axial loading-part 2

7/29/2019 Axial Loading-Part 2

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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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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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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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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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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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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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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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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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Fiber optic to measure tendon force


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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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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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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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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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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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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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Stress Concentration: Hole

Discontinuities of cross section may result in

high localized orconcentratedstresses. ave

max

=K


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Stress Concentration: Hole


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Stress Concentration: Fillet

ave

max

=K


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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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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=

axial loading-part 2

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