Strength of Materials

cypriotcamelUrban and Civil

Nov 29, 2013 (3 years and 8 months ago)

78 views

Strength of Materials


9/18/07

Topics to Cover


Pressure


Stress


Strain


Scale and Square
-
Cube Effect


Tension


Compression


Shear

Effect of force on non
-
rigid objects


Besides the force, what else is important?


Examples to consider:


Ice on a pond


Hammering a nail into wood


Pressure!


Area


The AREA over which it is applied; the larger
the area, the lower the pressure.


Utilizing pressure


Many simple devices


knives, picks, snowshoes, bulletproof vests, all simply
change the AREA over which a force is applied.


Distribution of Force
(
http://www.physics.unl.edu/outreach/football.html
)


Bed of Nails (
http://video.google.com/videoplay?docid=
-
1891333794864062153
)


Bed of Nails (wileyplus


ch. 6)


Pressure


What are the units of pressure?


Hint


force per unit area


Pounds per square inch, abbreviated psi.


Inflating a tire; a car tire typically takes ~ 30 psi.


Pressure of the air around us is about 14 psi.

Pressure


con’t


Pressure can take several forms in the material.


Pulling the material apart, material is in
TENSION, the pressure is called the stress.


Pushing together, material is in
COMPRESSION, and it’s again a stress.


Examples


strings can have tension, but not compression


Springs or rigid objects can do both

Think


Pair
-

Share


Consider a bed of nails with 600 nails


A person weighs 150 lbs lies on it


Will it puncture the skin? (It takes 2 lbs
per nail to puncture the skin)



How much weight must be applied to
puncture the skin?


http://phun.physics.virginia.edu/demos/nail_bed.html

Think


Pair
-

Share


High
-
heeled shoes can cause tremendous
pressure to be applied to a floor.


Suppose the radius of a heel is 6.00

×

10
-
3

m.


At times during a normal walking motion, nearly
the entire body weight acts perpendicular to the
surface of such a heel.


Find the pressure that is applied to the floor
under the heel because of the weight of a
50.0

kg woman.

Answer







kPa
m
s
m
kg
r
g
m
A
F
P
330
4
006
0
8
9
50
2
2
2
,
.
/
.










Think


Pair
-

Share


What exerts more pressure
-
per
-
square inch when
walking a 100

lb woman in high heels or a 6,000

lb
elephant in bare

feet?


[At the moment when only the heel rests on the ground.]


Stiletto heels have an area of about 1/16 of a square
inch.


Elephants, unlike humans, walk with two

feet on the
ground at a time.


Each foot is about 40

square

inches.


Answer

psi
in
lbs
P
psi
in
lbs
in
lbs
P
A
F
P
75
40
2
6000
1600
1600
16
1
100
2
2
2








Pressure



Woman’s heels




Elephant’s feet




The heels exert more pressure!

Stress


Internal distribution of forces within a body that
balances and reacts to the loads applied to it


Tension, compression, or shear


Strain


The amount of deformation in a material.
If we are stretching a bar, then the strain is
the percent change in its length.


Simple case of strain proportional to stress


Small forces


can’t model objects breaking, but it does
work for the amount a bridge sags when a
car drives on it



Equations

Strain
Stress
E
Strain
E
Stress













E is the modulus of elasticity


Units of pressure (N/m
2

or psi)


Stress and Strain


con’t


What is the amount a bar or wire can be
stretched or compressed in one dimension?
(Hooke’s Law).


Table 2.1 on page 21 of our book has a list
of constants for Moduli of Elasticity


Conversion from N/m
2

to Psi is 6830:1

L
L
Stress
E
Strain
Stress
E
L
L
Strain







Behavior of Materials


Elastic Behavior


Recovers shape


no deformation


Plastic Behavior


Range between elastic limit and breaking


Strain is not proportional to stress


The Ultimate Tensile (or Compressive) Strength is where
a stretched material breaks.


sample values in N/m
2

are


4e
7

for concrete or brick,


1e
8

for Al,


2.3e
8

copper,


5e
8

for steel


Bone 12e
7

tension, 17e
7

compression.

Stiffness


Ability to resist strain


Steels


Same E


different
yield points


different ultimate strengths

Stress and Strain


example 1


How far can you stretch a 20 ft steel cable
before it snaps?


E for steel is 2*10
11


Ultimate tensile strength for steel is 5.2*10
8



inches
ft
ft
L
E
Stress
L
L
L
L
Stress
L
L
Stress
Strain
Stress
E
6
0
05
0
10
2
10
2
5
20
11
8
.
.
.
*
*
*













Materials


Wood and metal pieces respond about the
same to compression and tension.


Stone pieces do NOT


strong in compression (hard to crush)


weak in tension (easy to pull apart)


Space Elevator


Discussion of Space Elevator
http://science.nasa.gov/headlines/y2000/ast07sep_1.htm


Basically a cable goes up to geosynchronous
orbit with ballast mass (asteroid) slightly above
it.


Best calculations are that a Tensile Strength of
6.2 x 10
10

N/m
2

= 9.1 x 10
6

psi (about 100 times
stronger than steel) is required.


Carbon nanotubes have a tensile strength 3
times this (Young’s Modulus is 10
12

N/m
2
),


Growing them long enough is a real concern.

How tall can a mountain

on Earth be?


Rock


Compressive Strength of 2 x 10
8

N/m
2
= 3 x 10
4

psi


Specific gravity ~ 3


Pressure at the bottom of a column is therefore weight/area


Weight = weight_water*specific gravity*height*area


areas cancel


Pressure at bottom = (0.036 pounds/in
3
)*(about 3)*height = Ultimate
Strength.


Height = (3 x 10
4

psi)/((.036 pounds/in
3
)*3) is about 3 x 10
5

inches, or
25,000 feet (over 4 miles)


Height of Mt. Everest is 29,000 feet


pretty good guess



If gravity was weaker, then the weight would be less, and mountains could
grow taller.


Mars has a gravity about 1/3 of Earth’s, and its tallest mountain (Mount
Olympus) is about three times as big as Everest.

Scale


Imagine a simple building, as a cube 10’ on a
side.


Volume is 10’ x 10‘ x 10’ = 1,000 ft
3


Area is 10’ x 10’ = 100 ft
2



If you double each dimension, what happens to
the VOLUME and AREA of the cube?


Volume is 20’ x 20‘ x 20’ = 8,000 ft
3


8 fold increase in volume


Area is 20’ x 20’ = 400 ft
2


4 fold increase in area

Scale


con’t


Weight of a structure scales with volume, and the strain
on the supports scale with area


Building larger and larger with the same materials leads
to more and more strain and eventually collapse.


Thus larger buildings need thicker supports:


If you double the size, for example, then AFTER
doubling you need to make the supports twice the area,
or 40% larger in each direction.


Of course eventually this becomes impractical, which is
why one needs stronger materials to build larger
structures: stone instead of wood, steel instead of stone.


Scale


con’t


If you could somehow make a mouse as big as a
gorilla, it would probably just collapse


Its bones (the support system) would be too thin
to support the increased weight at that size.


Any sci
-
fi movie with a “shrinking ray” is
nonsense.


Shear Modulus


Shear Modulus (ripping apart)


Shear refers to cases where we are not directly
pushing in or pulling out on a material, but are
applying forces perpendicular to the surface.


(Example: thick book)


Effectively shear is a combination of tension on
one axis and compression on another, causing
the material to deform or twist.



Given:


a 6400# weight is supported by an


8" x 8" post.

Determine:


a. the compression stress in the 8" x 8" post

b. the bearing stress between the post and the steel plate

c. the bearing stress between the steel plate and the concrete.

Solution:

a. The compressive force in the post : 6400 pounds/ 64 in2 = 100 psi


b. The stress at the point between the post and the steel bearing plate is dictated
by the smallest area being loaded. The stress between the two is again 6400
pounds divided by 64 square inches for a total stress of 100 psi.


c. The stress between the plate and the concrete floor is also determined by the
smallest area. In this case it is the 10" x 10" plate: 6400 pounds / 100 in2. The
resulting stress is 64 psi, a clear reduction in the stress.

Stress/Area Relationship Problem
-

Compresion

Think


Pair
-

Share

Stress/Area Relationship Problem
-

Tension


Given:



A 1000 lb load suspended from
a ceiling by a 1"x 1" member.

Determine:


a. the stress in the member

b. the stress if the size of the
member is increased to 2" x 2".


Solution:


a. The internal force in that member is 1000 lb tension and the tension stress
(intensity of the force per unit of area) is 1000 pounds divided by the area of 1
square inch. This is a stress of 1000 psi.


b. The 1000 lb load is now distributed evenly across an area of 4 square inches;
thus, 1000 lb divided by 4 in
2

is a stress of 250 psi. This clearly demonstrates the
inverse relationship between stress and area.


Solution


a. the stress in the member


The internal force in that member is 1000 lb tension
and the tension stress (intensity of the force per unit
of area) is 1000 pounds divided by the area of 1
square inch. This is a stress of 1000 psi.



b. the stress if the size of the member is
increased to 2" x 2".


The 1000 lb load is now distributed evenly across an
area of 4 square inches; thus, 1000# divided by 4 in^2
is a stress of 250 psi. This clearly demonstrates the
inverse relationship between stress and area.

Think


Pair
-

Share


Given:



A 5000 lb load on 1” diameter rod.

Determine:


a. Is the rod in tension or compression?


B. What is the stress in the rod?

b. What is the stress if the diameter of the rod is increased to 2".


Solution


Review


How materials respond to applied forces


When we apply a force over some area (a
pressure), we create stress in the material,
which then undergoes strain and changes
shape in response (tension, compression,
and shear).


Scaling


pressure scales as area, weight
as volume, so a larger object needs
proportionally thicker supports

Lecture for 2/8/07

Deformations from Temperature


Atomic chemistry reacts to changes in
energy


Solid materials


Can
contract

with decrease in temperature


Can
expand

with increase in temperature


Linear change can be measured per
degree

Thermal Deformation




-

the rate of strain per degree



UNITS:
°
F,
°
C



length change:



thermal strain:




-

no stress

when movement allowed


)
(
)
(
T
L
T
T
T








Changes in Size


Similar to the stress
-
strain relationship, assume:


temperature changes are not too extreme


change in the object is linear with temperature


Simple case is a solid bar of material with a length,
width, and depth.


When the object changes temperature by an
amount

T, the length of the bar becomes
increases.

degree
per

change

of

rate

is

hotter

is


ture,
in tempera

change

the
is

0









T
T
L
L

Expansion and Contraction


Example


a long bar of Aluminum 2” x 6” x 15 feet, with and


= 1.3 x 10
-
5

per
°
F



heat from room temperature of 70 F to 130 F, what are its new dimensions?



Solution


First step:

T=
130


70 = 60
°
F


Multiply


x

T =
1.3 x 10
-
5

/
°
F x 60
°
F = 0.00078


Consider each dimension


2” x
0.00078 = 2.002”


6” x
0.00078 = 6.005”


180” x
0.00078 = 180.14”


expands a bit more than an eight of an inch length
-
wise.


This is proportional to the length, so for a long bridge it could shift by an inch or even several
inches


enough to make a fixed connection untenable.


Hence a roller support can be an advantage for a material that expands and contracts


by
allowing one side to translate, there is no compression in the horizontal direction for the
bridge.



Demo


http://www.physics.umd.edu/lecdem/services/demos/demosi1/i1
-
11.htm


Superposition Method


Can remove a support


Replace the support with a reaction


Enforce the geometry constraint

Superposition Method


con’t


Total length change
restrained to zero


Constraint:

E
T
A
P
f
L
T
AE
PL
L
T
AE
PL
T
P
P
T
)
(


0
)
(

:
sub
)
(




0


























Masonry


Like all masonry, bricks are strong in compression but
weak in tension. You can see this with a single brick.
How?

Demonstration

Compression and Tension


DEMO: compressible foam


A force pushing down on a broad surface will clearly
cause compression beneath it.


If the force is applied in the middle, then clearly
everywhere will be compressed, though the sides less
than the middle.


However, if the force is applied at one edge, the top
will “see
-
saw” upwards, so that as we compress one
end we actually put the other side into tension.


How far out can I go and not cause any tension?

Middle Third Rule


Must stay within the “middle third” of the block to prevent
the opposite side from going into tension


this is the important MIDDLE THIRD RULE.


Why is it important?


Because bricks/stone/etc are much stronger in compression than
in tension; need to keep them in compression or they can break!





To get around this problem


reinforce concrete with a steel bar


the steel can take the tension when the concrete can’t.


We’ll go over this in more detail in Chapter 8.