WHAT IS A STRUCTURE?

A structure is something that will support an object or a

load.

A structure must be strong enough to support its own weight and whatever load is put on it !

There are three types of structure :

Mass StructuresFrame StructuresShell structures.

Mass Structures are solid structures which rely on their own weight to resist loads. A single brick is a mass structure but so is a large dam.

Mass Structures

Frame Structures

Frame structures are made from many small parts (called members (struts ties etc.)), joined together. Bridges, cranes and

parts of this oil rig are some examples.

Shell structures

•Shell structures are made or assembled to make one piece. Tin cans, bottles and other food containers are examples of shell structures. Larger things such as car and aeroplane bodies are examples of more complicated shell structures. Most shell structures are made from thin sheet material (which makes them light) and most have ridges or curves moulded into them (to make them strong).

Natural Structures

•Structures are not new, nature produced the first structures long before humans were able to. A leaf is a natural structure. Its veins provide support and carry nutrients. A tree has to carry the weight of its own branches as well as resisting strong winds.

Manufactured structures

• A manufactured structure is quite simply a structure built by human beings.

Many of Nature's structures have been copied by humans. The shell of a snail and the body of a modern car are both shell structures designed to protect their occupants.

Forces on Structures

• Force – push or pull that transfers energy to an object• External force – come from outside, act upon the

structure. An external force is a force placed on the structure from outside, by the wind perhaps or perhaps by someone sitting or standing on it.

• Internal force – force that parts exert on each other, act within structure. Internal forces are the forces which the structure must provide within itself to resist the external forces placed upon it. If the external forces are greater than the internal forces, a structure will collapse.

Forces acting on Structures

TYPES OF FORCES

Forces can be either static (stationary) or dynamic (moving).

Static forces are usually forces caused by the weight of the structure and anything which is permanently attached to it. Static Load – changes slowly or not at all, eg: bricks in a building, twigs in nest

Dynamic forces are caused by things such as wind, waves, people, and vehicles. Dynamic forces are usually much greater than static forces and are very difficult to predict. These are the most common reason for structural failures. Dynamic Load – move or change, eg: car crossing bridge, oil in pipeline

Forces can be internal or external

• 5 types of recognized forces:

compression, tension, torsion, shear & bending

• 1. Compression – shortens or crushes

• 2. Tension – stretches or pulls apart

• 3. Torsion – twists

• 4. Shear – pushes parts in opposite directions

• 5. Bending - stretches and squashes at the same time.

Forces acting on and within Structures

External forces or loads cause internal stresses to be set up in a structure. Not all forces or loads act in the same way. Forces can bend, pull, press, or twist. Each of these types of force are given special names.

Tension : Is the name given to a force which tries to pull something apart. A structural member in tension is called a tie. A Tie resists tensile stress. Tension forces stretch a material by pulling its ends apart Tensile strength measures the largest tension force the material can withstand before failing.

Compression : Is the name given to a force which tries to squash something together. A structural member in compression is called a strut. A strut resists compressive stress.. Compression forces crush a material by squeezing it together. Compressive strength measures the largest compression force the material can withstand before it loses its shape or fails.

Internal Forces Within Structures

Compression, Tension, Torsion and Shear •Shear : A shear force is created where two opposite forces try to cut tear or rip something in two. Shear forces bend or tear a material by pressing different parts in opposite directions at the same time. Shear strength measures the largest shear force the material can withstand before it rips apart.

•Torsion : Is the name given to a turning or a twisting force. Torsion forces twist a material by turning the ends in opposite directions. Torsion strength measures the largest torsion force the material can withstand and still spring back into its original shape.

•Bending : Bending is a word you will have met before. A structure which is subjected to bending is being stretched and squashed at the same time.

THE EFFECT OF FORCES ON MATERIALS

• When a material is bent, stretched or compressed but returns to its original size when the load is removed, we say it behaves in an elastic way. Elastic bands are good examples.

• When a material is bent, stretched or compressed but does not return to its original size when the load is removed we say it behaves in a plastic way.

• Moist clay is a good example of a 'plastic' material.

A simple experiment with a paper clip will show the difference between elastic and plastic behaviour. Up to a certain point, a paper clip will spring back into shape when you bend the end outwards and let go.

If you bend it too far, it springs back slightly but stays permanently bent. When this happens, it has been bent beyond its elastic Iimit. (see Hooke’s law for springs)

MEMBERS IN A STRUCTURE

• The different parts of a frame structure are called members. Each type of member has a different job to do in supporting the structure.

A Beam is a piece of material supported at either end.

When a beam is loaded the top is compressed and the bottom is in tension.

They are usually supported by two or more Columns. Ideally beams should be able to span a wide gap and support a load without deflecting.

Beams

• Beams used in larger structures take many different forms, some are simply solid, some are hollow, and others have special cross-sections to provide strength and rigidity.

A cantilever is a beam which is supported at one end only. Cantilevers are used where it is not possible to have a support at both ends (a diving board for instance).

When a cantilever is loaded, the top surface is in tension and the bottom is in compression.

Frame structures

• Frame structures achieve most of their strength and rigidity from the way they are assembled.

Most frameworks are built using a combination of struts and ties to make triangles. Triangles make very strong and rigid structures. Using triangles in this way is called Triangulation.

Shell structures

• Most shell structures achieve their strength and rigidity from the way they are shaped. Shell structures very rarely have large flat surfaces they tend to be designed and made with ribs to act as stiffeners.

Egg and light bulbs containers are good examples. Both eggs and light bulbs can withstand considerable static forces if they are applied carefully.

What they are not good at is resisting dynamic forces.

This is why their containers are designed to absorb impact.

Stress versus Strain• Mechanical Properties

– Deal directly with behavior of materials under applied forces. – Properties are described by applied stress and resulting strain, or applied

strain and resulting stress.• Example: 100 lb force applied to the end of a rod results in a stress

applied to the end of the rod causing it to stretch or elongate, which is measured as strain.

– Strength: ability of material to resist application of load without rupture. • Ultimate strength- maximum force per cross section area.• Yield strength- force at yield point per cross section area.• Other strengths include rupture strength, proportional strength, etc.

– Stiffness: resistance of material to deform under load while in elastic state.

• Stiffness is usually measured by the Modulus of Elasticity (Stress/strain)

• Steel is stiff (tough to bend). Some beds are stiff, some are soft (compliant)

21

Introduction

Mechanical properties that are important to a design engineer differ from those that are of interest to the manufacturing engineer.

• In design, mechanical properties such as elastic modulus and yield strength are important in order to resist permanent deformation under applied stresses. Thus, the focus is on the elastic properties.

• In manufacturing, the goal is to apply stresses that exceed the yield strength of the material so as to deform it to the required shape. Thus, the focus is on the plastic properties.

Testing Procedures

• Mechanical Testing

– Properties that deal with elastic or inelastic behavior of a material under load

– Primary measurements involved are load applied and effects of load application

– Two classification of tests; method of loading and the condition of the specimen during the test

• Primary types of tests

– Tensile– Compression

– Shear

– Torsion

– Flexure

Mechanical Test Considerations

• Principle factors are in three main areas– manner in which the load is applied– condition of material specimen at time of test– surrounding conditions (environment) during testing

• Tests classification- load application– kind of stress induced. Single load or Multiple loads– rate at which stress is developed: static versus dynamic– number of cycles of load application: single versus fatigue

• Primary types of loading

tension compressionshear

torsionflexure shear

24

Tensile Test- Basic Principles• The yield behavior of a material is

determined from the stress-strain relationship under an applied state of stress (tensile, compressive or shear).

• An axial force applied to a specimen of original length (lo) elongates it, resulting in a reduction in the cross-sectional area from Ao to A until fracture occurs.

• The load and change in length between two fixed points (gauge length) is recorded and used to determine the stress-strain relationship.

• A similar procedure can be adopted with a sheet specimen.

25

Basic Principles

• Step 1: Original shape and size of the specimen with no load.

• Step 2: Specimen undergoing uniform elongation.

• Step 3: Point of maximum load and ultimate tensile strength.

• Step 4: The onset of necking (plastic instability).

• Step 5: Specimen fractures.• Step 6: Final length.

26

Terminology

Engineering Stress and Strain:• These quantities are defined relative to the original

area and length of the specimen.• The engineering stress (σe) at any point is defined as

the ratio of the instantaneous load or force (F) and the original area (Ao).

• The engineering strain (e) is defined as the ratio of the change in length (L-Lo) and the original length (Lo).

Stress

Stress = Measure of force felt by material

Force is measured in Newtons (N): Area is measured using metres squared (m²)Therefore answer will be Newton/metres squared, denoted using N/m2

1 N/m2 = 1 Pa (same as pressure). Pa is the SI unit short for Pascals

Stress = ForceArea

Stress

• Stress: Intensity of the internally distributed forces or component of forces that resist a change in the form of a body.– Tension, Compression, Shear, Torsion, Flexure

• Stress calculated by force per unit area. Applied force divided by the cross sectional area of the specimen.

• Stress units– Pascals = Pa = Newtons/m2

– Pounds per square inch = Psi Note: 1MPa = 1 x106 Pa = 145 psi• Example

– Wire 12 in long is tied vertically. The wire has a diameter of 0.100 in and supports 100 lbs. What is the stress that is developed?

– Stress = F/A = F/πr2 = 100/(3.1415927 * 0.052 )= 12,739 psi = 87.86 MPa

A

F=σ

29

The engineering stress is:

P is the load in lbs. Kgs. Etc. on the specimen and A0 is the

original cross-sectional area near the center of the specimen. On the other hand, the true stress is the load divided by the true area, which continues to be smaller by the tensile load. The true stress continues to increase to the point of fracture, while the engineering stress decreases to the point of fracture due to the increasing load and the constant cross-sectional area.

0

P

Aσ =

Stress

Example •Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the stress that is developed? What is the Load?

– Stress = F/A = F/(width*thickness) = 100lbs/(1in*0.1in )=1,000 psi = 1000 psi/145psi = 6.897 Mpa

– Load = 100 lbs

•Block is 10 cm x 1 cm x 5 cm is mounted on its side in a test machine. The block is pulled with 100 N on both sides. What is the stress that is developed? What is the Load?

– Stress = F/A = F/(width*thickness) = 100N/(.01m * .10m )= 100,000 N/m2 = 100,000 Pa = 0.1 MPa= 0.1 MPa *145psi/MPa = 14.5 psi

– Load = 100 N

10cm 5cm

10in1 in

0.1 in

1 cm

100 lbs

Strain

Strain = Measure of deformation F

A∆L

L

Strain = ∆L

L

Strain

• Strain: Physical change in the dimensions of a specimen that results from applying a load to the test specimen.

• Strain calculated by the ratio of the change in length and the original length. (Deformation)

• Strain units (Dimensionless)– When units are given they usually are in/in or mm/mm. (Change in dimension

divided by original length)

• % Elongation = strain x 100%

0l

l∆=ε l0lF

33

The engineering strain is:

0

0

l l

lε −=

l is the gage length at a given load and l 0 is the original gage length with zero load

StrainExample

– Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the strain that is developed if the bar grows to 10.2in? What is % Elongation?

• Strain = (lf - l0)/l0 = (10.2 -10)/(10) = 0.02 in/in

• Percent Elongation = 0.02 * 100 = 2%

– Block is 10 cm x 1 cm x 5 cm is mounted on its side in a test machine. The block is pulled with 1000 kN on bone side. If the material elongation at yield is 1.5%, how far will it grow at yield?

• Strain = Percent Elongation /100 = 1.5%/100 = 0.015 cm /cm

• Strain = (lf - l0)/l0 = (lf -5)/(5) = 0.015 cm/cm

• Growth = 5 * 0.015 = 0.075 cm

• Final Length = 5.075 cm

10cm 5cm

10in1 in

0.1 in

1 cm

100 lbs

Strain• Permanent set is a change in form of a specimen once the

stress ends.• Axial strain is the strain that occurs in the same direction

as the applied stress.• Lateral strain is the strain that occurs perpendicular to the

direction of the applied stress.• Poisson’s ratio is ratio of lateral strain to axial strain.

Poisson’s ratio = lateral strain axial strain

– Example• Calculate the Poisson’s ratio of a material with lateral strain of

0.002 and an axial strain of 0.006• Poisson’s ratio = 0.002/0.006 = 0.333

AxialStrain

LateralStrain

Note: For most materials, Poisson’s ratio is between 0.25 and 0.5• Metals: 0.29 (304 SS) to 0.3 (1040 steel) to 0.35 (Mg)•Ceramics and Glasses: 0.19 (TiC) to 0.26 (BeO) to 0.31 (Cordierite)•Plastics: 0.35 (Acetals) to 0.41 (Nylons)

36

Stress strain diagramsStress-strain diagrams plot stress against the corresponding strain produced.Stress is the y-axisStrain is the x-axis

Stress-Strain Curve This stress-strain curve is produced from the tensile test.

37

Stress Strain for Different Materials

38

Modulus of Elasticity

• The slope of the stress-strain curve in the elastic deformation region is the modulus of elasticity, is known as Young's modulus

εσ=E

Y =F

A( )∆L

L( )

Young’s Modulus (Tension)

F

A∆L

L

•Measure of stiffness• Tensile refers to tension

tensile stress

tensile strain

S =F

A( )∆x

h( )

Shear Modulus

Sheer Stress

Sheer Strain

41

Shear Testing - Introduction

•Shear testing involves an applied force or load that acts in a direction parallel to the plane in which the load is applied. Shear loads act differently than, say, tensile or compressive loads that act normal or perpendicular to the axis of loading. Direct shear and torsional shear are important forces used to determine shear properties. Direct or torsional loading depends on the forces a material is expected to be subjected to during service.

Shear Testing - Procedure•Before testing, the specimen is accurately measured using proper instruments and the gage length is marked. The troptometer or a suitable replacement is attached to the specimen and zeroed out. Proper precautions should be taken to center the specimen in the machine or fixture. The grippers are tightened to insure against slippage, yet not so tight as to cause deformations which would affect test results.

•In general, shear testing involves either direct or torsional loading. In direct shear tests, the specimen is placed in the shear test fixture and a load is applied. This can be seen in the figure below. For plate specimens, a punch and die combination may be used. Plastics, generally, are square specimens with holes in either end to facilitate gripping. The applied load and resultant deformation are recorded and a suitable graph can be plotted.

42

B = −∆FA

∆VV

= − ∆P∆VV( )

Bulk Modulus

Change in Pressure

Volume Strain

B = Y 3

44

Tensile Testing - Procedure

•Tensile tests are used to determine the tensile properties of a material, including the tensile strength. The tensile strength of a material is the maximum tensile stress that can be developed in the material.

•In order to conduct a tensile test, the proper specimen must be obtained. This specimen should conform to ASTM standards for size and features. Prior to the test, the cross-sectional area may be calculated and a pre-determined gage length marked.

•The specimen is then loaded into a machine set up for tensile loads and placed in the proper grippers. Once loaded, the machine can then be used to apply a steady, continuous tensile load.

•Data is collected at pre-determined points or increments during the test. Depending on the material and specimen being tested, data points may be more or less frequent. Data include the applied load and change in gage length. The load is generally read from the machine panel in pounds or kilograms.

45

•The change in gage length is determined using an extensometer. An extensometer is firmly fixed to the machine or specimen and relates the amount of deformation or deflection over the gage length during a test.

•While paying close attention to the readings, data points are collected until the material starts to yield significantly. This can be seen when deformation continues without having to increase the applied load. Once this begins, the extensometer is removed and loading continued until failure. Ultimate tensile strength and rupture strength can be calculated from this latter loading.

•Once data have been collected, the tensile stress developed and the resultant strain can be calculated. Stress is calculated based on the applied load and cross-sectional area. Strain is the change in length divided by the original length.

46

Tensile Test

Tensile testing machine:

Load indicator

Clamps to hold the specimen

Unload Lever

Load Lever

Machine dial

Load indicator

Clamps to hold the specimen

Unload Lever

Load Lever

Load indicator

Clamps to hold the specimen

Unload Lever

Load Lever

Machine dial

47

Tensile Test

Test Specimen: • The tensile test can be conducted with either a round bar or sheet specimen.

• The round bar specimen used for the current test complies with the ASTM standards.

• A 2 inch gage length is marked on the specimen prior to testing.

• The specimen is held in the clamps at either end. Load and movement are applied to the bottom clamp.

Gauge mark ingsGauge mark ings

48

Comparison of f inal lengths (total elongation) of specimens at fracture with dif ferent n values using FE simulations

Fracture occurs after a certain amount of elongation that is influenced by the n-value

(a) n=0.2 (b) n =0.4 (c) n = 0.6

Simulation results- Fracture

49

Compression Testing - Introduction

•Simplistically, compression testing is the opposite of tensile testing. A compressive load tends to squeeze or compact the specimen. The choice of a compression test over other types of testing largely depends on the type of loading the material will see during application or service.

•Metals and many plastics, for example, are more efficient at resisting tensile loads. Therefore, they are more commonly tested using tensile loading, depending on the application, of course. Materials, such as concrete, brick, and some ceramic products, are more often used in applications for their compressive loading properties and are, therefore, tested in compression. Again, it is important to choose the test that best reflects the loads and conditions the material will be subjected to in application or service.

50

Compression Testing – Procedure

During a typical compression test, data are collected regarding the applied load, resultant deformation or deflection, and condition of the specimen. For brittle materials, the compressive strength is relatively easy to obtain, showing marked failure. However, for ductile materials, the compressive strength is generally based on an arbitrary deformation value. Ductile materials do not exhibit the sudden fractures that brittle materials present. They tend to buckle and "barrel out".

51

Barreling or Bulging of a Sample under Compressive Loads

52

•Prior to this and any test, the dimensions of the specimen should be measured with adequate precision using proper instruments. Once these measurements have been taken and recorded, the specimen should be loaded into the testing machine.

•In compression testing, and testing in general, care should be taken to insure that the axis of the specimen is centered and aligned with the axis of loading.

•Loading rates should be steady and continuous. Rates vary, but a general figure is 0.005 inches per minute strain rate. Loading rates typically range from 500-1000 lb/min.

53

Factor of safety (FoS), also known as (and used interchangeably with) safety factor (SF), is a term describing the structural capacity of a system beyond the expected loads or actual loads. Essentially, how much stronger the system is than it usually needs to be for an intended load. Safety factors are often calculated using detailed analysis because comprehensive testing is impractical on many projects, such as bridges and buildings, but the structure's ability to carry load must be determined to a reasonable accuracy.Many systems are purposefully built much stronger than needed for normal usage to allow for emergency situations, unexpected loads, misuse, or degradation

54

CalculationThere are several ways to compare the factor of safety for structures. All the different calculations fundamentally measure the same thing: how much extra load beyond what is intended a structure will actually take (or be required to withstand). The difference between the methods is the way in which the values are calculated and compared. Safety factor values can be thought of as a standardized way for comparing strength and reliability between systems.The use of a factor of safety does not imply that an item, structure, or design is "safe". Many quality assurance, engineering design, manufacturing, installation, and end-use factors may influence whether or not something is safe in any particular situation.                                       

EXAMPLE CALCULATION1) The tensile (material) strength of a specific metal bar is 766N/mm². the maximum permissible stress (design load) is 456N/mm². calculate the FoS.•FoS = 766N/mm² = 1.68 456N/mm²2) A column of pre-stressed concrete will be subjected to various loads and as such is required to have a permissible stress (design load) of 23N/mm², with a FoS of 3. Calculate the compressive stress (material strength) of the concrete.•FoS (3) = compressive stress 23N/mm²•Hence compressive stress = 3 x 23N/mm² = 69N/mm²3) Steel rebars are required to have a material strength of 524N/mm², with a FoS of 3. Calculate the design load for the steel.•FoS (3) = 524N/mm² design load•Hence design load = 524N/mm²÷ 3 = 174.7N/mm²