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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