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Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Dr. Konstantinos A. Sierros
Problem 1.2-4
A circular aluminum tube of length L = 400 mm is loaded in compression by
forces 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?
Problem 1.2-7
Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire
AB is at an angle α = 34° to the horizontal and wire BC is at an angle β = 48°.
Both wires have diameter 30 mils. (Wire diameters are often expressed in mils;
one mil equals 0.001 in.) Determine the tensile stresses AB and BC in the two
wires.
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 a
hook at a point 5.0 ft above the top of the slab. Each cable has an effective
cross-sectional area A = 0.12 in2 .
Determine the tensile stress σt in the cables due to the weight of the concrete
slab. (See Table H-1, Appendix H, for the weight density of reinforced concrete.)
1.3 Mechanical properties of materials
• In order to understand the mechanical
behaviour 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)
Stress (σ) – strain (ε) 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 contain
important information about
mechanical properties and
behaviour
FIG. 1-10 Stress-strain diagram for a
typical structural steel in tension (not to
scale)
Stress (σ) – strain (ε) diagrams
FIG. 1-10 Stress-strain
diagram for a typical
structural steel in
tension (not to scale)
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 hardening – changes in crystalline structure (increased resistance to
further deformation)
DE: Further stretching leads to reduction in the applied load and fracture
OABCE’: True stress-strain curve
Stress (σ) – strain (ε) diagrams
• The strains from zero to point A
are so small as compared to the
strains from point A to E and can
not be seen (it is a vertical line…)
• Metals, such as structural steel,
that undergo permanent large
strains before failure are ductile
• Ductile materials absorb large
amounts of strain energy
• Ductile materials: aluminium,
copper, magnesium, lead,
molybdenum, nickel, brass, nylon,
teflon
FIG. 1-12 Stress-strain diagram for
a typical structural steel in tension
(drawn to scale)
Aluminium alloys
•Although ductile…aluminium alloys
typically do not have a clearly definable
yield point…
•However, they have an initial linear region
with 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
FIG. 1-13 Typical stress-strain
diagram for an aluminum alloy.
Offset method
• When the yield point is not
obvious, like in the previous
case, and undergoes large
strains, an arbitrary yield stress
can be determined by the offset
method
• The intersection of the offset
line and the stress-strain curve
(point A) defines the yield
stress
Arbitrary yield stress determined by
the offset method
FIG 1-14
Copyright 2005 by Nelson, a division of Thomson Canada Limited
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
FIG. 1-15 Stress-strain curves
for two kinds of rubber in
tension
Measure of the amount
of necking
Brittle materials
• Brittle materials fail at relatively
low strains and little elongation
after the proportional limit
• Brittle materials: concrete,
marble, glass, ceramics and
metallic alloys
• The reduction in the crosssectional area until fracture (point
B) is insignificant and the fracture
stress (point B) is the same as the
ultimate stress
FIG. 1-16 Typical stress-strain
diagram for a brittle material
showing the proportional limit
(point A) and fracture stress
(point B)
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
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.
FIG. 1-17 Stress-strain
diagram for copper in
compression
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.
Have a good weekend…
Wednesday (23 January 2008): Quiz on Statics, I will send you
e-mail with further details…