Lectures 6 -7 - Lecture notes 6-7 PDF

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CEE 332

Statically indeterminate problems A problem is statically indeterminate if the equations of equilibrium are not sufficient to determine all unknowns. Rule of thumb: A structure is a mechanism if: A structure is statically determinate if: A structure is statically indeterminate if:

# unknowns < # equil. eqns. # unknowns = # equil. eqns. # unknowns > # equil. eqns.

Example: Trusses in 2-D (i.e., 2-force members).

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Statically indeterminate problems A problem is statically indeterminate if the equations of equilibrium are not sufficient to determine all unknowns. Rule of thumb: A structure is a mechanism if: A structure is statically determinate if: A structure is statically indeterminate if:

# unknowns < # equil. eqns. # unknowns = # equil. eqns. # unknowns > # equil. eqns.

Example: Trusses in 2-D (i.e., 2-force members). n  r  2j { ?

n  # members r  # reactions j  # joints 2 + 4 = 2(3)  statically determinate 32

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Statically indeterminate problems A problem is statically indeterminate if the equations of equilibrium are not sufficient to determine all unknowns. Rule of thumb: A structure is a mechanism if: A structure is statically determinate if: A structure is statically indeterminate if:

# unknowns < # equil. eqns. # unknowns = # equil. eqns. # unknowns > # equil. eqns.

Example: Trusses in 2-D (i.e., 2-force members). 2j n  r  { ?

n  # members r  # reactions j  # joints 2 + 4 = 2(3)   statically determinate

2j n  r  { ?

3 +6 < 2(4)   statically indeterminate 33

In problems of mechanics, equations from three physical categories must be satisfied: • Equilibrium: • Material model:

• Compatibility:

Up until now in MoM, compatibility has been a trivial issue. That is, all of our solutions have been fully compatible without the need to explicitly consider equations of compatibility.

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CEE 332 In problems of mechanics, equations from three physical categories must be satisfied: • Equilibrium:

∑F = 0, ∑M = 0.

• Material model:

Sometimes called “stress–strain relation”, “constitutive relation”, or “load–displacement relation”. E.g.,

 = E  , or  = PL/(AE). • Compatibility:

Sometimes called “kinematics.” Materials may not overlap (this is physically impossible) or separate (this is a crack), or leave the supports.

Up until now in MoM, compatibility has been a trivial issue. That is, all of our solutions have been fully compatible without the need to explicitly consider equations of compatibility.

Methods for solving statically indeterminate problems • Superposition: Your book uses this method a lot, but it is not very robust or methodical. • Fundamental approach: Write equations for equilibrium, material behavior and compatibility to produce a determinate system of equations.

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Example PROBLEM 2.35 - A 4-ft concrete post is reinforced with four steel bars, each with a 3/4-in. diameter. Knowing that ES = 29 × 106 psi and EC = 3.6 × 106 psi, determine the normal stresses in the steel and in the concrete when a 150-kip axial centric force P is applied to the post.

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PROBLEM 2.40 A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at both ends and supports two 6kip loads as shown. Knowing that E = 0.45 × 106 psi, determine (a) the reactions at A and C, (b) the normal stress in each portion of the rod.

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§2.9 Superposition, statically indeterminate problems

Superposition: If a structure is linear–elastic, superposition can be used to find stresses, strains, deflections, etc. • Superposition is useful for taking a problem with complicated loading, and replacing it by several problems having simpler loading. • Each of the simpler problems must have the same geometry, support conditions, and material properties.

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§2.10 Thermal effects Observation: When a material undergoes a change of temperature ∆T, it also undergoes a change of geometry: – expansion if ∆T > 0 – contraction if ∆T < 0 • If ∆T is not too large, then T =  ∆T T = “thermal strain” ∆T = change of temperature (+ for temperature increase)  = coefficient of thermal expansion  (units: 1/ºF or 1/ºC) • Temperature change produces expansion or contraction only – there is no distortion.

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Example PROBLEM 2.51 - A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel (Es = 200 GPa, αs = 11.7 × 10−6 / °C and portion BC is made of brass (Eb =105 GPa, αb = 20.9×10−6/ °C). Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 50°C. (Use superposition)

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§2.11-2.15 Poisson’s ratio, multiaxial loading Observation: When a material elongates in one direction, almost all materials will also contract in the other directions. This is called the “Poisson effect”. Consider the axial loading situation shown.

x 

x E

y  z

 

, y z  0 (for isotropic materials ... that is, for materials whose behavior does not depend on direction)

  lateral strain  y  z axial strain x x

 = Poisson’s ratio (Greek nu)

 is dimensionless

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Some values of  Material 2014-T6Al Ti Au Diamond Rubber Cork Lakes’ Foam

 0.33 0.33 0.42 0.08 0.49 ~0 – 0.7

Siméon Denis Poisson (1781-1840) All materials: –1 <  < 1/2. Many materials:  = 1/4 to 1/3. Materials with  < 0 are rare ... 49

Generalized Hooke’s law • For an element of material subjected to multiaxial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) Strain is linearly related to stress. 2) Deformations are small. 3) Material is isotropic (no direction dependence). • With these restrictions:

x   y   z  

x E

x E

 x

E 1 2 3

strains due to stress in x direction

 

 y E

y E

 y

 E 1 2 3

strains due to stress in y direction

 

z E

z E

z

 E 1 23

strains due to stress in z direction

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Rearranging terms, generalized Hooke’s law, for linear-elastic isotropic materials, is   x  x   y   z  y  z 

E

E

y



E

z E

 

E

 E

  T

 x   z 

  T



  T

 y 

x

include these terms for thermal effects

These relations can be inverted, and with ∆T=0, stresses in terms of strains are:





E (1   ) x    y  z  (1   )(1  2 ) E y  (1  ) y    x  z (1   )(1  2 )

x 

z 

E (1   )(1  2 )

(1  )    z

x



 y 

Note what happens when  —> 1/2!

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Dilatation and bulk modulus • Relative to the unstressed state, the change in volume is e  1 1  x 1 y 1  z   1 1  x  y  z  x  y  z 1 2

 x  y   z  E  dilatation (change in volume per unit volume)



• For element subjected to uniform hydrostatic pressure 3 1  2   p e  p  E k or p  ke. Note the similarity to   E E k  bulk modulus 3 1  2  

This is called Hooke’s Law for hydrostatic pressure • When material is subjected to uniform pressure, if 0   1 we require the dilatation to be negative (these are 2 sometimes called conventional materials) then

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Shear strain

• A cubic element subjected to a shear stress will deform into a rhomboid. The change of shape (distortion) is measured by the shear strain.

xy = shear strain

(Greek gamma) = change in angle between two initially perpendicular line segments.

units: radians (... actually, it is dimensionless). 53

Hooke’s law for shear stress – shear strain

 vs.  (torsion test) is similar to plots of  vs. , except that strength values

• A plot of

(shear stress where failure starts) are approximately half. For linear elastic isotropic materials: Hooke’s Law for shear

 xy  G  xy  yz  G  yz  zx  G  zx G = modulus of rigidity or shear modulus. units: force/area

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Example A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate displaces 0.04 in. under the action of the force, determine:

Strategy:

a) the average shearing strain in the material, and b) the force P exerted on the plate.

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Example A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate displaces 0.04 in. under the action of the force, determine: a) the average shearing strain in the material, and b) the force P exerted on the plate.

Strategy: • Determine the average angular deformation (or shearing strain) of the block. • Apply Hooke’s law for shear stress – shear strain to determine the shear stress. • Use the definition of shearing stress to find the force P. 56

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SOLUTION • Determine the average angular deformation (or shearing strain) of the block.  xy  tan  xy 

0.04 in. 2 in.

 xy  0.020 rad

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.

D

D’



xy

3  G xy  90 10 psi 0.020 rad  1800 psi

• Use the definition of shearing stress to find the force P. P   xy A  1800 psi 8 in. 2.5 in.   36  10 3 lb

P  36.0 kips

A

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Relationship between E, , and G • Before a load is applied, sketch an isosceles triangle on the bar. • After the axial load is applied, the isosceles triangle deforms into the shape of a general right triangle. • See §2.15 for derivation ... or ....    1  x — Start with tan    4 2  1 x — Use a trig identity. — Use x   x E where  x  F A 0 (§1.11 notes). — Use    G where   F 2A0 (§1.11 notes). — Algebra ....

assumes linear elastic isotropic material

E E  1   or G  2G 2 1  

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Summary: Hooke’s law for a linear elastic isotropic material x  y  z 

x E

y E

z

 yx   yz   xz 

E

  

 E

 E

 E



y

 z

  T

 x   z 

  T



  T

x

 y 

include these terms for thermal effects

 yx G

 yz G

 xz

E = elastic modulus G = shear modulus  = Poisson’s ratio

G

G

E 2(1  ) 59

Example

A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z = 20 ksi. For E = 10 x 106 psi and  = 1/3, determine the change in: a)

the length of diameter AB,

b) the length of diameter CD, c)

the thickness of the plate, and

d) the volume of the plate.

= 12 ksi = 20 ksi 60

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SOLUTION: 1) Apply generalized Hooke’s Law to determine the three components of normal strain. x  

2) Evaluate the deformation components.  B A   x d  0.533103 in./in. 9 in. B

 x  y  z 

E



E

E

1

1    12 ksi  0  3  20 ksi 10 10 6 psi    0.533 103 in./in. E

C

3

 14.4  10 3 in.

 t   0.800 10 3in.

  y  z z   x   E

D

t  yt  1.067 103 in./in.0.75 in.

E

  1.067 10 3 in./in.

E

 4.8  10 3 in.

C D   z d  1.600 103 in./in.9 in. 

   y   x  y  z E

A

3) Determine the change in volume 3 3 3 e  x y   z  1.067  10 in /in

V  eV  1.067  103 15  15  0.75 in 3

E

 1.600 10 in./in.

V  0.1800 in3

Or, V  V final  Vorig  (15 in)(1  x ) (15 in)(1  y ) (3 /4 in)(1 z )  (15 in)(15 in)(3 /4 in)  same result

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§2.17-2.20 St.Venant’s principle, stress concentrations, plasticity Adhémar Jean Claude Barré de Saint-Venant (1797- 1886) Q: Is the method of load application important? Loads applied to rigid end plates

Uniform state of stress



P A

Loads applied to sharp wedges

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Q: Is the method of load application important?

A: Yes, but ....

• Loads applied by the rigid frictionless plates results in a uniform distribution of stress and strain. • Concentrated loads result in very large stresses in the vicinity of the load application point. • However, the stress and strain distributions become uniform at a relatively short distance from the load application points. • Saint-Venant’s Principle (1855) describes this effect. • For all of the stress distributions shown, ∫  dA is the same.

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Saint–Venant’s Principle • Two statically equivalent force systems applied to the same region of a body produce essentially the same state of stress and deformation at sufficiently large distances from the region of loading (i.e. distances from the loaded region that are larger than the size of the loaded region).

Examples of statically equivalent force systems

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Stress Concentrations

• •

•

A Stress Concentration is a localized increase in stress due to a geometric or compositional change in the material. Stress concentration factors can be determined many different ways: analytically, experimentally, finite elements. For design, the maximum stress is usually most important (as opposed to the full stress distribution)

 max  max

 K ave

K

 stress concentration factor

Anet

 x-sectional area thru the plane w/ max. stress  average stress supported by the plane w/ max. stress  Fapplied Anet

 ave

 actual max. or peak stress

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Stress Concentration factors for a hole in a long plate

Remark: about D away from the hole, according to St. Venant’s principle, the stress is close to P/A.

 ave 

P (D  2r)t

 max  K ave

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Rather than using a plot for stress concentration factors, some references use a polynomial best fit*. E.g.,

2

3

K

 2r  2r   2r   3  3.13   3.66   1.53   ... D D   D

Anet

 D  2rt

 ave



t

 thickness of the plate

P P  D  2rt Dt

* Warren C. Young, Roark’s Formulas for Stress and Strain, McGraw-Hill.

Some extremes .... Really small hole: 2r 0 D

K 3

Really big hole: 2r 1 D

K2

(Note that although K–>2, ave –>∞)

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Stress Concentration factors for symmetric fillets in a bar

Q: What happens as r —> 0 ? 69

Example

The bar shown is made of steel with E = 30(10)6 psi and Y = 30 ksi.

a) Determine the value of the axial force P which will cause the bar to begin yielding. b) Determine the largest value of the axial force P the bar can support (the collapse load).

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Plastic deformations • Previous analyses assumed linear, elastic stress-strain behavior. Provided stresses remain below the yield stress, all deformations are reversible. • For brittle materials, linear elastic behavior can often be used all the way to rupture. • For ductile materials, linear elastic behavior can be used only until yielding begins. Once yielding starts, permanent deformations occur.

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Some popular models for plastic deformations

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Stages of plastic deformation

Assume elastic–perfectly plastic material.

A  P   ave A  max K

PY 

• Elastic: maximum stress is less than the yield stress. • Initial yield: Maximum stress just reaches the yield stress.

Y A K

• Continued yield: a region of plastic deformations develops near the hole and grows in size. • Fully plastic: plastic region expands until the section is at a uniform stress equal to the yield stress.

PU   Y A  K PY

collapse load!

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Residual stresses • If a member is loaded into the plastic range, after the external forces are removed, residual stresses will often remain in the member. • Example: Consider the strip with a hole from the previous slide. • Apply force P until the cross section by the hole is fully plastic.

• Remove P. The cross section by the hole has the residual stress field shown (assuming re-yielding in compression does not occur). 74

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Residual Stresses 1. Thermal Residual Stresses: W /I shape

Distribution of Residual-Stress in W section CE-409: Lecture 07

Prof. Dr Akhtar Naeem Khan

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Residual Stresses 1. Thermal Residual Stresses: W /I shape 20 W’s shapes were investigated: • • •

It Revealed that flange-tip stress frc varied from 4.1 to 18.7 Ksi, the average being 12.8 ksi Residual stresses in web center varied from 41Ksi compression to 18.2Ksi tension. Showing some W’s develop residual tension over entire web, instead the pattern shown.

CE-409: Lecture 07

Prof. Dr Akhtar Naeem Khan

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PROBLEM 2.102 Rod ABC consists of two cylindrica...