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
Two Conditions for Equilibrium
• 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
Two Conditions for Equilibrium
• 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
Work through Exercise 11-14Work through 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
Work through Exercise 11-22Work through Exercise 11-22
Compression vs. Tension
• tension (shown): pull on object
• compression: push on object(reverse directionof F shown at left)
• Ycompressive = Ytensile
Work through Exercise 11-26Work through Exercise 11-26
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
Work through Exercise 11-30Work through Exercise 11-30
Shear Stress and Strain
Shear Stress and Strain
• shear stress = F/A
• shear strain = x/h = tan• shear modulus = S
Shear Stress and Strain
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