Chapter 11 Equilibrium and Elasticity. Equilibrium.

Chapter 11

Equilibrium and Elasticity

Equilibrium

Two Conditions for Equilibrium

• To motivate these, recall:

dt

pdF cm

ext

dt

Ld

ext

0ext F

)point! (about

0ext

any

Defining Equilibrium

• Equilibrium= no net external force or torque = no change in translation or rotation)

• your text says L=0; others allow nonzero L:

constant

0

cm

cm

pdt

pd

constant

0

L

dt

Ld

Defining Static Equilibrium

• ‘Static’ Equilibrium= the special case of no translation or rotation at all

0constant

0

cm

cm

pdt

pd

0constant

0

L

dt

Ld


• When applying these, we must consider all external forces

• But the gravitational force is rather subtle

0ext F

)point! (about

0ext

any

Center of Gravity (cg)

• Gravity acts at every point of a body

• Let = the torque on a body due to gravity

• Can find by treating the body as a single particle (the ‘cg’)

Center of Mass (cm)

• it can be shown: if g = constant everywhere, then:

• center of gravity =center of mass

ii

iii

m

rmr

cm

Using the Center of Gravity

Pressent some more explanatory notesPressent some more explanatory notes

SolvingEquilibrium Problems


• From now on, in this chapter/lecture:

• center of mass = center of gravity

• ‘equilibrium’ means ‘static equilibrium’

• write: F and for Fext and ext

0ext F

0ext

First Conditionfor Equilibrium

0ext F

0

0

0

z

y

x

F

F

F

Second Condition for Equilibrium

0ext

0

0

0

z

y

x

Exercise 11-11

Work through Exercise 11-11Work through Exercise 11-11

Exercise 11-14


A different version of Example 11-3

The ‘Leaning Ladder’ Problem

Work through the variation the the text’s leaning ladder problemWork through the variation the the text’s leaning ladder problem

Problem 11-62

‘Wheel on the Curb’ Problem

Work through Problem 11-62Work through Problem 11-62

Elasticity

Elasticity

• Real bodies are not perfectly rigid

• They deform when forces are applied

• Elastic deformation: body returns to its original shape after the applied forces are removed

Stress and Strain

• stress: describes the applied forces

• strain: describes the resulting deformation

• Hooke’s Law: stress = modulus × strain

• modulus: property of material under stress

• (large modulus means small deformation)

Hooke’s Law and Beyond

• O to a :• small stress, strain• Hooke’s Law:

stress=modulus×strain

• a < b :• stress and strain are

no longer proportional

Units

• stress = modulus × strain

• stress (‘applied force’): pascal= Pa=N/m2

• strain (‘deformation’): dimensionless

• modulus: same unit as stress

Types of Stress and Strain

• Applied forces are perpendicular to surface:

• tensile stress

• bulk (volume) stress

• Applied forces are parallel to surface:

• shear stress

Tensile Stress and Strain

• tensile stress = F/A

• tensile strain = l/l0

• Young’s modulus = Y

Tensile Stress and Strain

0

strainstress

l

lY

A

F

Y


Compression vs. Tension

• tension (shown): pull on object

• compression: push on object(reverse directionof F shown at left)

• Ycompressive = Ytensile


Tension and Compression at once

Bulk Stress and Strain

• pressure: p=F/A

• bulk stress = p

• bulk strain = V/V0

• bulk modulus = B

Bulk Stress and Strain

• B > 0• negative sign above:

p and V have opposite signs

0

strainstress

V

VBp

B


Shear Stress and Strain


• shear stress = F/A

• shear strain = x/h = tan• shear modulus = S


SSA

F

h

xS

A

F

S

tanor

strainstress

| || |

Do Exercise 11-32Do Exercise 11-32

Regimes of Deformation

• O to a :• (small stress, strain)• stress=modulus×strain• elastic, reversible

• a < b :• elastic, reversible• but stress and strain

not proportional

Regimes of Deformation

• From point O to b :• elastic, reversible

• from point b to d:• plastic, irreversible • ductile materials have

long c–d curves• brittle materials have

short c–d curves

Demonstation

Tensile Strength and Fracture

Chapter 11 Equilibrium and Elasticity. Equilibrium.

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