TS ch04 Structures - HKEdCity · Technological Studies Structures 2 Structures 1 Introduction to...

Technological Studies Structures

1

Structures 1 Introduction to structures ................................................................................ 2 2 Forces in structure ............................................................................................ 2

(a) Compression ............................................................................................................................................. 2 (b) Tension ....................................................................................................................................................... 3 (c) Bending ...................................................................................................................................................... 3 (d) Torsion ....................................................................................................................................................... 3 (e) Shearing ..................................................................................................................................................... 4

3 Balance of structure ......................................................................................... 4 (a) Balance of force ...................................................................................................................................... 4 (b) Addition of forces ................................................................................................................................... 5 (c) Resolution of forces ............................................................................................................................... 6 (d) Forces in equilibrium ............................................................................................................................. 7 (e) Balance in moment of force ................................................................................................................. 9 (f) Conditions of equilibrium ................................................................................................................... 12

4 Materials and cross section ............................................................................ 14 (a) Characteristics of materials ................................................................................................................ 14 (b) Cross sectional area ............................................................................................................................. 15

5 Characteristics of structures ......................................................................... 15 (a) Limit deformation ................................................................................................................................. 15 (b) Stability ................................................................................................................................................... 16 (c) Strength .................................................................................................................................................... 17

(i) Enhancement of the strength of materials .......................................................................................... 17 (ii) Enhancement of the strength of the structure ..................................................................................... 18

(d) Different types of structure ................................................................................................................ 20 (i) Arch ................................................................................................................................................... 20 (ii) Frame ................................................................................................................................................. 20 (iii) Beam .................................................................................................................................................. 20 (iv) Box ..................................................................................................................................................... 21

6 Joining methods .............................................................................................. 22 (a) Permanent joining ................................................................................................................................. 22 (b) Semi-permanent joining ..................................................................................................................... 22 (c) Movable joining .................................................................................................................................... 22

7 Designing simple structures ........................................................................... 23 (a) Forces on the components .................................................................................................................. 23 (b) Cross sectional areas of components .............................................................................................. 24 (c) Factor of safety ...................................................................................................................................... 24

(i) Meaning of the factor of safety .......................................................................................................... 24 (ii) Cost efficiency ................................................................................................................................... 25

(d) Ways on strengthening structures..................................................................................................... 25

8 Tests for structure ........................................................................................... 25 (a) Loading of structure ............................................................................................................................. 26 (b) Strength of structure............................................................................................................................. 26 (c) Stability of structure ............................................................................................................................. 27 (d) Flexing of structure .............................................................................................................................. 28

Exercise………………………………………………………………………...30


2

Structures 1 Introduction to structures

A structure is composed of several light parts. It can be used to bear a large load. The parts forming the structure are called members. Many buildings and machines are made under the application of structure, for instance, pylon, house, bridge and crane.

(a) Pylon (b) Crane

Fig. 1

There are many advantages of using structure. Take the crane in Fig. 1b as an example. It is made up of rods of light metal. Therefore, it is lighter, easier to install or dismantle. Furthermore, it can support a very large load, so it is often used in constructions.

2 Forces in structure The members in a structure experience compression, tension, bending, torsion and shearing

when the structure is withstanding a load.

(a) Compression Compression shortens the axial length of the member when efforts are applied on both sides

along their axes of application. For example, a support is distorted by compression.

(a) Compression (b) Compressed spring

Fig. 2


3

(b) Tension Tension lengthens the axial distance of the member when efforts are applied on both sides

along their axes of application. The endurance of force on the cable of suspension bridge and the chains of swings are the examples of tension.

(a) Tension (b) Released spring

Fig. 3

(c) Bending When a force is applied to only one side, the member may bend. Bending makes the axial

distance of the member to experience the compression on the same side and tension on the opposite side, e.g. the bending of shelf in the bookshelf by compression.

Fig. 4 (a) Bending (b) Bending of shelf in the bookshelf by compression

(d) Torsion A torsion is formed by a couple of forces of the same magnitude but in opposite direction.

For examples, the turning of steering wheel in an automobile, opening of the bottle cap by hand, etc.

(a) Torsion (b) Turning of steering wheel in an automobile

Fig. 5


4

(e) Shearing Shearing is formed by a pair of forces with the same magnitude but in opposite direction.

The two forces are directed perpendicularly to the axis of the member, for instance, paper cutting by scissors, and steel plates are riveted by rivets.

(a) Shearing (b) Paper cutting by scissors

Fig. 6

3 Balance of structure (a) Balance of force

A force can change the motion state of objects. A force has magnitude and direction, so it is a vector and can be represented by a straight line and an arrow. The length represents the magnitude, while the arrow represents the directions. Fig. 7 shows various forces with different magnitudes and directions.

(a) Horizontal forces (b) Vertical forces (c) Inclined forces

Fig. 7

If a stationary object is pulled by a force F, it starts to move. F is called the unbalanced force.

Fig. 8 (a) Unbalanced force (b) Balanced forces

If both forces applied to the stationary object F and F’ have the same magnitude but in opposite directions, the object will remain stationary. The net force on the object equals to zero.


5

(b) Addition of forces The unit of force is Newton, denoted as N. For instance, an object of 1 kg mass has an

approximately weight of 10 N. Since a force is a vector quantity, during the addition of forces, magnitudes and directions should be noted.

Fig. 9 (a) (b)

Addition of the forces with 3 N and 4 N respectively in Fig. 9 can be made by drawing the two forces, using an appropriate scale (like 1 N is represented by 1 cm) and direction represented by arrows, as shown in Fig. 9b. Then construct a straight line from the starting point O to B, the straight line OB is the resultant force of 3 N and 4 N. For clear identification, the resultant force in the figure is represented by double arrows. This method is called the vector sum or graphical method. The length of the resultant force OB is 6.1 cm, i.e. 6.1 N in this case. The angle between the resultant force and the horizontal line, which can be represented as

In fact, the resultant force can be found by drawing a parallelogram. As shown in Fig. 10, the resultant force OA is the diagonal of the parallelogram OXAY. Therefore, this method is also known as parallelogram method.

Fig. 10 Fig. 11

The addition of vectors can be applied to the parallel forces. For instance, in Fig. 11, the resultant force is 4 –1 = 3 N to the right. We have used this method before to find the net force.

Addition of vectors can be applied to more than two forces. Fig. 12a shows the four forces F1, F2, F3 and F As shown in Fig. 12b, the forces are joined in scale by tip-to-tail, and their resultant force is F. The forces formed a polygon afterwards. This method is also called polygon method.

25.


6

Fig. 12 (a) (b)

(c) Resolution of forces Several forces can form a resultant force. Conversely, a force can also be resolved into

several component forces. This method is called resolution of force. In Fig. 13, the angle between the force F and x axis is . When the force F is resolved into Fx, which lies on the X-axis and Fy, which lies on the Y-axis respectively, we get

cosF

Fx sinF

Fy

Fx = F cos Fy = F sin

If in fig 13, F = 10 N， = 30，then

Fx = 10 cos 30 = 8.7 N

Fy = 10 sin 30 = 5 N

Fig. 13


7

(d) Forces in equilibrium When the resultant force is on an object is zero, it is said to be in equilibrium.

Fig. 15 (a) Three balanced forces (b) The balanced forces form a closed force diagram

Example 1

In a frame structure, two forces are applied to the same point O in the two members, as shown in Fig. 14a. Use the method of resolution of forces to find the resultant forces.

Fig. 14 (a) (b) (c)

Solution

Using the method of resolution of forces, the 3 N force is resolved into the horizontal and vertical forces Fx and Fy (Fig. 14b),

Fx = 3 cos60 = 1.5 N

Fy = 3 sin60 = 2.6 N

Net force in the direction of x axis = 1.5 + 4 = 5.5 N

Net force in the direction of y axis = Fy = 2.6 N

The resultant force as shown in Fig 14c is

N1.66.25.5 22 F

473.05.5

2.6tan

3.25

The resultant force is 6.1 N, which makes an angle 25.3 with the horizontal direction.


8

As shown in Fig. 15a, three applied forces lie on the same point. If their component forces make a closed triangle, that means that their resultant force is zero, i.e. in equilibrium (Fig. 15b). From this we know that when the forces are in equilibrium, the component forces will formed a closed force diagram. As in Example 2, the force diagram can be used to solve the problems in the balance of force. This is called the graphical method.

Example 2

Fig. 16 (a) (b) (c) Force diagram

In Fig. 16a, a cargo of 50 N is suspended by using two ropes.

(a) Construct a force diagram with all the tensions in the ropes.

(b) Find the tension T1 and T2.

Solution

(a) Fig. 16b shows the relationships between different forces. With the scale of 1 cm = 10 N. Draw the object which weighs 50 N. After that, draw T1 and T2 with appropriate angles and complete the force diagram (Fig. 16c). When constructing the force diagram, it is a common practice to construct the forces in a clockwise direction.

(b) Measure the force diagram in Fig. 16c, we can get:

T1 = 5.7 cm = 5.7 10 = 57 N

T2 = 2.9 cm = 2.9 10 = 29 N


9

(e) Balance in moment of force

Fig. 17 (a) Lesser effort (b) Larger effort

If we use the spanner to release a bolt, the shorter the distance between the bolt and the effort is, the greater force will be needed (Fig. 17). It is because the rotational motion of force is determined by the moment. In the rotation of spanner of Fig. 17, the pivot lies on the bolt, the definition of moment of force is:

Moment of a force = Force Perpendicular distance from that point

The distance between the applied force F and the pivot is called moment arm d. Therefore, the moment of force can be represented as:

Moment of force = Force Moment arm

or

M = F d

The unit of moment is Nm. In Fig. 17b, the moment arm is shorter than that of Fig. 17a, therefore a larger effort is needed.

Example 3

With reference to Fig 17a, if the applied force is 50 N, the moment arm is 0.3 m, find the

moment of force at the centre of the bolt.

Solution

Moment of force M = F d = 50 0.3 = 15 Nm (anticlockwise)


10

Fig. 18 (a) Clockwise moment (b) Zero moment

In Fig. 17, the anticlockwise movement of the spanner is caused by the moment of force. It is a common practice to set the anticlockwise moment as positive. In Fig. 18a, the clockwise movement of the spanner is caused by the moment of force, however, the movement is set to be negative. In Fig. 18b, the moment arm is zero as the effort pass through the pivot. Therefore, the moment of force is zero.

Fig. 19 (a) (b)

In Fig. 19a, if the force F is not perpendicular to OA, the moment arm of F about O (perpendicular distance) OX is

OX

d sin or OX = d sin

where d is the length of OA, and is angle between the extended line of F and OA, therefore,

Moment of force = Force d sin

The result is applicable to Fig. 19b


11

Fig. 21 (a) Unequal moment of force (b) Net moment of force = 0

In Fig. 21a, clockwise moment of force = 30 × 0.5 = 15 Nm, the anticlockwise moment of force = 50 × 0.5 = 25 Nm. As the moment of force in anticlockwise direction is larger, the plank rotates in anticlockwise direction. In Fig. 21b, clockwise moment of force is = 30 × 0.5 = 15 Nm, while the anticlockwise moment of force is = 50 × 0.3 = 15 Nm. Therefore the clockwise and anticlockwise moment of force is the same, the net moment = 0. The plank remains stationary. From this, it can be deduced that:

If an object does not rotate, the total clockwise moment = total anticlockwise moment

This is the principle of moment.

Example 4 In Fig. 20, a force of 80 N is needed to pull up a plank with 2 meter long. Find the moment of the tension about point O.

Solution

Moment = F d sin = 80 2 sin 40º = 103 Nm

Fig. 20


12

(f) Conditions of equilibrium If an object remains stationary and does not rotate, it is said to be in equilibrium. Therefore

the condition for equilibrium is:

Net force = 0

and

Net moment = 0

Example 5

In Fig. 22, a plank is used to carry goods with 50 N. If the

plank remains balanced, find the tension of the string T.

Solution

Take O as pivot.

Clockwise moment of force = 50 2 = 100 Nm

Anti-clockwise moment of force= T (6 sin 60 º)

According to the principle of moment , we know that

T (6sin60 º) = 100

T = 19.2 N

Fig. 22


13

Example 6 A 40 N cargo is put on a plank with 2 M long, as shown in Fig. 23. Find the reaction force RA and RB on the plank of the supports at A and B. Ignor the weight of the plank.

Fig. 23

Solution

Since the plank is in equilibrium, the conditions of equilibrium can be applied.

Equilibrium condition 2: Net moment = 0 or

Total clockwise moment = Total anti-clockwise moment

Moment about A:

40 0.4 = RB 2

RB = 8 N

Equilibrium condition 1: Net force = 0 or Total upward forces = Total downward forces

RA + RB = 40

RA + 8 = 40 RA= 40 8 = 32 N


14

4 Materials and cross section (a) Characteristics of materials

Fig. 24 Slings in Tsing Ma Bridge Fig. 25 Wooden structure (cover and frame of a building)

Loads in structures are mainly compression, tension, bending, torsion and shearing. In designing and constructing structures, appropriate materials should be used. The use of wrong materials may cause the damage or permanent deformation of the structures by external forces. For instance, steel cable can withstand a larger tension, but it cannot withstand any compression (Fig. 24). On the other hand, concrete can withstand a very large compression but can be broken down by tension easily. Table 1 lists the examples of appropriate materials for loads.

Load Applications Criteria for selection of materials

Common examples

Compression Bridge pier, building pier, etc

Better in hardness Metal, concrete

Tension Steel wire, slings in suspension bridge, etc

Better in toughness Metal, fibre (rope)

Torsion Rotational shaft, screw, etc

Better in hardness and toughness

Metal

Bending Bridge surface, shelf in bookshelf, beam, propeller shaft, etc

Better in hardness, toughness, elasticity and malleable

Wood, plastics and metal

Shearing Rivet, screw, etc Better in hardness Metal

Table 1 Example of appropriate materials for loads

Since different parts of a structure may withstand different loads, each part may need to be made of different materials. Take the structure of Fig. 25 as example, its frame is made of wood joined by metal screws.


15

(b) Cross sectional area If the cross sectional area of a material increases, it can withstand a larger force. Stress can

be calculated to compare the force exerted on two different cross sectional areas.

areasectionalCross

ForceStress

Fig. 26 Cylinders with different cross sectional area

In Fig.26,

Stress on cylinder A = 22

mN400m0.5

N200

Stress on cylinder B= 22

mN250m4

N1000

Therefore, in practice , the average force exerted on cylinder A is larger.

5 Characteristics of structures (a) Limit deformation

When a force is applied to members in a structure, the length will change. If the force applied is not too great, it will restore to its original length after the removal of the force. This phenomenon is called elastic deformation. However, if the applied force is larger than the elastic limit, when the applied force is removed, only part of the length will change. This is called plastic deformation. Plastic deformation changes the shape of the structure permanently. Therefore, plastic deformation should be avoided when designing structures.

Limiting deformation refers to the changes in the components of a structure within the range of elastic deformation. In this case, the components withstand the largest applied force, but will not lead to permanent deformation. This is the largest scope of change in the length and shape of a structure. The forces of each of the components are then calculated to find the limit deformation of the whole structure. In designing a structure, its limit of bearing the loads should be taken into consideration.


16

(a) non-obvious deformation (b) Limit deformation (c) Improved design

Fig. 27

Fig. 27 shows the effect of limit deformation to the structure. In Fig. 27a, a small load is put on the frame structure. The deformation is not obvious. In Fig. 27b, a load with maximum loading is placed on the plank that causes a very large limit deformation and makes the plank tilt very obviously. Although there is no damage in the structure, the load may slip on the floor. In Fig. 27c, after finding the limit deformation, a component with a larger cross sectional area is used or the plank is tilted upward to make the plank horizontal when a larger load is put on it.

(b) Stability

(a) Unstable frame structure (b) More vulnerable to deformation (c) More stable structure by external forces

Fig. 28

Structures joined by rods may not be stable. For example, in Fig. 28a, the rod are joined by the screw bolt and nut. Its shape is not stable. The whole structure can be changed to a shape as shown in Fig. 28b due to the external force. If a rod is added to the structure, the whole structure will becomes more stable as shown in Fig. 28c.

When a force is applied to a structure without permanent deformation, it is said to be a structure with high stability. Among various structures, triangular structure has a high stability. Therefore, it is widely used in machinery and architectural structures (Fig. 29). In designing structures, the use of the triangular structures can raise the stability.


17

(a) (b)

Fig. 29 Examples of triangular structure

Another way to enhance stability is to join the rods into 3-dimensional shapes. For example, trolleys in the supermarket are made of rods perpendicular to each other. Since each rod is bent into U-shaped and is fixed in the case of the basket, it will not deform easily. This kind of trolley is very strong, light, easy to clean and can be stacked up easily.

Most of the prams are made of tough frame structures. Each rod is fixed by the enclosure of the pram that can carry the baby safely. Besides the pram is not in use, it can be folded conveniently.

(c) Strength An object is said to be stiff if it can withstand a load without deformation. Therefore, the

strength of a metal is its ability to withstand external force without being damaged or deformed permanently. The strength of a structure is determined by its materials and structure.

(i) Enhancement of the strength of materials There are three ways to increase the strength of a material: (i) Heat treatment, (ii) Mixing, and (iii)

Laminating.

Heat treatment

Heat treatment is a method to improve the physical property by heating. Materials are heated to the required temperature and cool down so as to change the molecular structure or compositions in the surface area. Common methods are annealing, quenching, hardening, tempering and case hardening.

Fusion

Fusion is to make different substances with different properties into a single substance according to assigned proportion. In this process, chemical reaction occurs and the strength of the final product will be increased. For example, concrete is made from small stones, cement, sand and water. After these substances are fused and hardened, the final product becomes a strong material.


18

Mixing

Alloy is mixed from two or more metals in assigned proportions. In this process, no chemical reaction occurs. For example, brass is an alloy of copper and zinc. When zinc is added into copper, the strength of copper increases, but its elasticity will decrease. Brass is used widely to make cheap jewellery, key locks, water tap, pipe, navigation tools and simple bearings.

Lamination

Mixing up materials layer by layer forms lamination. For instance, the anti-compression ability of concrete is very high but it has a very poor tensile strength. If concrete is used to build a bridge, the bottom part of the bridge encounters a larger tension, and easily forms cracks (Fig. 30a). Addition of reinforcing steel to the concrete can increase its tensile strength.

Fig. 30 (a) Reinforcing steel (b) Reinforced concrete

Another example is the glass fibre plate. Glass fibre are bundled or made into yarn. Appropriate adhesion agent is then applied, such as polyester resin, to produce glass fibre plate. There is a variety of glass fibre plates. They are generally tougher than plastic materials. Glass fibre plates can be used to make canoe, the body of a speedboat or cars.

(ii) Enhancement of the strength of the structure

To make a strong and hard structure, besides the choice of materials, the cross sectional shape is an important factor. Take the cylindrical pillar and the beam as examples. They can withstand axial force without being bent (Fig. 31a). However, they can be easily bent with radial force (Fig. 31b).

(a) Withstanding axial force (b) Withstanding radial force

Fig. 31


19

Hollow cross section To prevent the inner column being bent, the design with a hollow cross section can be used.

Relative to the solid column with the same material, hollow column has a larger radius, and hence a larger radial force is needed to bend the column (Fig. 32). For instance, the column in the street lamp, the flagpoles in schools are usually made of hollow columns.

Fig. 32 Fig. 33

Folded cross section Another way to prevent the structure being easily bent is to fold the cross section into I-shaped,

L-shaped or T-shaped, as shown in Fig. 33. These cubic structures are not easy to bend. Many metal beams use these designs to construct flyovers, rooftops, shelves, air-conditioner support, crane frames, etc.

Fig. 34 Air-conditioner support

Folded cross section can be used to increase the strength of a plate. For instance, metal foil folded into cross section of triangular, rectangular or wave shapes can be used as a structure to build walls, containers and rooftop, etc.

(a) Folded metal foil (b) Container made of folded metal foil

Fig. 35


20

(d) Different types of structure

(i) Arch An arch is a structure to support load. The downward force of a load is transmitted to both

left and right through the design of arches (Fig. 36a). There are no joining in the structure, only putting components onto suitable positions can produce a strong structure. Materials commonly used for constructing an arch are stones, bricks and concrete. Arch is often used in Ancient Chinese and European architecture (Fig. 36b).

Fig. 36 (a) Structure of arch (b) Example of arch structure

(ii) Frame The frame is linked by the independent objects in a rod or in a

cylindrical shape. The advantages of the frame structure are that they are simple and light. Usually, they are used to carry heavy loads. The major components of a frame are the lever and joints (Fig. 37a). There are various examples for frames. Many buildings and machines are made of frames, for instance, pylon, building, beam and crane, etc (Fig. 37b).

(a) Structure of frame (b) Example of frame

Fig. 37

(iii) Beam Beam is used to bear pressure from top to bottom. The forces exerted on its cross section are

classified as shearing and bending. Commonly used beams include I-shaped beam, box shaped beam and L-shaped beam, etc. Table 2 shows the characteristics and applications of some beams.


21

Beam Diagram Characteristics Applications

I-shaped beam

both upper and lower surfaces can bear strong pressure

not easy to bend

Construction, highway collision rail

Box shaped beam

Stronger than I-shaped beam

Can withstand a larger load, especially torsion.

Construction,

pile

L-shaped beam

Weakest strength

use less materials

Support of air conditioner

Table 2 Examples of beam

(iv) Box Boxes are rectangular structures combined by lighter materials like paper, wood and metal foil,

etc (Fig. 38a). They can withstand a larger force from up to down, but can withstand a smaller horizontal compression force. Examples commonly seen are paper boxes in office, wooden boxes, steel shelves and wooden shelves (Fig. 38b).

(a) Rectangular structure (b) Wooden shelves

Fig. 38 (c) A container of a container car is an example of the box structure


22

6 Joining methods There are three types of joining, permanent joining, semi-permanent joining and movable

joining. (For all detailed joining methods, please refer to ‘2.4 Material joining’ of Chapter 2 Material Processing Technology.)

(a) Permanent joining The component is extremely difficult to separate if it is joined. It has a specific shape and

cannot move. Methods of permanent joining include adhesive, welding, soft soldering and riveting, metal welding, and jointing of wood by adhesive agent. Adhesive agents or rivets can join different materials.

(b) Semi-permanent joining Components have fixed shape if they are joined semi-permanently, but it is easier to separate

them. Common applications of semi-permanent joining include: screw, bolt , nut and knock-down fitting, etc. The characteristics of semi-permanent joining method include: wide areas of application; can be used to join materials from different components (such as wood and metal; timber and plastics.). For instance, use of screw bolt and nut to fix the components of the electrical appliances, use of screw to join the plank and metal shelves. If two or more components are to be jointed, the thickness should be considered.

(c) Movable Joining The components have a fixed movable space after joining. For example, the door can be

opened or closed after the screws are screwed at the back of hinges. Besides that, movable joining can be found in folded tables and door hinges (Fig 39a). Furthermore, tools like scissors and tongs are also examples of movable joining (Fig 39b). This joining can be used to join same or different materials, but it is better to join two components only.

(a) Door hinges (b) Scissors

Fig. 39 Examples of movable joints


23

7 Designing simple structures (a) Forces on the components

In a structure for carrying loads, different components experience different stress. To design the structures accurately, the forces experienced by each of the components need to be found. The force diagram shows all the forces on a frame.

Fig. 40 Frame structure with the load.

Fig. 40 shows a frame structure with the load. Its rods experience compression and tension respectively. Since the whole frame is in equilibrium, therefore

Load = Sum of two reaction forces = R1 + R2

Since forces applied to each of the joints in the frame are also in equilibrium. A force diagram can be used to find the magnitude of the forces. For instance, in the upper joint, the load and compression are balanced (Fig. 41a). If the load is known, using an appropriate scale to draw a force diagram, the compression force T1 and T2 can then be found.

(a) Balanced forces in the joint (b) Force diagram

Fig. 41


24

(b) Cross sectional areas of components When the force exerted on each component is found, its cross sectional area can be calculated.

The force exerted on the component per unit area is called stress.

Stress )( areasectionalCross

)( exertedForce

A

F (The unit of stress is Nm2)

The stress for the components should not be too large, or else plastic deformation is caused and it will break. Therefore, knowing the limitation of the stress on the components, the cross sectional area of a component can be calculated by using the formula above.

(c) Factor of safety (i) Meaning of the factor of safety

When designing the structures, the largest load that the structure can withstand must be considered. However, in practice, the load may increase accidentally. For instance, during a strike, a lot of trucks will stay on the flyover. This produces a heavy load. Therefore, when designing the structures, the increase in load should be in accordance with the factor of safety.

The definition of the factor of safety is as follows:

components on the exerted load actual the

damage toleads load thesafety ofFactor

Different materials have different factors of safety, for instance: the factor of safety of steel is 1.5 and the factor of safety of brick is 5.5, etc.

Example 7 A component in a structure can withstand a tensile force of 20 106 N. The bearable

stress is 5 106 Nm–2. Find the minimum cross sectional area of the component.

Solution

Stress = A

F

A

66 1020

105

26

6

m4105

1020

A

Therefore the minimum cross sectional area of the component should be 4 m2.


25

(ii) Cost efficiency The larger the factor of safety, the stronger the material and structure are needed. The

construction cost is then raised accordingly. However, when it comes to the matter of human life, the safety level should be of a high priority. Therefore, the cost is higher. For instance, the structures of nuclear power station, dam and loading lift, etc. should be strong and safe. Conversely, if the structure does not affect human life and properties, a lower safety level to save cost can be applied.

(d) Ways to strengthen structures When building structures, part of the components may withstand a large pressure.

Strengthening of the structures can be made by improvement in designs, for instance, the structure of the carrying baskets in supermarket can be made more stable with the strengthened edge (Fig. 42a).

(a) Carrying basket (b) Trolley

Fig. 42

Another example is the antenna support. In order to increase the strength of the structure, steel wire can be used to fix the antenna on the roof.

8 Tests for structure After designing and constructing the structures, it should be tested to see if it could meet the

criteria of the design. For massive structures, like a flyover, a model should be made for the experiment. For example, Fig. 43a shows a model of Tsing Ma Bridge, which is used to test for the changes on the surface of the bridge under different loadings.

Fig. 43 (a) Model of Tsing Ma Bridge (b) Tsing Ma Bridge

Common items in the testing for the structures include: the measurements of loading, strength, stability and flexing.


26

(a) Loading of structure When a frame structure withstands loading on itself, the loading of all components can be measured by using spring balance (Fig. 44a).

Fig. 44 (a) Turntable-type spring balance (b) Using of a turntable-type spring balance

In practice, several turntable-type spring balances can be mounted firmly on the suitable components as shown in Fig. 45. Afterwards, appropriate weights representing loading are added on the structure (or model) to exert forces on the components. By observing the readings of the turntable-type spring balances, the magnitude of the loading on each of the component is then found.

Fig. 45 A model for the measurement of the loading in a structure

(b) Strength of structure Excessive compressive force or tensile force exerted on the components of the structure may cause breakage of these components. However, the larger the cross sectional area of the component is, the greater the loading it can withstand. Thus stress (i.e. force applied on each unit area) is more suitable for representing the condition of loading on the component. Strength of a component is the maximum stress it can withstand, and the strength of a structure is then determined by the strength of all components.

Strength can be further categorized as tensile strength and compressive strength. Tensile strength represents the ability of a material to withstand the deformation under tensile force. It can be measured by using tensile force testing machine. The machine will continuously pull the testing component till it deforms and breaks with results recorded in the whole process. Fig. 46 shows a tensile force testing machine and the steel rod to be tested.


27

Fig. 46 (a) Tensile force testing machine (b) Steel rod under the tensile force test

Compressive force represents the ability of a material to withstand compressive force, and it can be measured by using compression testing machine. Compression testing machine operates by exerting compressive force on the component till deformation or breakage, with results recorded in the whole testing. Fig. 47 shows a compressive force testing machine and compressive force testing of concrete.

Fig. 47 (a) Compression testing machine (b) Concrete under the compression test

(c) Stability of structure If a structure is easily deformed or toppled, then the structure is said to be unstable. Fig. 48 shows a frame structure that is joined by bolts and nuts. It is an example of an unstable structure as it is deformed if a horizontal force is applied to it. If a rod is added to the structure to make it as a triangular structure, the newly formed structure becomes more stable. However, if an excessive rod (in dotted lines) is added to the structure, then it is considered to be redundant.

(a) Original shape (b) Deformed shape (c) More stable structure

Fig. 48 Unstable structures are easily subject to deformation


28

To test whether structure can easily deform, possible point of application of the force with its estimated direction and magnitude should be found. After estimation, appropriate force is then exerted on this point along the direction of application. Observe whether the structure is deformed or not. The stability of the structure is then estimated in accordance with the results of the testing.

Another testing for the stability is by using a vibrator. As shown in Fig. 49, a model is put on the vibrator. After switching on the power, it will continuously vibrate while the amplitude and frequency of the vibration can be adjusted. This method of testing can simulate the different levels of earthquake, and can be used to test the effects of earthquake on the stability of a building.

Fig. 49 A vibrator for the simulation of earthquake

(d) Flexing of structure When a structure is subject to the force applied to it, different degrees of deflection may occur among the components of the structure. This is known as flexing. A dial indicator can be used to measure the flexing of a simple structure. Fig. 50 shows a dial indicator which is a delicate device for precise measurement of flexing. When the probe of the dial indicator is pressed, it will then drive the mechanical parts inside the dial indicator to cause the pointer pointing at the right reading of flexing on the marking disc. By observing the reading indicated, we could know the degree of flexing of the structure.

(a) Dial indicator (b) Use of the dial indicator (c) Digital dial indicator

Fig. 50


29

As shown in Fig. 51a, the material tested is arranged to put on the pivot. A dial indicator is put at the bottom of the beam and is adjusted to zero. Put the weight on the beam and record the deflection of the dial indicator. The value is the degree of tilting. The test can be done as shown in Fig. 51b.

(a) Test for load (b) Test for the suspension arm

Fig. 51


30

Exercise

1. Give one example each of the following five types of forces (compression, tension, bending, torsion and shearing) exerted on a structure:

Forces Example

Compression force

Tensile force

Bending force

Torsion

Shearing

Table 3

2. In a pasture, a horse is tightened by two ropes. The forces exerted on the ropes are shown in Fig. 53. Find the resultant force by using resolution of forces.

Fig. 52 Fig. 53

3. After reaching the pier, a sailor on a ferry uses 60 N to pull up the plank for passengers to land. The length of the plank is 2 m long, find the moment of tension in the rope about the hinge at point O.

Fig. 54


31

4. List the materials used in five common types of the load, and give one example of each of the loading.

Load Example Material used

Compression

Tension

Torsion

Bending

Shearing

Table 4

5.

(a) State and elaborate three methods used for material reinforcement.

(b) Siu Ming and his teacher go to observe a bridge under construction. Siu Ming finds that there are numerous steel rods embedded in the concrete. He does not understand the function of these steel rods in the concrete. Can you explain to him?

6.

(a) In considering the strength of the structure, explain why a flag pole in school is made of a hollow column.

(b) Give two examples for enhancing the strength of the structure by changing its cross section.

7. State and explain the features of an arch structure with simple illustration.

8. In the following metallic box, identify the type of joints used at the three zones A, B and C (permanent, semi-permanent or movable joints). Explain your answers.

Fig. 55


32

Zone Joining method

Reason

A

B

C

Table 5

9. A boat anchors along the shore by using a rope. A tension of 10 105 N develops in the rope and the maximum stress of the rope is 15 105 Nm-2. Find the radius of that rope (assuming the cross sectional shape of the rope is circular).

10.

(a) Briefly describe a test for loading in a beam.

(b) If the beam breaks at the loading of 400 kg in the test, while its components can only bear a loading of 80 kg, find the factor of safety of this structure.

TS ch04 Structures - HKEdCity · Technological Studies Structures 2 Structures 1 Introduction to...

Documents

Transcript of TS ch04 Structures - HKEdCity · Technological Studies Structures 2 Structures 1 Introduction to...