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Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
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)
FIG. 1-10 Stress-strain diagram for a
typical structural steel in tension (not to
scale)
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
Stress (σ) – strain (ε) 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 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
FIG. 1-10 Stress-strain
diagram for a typical
structural steel in
tension (not to scale)
FIG. 1-12 Stress-strain diagram for
a typical structural steel in tension
(drawn to scale)
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. 1-13 Typical stress-strain
diagram for an aluminum alloy.
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
Copyright 2005 by Nelson, a division of Thomson Canada Limited
FIG 1-14 Arbitrary yield stress determined by
the offset method
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
FIG. 1-15 Stress-strain curves
for two kinds of rubber in
tension
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
FIG. 1-16 Typical stress-strain
diagram for a brittle material
showing the proportional limit
(point A) and fracture stress
(point B)
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 cross-
sectional area until fracture (point
B) is insignificant and the fracture
stress (point B) is the same as the
ultimate stress
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
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.
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.
Wednesday (23 January 2008): Quiz on Statics, I will send you
e-mail with further details…
Have a good weekend…

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Mechanics of Materials

  • 1. Mechanics of Materials – MAE 243 (Section 002) Spring 2008
  • 2. 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?
  • 3. 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.
  • 4. 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.)
  • 5. 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)
  • 6. FIG. 1-10 Stress-strain diagram for a typical structural steel in tension (not to scale) 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
  • 7. Stress (σ) – strain (ε) 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 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 FIG. 1-10 Stress-strain diagram for a typical structural steel in tension (not to scale)
  • 8. FIG. 1-12 Stress-strain diagram for a typical structural steel in tension (drawn to scale) 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
  • 9. FIG. 1-13 Typical stress-strain diagram for an aluminum alloy. 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
  • 10. Copyright 2005 by Nelson, a division of Thomson Canada Limited FIG 1-14 Arbitrary yield stress determined by the offset method 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
  • 11. FIG. 1-15 Stress-strain curves for two kinds of rubber in tension 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
  • 12. FIG. 1-16 Typical stress-strain diagram for a brittle material showing the proportional limit (point A) and fracture stress (point B) 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 cross- sectional area until fracture (point B) is insignificant and the fracture stress (point B) is the same as the ultimate stress
  • 13. 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
  • 14. 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.
  • 15. 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.
  • 16. Wednesday (23 January 2008): Quiz on Statics, I will send you e-mail with further details… Have a good weekend…