•  Main points of today’s lecture: •  Tensile stress and strain

•  Bulk stress and strain:

•  Pressure in fluids:

•  Buoyancy

Physic 231 Lecture 23

ghPPAFP topbot ρ+== ;

0LLY

AF Δ=Δ

VVBP

AF Δ=Δ=Δ

gVWF displacedfdisplacedB ρ==

Stress and strain in solids •  All materials deform when under stress, ie.

when a force is applied to them. The atoms in a solid are held in place by forces that act like springs.

FA= Y ΔL

L0

; FA= Stress; ΔL

L0

= Strain

•  If you pull on a bar of cross-sectional area A with a force F, the bar will stretch a distance ΔL. In this case, the force is perpendicular to the surface, and the force F and distance ΔL are related by Young’s modulus Y:

•  The force will be linearly proportional to the strain up to the elastic limit. Beyond this point, the bar will be permanently deformed. Stress units Pa=N/m2.

FA

ΔL L 0

Note: In elastic region: F= -k ΔL, k=A/L0

Reading Quiz 1.  The SI unit of pressure is

A.  N B.  kg/m2

C.  Pa D.  kg/m3

Slide 13-6

Example

•  Bone has a Young’s modulus of about 18 x 109 Pa. Under compression, it can withstand a stress of about 160 x 106 Pa before breaking. Assume that a femur (thigh bone) is 0.50 m long and calculate the amount of compression this bone can withstand before breaking. –  a) 4.4 mm –  b) 6.4 mm –  c) 1.6 cm –  d) 2.0 cm

6 2F 160x10 N / mA

=0

LYLΔ=

0LFLA Y

⇒ Δ = 6 29 2

0.5m160x10 N / m 4.4mm18x10 N / m

= =

If this force is uniformly applied, then F/A=P, where P is the pressure or force per unit area applied to the femur. Units for pressure are N/m2 = Pa. (Pa = Pascals). Atmospheric P=1.01x105 Pa

Y 18 x 109 Pa F/A 160 x 106 Pa

L0 0.50 m

ΔL ?

Bulk modulus

•  If you drop a metal block in the ocean, it will sink to bottom where the pressure is much greater than that of the atmosphere. This will compress the block; the amount of compression is governed by the bulk modulus B.

strain Volume ;change pressure ; =Δ=ΔΔ=ΔVVP

VVBP

Magdeburg’s Hemispheres Demonstration of atmospheric pressure

O"ovonGuericke(MayorofMagdeburg,17thcentury)

!F !F !F !F

T T

!F !F !F !F

tree

Conceptual question

•  In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and removed the air from the resulting sphere with a pump. Two four-horse teams could not pull the halves apart even though the hemispheres fell apart when air was readmitted. Suppose von Güricke had tied both teams of horses to one side and bolted the other side to a heavy tree trunk. In this case, the tension on the hemispheres would be –  a) twice –  b) exactly the same as –  c) half

what it was before.

!F !F !F !F

T T !F !F !F !F

How to calculate the force? •  Pressure force is perpendicular to surface. •  Vertical components of forces on each half add up to zero. •  Force on left half points to the right. Force on right half points left. •  Imagine the situation where the left hemisphere is replaced with a flat

plate: Then Fleft=P�A=P�πR2=Fright, where R is radius of the sphere.

P P

!Fright

!Fleft

!Fleft

Density

Slide 13-12

Nuclear matter 2.66x1017

Pressure in fluids, Pascal’s principle •  Consider a vessel filled with liquid and a smaller rectangular

volume V of the same liquid contained within the vessel. If the liquid is at rest, there is a pressure force F=P�A perpendicular to all 6 surfaces.

y

z x

front backz: F F 0; − =

•  More detailed considerations show that pressure always exerts a force perpendicular to a surface and that pressure is the same everywhere at the same height, i.e. Pfront =Pback, and Pleft =Pright and Pfront= Pright, etc.

•  Defining density ρ of the liquid by ρ=M/V , and using V=Ayh we can simplify the dependence of P on height as follows:

yW Mg Vg A hgρ ρ= = =

In equilibrium: F 0=∑v

bot topy: F F W 0; − − =

left rightx : F F 0; − = left left x right right xF P A ;F P A= = left rightP P⇒ =

back back zF P A= front backP P⇒ =front front z F P A ;=

bot bot y F P A ;= top top yF P A= bot topy

WP PA

⇒ = +

ybot top

y

A hgP P

Aρ

= + bot topP P ghρ⇒ = +

Pressure and Depth equation

•  If the top is open, Pbot is normal atmospheric pressure –  1.013 x 105 Pa = 14.7 lb/in2

•  The pressure does not depend upon the shape of the container

•  Gauge pressure is the difference between the pressure in the fluid and atmospheric pressure. If you connected a gauge which measures the pressure relative to atmospheric pressure to the fluid at depth h, then

•  for this specific case:

ghPP topbot ρ+=

guage bottom,h _ below _surface atmosphereP P P ghρ= − =

⇒ PA = PB = PC = PD

•  Because F=PA, one can make hydraulic machines that can generate large forces.

•  Units: One atmosphere (1 atm) = –  76.0 cm of mercury –  1.013 x 105 Pa = 1.013 x 105 N/m2 –  14.7 lb/in

Pascal’s principal and hydraulic machines

•  Quiz: A simple hydraulic lift is employed to lift a truck that weighs 30000N. If A1=1 cm2 and A2= 1000 cm2, what force F1 is needed to lift the truck? (Hint: Are the pressures equal under each piston?) –  a) 3N –  b) 300N –  c) 30N –  d) .0.33N

1 2P P ; this is pascal's principal=

21 1

2

FF AA

⇒ = 22

30000N1cm 30N1000cm

= =1 2

1 2

F FA A

⇒ =

Reading Quiz 2.  Gauge pressure is true pressure. Explain.

A.  larger than B.  smaller than C.  the same as D.  None of the above

Slide 13-8

Bulk modulus example

V2) P BVΔΔ =

•  A steel block with a volume of 8 cm3 is dropped in the ocean and comes to rest at the bottom, which is 3 km below the surface. 1) What is the pressure at the bottom of the ocean? 2) By what amount ΔV is the volume decreased. (The Bulk modulus for steel is 1.6x1011 N/m2. The density of water is 1000kg/m3. 1 atm. =1.01x105 N/m2) –  a) 2x105 N/m2

–  b) 4x105 N/m2

–  c) 2x106 N/m2 –  d) 3x107 N/m2

bot top1) P P ghρ= +

( )5 3 2botP 1.0x10 Pa 1000kg / m 9.8m / s 3000m= +

7 2botP 3.0x10 N / m=

( )bot topV VV P P PB B

Δ = Δ = −

( )3

7 2 5 211 2

8cm 3x10 N / m 1x10 N / m1.6x10 N / m

= −

3 31.5x10 cm−=