lecture 6 –axial deformation

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Lecture 6 – Axial deformation Instructor: Prof. Marcial Gonzalez Fall, 2021 ME 323 – Mechanics of Materials Reading assignment: 3.1 – 3.4, 3.6 Last modified: 8/16/21 9:12:58 AM

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Page 1: Lecture 6 –Axial deformation

Lecture 6 – Axial deformation

Instructor: Prof. Marcial Gonzalez

Fall, 2021ME 323 – Mechanics of Materials

Reading assignment: 3.1 – 3.4, 3.6

Last modified: 8/16/21 9:12:58 AM

Page 2: Lecture 6 –Axial deformation

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Generalized Hooke’s law (review)- Isotropic material: material properties are independent of the

orientation of the body- The principle of superposition can be used for linear elastic materials

and small deformations (i.e., strains may be added linearly)

Generalized Hooke’s law

Q: Plane stress and plane strain configurations?!

Page 3: Lecture 6 –Axial deformation

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Axial deformation- Why axial deformations and not three-dimensional deformations?

As engineers, we will always try to identify simplifications in the problem description that will produce idealized models amenable to analytical solutions. These simplifications are nothing but hypotheses that will have to be verified experimentally or with more descriptive models.

Note: knowing the procedure for finding the analytical solution of an idealized model is as important as knowing the hypotheses behind the idealized model!!

Axial deformation

Page 4: Lecture 6 –Axial deformation

Theory of deformable bodies

ExperimentalmechanicsMathematics

reference configuration

thermo-mechanical loads

KINEMATICS OF DEFORMATIONS

deformed configuration

CONTINUOUSMEDIA

atomic/micro/meso

structureis revealed

laws of nature .CONSERVATION OF MASS

BALANCE OF LINEAR MOMENTUMBALANCE OF ANGULAR MOMENTUM

LAWS OF THERMODYNAMICSCONSTITUTIVE

EQUATIONS

Empirical observation

Multi-scaleapproaches

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materialbehavior

J.L. Lagrange

Page 5: Lecture 6 –Axial deformation

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Axial deformation (@ ME 323)- Geometry of the solid body: straight, slender member with cross

section that is either constant or that changes slowly along the length of the member.

- Kinematic assumptions: cross sections, which are plane and are perpendicular to the axis before deformation, remain plane and remain perpendicular to the axis after deformation. In addition, cross sections do not rotate about the axis.

- Material behavior: isotropic linear elastic material; small deformations.

- Equilibrium: the above assumptions reduce the problem to a one-dimensional problem!!

Axial deformation

Page 6: Lecture 6 –Axial deformation

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Axial deformation- Geometry of the solid body: straight, slender member with cross

section that is either constant or that changes slowly along the length of the member.

Axial deformation

Q: Yes/No? Q: Yes/No? … “changes slowly”?!

Page 7: Lecture 6 –Axial deformation

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Axial deformation- Kinematic assumptions: cross sections, which are plane and are

perpendicular to the axis before deformation, remain plane and remain perpendicular to the axis after deformation. In addition, cross sections do not rotate about the axis.

Axial deformation

undeformed

deformed

The strain-displacement relationship is …(recall lecture 2)

Q: Shall we go back to thethree-dimension problem?

Page 8: Lecture 6 –Axial deformation

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Axial deformation- Kinematic assumptions: cross sections, which are plane and are

perpendicular to the axis before deformation, remain plane and remain perpendicular to the axis after deformation. In addition, cross sections do not rotate about the axis.

Axial deformation

undeformed

deformed

Q: what is the elongation of the member?

Sanity check:

Page 9: Lecture 6 –Axial deformation

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Axial deformation- Material behavior: isotropic linear elastic material;

small deformations.Generalized Hooke’s law reduces to …

Q: Can the Young’s modulus depend on the axial coordinates?

- Can the Young’s modulus change across the cross-section?

If yes, would it still be a case of axial deformations?

Axial deformation

Page 10: Lecture 6 –Axial deformation

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Axial deformation- Equilibrium: the above assumptions reduce the problem to

a one-dimensional problem!!+ Resultants for homogeneous materials (recall Lecture 5)

Axial deformation

Q: Shall we use the centroid of the cross section as the axis of the member and thus as the x-axis of the idealized model? Yes!

Page 11: Lecture 6 –Axial deformation

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Axial deformation- Equilibrium: the above assumptions reduce the problem to

a one-dimensional problem!!

+ Let’s consider an axisymmetric member loaded with axisymmetric body forces

Equilibrium:

… rearrange and take the limit for a differential volume

Axial deformation

Q: Units of the body forces p(x)?

Page 12: Lecture 6 –Axial deformation

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Axial deformation (simplest case)

- Stiffness coefficient: force required to produce a unit elongation

- Flexibility coefficient: elongation produced when a unit force is applied

Axial deformation

Homogeneous, constant cross section, no body forces:

Note: recall the behavior of a linear spring

Page 13: Lecture 6 –Axial deformation

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Axial deformation (thermal effects)

- Principle of superposition

- For material properties and thermal loads that only depend on x

- Resultant force:

- Elongation (force-temperature-deformation)

- Homogeneous, constant cross-section, thermal forces (no body forces)

Axial deformation

+" +" +"

Page 14: Lecture 6 –Axial deformation

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Axial deformation (summary)- Geometry of the solid body: straight, slender member with cross section that

is either constant or that changes slowly along the length of the member.- Kinematic assumptions: cross sections, which are plane and are perpendicular

to the axis before deformation, remain plane and remain perpendicular to the axis after deformation. In addition, cross sections do not rotate about the axis.

- Material behavior: isotropic linear elastic material; small deformations.

- Equilibrium:

Axial deformation

Strain:

Elongation:

Homogeneous:

Homogeneous:

Homogeneous, constant cross section, no body forces, thermal load:

Homogeneous, loaded with body forces:

Page 15: Lecture 6 –Axial deformation

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Example 5 (Example 3 from Lecture 2):Two solid circular rods are welded to a plate at B to form a single rod, asshown in the figure. Consider the 30-kN force at B to be uniformly distributedaround the circumference of the collar at B and the 10 kN load at C to beapplied at the centroid of the end cross section. Determine the axial stress ineach portion of the rod. L1=300mm, L2=200mm, E1=600 GPa, E2=400GPa.

Determine the displacement of end C.…. easy!

Strain energy density – Axial deformation

Answer: uC = �3.6µm

Page 16: Lecture 6 –Axial deformation

Any questions?

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Axial deformation