Introduction Aim To introduce the basic concepts in mechanics, especially as they apply to civil engineering.

Transcription

1 Mechanics Introduction im To introduce the basic concepts in mechanics, especially as they apply to civil engineering. Mechanics Mathematical models describing the effects of forces and motion on objects (structures, machines, etc). In constructing mathematical models we make assumptions and approximations. It is important that you are clear about the assumptions being made and the limitations of the approximations. Outline Statics B Kinematics Dynamics D Energy E Rotation Oscillations In this unit we will concentrate on the behaviour of particles and rigid bodies in two dimensions (typically height and one horizontal dimension). We will use SI units throughout but it is useful to have some general knowledge of other systems of units. Staff Dr G Lane-Serff: sections - and unit coordinator. Extn 64602, room P/B20, gregory.f.lane-serff@manchester.ac.uk Dr D D psley: sections D- Extn 63732, room P/15, d.apsley@manchester.ac.uk ssessment oursework (laboratory work and problems): orces and moments 3% Structures 3% Momentum 4% Energy 5% Rotation and oscillation 5% Exam: our out of six questions 80% (Two from three on sections -, two from three on sections D-.) It is important that you develop the habit of working in a clear and logical way, setting out your calculations so that they are easy to follow and check. Marks will be deducted for inadequate working (even if you get the correct answer). Books Meriam JL and Kraige LG, Engineering Mechanics (volume 1 Statics, volume 2 Dynamics) 531 Beer P and Johnston ER, Mechanics for Engineers (volume 1 Statics, volume 2 Dynamics) Both have lots of example questions: there s no substitute for plenty of practice! These and many other useful books (e.g. Young HD, undamentals of mechanics and heat) can be found in the Joule Library, classmarks: mechanics: applied; 531 mechanics: physics. G Lane-Serff 1 10-Sep-10

2 Statics 1 orces orce: action on a body Newton s irst Law body remains at rest or moves with constant velocity unless acted upon by a net force. orces are vectors, having magnitude (size) and direction; written in various ways, eg or or. To completely describe a force acting on a body we need to know its magnitude, direction and point of application (where the force is applied). Internal orces: orces within the body. Important in considering strength (failure) elasticity and plasticity (bending and deformation) but not studied in this lecture course. External forces: orces on the body under consideration. Rigid bodies: Bodies that stay the same shape (don t deform) under the action of forces. Principle of Transmissibility (Line of ction) = or a rigid body, a force can be treated as being applied at any point along its line of action (this doesn t have to be within the body). dding forces orces applied at the same point 1 R = R parallelogram rule 2 orces applied at different points (on a rigid body) Use principle of transmissibility to move forces along their lines of action to the point where they meet. G Lane-Serff 2 10-Sep-10

3 dding parallel (or nearly parallel) forces. irst add two equal and opposite forces (acting along a common line) 1 - R 1 2 R 2 Then find the resultant force (this can be moved back along its line of action if desired). R 1 R 2 R 2 R = R 1 + R 2 R 1 omponents force can be decomposed into two components along any two non-parallel directions by constructing the appropriate parallelogram. It is often convenient to use rectangular components. = x + y y θ x = cosθ y = sinθ = 2 2 x + y x or = x i + y j i unit vector in x-direction j unit vector in y-direction (k unit vector in z-direction) tanθ = y / x θ = tan -1 ( y / x ) The resultant of two forces can be found be summing the components separately: y R = R R x = 1x + 2x R y = 1y + 2y x e.g. 1 = (2,3), 2 = (4,-2) 2 R = (6,1). Provided the axes are perpendicular, you can make your own choice of axes to simplify the calculations. G Lane-Serff 3 10-Sep-10

4 2 Moments and couples Moment of a force about an axis is the tendency of the force to rotate an object about that axis. M The force is applied at right angles to the spanner handle. The magnitude of the moment about the centre of the bolt is: d M = d If the force is not perpendicular to the line joining the point of application to the axis, we need to find the perpendicular distance from the line of action to the axis. d M r α M = d or M = r sinα The usual sign convention follows the right hand rule, with a positive moment for anticlockwise motion. The moment can be represented as a vector, with the direction of the vector being the axis of rotation. or the example above M = d k, where k is the unit vector in the z-direction. or 2D problems the rotation axis is always perpendicular to the plane we are working in, and so the sign is all that is needed to define the sense of the rotation. or 3D problems the moment vector is given by M = r or r (Not r which equals -r.) s with forces, where we needed to state the point of application, a complete description of a moment requires its magnitude, direction and position of the axis (this is simply a point in 2D problems). Units of moment are N m (force distance). Varignon s Theorem The moment of a force about a point (axis) is equal to the sum of the moments of the components of the force about the same point. (The components need not be rectangular components.) R = r R = r ( ) = (r 1 ) + (r 2 ) Moment about d 2 r 1 R d R = d d 2 2 d d 1 2 G Lane-Serff 4 10-Sep-10

5 ouples onsider a pair of equal and opposite forces acting on points separated by a perpendicular distance d. (It is not possible to calculate a single resultant force from such a pair: try it!) - B d Total moment about is 0 + d = d Total moment about B is d + 0 = d Total moment about is -x + (x + d) = d x The moment of a couple is the same about all points. couple is a free vector. orce-couple systems ny force acting on a body tends to push/pull in the direction of the force and tends to rotate about any axis not on the line of action of the force. We can replace any force at one point by a force and couple applied at a different point. This can be useful when finding the reaction at a support. d B 1) dd equal and opposite forces at B. 2) Then these two forces can be treated as a couple - B M = -d (clockwise) M B So the force applied at is equivalent to a force and couple applied at B. G Lane-Serff 5 10-Sep-10

6 3 Example session: orces and moments Resultant of several forces at a point. Moments. G Lane-Serff 6 10-Sep-10

7 4 Equilibrium and free body diagrams Equilibrium or a body to be in equilibrium, the net effect of the forces on the body must neither tend to make the body move nor rotate. Thus the resultant force R and resultant couple N must both be zero. R = = 0 N = M = 0 In two dimensions, with forces in the x- and y-directions and all moments are about an axis perpendicular to the plane, this can be written as x = 0 y = 0 M = 0, where the M are all the moments about any point (which may be on or off the body). Other versions of these equilibrium equations are x = 0 M = 0 M B = 0, where the points and B must not lie on a line perpendicular to the x-direction, and M = 0 M B = 0 M = 0, where the three points, B and must not lie on a straight line. In tackling a given problem, it is worth considering which choice of equations, coordinate system and points will give the simplest set of equations to solve. ree Body Diagrams 1) Define: In considering a body in equilibrium it is first necessary to define the body under consideration. This may be a composite body consisting of several elements. 2) Isolate: Draw a diagram that represents the complete external boundary of the body. 3) orces: Represent all the forces acting on the body. Supports and connections Support or connection pin orces/couples supported forces, two components R y (no couple) R x rollers normal force only N smooth contact normal force only N rough contact normal and friction N cable tension (in direction of cable at body) T fixed forces and couple y x M mass under gravity m weight W = mg (down) W G Lane-Serff 7 10-Sep-10

8 Examples Mass at mid-point on beam (length L) ree body diagram m B x W = mg x-component forces y-component forces moments about mid-point (or use or B) x = 0 y + By W = 0 -½L y + ½L By = 0 y By inal result x = 0, y = By = ½ W = ½ mg Simple structure with a cable T=10N T=10N 2m 2m B x 60 y By x-component forces y-component forces moments about x + T = 0 y + By = 0 2 By 2 sin 60 T= 0 inal result x = -10 N, y = - By = -10 sin 60 = N Structure with cable and mass cable T 60 2m 2m x mass m y 4m W=mg moments about x-component forces y-component forces 2T 4W = 0 x T sin 60 = 0 y + T cos 60 - W = 0 inal result T = 2W = 2mg x = T sin 60 = 1.73 mg y = W - T cos 60 = 0 G Lane-Serff 8 10-Sep-10

9 5 Structures Statically determinate structures These are structures where the equilibrium equations are sufficient to calculate all the forces. We will concentrate on statically determined structures in this course, but you should be aware that not all structures are statically determinate. Statically indeterminate x m Bx B With the beam pinned at both ends, the x-component of the forces gives x = - Bx y Inadequate constraints By and there is no way to calculate the undetermined compression/tension in the beam. m B In this case there is still an unknown tension in the beam but also there is no way of counteracting the moment about. (Plane) Trusses Trusses are made up of simple two-force members (Equal and opposite forces.) orces on the members: T Tension or ompression T We will ignore (for now) any bending or deformation of the members and we will usually consider them to have negligible mass (weight). Members under tension could be replaced by cables. The basic unit for plane trusses is the triangle. Method of Joints We consider the forces acting on the connecting pins (these are in the opposite direction to the forces acting on the members). T forces on the connecting pin The resultant force on each pin must be zero. G Lane-Serff 9 10-Sep-10

10 Example of a plane truss 2m 2m L 2m B 1) Draw a free body diagram alculate the external forces Bx = 0 y = By = Bx y L 2) Use method of joints Work round the truss joint by joint, finding equilibrium equations (in e.g. x- and y-directions) for each joint. By using a sensible order, the number of simultaneous equations to be solved can be minimized or eliminated. By orces on the pins (joints) L 0 dd the forces to your diagram as you calculate them. Equations at joint B gives compression in vertical member, no force in horizontal member. L/ 2 L/ 2 t joint, the y-direction equation implies a compressive load in the diagonal member, which in turn implies tension in the horizontal member (from the x-equation). L/ 2 T L 0 T L/ 2 L/ 2 Most of the rest of the forces can be calculated from the joint above B. L/ 2 The force in the final member can be calculated from the joint where the load is applied, and checked with the joint vertically above this. L/ 2 T T L 0 T L/ 2 G Lane-Serff Sep-10

11 6 Example session: Statics Structures G Lane-Serff Sep-10

12 B Kinematics B1 Velocity and acceleration: rectilinear motion Rectilinear motion: motion in a straight line (1D) s s=0 s s+ s t t+ t Position of object at time t is s(t), at time t+ t the object is at s+ s. lim s Velocity, v = t 0 t = ds dt = ṡ acceleration, a = lim v t 0 t = dv dt = v. = d2 s dt 2 = ṡ. s v t s t v t an also make use of a = d dt (v) = ds dt t d ds (v) = v dv ds v t 1 t 2 t s 2 t 2 ds = vdt s 1 t 1 Distance travelled is equal to the area under a graph of velocity versus time. example s = 5t 3 + 2t + 7 v = ds dt = 15t2 + 2 a = dv dt = 30t onstant acceleration If a is constant (and v = v 0 and s = s 0 at t=0) then v = v 0 + a t s = s 0 + v 0 t + ½a t 2 v 2 = v a (s - s 0 ) Example: a = -5 m s -2, v 0 = 30 m s -1, s 0 = 0 m. 30 v (m s 1 ) Graphically, area = 90 m Mathematically, time to stop, t 1 0 = 30 5 t 1, t 1 = 6 seconds s = ½ = 90 m t (seconds) 6 G Lane-Serff Sep-10

13 The graphical representation of distance versus time (where the slope of the graph is the velocity) is useful when considering problems involving two bodies. Example train leaves Station travelling at 15 m s -1. One minute later a train leaves Station B travelling at 20 m s -1. The two lines meet at Junction, 3 km from and 1.5 km from B. Which train gets to the junction first and by how long? distance 3 km Junction 65 s 20 m s m s km Station B 0 km time (seconds) You may wish (or need) to do some of the calculations mathematically but the graphical representation is at least a very useful tool in clarifying the problem. It is sometimes also useful to plot a velocity-time graph (the integral of which is a distance-time graph). E.g. acceleration 2 m s -2 for 6 s, then constant velocity. 15 velocity (m s -1 ) time (s) distance (m) time (s) G Lane-Serff Sep-10

14 B2 Plane curvilinear motion Motion in two dimensions y Position is now a vector r(t), with position moving to r+ r at time t+ t. r lim r Velocity, v = t 0 t = dr dt = ṙ, r x acceleration, a = lim v t 0 t = dv dt = v. = d2 r dt 2 = ṙ. In rectangular coordinates: r = (x,y) = xi + yj v = (v x,v y ) = ẋi + ẏj magnitude (speed): v = v 2 2 x + v y a = (a x,a y ) = ẋ. i + ẏ. j tanθ = v y /v x Projectiles onstant acceleration (due to gravity) a x = 0, a y = -g (where g 9.81 m s -2 ). (Neglect air resistance.) Initial conditions: x = y = 0 at t = 0 (or r = 0 at t = 0). y v 0 θ x v x = v x0 = v 0 cos θ (constant horizontal velocity) v y = v y0 gt = v 0 sin θ gt (constant downward acceleration) Initial conditions: x 0 = y 0 = 0 Where does it hit the ground? x = v x0 t y = v y0 t ½gt 2 y = 0, v y0 t = ½gt 2, and so t = 2v y0 /g (t = 0 is also a solution). 2 2v y0 x = v x0 t = v x0 g = 2v 0 g sin θ cos θ = v 2 0 g sin 2θ G Lane-Serff Sep-10

15 Example iring a projectile over an obstacle, with initial speed 100 m s -1. y v 0 θ 3 m target 100 m 50 m x irst find the angle rom solving where does it hit the ground? we have: x = v 0 2 g sin 2θ, and here x = 150 m. Putting in v 0 = 100 m s -1 and g = 9.81 m s -2 gives sin2θ = 0.147, so 2θ = 8.5 or and so, θ = 4.25 or (it is important to notice that there are usually two solutions). Will it clear the obstacle? θ = 4.25 v x0 = v 0 cos4.25 = 99.7 m s -1, v y0 = v 0 sin4.25 = 7.4 m s -1., t x = 100 m, t = x/v x0 = s. Thus y = ½ = 2.49 m. So the projectile will not clear the obstacle at this angle. θ = v x0 = v 0 cos85.75 = 7.4 m s -1, v y0 = v 0 sin85.75 = 99.7 m s -1., t x = 100 m, t = x/v x0 = s. Thus y = ½ = 452 m. So the projectile will clear the obstacle at this angle by a considerable height! G Lane-Serff Sep-10

16 B3 Example session: kinematics car accelerates from 0 to 144 kmh (40 m s -1 ) in 8 seconds (with constant acceleration). It then continues travelling at this speed. There is a police car 90 m from the position where the first car started. Immediately the police car is passed, it begins to accelerate to 180 kmh (50 m s -1 ) in 10 seconds. With the aid of graphical representations, find out when and where the police car overtakes the first car? G Lane-Serff Sep-10

17 Dynamics 1 orce and acceleration: Newton s Laws of Motion Newton s Laws Newton s irst Law body remains at rest or continues to move at constant velocity unless acted upon by a net force. (Galileo s Principle of Inertia) Newton s Second Law The rate of change of momentum of a body is proportional to the force acting on it. or bodies of fixed mass: = ma m is mass (kg), a is acceleration (m s -2 ), is the force (Newtons N). So 1 N = 1 kg m s -2 [We are using a system of units such that the constant of proportionality is one.] More generally, = d dt (mv) Momentum is a vector, p = mv. Newton s Third Law ction and reaction are equal (in magnitude) and opposite (in direction). These opposing forces act on the different interacting bodies. This includes contact forces and body forces. ontact forces - B orce by on B orce by B on Body forces, e.g. gravity - B Gravitational force of on B Gravitational force of B on If has mass m, B has mass M and r is the distance between the centres of mass, the magnitude of the force is = GmM r 2 G = N m 2 kg -2 On the surface of the earth r is approximately constant, (universal constant) = GM 2 r m = g m, g 9.81 m s-2 G Lane-Serff Sep-10

18 Examples 1) If a car has a mass of 500 kg, what is the force required to accelerate it from 0 to 30 m s -1 in 5 s (at a constant rate)? a = 30/5 = 6 m s -2, = ma = = 3000 N (or 3 kn) 2) The engines of a ship provide a maximum thrust T (N) which depends on the ship s speed V (m s -1 ) T = 10 5 (1-0.01V) The drag on the ship as it moves through the water is D = 200V 2 If the mass of the ship is 5000 tonnes, calculate the initial acceleration if the ship starts with full engine power from rest and the maximum speed of the ship. Draw up a table of acceleration as a function of velocity and sketch velocity as a function of time for the ship as it accelerates from rest to 95% of maximum speed. t rest (V=0) a= m s -2 orce = 10 5 (1-0.01V) - 200V 2 = ma a = /m = = (1 0.01V) V 2 Maximum speed when a=0, solving the quadratic gives Vmax=20 m s -1 (the other, unrealistic, solution is 25 m s -1 ) Draw up table of acceleration as a function of velocity, then assume that during acceleration we can approximate the time taken to accelerate from v to v+ v as t v/a. (In the table below v = 1 m s -1.) v (m s -1 ) a (m s -2 ) t (s) t = Σ t (s) v (m s -1 ) analytic approx time (s) [It is also possible to solve the equation for v as a function of t analytically by integrating the equation for dv/dt (=a), after some rearrangement (actually integrating dt/dv).] G Lane-Serff Sep-10

19 2 onservation of momentum onservation of momentum follows from Newton s Second and Third Laws. onsider two interacting bodies: The rate of change of momentum of the first body is m 1 - v 1 m 2 d dt (m 1v 1 ) = d dt (p 1) = - v 2 and of the second body d dt (m 2v 2 ) = d dt (p 2) = dding these gives d dt (p 1) + d dt (p 2) = - + = 0, or d dt (p 1 + p 2 ) = 0, so (p 1 + p 2 ) is constant. Principle of conservation of momentum: The total momentum of an isolated system of two (or more) interacting bodies doesn t change. Examples 1) Two masses (m 1 =100 kg and m 2 =50 kg) are joined together and moving at 10 m s -1. spring is released which separates the two masses such that the final speed of the second mass is 20 m s -1. What is the final speed of the first mass? m 1 m 2 10 m s m s -1 Initial momentum ( )10 = 1500 kg m s -1 inal momentum 100 v = 100 v , Initial momentum = inal momentum, so the final momentum of the first body is 100 v 1 = 500 kg m s -1, nd thus v 1 = 5 m s -1 v 1 2) small mass (0.1 kg) travelling at 250 m s -1 strikes a stationary larger mass (500 kg) suspended on a cable. fter the impact, the small mass remains imbedded in the larger mass. What is the final speed of the combined masses? Initial momentum inal momentum = 25 kg m s v = 25 kg m s -1 v = 25/500.1 = m s -1 G Lane-Serff Sep-10

20 orce and rate of change of momentum 3) car of mass 700 kg is travelling eastward at 20 m s -1. The road bends 45 and the car slows to 16 m s -1, now travelling north-east. ssuming that the change in speed and direction of the car is achieved by a constant applied force acting over 3 seconds, find. 16 m s m s Use rectangular components with the initial velocity in the x-direction. Initial velocity, v i = (20, 0), momentum p i = (20 700, 0) = (14000, 0) inal velocity, v f = (16 cos45, 16 sin45 ) = (11.31, 11.31), momentum p f = ( , ) = (7917, 7917) hange in momentum (final - initial) p f - p i = (7917, 7917) - (14000, 0) = (-6083, 7917) If this change happened in a time of 3 seconds, then the rate of change of momentum (which is equal to the force) is given by, = (p f - p i )/ t = (-6083, 7917)/3 = (-2028, 2639) N Note the total change of momentum (p f - p i ) = t is given by force multiplied by time. G Lane-Serff Sep-10

21 3 Impacts and impulse In the first two example problems in the previous lecture we didn t calculate the forces involved (to do so we would need information about the length of time of the interactions). rom the final problem we saw how the total change in momentum of a body was given by force time (when the force acting on the body is constant). Impacts (collisions) v 1 v 2 m 1 m 2 When two bodies collide the interaction between the bodies involves deformation of the bodies and a complicated force-time history (with rapidly changing and very large forces). v 1 ' v 2 ' We will begin with cases where the motion is in a straight line, so that we only have to consider one component of the velocities and momentum. or angled collisions both the x- and y-components of the total momentum must be conserved (use separate equations for the x and y momentum). Impulse The total change of momentum that a body experiences in such an interaction is known as an impulse. I = (p' - p) impulse = (final momentum initial momentum) or a steady force I = t, in general we would need to integrate the varying force over the duration of the interaction. I = dt. oefficient of restitution In an impact between two bodies, the elasticity of the collision can be described by the coefficient of restitution, e, given by In terms of the velocities this is: speed of separation e = speed of approach. e = v 1' - v 2 ' v 1 - v 2. or motion in two-dimensions, note that you find the difference in velocities first, then the magnitude of this difference. E.g. if v 1 = (3, -1) m/s and v 2 = (4, 2) m/s, then v 1 - v 2 = (-1, -3) = 10 = 3.16 m/s. The coefficient of restitution is also the ratio of restorative forces (as the bodies resume their shape) to deformative forces (as the bodies deform) and e is always between 0 and 1. Perfectly inelastic or plastic impacts have e=0 (the bodies stick together after impact). Perfectly elastic impacts have e=1 (the bodies fly apart at the same relative speed as they approached). G Lane-Serff Sep-10

22 Example ind the final velocities after a collision between two bodies travelling along a straight line, with: for the three cases e=0, e=0.5 and e=1. Total initial momentum a) e=0 (perfectly plastic) m 1 =1 kg, m 2 = 10 kg, v 1 = 15 m s -1 and v 2 = 4 m s -1 ; m 1 v 1 + m 2 v 2 = = = 55 kg m s -1 In this case the bodies stick together after collision with v 1 ' = v 2 '. Total momentum b) e=0.5 m 1 v 1 ' + m 2 v 2 ' = 11 v 1 ' = 55 kg m s -1, so v 1 ' = 5 m s -1 = v 2 '. v 1 ' - v 2 ' = 0.5 v 1 - v 2 = 0.5 (15 4) = 5.5 m s -1. We expect v 2 '>v 1 ', so this implies that rom the final momentum we have v 2 ' - v 1 ' = 5.5 m s -1. (So, v 1 ' = v 2 ' ) m 1 v 1 ' + m 2 v 2 ' = 1v 1 ' + 10v 2 ' = 55 v 2 ' v 2 ' = 55 11v 2 ' = 60.5 v 2 ' = 60.5/11 = 5.5 m s -1. nd so v 1 ' = 0 (i.e. the smaller mass is stationary after the collision). c) e=1 (perfectly elastic) v 1 ' - v 2 ' = 1.0 v 1 - v 2 = 1.0 (15 4) = 11 m s -1. We expect v 2 '>v 1 ', so this implies that rom the final momentum we have v 2 ' - v 1 ' = 11 m s -1. (So, v 1 ' = v 2 ' - 11.) m 1 v 1 ' + m 2 v 2 ' = 1v 1 ' + 10v 2 ' = 55 v 2 ' v 2 ' = 55 11v 2 ' = 66 v 2 ' = 66/11 = 6 m s -1. nd so v 1 ' = -5 m s -1 (i.e. the smaller mass bounces back in the opposite direction after the collision). G Lane-Serff Sep-10

23 4 Example session: Momentum Plastic/elastic collisions G Lane-Serff Sep-10