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Chapter 2 /1

DEFORMABLE BODIESChapter 2

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Chapter 2 /2

ANALYSIS STEPS

Three Steps for Analysis:1. Forces and equilibrium requirements

2. Deformation and geometric compatibility

3. Application of force-deformation relations

EXAMPLE 2.1:Piston compressing

two coaxial springsFree Body Diagrams Equilibrium requirement for

each free body:

Equilibrium requirement forpiston:

Geometric compatibility

requirement:

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Chapter 2 /3

EXAMPLE PROBLEM Application of force-displacement relations:

Solution now can be obtained

Example 2.2: Problem definition:

1. Very light and stiff  wood plank.

2. Two identical springs of constant k.

3. Plank just touches ground when

man reaches a distance b.

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Chapter 2 /4

Example 2.2….. 

Force equilibrium conditions:

Geometric compatibility:

Force displacement relations:

Unknowns: 7

Number of equations: 7

Solution can be obtained.

FBD of Plank

Results:

FBD of springs and geometric compatibility

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Chapter 2 /5

UNIAXIAL LOADING

Three rods of identical material loaded gradually by two balanced loads

Observation: Uniaxial load-elongation

relation of the material is linear.

Define Modulus of Elast ic i ty :

Plot of results: Results re-plotted:

Expression for δ: Expression for k:

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Chapter 2 /6

MODULUS OF ELASTICITY

Note: Units of P/A and of E (modulus of elasticity) are the force per unit area,

which are also the units of stress. We will define stress and modulus of

elasticity, more accurately, later.

MODULUS OF ELASTICITY OF DIFFERENT MATERIALS

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Chapter 2 /7

EXAMPLEThis is the same frame that we

had discussed earlier  

Forces in members from

equilibrium conditions: 

Deformations of members BD and CD: 

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Chapter 2 /8

Example…. 

Geometric compatibility 

Obtain location of D afterdeformation of links:

Horizontal and vertical

displacements of point D: 

Geometric compatibility: Simplified method

1. Replace arcs by tangents.2. Possible only for small deformations.

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Chapter 2 /9

THIN RING

Radius: r,

Thickness t,

Width b 

 Axial view  Half ring

FBD 

Force Analysis:

Symmetry requirement: FR = 0Radial force on arc length r  Δθ = pb(r Δθ)

Component of force in y direction = pb(r Δθ)sinθ 

Force equilibrium of half ring in y-direction:

Thus FT = prb

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Chapter 2 /10

THIN RING…... 

Geometric Compatibility:

Change in radius:

or

Neglecting (t/2r)

Deformation: Since hoop is thin, it may be thought of as a flat plate:

Length = 2π(r + t/2),

Width = b. and

Thickness = t

Increase in circumference of the hoop:

.

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Chapter 2 /11

EXAMPLE:FRICTION-BRAKE BAND

Schematic view Free Body Diagrams of Band in contact 

Problem definition: 

1.Steel band (1.6 mm thick, 50 mm wide)

used as a brake for motor.

2. Kinetic coefficient of friction = 0.4.

3. Tension in section BC = 40 kN.

4. Predict elongation of band.

Will tension in section AD be equal, lower orhigher than 40 kN?

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Chapter 2 /12

FRICTION-BRAKE BAND….. 

Solution:

FBD of band small section: Force equilibrium equations:

Useful

approximations:

Equations with

approximations:

In the limit as ∆θ → 0 

By integration from θ = 0 to θ = θ:

This is an important result in mechanics. Many machines employ this frictional

behavior. Sailors use it to halt the motion of a large ship by taking a few turns

of a rope around a piling.