engineering structures mechanics of material earthquakes · and materials the historical...

Engineering Structures and Materials

Mechanics of materials is a branch of applied mechanicsthat deals with the behavior of solid bodies subjected tovarious types of loading

A thorough understanding of mechanical behavior isessential for the safe design of all structures

Mechanics of materials is a basic subject in many engineering fields

Mechanics of Material Earthquakes

The 2017 Central Mexico earthquake struck at 13:14 on September, 19 2017 with a magnitude estimated to be Mw 7.1 and strong shaking for about 20 seconds.

Mechanics of Material Earthquakes

Mechanics of Material Space Shuttle Columbia

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

Mechanics of Material Space Shuttle Columbia

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.

Mechanics of Material I-35W Mississippi River Bridge

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

CIVL 1101 Introduction to Mechanics of Materials 1/10





Mechanics of Material FIU Pedestrian Bridge

On March 15, 2018, a 175-foot-long (53 m), recently-erected section of the FIU Sweetwater University City pedestrian bridge collapsed onto the Tamiami Trail (U.S. Route 41). Eight vehicles were crushed underneath, which resulted in six deaths and nine injuries.

Engineering Structuresand 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


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.


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 digitsprovides enough accuracy.


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

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

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

Stress and Strain

P

a a

P

a a

P

A

P

L

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

P

A

STRESS

(commonly denoted by the Greek letter ).

Stress

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

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

P A

P

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

Stress

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

a a

P

A

Sign convention for normal stresses is: (+) for tensile stresses and (-) for compressive stresses

Stress

-P+P

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


Stress

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).

7,000 Pa to make 1 psi

Normal Strain

P

P

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

Normal Strain

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

L

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

P

L

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 (-).

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

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

40.00070 7.0 10 31.4 10

2.0

m

m


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.

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

100 kN

3.5 m

Circular cross-section

Diameter = 25 mm

Example

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

100 kN

3.5 m


Diameter = 25.0 mm

2100

25.0

4

kN

mm

2203.718327

N

mm

1,000 1mm m

204MpaP

A

44.3 10

Example

The strain is

100 kN

3.5 m


Diameter = 25 mm

L

1.5 1

3.5 1,000

mm m

m mm

0.0004286.....

Group Problems

TopHat Problems

Group Problem 1

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

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?

Group Problem 4

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

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?

Stress–Strain Diagrams

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

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

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

Tension Test

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

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

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) J. V. Poncelet (1788-1867)


Developing a Stress–Strain Diagram 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:

Elastic Plastic StrainHardening

Necking

FractureUltimate Stress

Yield Stress

Stress–Strain for Steel

0.05 0.10 0.15 0.20 0.25

20

40

60

Strain

Stress–Strain for Aluminum

0.05 0.10 0.15 0.20 0.25

20

40

60

Strain

Steel

Aluminum

Stress–Strain for Rubber

2 4 6 8

1

2

3

Strain

Hard Rubber

Soft Rubber

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.


Linear Elasticity

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

where E is a constant known as the

modulus of elasticity

(units are either psi or Pa)

E

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.

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”

Hooke’s Law

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

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

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

0.005

psiE 5,000,000 psi

65 10 psi

Linear Elasticity

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

E P

A

L

PE

A L

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).


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

Elastic Plastic StrainHardening

Necking

FractureUltimate Stress

Yield Stress

s

Stress-Strain Diagram

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

Examples are concrete, stone, cast iron, glass, ceramic materials, and many common metallic alloys.

Strain

Ordinary glass is a nearly ideal brittle material

Compression Test

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

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)

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

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.

Compression Test

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

0.2 0.4 0.6 0.8 1.0

20

40

60

Strain

1.2

Elasticity

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

Elastic

Loading

Unloading


Plasticity

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

Elastic

Loading

Unloading

ResidualStrain

ElasticRecovery

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

P

Creep

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

Timet0

d0

Relaxation

0

t0

Tight cable

Time

End of Mechanics of Materials

Any Questions?


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