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MECHANICS OF MATERIALS
2 - 1Nazarena Mazzaro, AAU
Static Indeterminacy
Cases in which internal forces and reactions cannot be
determined from statics are said to bestatically indeterminate.
We have more unknowns than equations
We introduce other relationships: deformations
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MECHANICS OF MATERIALS
2 - 2Nazarena Mazzaro, AAU
Static Indeterminacy
0=+=RL
Deformations due to actual loads and redundant
reactions are determined separately and then added
orsuperposed.
Superposition Method: Redundant reactions
are replaced with unknown loads which along
with the other loads produce deformations.
A structure will be statically indeterminate
whenever it is held by more supports than are
required to maintain its equilibrium.
Statics:
RA + RB = PK+ PD
One equation, 2 unknowns
Statically indeterminate
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MECHANICS OF MATERIALS
2 - 3Nazarena Mazzaro, AAU
Example 2.04
Determine the reactions atA andB assuming a close
fit at both supports before the loads are applied.
Solve for the reaction atA due to applied loads
and the reaction found atB.
Require that L + R= 0
b) Solve for the RatB due to the redundant
reaction atB.
SOLUTION:
a) Consider the reaction atB as redundant,
release the bar from that support, and solve the
L atB due to the applied loads.
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MECHANICS OF MATERIALS
2 - 4Nazarena Mazzaro, AAU
SOLUTION: Solve L atB due to the applied loads with the
redundant constraint released -> internal forces
EEA
LP
LLLL
AAAA
PPPP
i ii
ii9
L
4321
2643
2621
34
3321
10125.1
m150.0
m10250m10400
N10900N106000
==
====
====
====
Solve RatB due to the redundant constraint,
( )
==
==
==
==
i
B
ii
ii
R
B
E
R
EA
LP
LL
AA
RPP
3
21
262
261
21
1095.1
m300.0
m10250m10400
Example 2.04
P2
P3
P4
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MECHANICS OF MATERIALS
2 - 5Nazarena Mazzaro, AAU
Require that the displacements due to the loads and due to
the redundant reaction be compatible,
( )
kN577N10577
01095.110125.1
0
3
39
==
=
=
=+=
B
B
RL
R
E
R
E
Example 2.04
Find the reaction atA due to the loads and the reaction atB
kN323
kN577kN600kN3000
=
+== Ay
R
RF
kN577
kN323
=
=A
R
R
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MECHANICS OF MATERIALS
Nazarena Mazzaro, AAU
Thermal Stresses
If the temperature increased by T the rod elongates by T
which is proportional to the temperature change and the
length of the rod.
T=(T )L
: coeficient of thermal expansion [1/C]
Thermal strain: T= T
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MECHANICS OF MATERIALS
2 - 7Nazarena Mazzaro, AAU
Thermal Stresses
A temperature change results in thermal strain.
There is no stress associated with the thermal strain
unless the elongation is restrained by the supports.
( ) AEPL
LT PT == ;
Treat the additional support as redundant and apply
the principle of superposition. P represents the
redundant action at B
0=+= PT
The thermal deformation and the deformation from
the redundant support must be compatible.
( )
( )
( )TEP
TAEP
AE
PLLT
==
=
=+
0
ANIMATION
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MECHANICS OF MATERIALS
2 - 8Nazarena Mazzaro, AAU
Poissons Ratio
For a slender bar subjected to axial loading:
0=== zyx
xE
The elongation in the x-direction is
accompanied by a contraction in the other
directions. Assuming that the material is
isotropic (no directional dependence),0= zy
Poissons ratio is defined as
E
xzy
xx
x
z
x
y
===
===
;
strainaxial
strainlateral
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MECHANICS OF MATERIALS
Nazarena Mazzaro, AAU
Fiber optic to measure tendon force
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MECHANICS OF MATERIALS
Nazarena Mazzaro, AAU
Ft =[0.08/(0.00025)] Fc = 320 Fc
PMMA: Compression Strength = 70 131 MPa
Ifc = 70 MPa => t 70 [MPa] (320)= 11.2 GPa
Maximum stress measured in the Achilles Tendon during running using a buckle
transducer = 0.11 GPa [Komi 1990]
11.2 /0.11 200 !!!
Fiber compression theory
(Decrease in light transmission)
Fiber optic to measure tendon forces
The Achilles tendon has to be loaded 200 times the peak load measured during running to
compress the fiber
8cm
/2= 0.25 mm
Fc
Ft
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MECHANICS OF MATERIALS
2 - 11Nazarena Mazzaro, AAU
Generalized Hookes Law
Multi-axial loading: normal components
resulting from components may be determined
from theprinciple of superposition. This
requires:
1) strain is linearly related to stress
2) deformations are small
E
EEE
EEE
zyxz
zyx
y
zyxx
+=
+=
+=
With these restrictions:
ANIMATION
Generalized
Hooks law
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MECHANICS OF MATERIALS
2 - 12Nazarena Mazzaro, AAU
Shearing Strain
A cubic element subjected to shear stress will deform
into a rhomboid. Theshear strain is the change in
angle between the sides.
[rad] represents theshearing strain
xyxy f =
A plot of- is similar to plots of-. For small strains:
zxzx
yzyz
xyxy
G
G
G
=
=
=
G [Pa] is Modulus of rigidity or shear modulus.
xy [rad] angle corresponding to the x and y directions
Hook Law
for-
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MECHANICS OF MATERIALS
2 - 13Nazarena Mazzaro, AAU
Example 2.10
A block of material with modulus of
rigidity G = 630 GPa is bonded to two
rigid horizontal plates. The lower plate
is fixed, while the upper plate is
subjected to a horizontal forceP. The
upper plate moves 1mm under theaction of the P; determine a) the
average shearing strain, and b)P.
SOLUTION:
Determine the angular deformation
or shearing strain of the block.
Use the definition of shearing stress to
find the forceP.
Apply Hookes law to find the
corresponding shearing stress.
50 mm
62 mm
200 mm
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MECHANICS OF MATERIALS
2 - 14Nazarena Mazzaro, AAU
Determine the angular deformation orshearing strain.
rad020.00mm5
1tan == xyxyxy
mm
Apply Hookes law to find the shearing
stress.
( )( ) MPaMPaG xyxy 6.12rad020.0630 ===
Use the definition of shearing stress to find
the forceP.
( )( )( ) kNmmmmMPaAP xy 2.156622006.12 ===
kNP 2.156=
1mm
50 mm
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MECHANICS OF MATERIALS
2 - 15Nazarena Mazzaro, AAU
Saint-Venants Principle
Loads transmitted through rigid plates
result in uniform distribution of stress
and strain.
Saint-VenantsPrinciple: distribution may be assumed
independent of the mode of load
application except in the immediate
vicinity of load application point.
and distributions become uniformrelatively close the load application
points.
Concentrated loads cause large stresses in
the vicinity of the load application point.
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MECHANICS OF MATERIALS
2 - 16Nazarena Mazzaro, AAU
Stress Concentration: Hole
Discontinuities of cross section may result in
high localized orconcentratedstresses. ave
max
=K
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MECHANICS OF MATERIALS
Nazarena Mazzaro, AAU
Stress Concentration: Hole
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MECHANICS OF MATERIALS
2 - 18Nazarena Mazzaro, AAU
Stress Concentration: Fillet
ave
max
=K
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MECHANICS OF MATERIALS
2 - 19Nazarena Mazzaro, AAU
Example 2.12
Determine the largest axial loadP
that can be safely supported by a
flat steel bar consisting of two
portions, both 10 mm thick, and
respectively 40 and 60 mm wide,
connected by fillets of radius r= 8
mm. Assume an all 165 MPa.
SOLUTION:
Determine the geometric ratios and
find the stress concentration factor K
Apply the definition of normal stress to
find the allowable load.
Find the allowable-average normal
stress ave using the material allowable
normal stress all and K.
ave
max
=K
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MECHANICS OF MATERIALS
2 - 20Nazarena Mazzaro, AAU
Determine the geometric ratios and
find K
82.1
20.0mm40
mm850.1
mm40
mm60
=
====
K
d
r
d
Find the average normal stress ave
using the material allowable normal
stress all and K.
NOTE: max all
MPa7.9082.1
MPa165maxave ===
K
Apply the definition to find the allowable P.
( )( )( )MPa7.90mm10mm40== aveAP
kN3.36=
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