Unit 4: Science and Materials in Construction and the Built Environment. Chapter 14. Understand how Forces act on Structures


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1 Chapter 14 Understand how Forces act on Structures 14.1 Introduction The analysis of structures considered here will be based on a number of fundamental concepts which follow from simple Newtonian mechanics; it is necessary that we first review Newton s Laws of Motion. The word Laws is often replaced with Axioms, as they cannot be proved in the normal experimental sense but are selfevident truths which are believed to be correct because all results obtained assuming them to be true agree with experimental observations. In 1687 Sir Isaac Newton published a work that clearly set out the Laws of Mechanics. He proposed the following three laws to govern motion: Newton s Laws of Motion Law 1: Law 2: Law 3: Everybody will continue in a state of rest or uniform motion in a straight line unless acted on by a resultant force. The change in momentum per unit time is proportional to the impressed force, and takes place in the direction of the straight line along the axis in which the force acts. Action and reaction are equal and opposite. Based on these laws we are able to define some basic concepts which will assist us in our analysis of structures. Chapter 14: Understand how Forces act on Structures Page 1
2 14.2 Equilibrium Equilibrium is an unchanging state; it is a state of balance. In the analysis of structures this will be achieved when the total of all the applied forces, reactions and moments equate to zero. In this condition, the structure will be in balance and no motion will occur. We will now consider three types of static equilibrium. Figure 1 shows an object, in this case a ball, placed on three differently shaped surfaces. i Potential Energy Gained ii Potential Energy Lost iii Potential Energy Constant Figure 1: A ball, placed on three differently shaped surfaces (Note: Potential Energy = change in energy if the body is displaced. Potential energy is the energy the body possesses by virtue of its position above a known datum, in this case the apex of the surface on which it rests.) In (iii) we have a neutral equilibrium position. The ball will remain at rest unless acted on by a force. The potential energy of the system is constant. In system (i), any movement of the ball will require a gain in potential energy. When released, the ball will try to achieve equilibrium by returning to its original position. In (ii) any movement will cause the ball to move. The shape of the surface on which it rests will further promote this movement. Relative to its original position, potential energy will be lost. Static equilibrium is achieved by having a zero force resultant. It is perhaps worth noting here that our early analysis of structures will be based wholly on the principles of statics alone. That is, force will be constant with respect to time. Hence we will consider the static analyses of structures. The study of structures subject to forces that vary with time is known as dynamic analyses. Chapter 14: Understand how Forces act on Structures Page 2
3 14.3 Force From Newton s Second Law, and since: Momentum = mass velocity Equation 1 We can derive an expression for force such that: Force = change in momentum per unit time = mass acceleration or F = m a Equation 2 where F = force (N) m = mass (kg) a = acceleration = (ms 2 ) Acceleration is a vector quantity since it has both direction and magnitude. It is a measure of change of speed (velocity) over time taken. We commonly look at the acceleration rates of cars as the time it takes to go from 0 to 60 miles per hour (mph) or 0 to 96 kilometres per hour (km/h), around 5 seconds for a Ferrari. This can be represented differentially as: Equation 3 where a = acceleration dv = change in velocity (a vector quantity, i.e. one that has both direction and magnitude) dt = change in time We also know that velocity is a measure of distance covered over time, e.g. at a velocity (or more familiarly a speed) of 96 km/h a car would cover 96 km in 1 hour or m/s. This can also be represented differentially as: Equation 4 where v = velocity ds = change in distance dt = change in time Chapter 14: Understand how Forces act on Structures Page 3
4 So acceleration can be expressed in terms of change in distance over time taken as, the second derivative of a distancetime graph. Equation 5 Acceleration (and thus velocity) can be rectilinear (in a straight line) or rotational. In order to determine forces on a structure we first need to consider the differences between weight and mass. The weight of an object is defined as the force acting on it due to the influence of a gravitational attraction, or gravity. Thus, attaching an object to a spring balance and noting the extension will enable us to determine its weight. From our knowledge of physics we know that within the elastic range, extension is proportional to force (Hooke s Law), and most spring balances are calibrated to read weight directly. Consider an object of mass m and weight W. If the object is held at a certain height above the earth s surface and is released it will fall to the ground. Its acceleration in this case, will be the acceleration due to gravitational force; this is normally denoted by g and taken to be 9.81 m/s 2. Since from equation 2: Force = mass acceleration The force acting on the object, that is its weight, will be: W = m g Equation 6 where W = weight m = mass g = acceleration due to gravity. Hence, we can derive the force (weight) of various objects by multiplying its mass by its acceleration due to gravity, e.g. an object of mass 1 kg will have a force of 1 kg 9.81 m/s 2 = 9.81 kgm/s 2 or 9.81 Newtons. The units of force are the Newton and are normally denoted as N. (Note: 100g is approximately 1 N The weight of an average apple). It is important to note that the acceleration due to gravity is not actually constant over the whole Earth s surface. This is due to the earth being ellipsoidal in shape. It is also interesting to note that the weight of a body on the Moon will be approximately onesixth of that on the Earth. The Chapter 14: Understand how Forces act on Structures Page 4
5 Unit 4: Science and Materials in Construction and the Built Environment mass of an object is therefore constant, whereas its weight will vary in magnitude with variations in gravitational intensity g. Having determined the force exerted by an object, some basic geometric properties may be defined. All forces are vector quantities, which means, that they have both magnitude and direction. They may therefore be the subject of vector addition. Consider the situation shown in figure 2. Tractor a 500N Post Figure 2: Force Vectors Tractor b Two tractors (seen here in plan) are used to remove (pull out) a post from the ground by exerting horizontal forces as shown. Both tractors are attached by cables to the post and exert forces of 500N in the directions indicated. In force vector terms we can represent our system as shown in figure 3. The forces are represented by straight lines, which can be drawn to scale, denoting both the direction and the magnitude of the force. Any suitable scale can be used to construct the diagram, however in most cases such a diagram, will not be necessary. (i) (ii) (iii) y x Post Figure 3: Graphical Representation of Forces Chapter 14: Understand how Forces act on Structures Page 5
6 It is possible to achieve the same overall result by replacing the forces exerted by tractors a and b with a single tractor c and therefore replacing the system represented in figure 3(i). In order to determine the direction and magnitude of force required by the single tractor we analyse our initial system. Using elementary geometrical relationships we can determine the magnitude and direction of the vector. This can be calculated using the Pythagoras theorem and standard Sine, Cosine and Tangent relationships. The analysis will be completed using normal Cartesian coordinates as shown graphically in figure 3(ii) and (iii). (Note the bar on top of the letters, for example, indicates a vector quantity.) Therefore to calculate the direction and magnitude of the new vector c we can use simple vector algebra: Direction: Magnitude: Thus we can replace our original system with that shown in figure 4. Tractor c Post 707.1N 45 o Figure 4: Single Force System Similarly we can break down a single force c into its mutually orthogonal components, a and b. From figure 3(iii) we find the vector: in the x direction in the y direction Force vectors can therefore be resolved into a resultant, or broken down into horizontal and vertical components. For this year, we will only consider 2D structures; that is, structures with both breadth and height only. The proposed coordinate system is shown in figure 5. Chapter 14: Understand how Forces act on Structures Page 6
7 y x z Figure 5: TwoDimensional coordinate system We will apply this coordinate system across the entire structure. The coordinate system will therefore be considered as global. In some methods of analysis the coordinate may be oriented to the individual member axis and will then be considered as a local system. For equilibrium in a 2D system we must ensure that the summation of the forces in the x direction, the summation of the forces in the y direction and the moments about the z axis equate to zero. Or: Equation 7 However, real structures exist in three dimensional (3D) space. In three dimensions our coordinate system will be that shown in figure 6. For stable equilibrium of a rigid 3D body at the origin, we must now consider six equations. Equation 8 Chapter 14: Understand how Forces act on Structures Page 7
8 y x Note: Clockwise moments taken as positive z Figure 6: Three dimensional Coordinate System Therefore, in two dimensional analysis we are only required to solve for three equations in order to ensure static equilibrium of a rigid body. For three dimensional structures the analysis is much more complex, requiring the solution of six equations, and will not form part of this years studies. Practice has shown that, in the formative years of study, it is easier to analyse structures using the global Cartesian coordinates concept and resolving forces into horizontal and vertical components in order to check for equilibrium. This may require the resolution of a number of concurrent forces in order to determine the total horizontal and vertical forces applied at a particular position on the structure. Consider the two forces applied at point T as shown in figure 7. Chapter 14: Understand how Forces act on Structures Page 8
9 + y  x Point T R horizontal R vertical + x P Vertical P horizontal  y Figure 7: Concurrent Forces applied to a joint In order to simplify our analysis we will resolve each force into its horizontal and vertical components and then sum the results to find the resultant horizontal and vertical forces. Hence: For vector (force) R For vector (force) P R vertical = R Sin P vertical = P Cos R horizontal = R Cos P horizontal = P Sin Hence resultant forces are: Resultant vertical force Resultant horizontal force = + R Sin  P Cos = + R Cos  P Sin Note that the positive and the negative signs are generated in normal Cartesian coordinates by the force directions shown in figure 7. Also note that we have dropped the bar convention on the vectors to simplify the equations. We can therefore resolve any number of forces into a single horizontal and a single vertical component by adding all horizontal and vertical forces respectively acting at the point under consideration. Chapter 14: Understand how Forces act on Structures Page 9
10 Practical Example 1 Consider the point J shown in figure 8 and the forces applied to it. Determine the magnitude and direction of the horizontal and vertical forces H and V required to ensure equilibrium. Hence or otherwise, calculate the single force and direction required to ensure equilibrium. Force N 30kN Force K 15kN 45 o Point J 50 o 20 o Force H Force M 10kN 40 o Force V Force L 25kN Figure 8: Forces applied at point J Such problem can be solved mathematically or graphically. Mathematically will involve, solving each component by its horizontal and vertical component and hence add all vertical components and horizontal components by using the Cartesian properties. Graphically will involve drawing two components at a time up to scale and by using the parallelogram of forces one will find the resultant force and direction. For such methods and answers, follow the teachers work during the lesson. Chapter 14: Understand how Forces act on Structures Page 10
11 Exercise 14a Choose the correct answer. Mark a very good in the appropriate box. 1 a) stable A funnel that is balanced upright on a table on its narrow tip is in equilibrium. b) unstable c) neutral 2 a) stable An inverted funnel placed on a table is in equilibrium. b) unstable c) neutral 3 a) stable A funnel lying horizontally on its side at rest on a table is in equilibrium. b) unstable c) neutral 4 a) True b) False Two equal and opposite forces acting at a point are in equilibrium. 5 a) True b) False The resultant of two forces acting at a point can be determined by using the Parallelogram of Forces. 6 There are fundamentally kinds of equilibrium. a) four b) two c) three Chapter 14: Understand how Forces act on Structures Page 11
12 7 Three forces in equilibrium acting at a point can be represented in magnitude and direction by the sides of a triangle taken in order. a) true b) false 8 A pendulum at rest is in equilibrium. a) stable b) unstable c) neutral 9 A force of 3 N acts on a small body pushing it vertically upwards. Simultaneously, another force of 8 N acts on it pushing it vertically downwards. The resultant of the two forces is. a) 5 N downwards b) 5 N upwards c) 11 N downwards d) 11 N upwards 10 Two forces acting at right angles, which have the same effect as a single force, are called. a) resultants b) components c) constituents 11 a) stable A sphere that is at rest on a table is in equilibrium. b) unstable c) neutral 12 a) stable A chalkstick that is balanced upright on a table is in equilibrium. b) unstable c) neutral Chapter 14: Understand how Forces act on Structures Page 12
13 Exercise 14b 1. Determine whether or not a net force exists in the following situations. Description of Motion Net Force: Yes or No? 2. Freebody diagrams for four situations are shown below. For each situation, determine the net force acting upon the object. Chapter 14: Understand how Forces act on Structures Page 13
14 3. Freebody diagrams for four situations are shown below. The net force is known for each situation. However, the magnitudes of a few of the individual forces are not known. Analyse each situation individually and determine the magnitude of the unknown forces. Then click the button to view the answers. Exercise 14c Choose the correct answer. 1 Which of Newton's Three Laws does the following statement satisfy? The relationship between an object's mass (m), its acceleration (a), and the applied force F is F = ma. Acceleration and force are vectors. This law requires that the direction of the acceleration vector is in the same direction as the force vector. a) Newton s first Law b) Newton s second Law c) Newton s third Law d) All of the above 2 Which of Newton's Three Laws does the following statement satisfy? For every action there is an equal and opposite reaction. a) Newton s first Law b) Newton s second Law c) Newton s third Law d) All of the above Chapter 14: Understand how Forces act on Structures Page 14
15 3 Which of Newton's Three Laws does the following statement satisfy? Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. a) Newton s first Law b) Newton s second Law c) Newton s third Law d) All of the above 4 Which of Newton's three laws does the following example illustrate? If you have a hockey puck sliding along a table, it will eventually come to a stop. a) Newton s first Law b) Newton s second Law c) Newton s third Law d) All of the above 5 Which of Newton's Laws does this situation represent? Imagine a ball moving in a straight line directly toward when another ball collides with it. The moving ball exerts a force on the ball at rest. This causes the ball at rest to accelerate. However, the ball at rest also exerts the same magnitude of force (in the opposite direction) of the moving ball. This will cause the moving ball to decelerate or even move in another direction. a) Newton s first Law b) Newton s second Law c) Newton s third Law d) All of the above Chapter 14: Understand how Forces act on Structures Page 15
16 6 In the following example, what are the forces that are acting on the ball? Check all that apply. If a ball is thrown in the air, it will keep going the same velocity unless a force changes the velocity (speed and direction). a) Air friction b) Gravity c) Resistance of the ground d) Mass of the ball 7 Which law states the need to wear seatbelts? a) Newton s first Law b) Newton s second Law c) Newton s third Law d) All of the above 8 was the scientist who gave us the Laws of Motion a) Albert Einstein b) Michael Faraday c) Isaac Newton d) None of the above 9 What is another name for the Newton's first law of motion? a) Law of acceleration b) Law of velocity c) Law of inertia d) Law of mass Chapter 14: Understand how Forces act on Structures Page 16
17 Exercise 14d 1. A block of mass 2kg is pushed along a table with a constant velocity by a force of 5N. When the push is increased to 9N what is; a) the resultant force b) the acceleration? 2. How much net force is required to accelerate a 1000 kg car at 5m/s 2? 3. If you apply a net force of 1 N on 200g book, what is the acceleration of the book? 4. What is the net force on 200 g ball when it hits a wall with acceleration of 10 m/s 2? 5. What is the mass of an object that has a weight of 115 N on the Moon? The gravity of the Moon is 1/6 of g (which is 9.8 m/s 2 ). 6. What is the normal force acting on a 70kg person on the Moon? 7. A car is moving at a constant velocity of 20 km/h (5.56 m/s). How much net force is required to raise its velocity to 50 km/h (13.89 m/s) in 30 seconds? Suppose the car has a mass of 150 kg. 8. On Planet X, a 70 kg object can be lifted by a force of 400 N. a) What is the acceleration of gravity on Planet X? b) How much force is required if the same object is lifted on Earth? c) Suppose your car was taken to Planet X. If the car has a mass of 1500 kg, what would its weight be? 9. Calculate the weight, in kn, of each of the following two people. a) A young woman with a mass of 70kg. b) A middleaged man with a mass of 95kg. What would be the weights of each of these people on the moon if the gravitational acceleration on the moon is onesixth of that on earth? 10. Calculate the mass of a brick of length 215 mm, breadth mm and height 65 mm if its density is 1800 kg/m 3. What would be the weight of this brick? Chapter 14: Understand how Forces act on Structures Page 17
18 11. Calculate the weight of a 9 metre long reinforced concrete beam of breadth 200 mm and depth 350 mm if the unit weight of reinforced concrete is 34 kn/m As we will see in later chapters, the term live load is used to describe nonpermanent load within a building that is, those loads due to people and furniture. If a university classroom is 12 metres long and 10 metres wide and is designed to accommodate up to 60 students, calculate the live load in the classroom when full. (Note that you will have to make an assessment of the weight of an individual student, desk and chair.) Compare your answer with the British Standard value of live load (3.0 kn/m 2 ) for classrooms. 13. An international hotel chain plans to upgrade its hotel in a particular glamorous and exotic location by installing a rooftop swimming pool on top of its existing highrise bedroom block. The swimming pool will be 25 metres long and 10 metres wide and will vary uniformly in depth from 1 metre to 2 metres. Calculate the volume of water in the pool. If the unit weight of water is 10 kn/m 3, calculate the weight of water in the pool, in tonnes. If a small modern car weighs 1 tonne, calculate the number of cars that would be equivalent, in weight, to the water in the proposed swimming pool. If you were appointed as structural engineer for the project, what would be your initial advice to the architect and client? Chapter 14: Understand how Forces act on Structures Page 18
19 Exercise 14e 1. For the system of forces shown in the figure below: a) Calculate the horizontal and vertical components for each of the forces. b) Determine if the given system of forces are in static equilibrium. c) Solve this problem graphically to verify if the answers produced in (a) and (b) match. Force N 19.8kN + y Force K 30.23kN 25 o Point T 40 o 13.5 o 20 o + x Force M 72.8kN Force L 59.6kN Chapter 14: Understand how Forces act on Structures Page 19
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