Structures and Materials Objectives

Free Web Hosting Provider - Web Hosting - E-commerce - High Speed Internet - Free Web Page

The method used to add or take vectors includes drawing a diagram using the head-to-tail method then using Pythagorean theorem and trigonometric functions to find out the figures.

The head-to-tail method involves drawing a vector diagram; where the head of this vector ends the tail of the next vector begins (thus, head-to-tail method). The process is repeated for all vectors which are added. Once all vectors have been added head-to-tail, the resultant is drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be calculated and converted to real units using the given scale. The magnitude and direction of vector R in the diagram above can be determined by use of trigonometric functions.

2.
When calculating the torque or rotational force caused by the weight of an object, the object may be treated as a rigid, massless shell with all its mass or weight concentrated at the center of mass. For a regular, uniform object, this center of mass or center of gravity is located at the ordinary center of the object.

Center of Mass (CM) The point where all the mass of an object seems to act. For example, if you look at the motion of a high jumper, there is one special spot that moves in a parabolic path. It would be the same point where you could balance that person. An object is in equilibrium as long as its CM is over its base level. An object is considered uniform when the CM is its geometric center.

When we throw a baseball, we expect it to follow a smooth, parabolic trajectory.

How can we locate the Center of Mass in an extended object?
We can suspend it. The Center of Mass will always be directly under the point of support. Then we can suspend it from some other point and know that the Center of Mass will always be directly under this point of support as well.

3. Torque is a force applied in such a way that it creates a change in the rate of rotation (angular velocity) of an object. It is defined as the product of the applied Force and the perpendicular Length from some point (typically measured from the pivot point). This length measurement is often called the "lever arm".

A torque produces angular acceleration. A torque is required to start the rotation of a body. A rotating rigid body can be considered as consisting of many particles located at various distances from the axis of rotation. The sum of the torques due to each of these particles is just the total torque. A rotational force depends upon the force and where that force is applied; torque = moment arm x force. The lever arm is the perpendicular distance from the force to the axis of rotation.

A torque is a "rotational force" or the rotational equivalent of a force,
Torque = moment arm x force

A force F acts at at an angle at a point P on a rigid body that is free to rotate about an axis through O a distance r from the axis of rotation. Only the tangential component of F or F_T can have any effect on the rigid body. F_R (the radial component) passes through the axis of rotation and cannot cause the object to rotate.

An object is said to be in rotational equilibrium when all the torques acting on it are balanced (or, S t = 0). A torque can cause a counterclockwise (cc) or a clockwise rotation (cw).
St_cw = St_cc

Static Equilibrium: An analysis of static equilibrium is very important in engineering. The design engineer must identify and isolate all external forces and torques acting on the structure. With a good design and a correct choice of materials, structures can support loads. Landing gear of aircraft survive the shock of rough landings and bridges don't collapse under traffic loads and the wind.
Translational Equilibrium: An object is in translational equilibrium (its momentum is constant) if the sum of the forces acting on it is zero. S F_x = 0, S F_y = 0, SF_z =0

Rotational Equilibrium An object is in rotational equilibirum (its angular momentum is constant) if the sum of the torques acting on it is zero.
An object will be in equilibrium if it is suspended from its center of gravity or its center of gravity is below the suspension point.

Newton's First Law (adapted to torque)
In the absence of a net torque (

) an object will continue with its present rotational (angular) velocity. In other words, if it is not rotating, it will continue to not rotate. If it is rotating it will keep the same rate of rotation. It will have no rotational (angular) acceleration.

Newton's Second Law (adapted to torque)
If there is a net torque (

), the object will have a rotational (angular) acceleration. It will either be increasing or decreasing its rate of rotation. The amount of the object's acceleration will be related to its rotational inertia (moment of inertia) which is the rotational version of mass (regular inertia).

Newton's Third Law (remains unchanged for torque)
When one object applies a force on another object the second object applies an equal and opposite force back on the first object.

What forces are exerted by the other two ropes?
To answer this, look at the forces exerted on the knot where the three ropes are joined. The knot at rest so the sum of the forces acting on it must be zero.

T is the magnitude of the force on the knot from the rope attached directly to the weight. T_L is the magnitude of the force on the knot from the rope on the left that makes an angle of 45° with the horizontal. And T_R is the magnitude of the force on the knot from the rope on the right that makes an angle of 30° with the horizontal. These three forces must add to zero. Graphically, they can be added as shown here; the force vectors form a closed triangle.

6.
First Condition of Equilibrium
We may say that an object at rest is in equilibrium or in static equilibrium. An object at rest is described by Newton's First Law of Motion. An object in static equilibrium has zero net force acting upon it.

The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as

Second Condition of Equilibrium
An object in equilibrium does not move along a straight line -- it does not translate -- that means the sum of all the forces on it is zero. That was the first condition of equilibrium.
But an object in equilibrium also does not rotate. That means the sum of all the rotational forces on it is also zero. The sum of all the torques on an object is equilibrium is zero. This is the Second Condition of Equilibrium.

Torques that would rotate an object counter clockwise may be taken as positive and torques that would rotate an objectclockwise may be taken as negative. Then we can write this Second Condition of Equilibrium as

or we can calculate the sum of the clockwise torques and set them equal to the sum of the counterclockwise torques. Then we can write this Second Condition of Equilibrium as

An object in equilibrium is at rest. It is not moving (or "translating") and it is not rotating.
There are two conditions of equilibrium. That a body is not moving (or "translating") means the sum of the forces is zero. That a body is not rotating means the sum of the torques is zero.

7. An object will topple if its center of mass is pushed outside of its base. A body is more stable the lower the center of gravity and the wider the base. Hence it is harder to push the center of mass outside the base.

The Centre of mass of theses birds is over the pivot point of the beak.

8.
Consider a bar subjected to a load P as shown in Fig. 1. Due to the load, the bar will deform and the right end of the bar will move in the direction of the load by a distance d. We note that d also denotes the change in length of the bar. Let L and A be the length and cross-sectional area of the bar respectively. As the load P is increased, d also increases. For most engineering materials, the plot of P against d appears linear as shown in Fig. 1.b. Here we assume that the load P is small enough that no permanent deformation occurs in the bar (more on this later). When a given material deforms linearly under an applied load as shown in Fig. 1.b, the material is said to behave linear elastically.

Next consider three bars made of the same material subjected to a load P as shown in Fig. 2.a. In case (1), the bar has a length L and cross-sectional area A, in case (2), the bar has a length 2L and cross-sectional area A whereas in case (3), the bar has a length L and cross-sectional area 2A. The load-displacements curves for these three cases are shown in Fig. 2.b. As the figure indicates, each bar is represented by a different curve even though all the bars are made of the same material. Clearly, force-displacement curves are not very useful for characterizing the material behavior since these curves depend on the length and cross-sectional area of the bar under consideration.

Elasticity is the branch of physics which deals with how objects deform when forces are applied to them.
The Elastic Limit is the point where the material being deformed suffers permanent deformation and will not return to its original shape.

There are three ways an object can change its dimensions when forces act on it:
- An object can be deformed by shearing forces. It will behave like the pages of a book when under shear. An example: the movement of layers of rock in an earthquake.
- An object can be deformed by stretching or compressing forces. An example: tensile forces that stretch the length of a string until it breaks. An example: Piling weights on a cylinder until it breaks.
- An fluid can be deformed by bulk forces. An example: A fluid under high pressure can be compressed on all sides resuliting in a volume change.

9. A strain on a material can be defined as any change in the materials dimension, and any force acting on a material produces a stress. With tensile materials, strain (e ) is the same as stretch, and is simply the ratio of the change in size to some basic (or original) size (often given as a percentage; e = 0.1 indicates that each unit of length has extended by 10%). The unit for stress (s ) is the pascal (Pa) or megapascal (MPa), which is the force per unit area (a newton or meganewton, respectively, per square meter).

10.

A stress/strain curve shows the amount of force or tension (stress) that is required to produce a given stretch or elongation (strain). A stress/strain diagram can be developed for any metal. A simple stress/ strain diagram for carbon steel is shown above. The two most often used mechanical specifications are yield point and ultimate tensile strength. The yield point represents the force or elongation at which a permanent change in the metal begins to occur. It generally is expressed in pounds of force per square inch of cross section area; although, it also could be expressed as a percent elongation. Ultimate tensile strength describes the point at which failure of the metal occurs. Failure might be fracture during a shearing operation, which is desired, or failure of a structural component in a bridge. It also generally is expressed in pounds per square inch (psi) of cross section.

The straight section of the stress/ strain curve, up to the yield point, is called the elastic range. This is a very descriptive term because material in this range acts in an elastic manner--like a rubber band. Elongation or stretch is directly proportional to force. If a force is exerted, the material deflects or elongates, and then returns to its original form when that force is removed. Doubling the force will double the elongation or strain.

Within the elastic range, the ratio of force or stress divided by the elongation or strain is called the modulus of elasticity. It is essentially the same for all carbon steels. If the steel has been hardened or if it includes more points of carbon, the yield point will be increased; but the ratio of force to elongation, the modulus of elasticity, or slope of this part of the stress/strain curve will not change.

Aluminum is less stiff or rigid and more springy or elastic than steel. For any given loading, aluminum having the same cross section will deflect more than steel. For a coil of aluminum with the same thickness and yield strength as steel, a real possibility, the necessary structural integrity or strength of the equipment would be the same. But, since the aluminum would have to be stretched farther for that same loading, it requires more horsepower than needed for steel.

The section of the curve following the yield point is called the plastic range, as shown below. This is another descriptive term because the material acts in a plastic manner when it is forced to stretch past its yield point. In this range, the metal begins to achieve a positive and permanent change in form.

This is the working range of stress or strain for leveling, flattening, drawing, roll forming, bending, or any other forming application. You must be somewhat beyond the yield point but not up to the ultimate tensile strength or failure point to form metal. Below the yield strength, there is no permanent change. Exceeding the tensile strength causes the metal to fracture. If the metal is subject to a force causing an elongation somewhere in the plastic range and that force is removed, the material will relax slightly. Some permanent change in shape will remain. For example, as shown below, stretch the material with a given force to a certain point (a) in the elastic range. When that force is returned to zero, the strain or amount of elongation (b) will follow a straight line from the maximum point, parallel to the original elastic line and down to the appropriate remaining "set" (c).

The difference between the elongation when the force is placed on it and the elongation after that force has been removed is referred to as spring back. In order to achieve the desired change in shape, we must 1) exceed the yield point and 2) stretch the material farther than we want it to be when we are finished in order to allow for spring back.

If we have two metal strips of unequal length and we want to make them equal, it is necessary to stretch the short one past the yield point, so that it is longer than the longer strip in order to allow for spring back.

Therefore, it is important that we consider quantities that characterize a material response to axial loads without depending on the geometry of the bar. This is done by considering force per unit area of the cross-section and change in length per unit length of the bar. For the bars shown in Fig. 2, if we now plot force per unit area versus displacement of the right end divided by the total length, it will be found that for all the three cases, the curves fall on the same straight-line as shown in Fig. 3.

Stress-Strain Curve

As long as the applied load is sufficiently small, the load-displacement curve and the stress-stress curve are linear. To study the response of a bar when subjected to an arbitrary load P, let us consider a bar as shown in Fig. 5 subjected to a load P. As the load is gradually increased, the bar elongates. If the normal stress is plotted against normal strain (defined in Fig. 5), the resulting curves appear as shown in Figs. 6 and 7 for ductile materials.

Some examples of ductile materials are structural steels and aluminum alloys. As the load P is increased, the bar elongates and initially the stress-strain curve follows a straight line and the material behaves elastically (i.e., if the material is unloaded, no permanent deformation remains). When the stress reaches a certain value called the proportional limit

, the stress-strain curve ceases being a straight line. Under further loading, the stress reaches a critical value called the elastic limit at which the material ceases being elastic and the material begins to deform plastically. During this stage of deformation, if the bar is unloaded, only a certain part of the total deformation is recovered and the bar does not return to its original length. As the load is increased, the stress also increases until a critical value called the ultimate stress

is reached. After this, the diameter and hence the cross-sectional area begins to decrease in some region of the bar and failure by fracture occurs eventually.

The proportional limit and elastic limit are usually very close to each other and it is very difficult to determine these two precisely. Therefore, a quantity called the yield strength of the material is used to determine the onset of plastic deformation. This is defined by the stress that is required to produce a 0.2% permanent strain and obtained by passing through the strain=0.02 point on the strain-axis, a line parallel to the initial straight-line portion of the stress-strain curve.

The stress-strain curve for brittle materials is shown in Fig. 8. Some examples of brittle materials are concrete, glass, ceramics and cast iron. These materials exhibit little or no yielding before failure.

The distinction between ductile and brittle materials is somewhat arbitrary. Usually, a quantity called the percentage elongation is used to distinguish between brittleness and ductileness. Let

and

be the initial and final lengths of the specimen on which the tensile test is being conducted. Then the percent elongation is defined as

If the percent elongation exceeds 10%, then the material is called ductile. If the percent elongation is less than 10%, it is called brittle.

Since stress is force per unit area, it has the dimensions of newton per square meter (N/m2) in SI system and pounds per square inch (psi) in FPS system. Since stress is a frequently occurring quantity, the base unit Newton per square meter is given the name Pascal. Thus, in SI system, stress is specified in the units of Pascals. Since this quantity is quite small, stress is typically specified in terms of KPa (Kilopascals = 1000 pascals), MPa (megapascals = 10^6 Pascals) or GPa (gigapascals = 10^9 pascals). Similarly, in FPS system, stress is often expressed in terms of ksi (kilo pounds per square inch).

Since strain is a ratio of two length dimensions, it is dimensionless. Therefore, Young's modulus E has the same dimensions as normal stress.

13.
A stress due to forces produces a strain or a deformation. Stress is proportional to strain and that proportionality cosntant is called its modulus. Stress is the product of modulus and strain. Stress is the ratio of the force exerted on the object to the cross-sectional area over which the force is exerted. Strain is the resulting deformation, whether it is the ratio of change in length to original length, change in height to original height, or change in volume to original volume.

where F is the force, A is the cross-sectional area in m², E is Young's modulus, L is the original length, and D L is the change in length.

where F is the force, A is the cross-sectional area in m², G is shear modulus, h is the original height, and D h is the change in height.

where F is the force, A is the cross-sectional area in m², B is bulk modulus, V is the original volume, and D V is the change in volume.
Remember, the ratio of F to A is the pressure (P) of the fluid.

Young's modulus of elasticity can be thought of as a measure of how well a substance stands up to tension.

15.
Young's modulus of elasticity (E), also known as the elastic modulus, is the ratio between stress and strain: E = s /e ,
and has the same units as stress. E is the slope of the stress-strain graph: the steeper the slope, the stiffer the material.

Since ancient times, engineers have designed three major types of bridges to withstand all forces of nature.In the below diagrams the places where the materials is under compression is shown by the Red arrows and the places of tension are shown by the Blue arrows.

When something pushes down on the beam, the beam bends. Its top edge is pushed together, and its bottom edge is pulled apart.

One feature of the suspension bridge which has given trouble in the past is the effect of wind on the roadway. One famous bridge started swinging in a strong wind some years ago and broke up completely. This was the bridge at Tacoma Narrows, in the northwest of the United Sates. Trouble can arise because the roadway, in section, is a relatively thin affair, flexible like the wing of an aeroplane. As the wind is funnelled past the bridge, between the banks, it is speeded up, and the roadway tends to lift, and fall back again, twisting as it is buffetted by the eddies cast off behind the sction alternately above and below. And as the wind speed is different for different parts of the roadway some parts lift and sag when others are moving the other way . This results in some twisting, which the road cannot endure, and failure occurs.

The classic failure of the Tacoma Narrows bridge is especially interesting because an amateur photographer happened to be there at the right time and took a film of the break-up. This film has been shown many times and I expect you will get a chance to see it one of these days. You will find it a stunning record, well worth the attention of a bridge-watcher. Planning a suspension bridge and building it can be to some extent a fairly straightforward matter in some ways and we have plenty of experience now, gathered over the past fifty years or so.

Although there are some truly tremendous suspension bridges to be seen you will like some of the smaller ones, too. The cables or chains make bold curves against the sky as they sweep up to the tops of the pylons and plunge down again to the anchorage points on the shore. When you walk across a small suspension foot-bridge you can often make it sway a bit, by swinging your weight this way and that and thus see just how the principle applies. It is probably better to do this when the only other people on the walkway are your friends, or you may get some black looks.

A Strut is a member that shortens under load. Struts must be designed to resist compression and buckling . You may sometimes see a strut composed of several slender members held wider apart in the middle than at the ends by a spacer, where lightness is important. Perhaps the most prominent struts of all are the pylons on suspension bridges which are really huge struts, resisting the heavy vertical loads imposed by the suspension cables. In framed structures the identification of struts by the bridge-watcher requires some thought.

Columns are vertical supports that carry a compressive load. Some of the Victorian columns in iron are bridge-watchers' gems, with fanciful decorations cast in to the shape. They are still to be seen in old bridges. The classic columns in masonry are associated with the Greek and Roman eras. The plainest of these was the Doric, a sturdy column capable of carrying large loads. These could therefore be spaced more widely than the more delicate and decorative styles such as the Corinthian. You can identify the Corinthian style from the leaves at the top, and this is the kind most often chosen by the old bridge designers working in iron. Wonderful brick columns were used in some of the early railway bridges, those of long span in Cornwall being particularly impressive. The modern concrete column tends to be plain and severe in treatment. The principal requirement of a column is that it shall neither crush nor buckle under load.

Arches are curved members or structures, using material in compression, the centre part rising above the level of the ends. They may be pinned at one end, at both ends, or at both ends and in the centre. The pinning of an arch rib ensures that there is no bending effect at that point, and permits movement to accommodate alterations of dimensions due to temperature changes. The chief feature of an arch bridge, apart from it being a handsome structure, is that at the abutments there are horizontal forces as well as vertical ones, so that the resultant reactions at the abutments are inclined to the vertical. The invention of the arch was a wonderful forward step in structural design.

18.
What you build a structure out of is just as important as how you build it! Different materials have vastly different properties.

Wood Properties
Type: Spruce (softwood)
Wood Pros+Cons
Strengths: Cheap, lightweight, moderately strong in compression and tension.
Weaknesses: Rots, swells and burns easily
Wood Applications: Bridges, houses, two- to three-story buildings, roller coasters
Compression: You squeezed this block easily, but it took a lot of effort to make it break. Wood is cheap and pretty strong in compression. That's why people build houses out of wood!
Tension: It wasn't easy to break this block of wood because wood is strong when you pull it in the direction of its fibers. It would have been three times easier for you to break this block if you'd stretched it from the top and bottom, across the direction of its fibers.

Plastic Properties
Type: High-strength plastic fabric Ingredients: Long chains of molecules
Plastic Pros+Cons
Strengths: Flexible, lightweight, long-lasting, strong in compression and tension
Weaknesses: Expensive
Plastic Applications: Umbrellas, inflatable roofs over sports arenas
Compression: Compared to steel, you squeezed this plastic block easily, but it took a lot of effort to make it break. The long chains of molecules that make up plastic can be pulled and pushed in many directions without failing.
Tension: You stretched this plastic pretty far before it finally broke. The long chains of molecules that make up plastic can be pulled in many directions without snapping. That's one of the reasons why circus tents are made of plastic fabric!

Aluminum Properties
Type: Aluminum alloy Ingredients: Aluminum with magnesium & copper
Aluminum Pros+Cons
Strengths: Lightweight, doesn't rust, strong in compression and tension
Weaknesses: Expensive
Aluminum Applications: Airplane wings, boats, cars, skyscraper "skin"
Compression: It was pretty hard for you to break this aluminum block. That's because the magnesium and copper inside this block makes it almost as strong as
steel!
Tension: It wasn't easy to break this aluminum block. That's because aluminum, when combined with metals like magnesium and copper, is almost as strong as steel!

Brick Properties
Type: Ordinary brick Ingredients: Burned clay
Brick Pros+Cons
Strengths: Cheap, strong in compression
Weaknesses: Heavy, weak in tension
Brick Applications: Walls of early skyscrapers and tunnels, domes
Compression: You had to push this brick very hard to make it crumble. Bricks are very strong in compression. That's why early houses were made of brick!
Tension: You pulled this brick apart easily! That's because bricks are very weak in tension.

Concrete Properties
Type: Fine-grain concrete Ingredients: Cement, water, small stones
Concrete Pros+Cons
Strengths: Cheap, fireproof and weatherproof, molds to any shape, strong in compression
Weaknesses: Cracks with temperature changes, weak in tension
Concrete Applications Early arch bridges and domes Example: Pantheon - Rome, Italy
Compression: You had to squeeze this concrete block really hard to make it break. That's because concrete is very strong in compression.
Tension: You pulled apart the small stones and cement in this concrete block easily. That's because concrete is weak in tension.

Reinforced Concrete Properties
Type: Fine-grain concrete with high-strength steel Ingredients: Steel bars hidden in concrete
Reinforced Concrete Pros+Cons
Strengths: Low cost, fireproof and weatherproof, molds to any shape, strong in compression and tension
Weaknesses: Can crack as it cools and hardens
Reinforced Concrete Applications: Bridges, dams, domes, beams and columns in skyscrapers
Compression: You had to squeeze this block really hard to make it break. That's because concrete and steel are both very strong in compression.
Tension: It was hard to pull this concrete block apart because the steel bars inside make it very strong in tension. That's why some of the tallest
skyscrapers in the world are made of reinforced concrete.

Steel Properties
Type: High-strength steel
Ingredients: Iron with a touch of carbon
Steel Pros+Cons
Strengths: One of strongest materials used in construction, strong in compression and tension
Weaknesses: Rusts, loses strength in extremely high temperatures
Steel Applications: Cables in suspension bridges, trusses, beams and columns in skyscrapers, roller coasters
Compression: You had to push extra hard on this steel block to make it bend and break. Steel is stronger than any other material in compression. That's why engineers choose steel beams and columns to support most skyscrapers.
Tension: You had to pull this block incredibly hard to make it break because steel is stronger than any other material in tension. That's why the cables in the Golden Gate Bridge are made of steel.

19.
Pre-stressing: Pulling on the reinforcement in a concrete beam before the concrete has been placed. The rods are thus under stress even before the load comes onto the bridge. As for post-stressed work, there is usually no visible indication in the finished bridge. The pre-stressing process means that when the member is complete and the pre-stressing forces removed, the rods return towards their former length, improving compressive loading on concrete otherwise in tension. This reduces the tendency to the formation of cracks on the tension face.

Post-stressing: Pulling on the reinforcing rods in a concrete beam after the concrete has set. This puts a higher stress in the rods and a compressive stress in the concrete to avoid tension in the concrete when the bridge is loaded. You are not likely to be able to tell by just looking whether or not this technique has been used, but you might see the forces being applied during construction. Usually the arrangements for applying post-tensioning are not visible once the span has been completed.

Created : 4th February 2002
Last modified : 4th February 2002
Author : Chad Silver email:Chaddysi@start.com.au
Site maintained by : Chad Silver

F_x = 0 F_x = - T_L cos 45^o + T_R cos 30^o = 0 - 0.707 T_L + 0.866 T_R = 0 0.866 T _R = 0.707 T_L	F_y = 0 F_y = T_L sin 45^o + T_R sin 30^o - 98 N = 0 0.707 T_L + 0.5 T_R - 98 N = 0 0.707 T_L + 0.5 T_R = 98 N
Now we can substitute,
	0.707 T_L + 0.5 T_R = 98 N 0.866 T _R + 0.5 T_R = 98 N ( 0.866 + 0.5 ) T_R = 98 N 1.366 T_R = 98 N T_R = 98 N / 1.366 T_R = 71.7 N
0.866 T _R = 0.707 T_L T_L = (0.866 / 0.707) T_R T_L = 1.22 T_R T_L = 1.22 ( 71.7 N ) T_L = 87.8 N

	In this case the lever arm is drawn and measured so that it is perpendicular to the force.
	In this case the force is broken down into components with one of them being perpendicular to the lever arm. The value for the perpendicular component is used as the force.
	In this case the force being applied and the lever arm are naturally perpendicular to each other. Any of the three formulas would work fine in this case.