- 1. Stresses and Strains
- Solid materialsare deformable, not rigid.
- We will study thestressesandstrainsthat forces produce in a
body
2. A.)Axial Tensile and CompressiveStresses
- Consider a 2 x 4 piece of wood with a force P applied at each
end.
800 lb 800 lb 2 4 A B 3.
- Anywhere you cut this bar across its section, in order to keep
the board from moving,the 800 lb force must act on that
section.
- F x= 0 = - 800 lb + P A= 0
P A 800 lb A B 4.
- We assume that the force is distributed evenly throughout the
section so that an equal portion of the 800 lb force acts on each
square inch of the cross-section
800 lb 2 4 1 1 5.
- Since we have 8 square makes, the amount of force on each
square inch is:
6.
- Which is the definition ofstress :
- A= cross-sectional area over which thestress develops
7.
- t= Tensile Stress (produced by
- c = Compressive Stress (produced by
8.
- B.Examples of Tensile and Compressive Stresses
9.
- C.)TENSILE AND COMPRESSIVE
10.
- Example:Dock with wooden ladder for
- This is an example of deformation or
- deflection due to bending stress which
11.
- Similarly, when a steel rod is in Tension,
- it will deform, but it is not as noticeable.
- = deformation = the amount a body is
- lengthened by a tensile force and
- shortened by a compressive force.
L T T 12.
- To permit comparison with acceptable
- values, the deformation is usually
- converted to a unit basis, which is the
- = deformation that occurs over length L
- L = original length of member
13.
- Example:a 3/8 cable, 100 long stretches1 before freeing a truck
which is stuck in the mud.
- Find the strain in the cable.
100 14. 15.
- Well come back to see if this is will break the cable.
16.
- Reviewof Stress and Strain
17.
- E.) The Relationship Between Stress
- As you apply load to a material, the strain increases
constantly (or proportionately) with stress.
18.
- Example: In a tension test you apply a gradually increasing
load to a sample.You can determine the amount of strain ( that
occurs in a sample at any given stress level ( .
19. Stress , (ksi) Strain , in/in x 0.001) 20.
- Since the stress is proportional to the strain, ratio of stress
to strain isconstant .
- (ksi) (in/in x 0.001) (ksi x 1000)
21.
- This constant ratio of stress to strain is called theModulus of
Elasticity (E).
- The Modulus of Elasticity is always the same for a given
material. We call it amaterial constant .
22.
- Knowing E for a given material and :
- 1.) We can find how much stress is in thematerial if we know
the strain:
- 2.) We can find how much strain is in thematerial if we know
the stress:
23.
- If the tension test continues, the stress will reach a level
called theProportional Limit ( PL).If the stress is increased above
PL,the strain will increase at a higher rate.
24. Stress ), ksi Strain ( ), in/in PL 25.
- Ex. Given:Previous Truck cable strain
- Find: Stress in the steel cable
- E= 30,000,000 psi (for steel)
26.
- E = 30,000,000 psi (.0008333 in/in)
- = 24,990 psi (pretty high)
- = 24,990 psi< PL= 34,000 psi(OK)
27.
- D.) Material Properties found using the TensionTest
Stress ), ksi Strain ( ), in/in PL Y U E = =slope 28.
- D.) Material Properties found using the Tension Test
- 1.) Ultimate Strength ( U ) - The maximum stress a material
will withstand before failing.
- 2.) Yield Strength ( Y ) - The maximum stress a material will
withstand before deforming permanently.
- 3.) Proportional Limit ( PL ) - The maximum stress a material
will withstand before stress-strain relationship becomes
non-linear.
29.
- D.) Material Properties found using the Tension Test
- 4.) Modulus of Elasticity - the ratio of stress over strain in
thelinearregion of the stress-strain curve.
- 5. Percentage Elongation-the plastic deformation at failure, as
a percentage of the original length = (L f L o )/ L ox 100
30.
-
- Ductile Material - will undergo plasticdeformation before
failing
Stress Strain Ductile Material 31.
- Brittle Material - will fail without any plastic deformation
(opposite of ductile)
Stress Strain Brittle Material 5.) Percent Elongation: