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Mechanics of MaterialsMAE 243 (Section 002)
Spring 2008
Dr. Konstantinos A. Sierros
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Problem 1.2-4
A circular aluminum tube of lengthL
= 400 mm is loaded in compression byforces P(see figure). The outside and inside diameters are 60 mm and 50 mm,
respectively. A strain gage is placed on the outside of the bar to measure
normal strains in the longitudinal direction.
(a) If the measured strain is 550 x 10-6 , what is the shortening of the bar?
(b) If the compressive stress in the bar is intended to be 40 MPa, what should
be the load P?
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Problem 1.2-7
Two steel wires,AB and BC, support a lamp weighing 18 lb (see figure). Wire
AB is at an angle = 34 to the horizontal and wire BCis at an angle = 48.
Both wires have diameter 30 mils. (Wire diameters are often expressed in mils;
one mil equals 0.001 in.) Determine the tensile stressesAB and BCin the two
wires.
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Problem 1.2-11
A reinforced concrete slab 8.0 ft square and 9.0 in. thick is lifted by four cables
attached to the corners, as shown in the figure. The cables are attached to ahook at a point 5.0 ft above the top of the slab. Each cable has an effective
cross-sectional areaA = 0.12 in2 .
Determine the tensile stress t in the cables due to the weight of the concrete
slab. (See Table H-1, Appendix H, for the weight density of reinforced concrete.)
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1.3 Mechanical properties of materials
In order to understand the mechanicalbehaviour of materials we need to perform
experimental testing in the lab
A tensile test machine is a typical equipment
of a mechanical testing lab ASTM (American Society for Testing and
Materials)
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FIG. 1-10 Stress-strain diagram for atypical structural steel in tension (not to
scale)
Stress () strain () diagrams
Nominal stress and strain (in the
calculations we use the initial
cross-sectional area A)
True stress (in the calculations
we use the cross-sectional area A
when failure occurs)
True strain if we use a strain
gauge
Stress-strain diagrams containimportant information about
mechanical properties and
behaviour
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Stress () strain () diagrams
OA: Initial region which is linear and proportional
Slope of OA is called modulus of elasticity
BC: Considerable elongation occurs with no noticeable increase in stress (yielding)
CD: Strain hardeningchanges in crystalline structure (increased resistance tofurther deformation)
DE: Further stretching leads to reduction in the applied load and fracture
OABCE: True stress-strain curve
FIG. 1-10 Stress-strain
diagram for a typical
structural steel in
tension (not to scale)
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FIG. 1-12Stress-strain diagram fora typical structural steel in tension
(drawn to scale)
Stress () strain () diagrams
The strains from zero to point A
are so small as compared to thestrains from point A to E and can
not be seen (it is a vertical line)
Metals, such as structural steel,
that undergo permanent largestrains before failure are ductile
Ductile materialsabsorb large
amounts of strain energy
Ductile materials: aluminium,
copper, magnesium, lead,
molybdenum, nickel, brass, nylon,
teflon
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FIG. 1-13 Typical stress-strain
diagram for an aluminum alloy.
Aluminium alloys
Although ductilealuminium alloys
typically do not have a clearly definable
yield point
However, they have an initial linear regionwith a recognizable proportional limit
Structural alloys have proportional limits
in the range of 70-410 MPa and ultimate
stresses in the range of 140-550 MPa
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Copyright 2005 by Nelson, a division of Thomson Canada Limited
FIG 1-14 Arbitrary yield stress determined bythe offset method
Offset method
When the yield point is notobvious, like in the previous
case, and undergoes large
strains, an arbitrary yield stress
can be determined by the offsetmethod
The intersection of the offset
line and the stress-strain curve
(point A) defines the yieldstress
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FIG. 1-15 Stress-strain curves
for two kinds of rubber intension
Rubber (elastomers)
Rubber maintains a linear relationship
between stress and strain up to relatively,
as compared to metals, large strains (up to
20%)
Beyond the proportional limit, the
behaviour depends on the type of rubber
(soft rubber stretches enormously without
failure!!!)
Rubber is not ductile but elastic material Percent elongation = (L1-Lo)/ Lo %
Percent reduction in area = (Ao-A1)/ Ao %
Parameters that characterize ductility
Measure of the amount
of necking
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FIG. 1-16 Typical stress-strain
diagram for a brittle materialshowing the proportional limit
(pointA) and fracture stress
(pointB)
Brittle materials
Brittle materials fail at relatively
low strains and little elongationafter the proportional limit
Brittle materials: concrete,
marble, glass, ceramics and
metallic alloys
The reduction in the cross-
sectional area until fracture (point
B) is insignificant and the fracture
stress (point B) is the same as theultimate stress
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Plastics
Viscoelasticity
Time and temperature dependence
Some plastics are brittle and some are
ductile
COMPOSITES (glass fiber reinforced
plastics) combine high strength with light
weight
Polymer
matrix
Glass
fiber
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FIG. 1-17 Stress-strain
diagram for copper in
compression
Compression
Stress-strain curves in compression are
different from those in tension Linear regime and proportional limit are
the same for tension and compression for
materials such as steel, aluminium and
copper (ductile materials)
However, after yielding begins the
behaviour is different. The material bulges
outward and eventually flattens out (curve
becomes really steep)
Brittle materials have higher ultimate
compressive stresses than when they are
under tension. They do not flatten out but
break at maximum load.
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Tables of mechanical properties
Appendix H contains tables that list materials properties.Please make sure that you use these tables when solving problems
that require input of material properties data.
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Wednesday (23 January 2008): Quiz on Statics, I will send you
e-mail with further details
Have a good weekend