Engineering Structures and Materials Mechanics of Material ...

CIVL 1101

Introduction to Mechanics of Materials

Engineering Structures and Materials

Mechanics of Material

 Mechanics of materials is a branch of applied mechanics that deals with the behavior of solid bodies subjected to various types of loading

Early in the morning on Tuesday, August 17, 1999, a deadly magnitude 7.4 earthquake struck along the Anatolian fault in the northwestern region of Turkey

Earthquakes

 A thorough understanding of mechanical behavior is essential for the safe design of all structures  Mechanics of materials is a basic subject in many engineering fields



Earthquakes

Space Shuttle Columbia

The Space Shuttle Columbia disaster occurred on February 1, 2003, when the Space Shuttle Columbia disintegrated over Texas during reentry into the Earth's atmosphere.



Space Shuttle Columbia

I-35W Mississippi River Bridge

The loss of Columbia was a result of damage sustained during launch when a piece of foam insulation the size of a small briefcase broke off the Space Shuttle external. The debris struck the leading edge of the left wing, damaging the Shuttle's thermal protection system.

The I-35W Mississippi River bridge catastrophically failed during the evening rush hour on August 1, 2007, collapsing to the river and riverbanks beneath.

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Mechanics of Material Collapse of Can Tho Bridge


Engineering Structures and Materials The historical development of mechanics of materials is a fascinating blend of both theory and experiment  Leonardo da Vinci (1452–1519)  Galileo Galilei (1564–1642) performed experiments to determine the strength of wires, bars, and beams

On September 26, 2007 a 90 meter section of an approach ramp, which was over 30 meters above the ground, collapsed – possible due to rains weaken the foundation

Engineering Structures and Materials  Leonhard Euler (1707–1783) Developed the mathematical theory of columns and calculated the theoretical critical load of a column in 1744, long before any experimental evidence existed to show the significance of his results.

Engineering Structures and Materials  Numerical problems require that you work with specific units of measurements.  The two accepted standards of measurement are the International System of Units (SI) and the U.S. Customary System (USCS).  As you know significant digits are very important in engineering.  In our work in this section, three significant digits provides enough accuracy.

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Stress and Strain

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Stress and Strain

 The fundamental concepts of

stress

and

strain

can be illustrated by considering a prismatic bar that is loaded by axial forces P at the ends

P a

L

 A prismatic bar is a straight structural member having constant cross section throughout its length

P

 In this illustration, the axial forces produce a uniform stretching of the bar; hence, the bar is said to be in tension

a

P A a

 P

Stress

P

Stress

 The tensile load P acts at the bottom end of the bar  At the top of the bar are forces representing the action of the removed part of the bar

 The axial force is equal to the intensity s times the cross– sectional area A of the bar

P   A

 The intensity of force (that is, the force per unit area) is called the

STRESS

a



 When the bar is stretched by the force P, the resulting stresses are tensile stresses

P   A

 If the force P cause the bar to be compressed, we obtain compressive stresses

(commonly denoted by the Greek letter ).

P

Stress P   A

 a

a  The stress acting perpendicular to the cut surface, it is referred to as a normal stress  The equation  = P/A will give the average normal stress

P

Stress  Sign convention for normal stresses is: (+) for tensile stresses and (-) for compressive stresses  Because the normal stress s is obtained by dividing the axial force by the cross–sectional area, it has units of force per unit of area

+P

-P

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Stress

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Normal Strain

 In SI units: Force is expressed in newtons (N) and area in square meters (m2). A N/m2 is a pascals (Pa).

 In USCS units: Stress is customarily expressed in pounds per square inch (psi) or kips per square inch (ksi).

The change in length is denoted by the Greek letter  (delta).

L



7,000 Pa to make 1 psi

P

P

Normal Strain  An axially loaded bar undergoes a change in length, becoming longer when in tension and shorter when in compression

L 

P

 The concept of elongation per unit length, or strain, denoted by the Greek letter  (epsilon) and given by the equation

 



Normal Strain  If the bar is in tension, the strain is called a

tensile strain  If the bar is in compression, the strain is called a

compressive strain  Tensile strain is taken as positive (+), and compressive strain as negative (-).

L

Normal Strain  The strain  is called a normal strain because it is associated with normal stresses.  Because normal strain  is the ratio of two lengths, it is a dimensionless quantity; that is, it has no units.

Example Consider a steel bar having length L of 2.0 m. When loaded in tension, the bar might elongate by an amount  equal to 1.4 mm.



 L



1.4  103 m  0.00070  7.0  10 4 2.0m

The resulting state of stress and strain is called uniaxial stress and strain

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Example

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Example

A prismatic bar with a circular cross section is subjected to an axial tensile force. The measured elongation is  = 1.5 mm. Calculate the tensile stress and strain in the bar.

100 kN

Diameter = 25.0 mm

3.5 m

100 kN

Circular cross-section

Assuming the axial force act at the center of the end cross section, then the stress is: Diameter = 25 mm

3.5 m

The resulting state of stress and strain is called uniaxial stress and strain



P 100kN  A  25.0mm



4

Example 100 kN 3.5 m

Circular cross-section



2

 203.718327

N  mm 2

204Mpa

1,000 mm  1 m

Group Problem 1 Circular cross-section

If the applied load on the bar is P = 10,000 lb. and the crosssectional area is A = 0.50 in2, what is the stress?

Diameter = 25 mm

The strain is



 L



 1.5 mm  1m    0.0004286..... 3.5 m  1,000 mm 

 4.3  10 4

Group Problem 2 If the allowable stress at failure for the material is 35,000 psi and the applied load on the bar is P = 20,000 lb., what is the minimum area require to prevent failure?

Group Problem 3 A bar has 2 in.2 cross-sectional area; if the allowable stress at failure for the material is 25,000 psi, what is the maximum forced the bar would support?

CIVL 1101


Group Problem 4 If a bar elongates 2.5 in. and the original length is 10.0 ft., what is the strain?

Stress–Strain Diagrams

Group Problem 5 If the bar fails at strains greater than 0.15 and the original length of the bar is L = is 10 ft., what is the maximum allowable deformation before failure?

Tension Test

 The mechanical properties of materials are determined by tests performed on small specimens of the material

 The axial stress  in the test specimen is calculated by dividing the load P by the cross–sectional area A

 In order that test results may be compared easily, the dimensions of test specimens and the methods of applying loads have been standardized

 Strain in the bar is found from the measured elongation  between the gage marks by dividing  by the gage length L

Standards organizations: American Society for Testing and Materials (ASTM), American Standards Association (ASA) and the National Bureau of Standards (NBS)

Developing a Stress–Strain Diagram  After performing a tension or compression test and determining the stress and strain at various magnitudes of the load, we can plot a diagram of stress versus strain  Stress–strain diagrams were originated by: Jacob Bernoulli (1654–1705) and J. V. Poncelet (1788– 1867)

Developing a Stress–Strain Diagram

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Stress–Strain for Steel

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Stress–Strain for Steel

 The first material we will discuss is: structural steel  A stress–strain diagram for a typical structural steel in tension is shown:

 60

s 40 Fracture

Ultimate Stress Yield Stress

20

0.05

0.10

0.15

0.20

0.25

Strain Elastic

Plastic





Strain Necking Hardening

Stress–Strain for Aluminum

Stress–Strain for Rubber 



3

60

Steel

Hard Rubber 2

40

Aluminum 20

0.05

0.10

0.15

0.20

Strain

0.25



Linear Elasticity  When a material returns to its original dimensions after unloading, it is called elastic  When a material behaves elastically and also exhibits a linear relationship between stress and strain, it is said to be

linearly elastic.

Soft Rubber 1

2

4

6

8

Strain



Linear Elasticity The linear relationship between stress and strain for a bar in simple tension or compression can be expressed by the equation:

  E where E is a constant known as the

modulus of elasticity (units are either psi or Pa)

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Hooke’s Law

Hooke’s Law

The equation  = E  commonly known as Hooke’s law 

For the famous English scientist Robert Hooke (1635–1703).



Hooke was the first person to investigate the elastic properties of materials, and he tested such diverse materials as metal, wood, stone, bones, and sinews.



 Aluminum is approximately 10,600ksi or 70 GPa  Wood is 1,600 ksi or 11 Gpa

 The modulus of elasticity is often called Youug’s modulus, after another English scientist, Thomas Young (1773–1829)

Linear Elasticity

Linear Elasticity

If the material in the bar is considered linear-elastic and the tensile stress is 25,000 psi and the tensile strain is 0.005, what is the modulus of elasticity of the material?

E

  E 

25,000 psi 0.005

 The modulus of elasticity E has relatively large values for materials that are very stiff, such as structural metals  Steel has a modulus of 30,000 ksi or 200 GPa

He measured the stretching of long wires supporting weights and observed that the elongations “always bear the same proportions one to the other that the weights do that make them”

  E

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

5,000,000 psi



5  10 psi 6

Group Problem 6 Determine the cross-sectional area of a 100-ft. steel cable supporting a 20,000 lb. tensile force while not exceed the an allowable tensile stress of 50,000 psi or a maximum elongation of 0.050 ft. Assume the modulus of elasticity of steel is E = 29,000,000 psi (assume all value are “exact” measurements).

If you substitute the formulas for stress and strain into Hooke’s Law you get:

  E









P A



L

P  E A L

Stress–Strain Diagram  Materials that undergo large strains before failure are classified as ductile  Ductile materials include mild steel, aluminum and some of its alloys, copper, magnesium, lead, molybdenum, nickel, brass, bronze, nylon, teflon, and many others s Fracture

Ultimate Stress Yield Stress

Elastic

Plastic

Strain Necking Hardening



CIVL 1101


Stress–Strain Diagram

Compression Test

 Materials that fail in tension at relatively low values of strain are classified as brittle materials.

 Compression tests of metals are customarily made on small specimens in the shape of cubes or circular cylinders.

 Examples are concrete, stone, cast iron, glass, ceramic materials, and many common metallic alloys.  Ordinary glass is a nearly ideal brittle material

 Concrete is tested in compression on every important construction project to ensure that the required strengths have been obtained.



Strain

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

 The standard ASTM concrete test specimen is 6 in. in diameter, 12 in. long, and 28 days old (the age of concrete is important because concrete gains strength as it cures)

Compression Test

Compression Test

 Stress–strain diagrams for compression have different shapes from those for tension.  Ductile metals such as steel, aluminum, and copper have proportional limits in compression very close to those in tension.

 However, when yielding begins, the behavior is quite different. Consider compression of copper:

 60

40

20

0.2

0.4

0.6

0.8

1.0

1.2



Strain

Elasticity

Plasticity

 The stress–strain diagrams described in the preceding section illustrate the behavior of various materials as they are loaded statically in tension or compression.  Now let us consider what happens when the load is slowly removed, and the material is unloaded



 If the loading is too great a residual strain, or permanent strain, remains in the material  The corresponding residual elongation of the bar is called the permanent set. The material is said to be partially elastic

Loading

Unloading

Elastic

 Now let us suppose that we load this same material to a much higher level



 Loading

Unloading

Elastic Residual Elastic Strain Recovery



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Creep

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Relaxation

 Development of additional strains over long periods of time and are said to creep

 0



Tight cable

d0

P t0

t0 Time

End of Mechanics of Materials

Any Questions?

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