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

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:

800 lb = 100lb= 100 psi

8in 2 in 2

6.

Which is the definition ofstress :

=P

A

= stress = unit stress

= average stress

= engineering stress

P = applied force

A= cross-sectional area over which thestress develops

7.

t= Tensile Stress (produced by

Tensile Forces)

c = Compressive Stress (produced by

Compressive Forces)

8.

B.Examples of Tensile and Compressive Stresses

9.

C.)TENSILE AND COMPRESSIVE

STRAINS AND DEFORMATIONS

10.

Example:Dock with wooden ladder for

a footbridge.

This is an example of deformation or

deflection due to bending stress which

we will cover later.

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

strain .

=

L

= strain (= unit strain)

= 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.

=

L

= 1

L = 100 (12/1) = 1200

= 1= 0.0008333 in/in

1200

Well come back to see if this is will break the cable.

16.

Reviewof Stress and Strain

Axial Stress and Strain

=P

A

= L

Shear Stress

= V

A

17.

E.) The Relationship Between Stress

and Strain

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 ( .

(ksi) (in/in x 0.001)

0 0

3 1

6 2

9 3

12 4

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)

0 0 0

3 1 3

6 2 3

9 3 3

12 4 3

21.

This constant ratio of stress to strain is called theModulus of Elasticity (E).

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 :

E = /

1.) We can find how much stress is in thematerial if we know the strain:

= E

2.) We can find how much strain is in thematerial if we know the stress:

= E

23.

CAUTION !

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

=1

L = 1200

= 1= 0.0008333 in/in

1200

E( as long as PL )

E= 30,000,000 psi (for steel)

26.

E = 30,000,000 psi (.0008333 in/in)

= 24,990 psi (pretty high)

CHECK:is < PL?

= 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.

5.) Percent Elongation:

• 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: