknesstis small compared to other dimensions
= T
2ta
shear stress;
thin-walled hollow noncircular shaft
= T L
4a2G ds
t angle of twist;
thin-walled hollow noncircular shaft
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
IFn
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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
2024-T6aluminum alloy;
SI units
C1=30.9ksi C2=0.229ksi C3=55400ksi
2024-T6aluminum alloy;
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;
subjected to internal and external pressures
r= 1
E(r)
= 1E
( r)
z=E
2
a2pi b2peb2 a2
r=rz=z=0
strain; thick-walled cylinder;
subjected to internal and external pressures
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;
interference fit cylinders
=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;
interference fit cylinders
ASSUMESbiao=c a negative pressure is compressive
r(r) =
1
Er(r)
E ( (r) + z (r)) + T radial strain;
interference fit cylinders
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;
parallel cylinders in contact;
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;
parallel cylinders in contact;
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;
2D triangular element;
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;
2D triangular element;
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
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