3 Stress Strain Tension Test1233
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Stresses and Strains Stresses and Strains Solid materials are deformable, not rigid. We will study the stresses and strains that forces produce in a body
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Transcript of 3 Stress Strain Tension Test1233
Stresses and StrainsStresses and Strains
Solid materials are deformable, not rigid.
We will study the stresses and strains that forces produce in a body
800 lb800 lb
2”
4”
A
B
A.) Axial Tensile and Compressive Stresses
• Consider a 2” x 4” piece of wood with a force P applied at each end.
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.
Fx = 0 = - 800 lb + PA = 0
PA = 800 lb
PA800 lb
A
B
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”
Since we have 8 square makes, the amount of force on each square inch is:
800 lb = 100 lb = 100 psi
8in2 in2
Which is the definition of stress:
= P A
= stress = unit stress= average stress= engineering stress
P = applied force
A= cross-sectional area over which the stress develops
t = Tensile Stress (produced by
Tensile Forces)
c= Compressive Stress (produced by
Compressive Forces)
B. Examples of Tensile and Compressive Stresses
C.) TENSILE AND COMPRESSIVE
STRAINS AND DEFORMATIONS
Example: Dock with wooden ladder for a footbridge.
This is an example of deformation ordeflection due to bending stress whichwe will cover later.
Similarly, when a steel rod is in Tension, it will deform, but it is not as noticeable.
= deformation = the amount a body islengthened by a tensile force and shortened by a compressive force.
L
T T
To permit comparison with acceptablevalues, the deformation is usually converted to a unit basis, which is thestrain.
= L = strain (= unit strain) = deformation that occurs over length LL = original length of member
Example: a 3/8” cable, 100’ long stretches 1” before freeing a truck which is stuck in the mud.
Find the strain in the cable.
100’
= L
= 1”
L = 100’ (12”/1) = 1200”
= 1” = 0.0008333 in/in 1200”
We’ll come back to see if this is will break the cable.
Review of Stress and Strain
Axial Stress and Strain = P
A = L Shear Stress = V
A
E.) The Relationship Between Stressand Strain
As you apply load to a material, the strain increases constantly (or proportionately) with stress.
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 (.
(ksi) (in/in x 0.001)
0 0 3 1 6 2 9 3
12 4
Str
ess
, (
ksi)
Strain ,in/in x 0.001)
Since the stress is proportional to the strain, ratio of stress to strain is constant.
/
(ksi)(in/in x 0.001)(ksi x 1000) 0 0 0 3 1 3 6 2 3 9 3 3
12 4 3
This constant ratio of stress to strain is called the Modulus of Elasticity (E).
E = /
The Modulus of Elasticity is always the same for a given material. We call it a material constant.
Knowing E for a given material and :E = /
1.) We can find how much stress is in the material if we know the strain:
= E
2.) We can find how much strain is in the material if we know the stress:
= E
CAUTION !
If the tension test continues, the stress will reach a level called the Proportional Limit ( PL ). If the stress is increased above PL ,
the strain will increase at a higher rate.
Str
ess
), k
si
Strain (), in/in
PL
Ex. Given: Previous Truck cable strain Find: Stress in the steel cable
= 1”L = 1200”
= 1” = 0.0008333 in/in 1200”
E ( as long as PL)
E= 30,000,000 psi (for steel)
E = 30,000,000 psi (.0008333 in/in)= 24,990 psi (pretty high)
CHECK: is < PL ?
= 24,990 psi < PL = 34,000 psi (OK)
D.) Material Properties found using the Tension Test
Str
ess
), k
si
Strain (), in/in
PL
Y
U
E = =slope
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.
D.) Material Properties found using the Tension Test
4.) Modulus of Elasticity - the ratio of stress over strain in the linear region of the stress-strain curve.
5. Percentage Elongation-the plastic deformation at failure, as a percentage of the original length = (Lf – Lo)/ Lo x 100
5.) Percent Elongation: Ductile Material - will undergo plastic
deformation before failing
Str
ess
Strain
Ductile Material
Brittle Material - will fail without any plastic deformation (opposite of ductile)
Str
ess
Strain
Brittle Material
5.) Percent Elongation:
