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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 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.
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.
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
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 digitsprovides enough accuracy.
CIVL 1101 Introduction to Mechanics of Materials 2/10
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
CIVL 1101 Introduction to Mechanics of Materials 3/10
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
CIVL 1101 Introduction to Mechanics of Materials 4/10
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
Circular cross-section
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
Circular cross-section
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?
CIVL 1101 Introduction to Mechanics of Materials 5/10
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)
CIVL 1101 Introduction to Mechanics of Materials 6/10
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.
CIVL 1101 Introduction to Mechanics of Materials 7/10
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).
CIVL 1101 Introduction to Mechanics of Materials 8/10
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
CIVL 1101 Introduction to Mechanics of Materials 9/10
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?
CIVL 1101 Introduction to Mechanics of Materials 10/10
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