cub_surg_lesson02_presentation_v2_tedl_dwc.ppt

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This document discusses the fundamentals of mechanics of elastic materials. It defines key terms like stress, strain, and modulus of elasticity used to characterize how materials deform under force. Stress represents the force applied over an area, strain is the deformation normalized by length. Modulus of elasticity relates stress and strain and indicates a material's stiffness. Examples show how to calculate stress, strain, and modulus of elasticity for an elastic rod under tension. The stress-strain curve is also explained, distinguishing elastic deformation, permanent plastic deformation, yield point, ultimate strength, and failure point.

Why study mechanics?
Useful for the analysis and
design of load-bearing
structures, such as:
buildings
bridges
space shuttles
prosthetics
biological implants
Also used to
characterize materials

Stress
The force per unit area, or intensity of the forces
distributed over a given section. (units = Pascals
[Pa] or pounds per square inch [psi])
σ = F/A
• Stress is how engineers normalize the force that is
applied to a material to account for differences in
geometry.
• Useful for predicting failure conditions for materials.

Strain
Deformation per unit length (units: none [unitless])
ε = ΔL/L
Strain is how engineers normalize the deformation
that a material experiences to account for
differences in geometry.
Useful for determining how much a material can
deform before failure.

Modulus of Elasticity
A representation of the stiffness of a material that
behaves elastically (units: Pascals [Pa] or pounds per
square inch [psi])
E = σ/ε
What equation is this similar to?
k = F /Δx
Modulus of elasticity is how engineers characterize
material behavior.
Useful for knowing how materials behave, material
selection for device design, and calculating the stress in
a material since it is easier to measure deformation than
it is to determine the exact force on a material.

Example 1
This rod is exposed to a
tensile force of 20 N. What is
the stress in the rod?
20 N
20 N
0.5 m
0.5 m
σ =F/A
F = 20 N (given)
A = 0.5 m * 0.5 m = 0.25 m2
σ = 20 N / 0.25 m2
σ = 80 Pa
3 m

Example 2
The rod below is exposed to a
tensile force of 20 N and
elongates by 0.03 m. Calculate
the strain.
ε = ΔL/L
ΔL = 0.03 m (given)
L = 3 m
ε = 0.03 m / 3 m
ε = 0.01
20 N
20 N
0.5 m
0.5 m
3 m

Example 3
The rod below is exposed to
a tensile force of 20 N and
elongates by 0.03 m.
Calculate the modulus of
elasticity.
E = σ/ε
σ = 80 Pa (from first example)
ε = 0.01 (from second example)
E = 80 Pa / 0.01
E = 8000 Pa or 8 kPa
20 N
20 N
0.5 m
0.5 m
3 m

Next:
Complete the Solid Mechanics
Worksheet

$Elastic Behavior stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress F F tensile load direction neck steel tensile specimen$

$Understanding the Stress-Strain Curve elastic range – The linear portion of the stress-strain curve. When the force is released, the material returns to its original dimensions. plastic range – The region of permanent deformation. stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress$

$Understanding the Stress-Strain Curve yield stress – The minimum stress that causes permanent deformation. ultimate tensile strength – The maximum stress that the material can withstand. Also defines the beginning of necking. stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress$

$The Stress-Strain Curve necking – A localized decrease in cross sectional area that causes a decrease in stress with an increase in strain. fracture stress – Stress in which the material fails. stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress F F tensile load direction neck steel tensile specime n$

Image Sources
Line diagrams: 2011 © Brandi N. Briggs, ITL Program, College of Engineering, University of Colorado Boulder
tomruen, wikimedia.org http://sv.wikipedia.org/wiki/Fil:I-35W_bridge_collapse_TLR1.jpg
Glenn Research Center, NASA
http://www.nasa.gov/centers/glenn/moonandmars/med_topic_atomic_oxygen.html
NOAA http://www.photolib.noaa.gov/htmls/corp2239.htm

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cub_surg_lesson02_presentation_v2_tedl_dwc.ppt

1. Mechanics of Elastic Materials

2. Why study mechanics? Useful for the analysis and design of load-bearing structures, such as: buildings bridges space shuttles prosthetics biological implants Also used to characterize materials

3. Stress The force per unit area, or intensity of the forces distributed over a given section. (units = Pascals [Pa] or pounds per square inch [psi]) σ = F/A • Stress is how engineers normalize the force that is applied to a material to account for differences in geometry. • Useful for predicting failure conditions for materials.

4. Strain Deformation per unit length (units: none [unitless]) ε = ΔL/L Strain is how engineers normalize the deformation that a material experiences to account for differences in geometry. Useful for determining how much a material can deform before failure.

5. Modulus of Elasticity A representation of the stiffness of a material that behaves elastically (units: Pascals [Pa] or pounds per square inch [psi]) E = σ/ε What equation is this similar to? k = F /Δx Modulus of elasticity is how engineers characterize material behavior. Useful for knowing how materials behave, material selection for device design, and calculating the stress in a material since it is easier to measure deformation than it is to determine the exact force on a material.

6. Solid Mechanics In-Class Examples

7. Example 1 This rod is exposed to a tensile force of 20 N. What is the stress in the rod? 20 N 20 N 0.5 m 0.5 m σ =F/A F = 20 N (given) A = 0.5 m * 0.5 m = 0.25 m2 σ = 20 N / 0.25 m2 σ = 80 Pa 3 m

8. Example 2 The rod below is exposed to a tensile force of 20 N and elongates by 0.03 m. Calculate the strain. ε = ΔL/L ΔL = 0.03 m (given) L = 3 m ε = 0.03 m / 3 m ε = 0.01 20 N 20 N 0.5 m 0.5 m 3 m

9. Example 3 The rod below is exposed to a tensile force of 20 N and elongates by 0.03 m. Calculate the modulus of elasticity. E = σ/ε σ = 80 Pa (from first example) ε = 0.01 (from second example) E = 80 Pa / 0.01 E = 8000 Pa or 8 kPa 20 N 20 N 0.5 m 0.5 m 3 m

10. Next: Complete the Solid Mechanics Worksheet

11. Elastic Behavior stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress F F tensile load direction neck steel tensile specimen

12. Understanding the Stress-Strain Curve elastic range – The linear portion of the stress-strain curve. When the force is released, the material returns to its original dimensions. plastic range – The region of permanent deformation. stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress

13. Understanding the Stress-Strain Curve yield stress – The minimum stress that causes permanent deformation. ultimate tensile strength – The maximum stress that the material can withstand. Also defines the beginning of necking. stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress

14. The Stress-Strain Curve necking – A localized decrease in cross sectional area that causes a decrease in stress with an increase in strain. fracture stress – Stress in which the material fails. stress strain elastic range plastic range yield stress ultimate tensile strength fracture stress F F tensile load direction neck steel tensile specime n

15. Image Sources Line diagrams: 2011 © Brandi N. Briggs, ITL Program, College of Engineering, University of Colorado Boulder tomruen, wikimedia.org http://sv.wikipedia.org/wiki/Fil:I-35W_bridge_collapse_TLR1.jpg Glenn Research Center, NASA http://www.nasa.gov/centers/glenn/moonandmars/med_topic_atomic_oxygen.html NOAA http://www.photolib.noaa.gov/htmls/corp2239.htm

Editor's Notes

#2: Explain why we are studying tissue mechanics. First, we need to gain an understanding of mechanics in general. This ties into many different areas of engineering. [Mechanics of Elastic Materials presentation, part of the Mechanics of Elastic Solids lesson in TeachEngineering.org]
#3: Example: To design and build the World Trade Center, different types of steel were used, based on stress analysis of the building design. Did not use high performance steel if it was not needed. Collapse of WTC was due to too many beams being broken when the plane hit. The remaining beams could not handle the stress/weight from floors above and they buckled/failed. Example: For biological implants, engineers analyze the stresses in bones to make sure the implant can withstand the stress that the natural joint withstood. Bone degenerates with lack of stress, so we don’t want the implant to take away from the stress borne by surrounding bone.
#4: Explain normalizing the force by comparing two pieces of material that are the same length but different cross sectional areas. The force it takes to displace them the same amount (or break them) is different, but the stress is the same. If you know what the yield stress is in your material and you know the dimensions, then you can calculate how much force the material can withstand before failing.
#5: If you have a very long piece of material compared to a very short piece, then the same amount of displacement is more detrimental to the short piece, compared to the long piece. The displacement may be on the same order of magnitude as the small piece, but on a smaller order of magnitude than the large piece. Strain normalizes this and puts it on the same scale. Yield stress corresponds to a certain strain as well. If you can only measure strain or your system depends on being strained a certain amount then you can predict failure based on the strain.
#12: Engineers physically test materials to obtain this stress-strain curve graph. They record force and displacement data and then calculate stress and strain. Dog bone-shaped test specimens prevent sample from breaking at grips.
#13: Explain that modulus of elasticity is the slope of the elastic region of the line.
#15: Explain the difference between uniform deformation and necking. Even though the cross sectional area is changing, the stress is calculated using the initial area throughout the entire test because it is very difficult to measure area during testing.

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Editor's Notes