EgR 236Lecture03

EGR 236 Properties and Mechanics of Materials Spring 2014Lecture 03: Deformation and Strain

Today:-- Homework questions:-- Strain and Deformation -- Normal Strain -- Shear Strain-- Homework: Read Section 2.1 to 2.2 Work Problems from Chap 2: 5, 27, 34, 35

Following today's class you should be able to:-- Understand the difference between deformation and strain-- Understand the difference between normal and shear strain-- Be able to calculate strain from a body's deformation

In Statics, problems were solved assuming that they behaved as rigid bodies. No matter how large the forces were, the bodies always maintained their same shape.

Real materials are not rigid. Whenever an external force or moment acts on real material, there is a change in length or shape associated with the material. This change in shape is called deformation.

Deformation: a change in shape of material due to an applied load or other external stimuli.

Undeformed Deformed

Consider the case of two rods of the same cross sectional area and the same material, but one is twice as long as the other.

If you were to apply the same force to each rod, you find that the longer rod would deform more than the short rod. Apparently the geometry of a part affects the deformation.

One way to try to express deformation without needing to account for the size of the part is to calculate the deformation per unit length. Such a quantify of unit deformation is called Strain.

Units: mm/mm in/in none

The specific example presented here with the rods, is an example of Normal Strain because the deformation was caused by a stress which acting normal to the internal plane.

Deformation Normal Strain

ΔLB

P

P

ΔLA

LA

LB LB'

LA'

σ

x x'

σ

Example 1:The two wires are connected together at A. If the force P causes point A to be displaced vertically 3 mm, determine the normal strain developed in each wire.

Solution:

Geometry: no Load Geometry: with Load

Change of length:

Strain:

30o 30o400 mm

P

400 mm

B

A

C

30o

400 mm

346.4mm

200 mm

346.4 + 3= 349.4 mm

200 mm

402.6 mm

Shear Strain: Consider an infinitesimal element of material subjected to only shearing stresses. In order for the element to be in equilibrium, shear stresses must act along all four sides of the element.

Shearing stresses tend to deform the angular appearance of a body. Shear strain is measured by the change in the angular deformation that the body undergoes measured in radians.

Consider the block shown below fixed along the bottom plane and deformed by the horizontal force P.

The shear stress represents the change in the angular feature expressed in radians. If the original corner of the block was a 90o angle, following application of the force, P, the new corner angle can be found as

so

Strain where the angle dimension is decreased is considered + strain.Strain where the angle dimension is increased is considered - strain.

γ/2

τ

τ

τ

τ

γ/2

1.5mm

A'A

30 mm

γ

90o θ

1.5

30

θ

P

Example 2: A piece of plastic is originally rectangular. Loads cause it to distort as shown by the dashed lines. a) Determine the shear strain at corner Ab) Determine the shear strain at corner B. c) Determine the average normal strain that occurs along the diagonal DB.

Solution:

x

5

C

AD

B 42

3

2

2

300

400

y

all units in mm

Example 2: A piece of plastic is originally rectangular. Loads cause it to distort as shown by the dashed lines. a) Determine the shear strain at corner Ab) Determine the shear strain at corner B. c) Determine the average normal strain that occurs along the diagonal DB.

Solution: a) Originally the each of angles A and B were 90 degrees. Following deformation, angle A now appears to be obtuse and angle B is acute.

Find new angle A:

so

= 180o-89.6206o+0.2843o=90.6637o

therefore

b) Find new angle B:

so = 89.6206o - 0.2843o =89.3363o

therefore

x

5

C

AD

B 42

3

2

2

300

400

y

all units in mm

α

βθA'

302

22

403

δ

β

θB'

302

2

2

403

c) find normal strain along DB.

LDB

B

300

D 400

LDB''

304

405

Other practice problems:

Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm/mm, determine the displacement of point D. Originally the cable is unstretched.____________________________________________________________________________Solution:

The rectangular plate is subjected to the deformation shown by the dashed line. Determine the shear strain, γx'y' in the plate. The x' axis is directed from A through point B.____________________________________________________________________________Solution:

y x'

EGR236 Mechanics of Materials Homework Set 03 Spring 2014Problem 2:5 Bar ABC is originally in a horizontal position. If the loads cause the end A to be displaced downwards ΔA = 0.002 in and the bar rotates θ = 0.2 degrees, determine the strains in the rods AD, BE, and CF.-------------------------------------------------------------------------------------------------------------------

EGR236 Mechanics of Materials Homework Set 03 Spring 2014Problem 2.27The triangular plate is fixed at its base and its apex A is given a horizontal displacement of 5 mm. Determine the shear strain , γxy at A. ____________________________________________________________________________Solution:

EGR236 Mechanics of Materials Homework Set 03 Spring 2014Problem 2.34 and 35A square piece of material is deformed into the dashed parallelogram.a) (Problem 34) Determine the average normal strain that occurs along diagonals AC and BD. b) (Problem 35) Determine the shear strain γxy at corners A and D.------------------------------------------------------------------------------------------------------------------Solution: