Dynamics of Solid Mechanics – MEE 201 Lecture 2 PDF

Dynamics of Solid Mechanics– MEE 201 Lecturer: Selase Kwame Mantey Department: Civil Engineering Sept. 2024 September 2024 1 Dynamics of Solid Mechanics– MEE 201 Lecture 2 Kinematics of Particles Sept. 2024...

Dynamics of Solid Mechanics– MEE 201 Lecturer: Selase Kwame Mantey Department: Civil Engineering Sept. 2024 September 2024 1 Dynamics of Solid Mechanics– MEE 201 Lecture 2 Kinematics of Particles Sept. 2024 September 2024 2 Contents Kinematics of Particles ◆ Background ◆ Rectilinear Motion ◆ Rectilinear erratic motion ◆ Curvilinear Motion ◆ Curvilinear Motion: Rectangular 3-Dimension ◆ Curvilinear Motion : Projectile motion ◆ Curvilinear Motion: Normal and Tangential Components ◆ Curvilinear Motion: Cylindrical Components Sept. 2024 3  Kinematics of Particles Background Applications 1. Moving loads on bridges 2. Wind load bridges, wind Kinematics turbines etc 3. Earthquake loads on structures 4. Motions of vehicles, cranes, Geometry of motion Force inducing motion airplanes etc. 5. Motions of artificial satellites and projectiles and spacecrafts. Particles Sept. 2024 4  Kinematics of Particles Rectilinear Motion The straight line path is defined by single Particle in a straight path is called rectilinear coordinate axis or scalar quantity (s). motion. Particle have mass but with negligible size and shape. Particle motions are characterized by mass centre. Rotation is neglected Rectilinear kinematics at any time (t), the particles position, A fixed reference point O, from which velocity and the position s of the particle can be acceleration is known. determined at time (t). Sept. 2024 5  Kinematics of Particles Rectilinear Motion Magnitude- the magnitude of s is the Displacement: the change in position is distance from O to the particle, measured in displacement of particle. meters. The sense of direction is defined by the algebraic sign of s (+ or -). If a particle move from one point to another, the displacement is (1) Sept. 2024 6  Kinematics of Particles Rectilinear Motion The average speed is always a positive scalar and is the total distance traveled by a particle, ST, divided by the elapsed time ∆𝒕 ; 𝒊𝒆 sT (vavg ) sp = (3) Velocity: it is the change in displacement t with time. Displacement that occur with respect to the measure of time consumed. The particles average velocity is vavg = − s t (4) ds v= (2) dt The magnitude of the velocity is the speed, m/s and the speed is always positive scalar Sept. 2024 7  Kinematics of Particles Rectilinear Motion The instantaneous acceleration at time t is a vector that is found by taking smaller and smaller values of ∆𝒕 and corresponding smaller and smaller values of v , so that Acceleration: it is the change in velocity a = lim ( v / t ) t → 0 between two points or dv a= v dt (6) aavg = (5) t Second Substituting with the velocity equation derivative of displacement d 2s a= 2 dt (7) At a decreasing speed, the particle is said to be decelerating, the velocity will be negative and the acceleration too negative. Units m/s2 Sept. 2024 8  Kinematics of Particles Rectilinear Motion Constant acceleration: at constant acceleration, the three kinematic equations can be integrated to obtain formulas that relate ac, v, s and t Velocity as a function of time: integrating ac=dv/dt, assuming that initial v=v0 when t=0, we get Combined equation: v t  dv =  a dt Initial velocity is the c constant of integration v0 0 (9) Finally, a differential equation can be formulated for v = v0 + ac t (Constant Acceleration) displacement, velocity, and acceleration by eliminating the time differential dt Position as a function of time: integrating v = ds/dt = v0+act, assuming that initially s=s0 when t=0, yields s t  ds =  ( v + ac t )dt Initial displacement is ds dv 0 the constant of dt = = (8) s0 0 integration v a (10) a ds = v dv 1 2 s = s0 + v0t + act (Constant Acceleration) 2 Sept. 2024 9  Kinematics of Particles Rectilinear Motion Velocity as a function of Position: if we solve t in the velocity equation with respect to constant acceleration, and substitute it into displacement equation or integrate vdv = acds, assuming that initially v=v0, at s=s0, we get v s  vdv =  a ds v0 s0 c (11) v 2 = v02 + 2ac ( s − s0 ) (Constant Acceleration) The algebraic signs of s0, v0 and ac assumed to be positive direction of s axis. The equations are applicable when acceleration is constant, t=0, s=s0, v=v0. e.g. free fall of body by gravity. Neglecting air resistance, constant acceleration is 9.81 m/s2 Sept. 2024 10  Kinematics of Particles Rectilinear Motion Solution The origin extend positive 0 to the car at a distance s. Since v=f(t), the car’s position can be determined from v=ds/dt, since this equation relates v, s and t. granted that s=0 when t=0, we have The car shore bellow moves in a straight line such that for a short time its velocity is defined by v=(0.9t2+0.6t) m/s. Determine its position and acceleration when t=3 s. when t=0, s=0 Acceleration: since v=f(t), the acceleration is determined from a=dv/dt, since this equation relates a, v, and t. ****Because the acceleration is a function of time the constant acceleration formula is not used here. Sept. 2024 11  Kinematics of Particles Rectilinear Motion Example: a small projectile is fired vertically downward into a fluid with an initial velocity of 60 m/s. Due to the drag resistance of the fluid the projectile experiences a deceleration of a=(-0.4𝑣 3 ) , where v is in m/s. Determine the projectile’s velocity and position 4 s after it is fired. velocity = 0.559m / s s = 4.43m Example: During a test rocket as shown in the figure below, it travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height Sb reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s2 due to gravity. Neglect the effect of air resistance. sB = 327 m vC = 80.1m / s Sept. 2024 12  Kinematics of Particles Rectilinear Motion A particle moves along a horizontal path with a velocity of v = (3t 2 − 6t ) m/s , where t is in seconds. If it is initially located at the origin O, determine A metallic particle is subjected to the influence of a magnetic field as it the distance travelled in 3.5 s, and the particle’s average velocity and travels downward from plate A to plate B. if the particle is released average speed during the time interval. from rest at the midpoint C, s=100 mm, and the acceleration is a=(4s) m/s2, s is in meters, determine the velocity of the particle when it reaches plate B, s=200 mm and the time it takes to travel from C to B. vB = 0.346 m / s t = 0.658 s Sept. 2024 13  Kinematics of Particles Rectilinear Erratic motion The s-t, v-t and a-t Graphs: To construct the v-t graph given the s-t graph, fig. a, the equation v=ds/dt should be used , since it relates the variables s and t to v. This equation states that Changing Erratic motions v = ds / dt , slope of s - t graph = velocity The velocity , acceleration, and displacement a = dv / dt , slope of v-t graph = acceleration cannot be described by a single continuous mathematical function along the path of motion. A series of mathematical equations describes the path By measuring the slope on the s-t graph when t-t1, the velocity is v1, from fig. a. The v-t graph can be constructed by plotting this and other Any two variable (s, v, a, t) known can be used values at each instant as shown in Fig. b. similarly the a-t graph can be to estimate other unknown variables from the constructed. equations: v = ds / dt , a = dv / dt , or ads = vdv (12) Sept. 2024 14 Rectilinear Erratic motion Kinematics of Particles If the a-t graph is given, Fig. a, the v-t graph may be constructed using a = dv / dt , written as (13) v =  a dt change in velocity = area under a - t graph Therefore, to construct the v-t graph, we begin with the particle’s initial velocity vo, and then add to this small increments of area (∆𝑣) determined from the a-t graph. In this manner successive points , v1=v0+∆𝑣, etc., are determined as shown in Fig. b. s-t graph can be determined using similar method when the v-t graph is given. Sept. 2024 15  Kinematics of Particles Rectilinear Erratic motion The v-s and a-s graphs: if the a-s graph can be constructed, then points on the v-s graph can be determined by using v dv = a ds (14) Integrating this equation between the limits v=v0 at s=s0 and v=v1 at s=s1, we have s1 ( v1 − v0 ) =  ads, area under a-s graph 1 2 2 2 (15) s0 If the area shaded in Fig. a is determined , and the initial velocity v0 at s0=0 is known, then (fig. b) 1/2  s1  v1 =  2  a ds + v0  2   (16)  0  Sept. 2024 16  Kinematics of Particles Rectilinear Erratic motion If v-s is given, the acceleration a at any point s can be determined using a ds = v dv Written as  dv  a =v    ds  velocity x acceleration = slope of v - s graph On graph (a) at point (s, v), the slope dv/ds of the v-s graph is measured. Then with v and dv/ds known, the value of a can be calculated and shown in (b). Note v-s graph can also be constructed from the a-s graph. Sept. 2024 17  Kinematics of Particles Rectilinear Erratic motion Solution: v-t graph- v=ds/dt, the v-t graph can be determined by differentiating the equations defining the s-t graph. We have Work examples: A bicycle moves along a straight road such that its position is described by the graph shown below. Construct v-t and a-t graphs for 0  t  30 s a-t graph- a=dv/dt, the a-t graph can be determined by differentiating the equations defining the v-t graph. We have Sept. 2024 18  Kinematics of Particles Rectilinear Erratic motion 2). The v-s graph describing the motion of a motorcycle is shown in below. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s=120 m. Work examples: The car below starts from rest and travels along a straight track such that it accelerates at 10 m/s2 for 10 s, and then decelerates at 2 m/s2. Draw the v-t graph and determine the time t’ needed to stop the car. Sept. 2024 19  Kinematics of Particles Rectilinear Erratic motion 2). The s-t graph for a train has been experimentally determined. From the data, construct the v-t and a-t graphs for the motion; 0  t  40 s Work examples: For 0t  30 s , the curve is s = (0.4t 2 ) m, and then it becomes Due to an external force, the particle travels along a straight track such that its position is described by the s-t graph. straight for t  30 s. Construct the v-t graph for the same time interval. Take v=0, a=0 when t=0 Sept. 2024 20  Kinematics of Particles Rectilinear Erratic motion The boat is originally travelling at a speed of 8 m/s when it is subjected to the acceleration shown in the graph. Determine the boat’s maximum speed and the time t when it stops. Work examples: The rocket has an acceleration described by the graph. If it starts from rest, construct the v-t and s-t graphs for the motion for the time interval 0  t  14 s Sept. 2024 21  Kinematics of Particles Curvilinear Motion Position- Consider a particle located at a point on a space curve defined by the path function s(t), fig. (a). The position of the particle, measured from a fixed point O, will be designated by the position vector r=r(t). Displacement Particle Curve path The particle moves a distance ∆ 𝒔 along the curve to a new position r`=r+ ∆r Curve motion Displacement, ∆r=r`-r Three dimension of motion Vector analysis description Position, velocity, Velocity-if ∆r occurs during the time ∆𝐭, then the average velocity of the particle is and acceleration r v avg = t Sept. 2024 22  Kinematics of Particles Curvilinear Motion Since dr is tangent to the curve, the direction of v is also tangent to the curve , Fig. c. The magnitude of v, which is called the speed, is obtained by realizing that the length of the straight-line segment ∆r in Fig.(b) approaches the arc length ∆𝐬 as ∆𝐭 → 𝟎 , and so we have Instantaneous ∆𝐭 → 𝟎 velocity Direction ∆r Approach the curve tangent v = lim ( r / t ) = lim ( s / t ) or t → 0 t → 0 dr v= lim ( r/t ) or v= t → 0 dt ds v= dt Sept. 2024 23  Kinematics of Particles Curvilinear Motion To investigate the rate of change the two velocity vectors are plotted below (Hodograph). Use velocity as the Acceleration-a particle with velocity v at time t and a velocity curve radius (Arm) v`=v+ ∆𝐯 at t+ ∆𝒕, as shown in Fig. d, then the acceleration of the particle during the time interval ∆𝐭 is v aavg = t The instantaneous acceleration, let ∆𝐭 → 𝟎, and so a = lim ( v / t ) or Where ∆𝐯 =v`- v. t → 0 dv a= dt by substitution d 2r a= 2 dt Acceleration is tangent to hodograph but not tangent to the path of particle. Sept. 2024 24  Kinematics of Particles Curvilinear Motion: When the particle moves, the x, y, z components or r will be functions of time; i.e., x=x(t), y=y(t) and z=z(t), so that r= r(t). Rectangular 3-dimension At any instant the magnitude of r is determined by 3-dimensional The motion particle coordinate x +y +z path 2 2 2 system of x, y, z r= Particle position defined by vector: And the direction of r is specified by the unit vector r = xi + yj + zk u r = r/r (-1,1) unit vector Sept. 2024 25  Kinematics of Particles Curvilinear Motion: The derivative of the i component of r is Rectangular 3-dimension dr dx di Velocity-The time derivative of r yields the velocity of the particle. Hence, ( xi ) = i + x dt dt dt dr d d d v= = ( xi ) + ( yj) + ( zk ) The last term is zero, because the x, y, z, reference frame is fixed, and therefore the direction ( and the magnitude) of i does not change with time. dt dt dt dt Similarly , j and k components are differentiated and the final result yields For each vector component (x, y, z) the magnitude and direction must be dr accounted for. v= = vx i + v y j + vz k dt where Magnitu vx = x v y = y vz = z de of The velocity magnitude is velocity in the 3- direction v = v +v +v 2 x 2 y 2 z u v = v/v Direction vector Sept. 2024 26  Kinematics of Particles Curvilinear Motion: The acceleration magnitude is Rectangular 3-dimension Acceleration-The acceleration is calculated by taking time derivative of the velocity equation. Hence, a = a +a +a 2 x 2 y 2 z = ( a x i ) + ( a y j) + ( a z k ) dv a= u a = a/a Direction vector dt where a x = vx = x ay = vy = y a z = vz = z Sept. 2024 27  Kinematics of Particles Curvilinear Motion: Rectangular 3-dimension Work examples: 1). At any instant the horizontal position of the weather balloon in the fig. bellow is defined by x=2.4t (m) where t is in seconds. If the equation of the path is y=x2/3, determine the magnitude and direction of the balloon’s velocity and the acceleration when t=2 s. Sept. 2024 28  Kinematics of Particles Curvilinear Motion: Projectile Horizontal motion- Since ax=0, application of the constant acceleration equation yields motion A projectile launched at point (x0, y0), with an initial velocity v0, having components (v0)x and (v0)y, shown in the fig. bellow. Neglecting air resistance, the only force acting on the projectile is its weight, and this causes the projectile to have a constant downward acceleration of The first and last equations simply indicate that the horizontal component of ac=g=9.81 m/s2 velocity always remains constant during the motion Vertical motion- since ay= -g, then applying equations yields, Sept. 2024 29  Kinematics of Particles Curvilinear Motion: Projectile Solution: Coordinate system- The origin of coordinates is established at the beginning of motion the path, point A. the initial velocity of the sack has components (vA)x =12 m/s and (vA)y =0 m/s. Also, between points A and B the acceleration is ay=-9.81 m/s2. Since (vB)x =(vA)x =12 m/s , the three unknowns are (vB)y , R, and the time of Work example: flight tAB. 1). A sack slides off the ramp, shown bellow, with horizontal Vertical motion- The vertical distance from A to B is known, and therefore we velocity of 12 m/s. If the height of the ramp is 6 m from the floor, can obtain a direct solution for tAB by using the equation determine the time needed for the sack to strike the floor and the 0 range R where the sack strikes the ground. Horizontal motion- since tAB has been calculated, R is determined as follows: Sept. 2024 30  Kinematics of Particles Curvilinear Motion: Projectile 3) If the x and y components of a particle’s velocity are defined bellow, motion determine the equation of the path y=f(x), if x=0 and y=0 when t=0. Work example: 2). The chipping machine is designed to eject wood chips at v0=7.5 m/s as shown. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube. 4) A particle is traveling along the straight path. If its position along the x axis is x= (8t) m, where t is in seconds, determine its speed when t=2 s. Sept. 2024 31  Kinematics of Particles Curvilinear Motion: Projectile A particle travels along the curve from A to B in 2 s. it takes 4 s for it to go motion from B to C and then 3 s to go from C to D. Determine its average speed when it goes from A to D. Work example: 2). Determine the speed at which the basketball at A must be thrown at the angle of 30° so that it makes it to the basket at B. Sept. 2024 32  Kinematics of Particles Curvilinear Motion: Normal Planar Motion- Consider the particle in fig.a, at positon s, measured from O. and Tangential Components the t axis is the tangent to the curve at the particle and is positive in the direction of increasing s. We will designate this positive direction with the unit vector ut. The normal axis n is perpendicular to the t axis and its The path of particle travel is decomposed into the normal n and the positive direction is towards the centre of curvature O’. The direction is tangential components axis using the particle as the origin at the instant designated by un. The plane on which the n and t axis are located is referred considered. to as the embracing or osculating plane, and in this instance it is fixed in the plane of motion Velocity – The particle velocity v has a direction that is always tangent to the path, fig. below, and a magnitude that is determined by taking the time derivative of the path function s=s(t), i.e. , v=ds/dt v=vu t where v=s Sept. 2024 33  Kinematics of Particles Curvilinear Motion: Normal direction  u n and Tangential Components du t = d u n The time derivate is Acceleration- The acceleration of the particle is the rate of change of velocity with respect to time. Therefore, ut =  u n a=v=vu t + vu t ds =  d To determine the time derivative u , as the particle move along the arc  =s/ t ds in time dt, u t preserves its magnitude of unity; however, its direction changes, and becomes u t. As shown in fig.2, (RS), we require u t = u t + du t , where du acts between the arrowheads of therefore u t and u t. t s v u t = u t = 1, then du t = (1)d ut =  u n = un = un   Sept. 2024 34  Kinematics of Particles Curvilinear Motion: Normal Three –Dimensional Motion and Tangential Components A particle moving in space at a given instant the t axis is uniquely specified and infinite number of straight line can be constructed to the tangent axis. The normal n axis is directed towards the centre of curvature. The n axis is  → in a straight line path of a particle. Then an = 0 the principal normal axis. Knowing n and t , then v and a can be calculated. For spatial motions a third unit vector u b is introduced, which defines the Then a = at = v the tangential component represent the rate of binormal axis b which is perpendicular to u and u n. change of velocity. t If a particle moves along a curve with a constant speed, then Related by at = v = 0 and a = an = v  2 vector cross- product u b =u t  u n Sept. 2024  Kinematics of Particles Curvilinear Motion: Normal and Tangential Components Velocity:- ◆ The particle’s velocity is always tangent to the path Normal Acceleration:- ◆ The magnitude of the velocity is found from the time derivative of the path function ◆ The normal component of the acceleration represents the rate of change of the velocity’s direction with respect to time. And it is always directed towards the centre of curvature of the path. Tangential Acceleration:- ◆ Rate of change of velocity magnitude with time. It can be increasing of decreasing. Sept. 2024  Kinematics of Particles Curvilinear Motion: Normal Solution: and Tangential Components Since the path has been express in terms of x and y coordinates , we will establish the origin of the n, t axes at the Work example: fixed point A on the path and determine the components of v The skier has a speed of 6 m/s at point A, which is and a along these axes. increasing at 2 m/s2. Determine the direction of the velocity and the direction and magnitude of acceleration at this instant. Neglect the size of the Velocity: skier in the calculation By definition, the velocity is always tangent to the path. 1 2 1 Since y = x , dy dx = x, at x = 10 m, dy / dx = 1. 20 10 Hence, at A, v makes an angle of  = tan − 1 = 45o with the x axis Therefore, v A = 6 m/s 45o Sept. 2024  Kinematics of Particles Curvilinear Motion: Normal Solution: and Tangential Components Acceleration: The acceleration is determined from a=vu t + v  u n 2 ( ) The radius of curvature of the path at A (10 m, 5 m). d2y 1 = , then dx 2 10  = 28.28 m Acceleration a A = 2u t + 1.273u n  m/s 2 a = 22 + 1.2732 = 2.37 m/s 2 2  = tan −1 = 57.5o 1.273 Thus, 45o + 90o + 57.5o − 180o = 12.5o a = 2.37 m/s 2 12.5o Sept. 2024  Kinematics of Particles Curvilinear Motion: Cylindrical Components Sometimes the best way of describing a motion of a particle is by using cylindrical coordinates or restricted motions on planes using polar coordinates. Polar Coordinates:- the location of particle can be inferred by using the radial coordinate r, which extends outward from the fixed origin 0 to the particle , and a transverse coordinate 𝜃, which is the counterclockwise angle between a fixed reference line and the r axis. The angle is measure in degrees or radians, 1 rad=180°/π. The positive direction of r and 𝜃 coordinates are defined by the unit vectors u r and u , respectively. u r and u are the direction of increasing r and 𝜃 respectively when one variable is held fixed. Sept. 2024  Kinematics of Particles Curvilinear Motion: Cylindrical Components Position:- At any instant the position of the particle is defined by the position vector r=ru r Velocity:- the velocity v is obtained by taking the time derivative of r. v=r=ru r + ru r In time interval ∆𝑡 , a change in ∆𝑟 will not cause a change in direction of u r. A change in ∆𝜃 will cause u r to become u r ; u r = u r + u r For small angles ∆𝜃 , ur  1(  ) , u r =  u Sept. 2024  Kinematics of Particles Curvilinear Motion: Cylindrical Components u r    u r = lim =  lim  u t → 0 t t → 0 t   u r =  u The velocity can then be written in component form as The radial component Vr is a measure of the rate of increase or decrease in the length of the radial coordinate,. The transverse component V𝜃 is interpreted as the velocity Where along the circumference of a circle having a radius r. The angular velocity is therefore  = d dt rad/s. Velocity magnitude is therefore: Sept. 2024  Kinematics of Particles Curvilinear Motion: Cylindrical Components Acceleration:- taking time derivate of : The particle acceleration is obtained as Angular acceleration Rad/s2 u    u = lim =  lim  ur t → 0 t  t → 0 t  Magnitude of acceleration u = − u r Sept. 2024  Kinematics of Particles Curvilinear Motion: Cylindrical Components Cylindrical Coordinates:- If the particle moves along a space curve, then its location can be specified by the three cylindrical coordinates, r, 𝜃 ,z, where z coordinate is identical to that used for rectangular coordinates. Since the unit vector defining its direction, uz , is constant, the time derivatives of this vector are zero, and therefore the position, velocity, and acceleration of the particle can be written in terms of its cylindrical coordinates as 1. Polar coordinates specified as time parametric equations, r = r (t ) and  =  (t ), then the time derivatives can be found directly. 2. if the time parametric equations are not given, then the path r = f ( ) must be known. Using chain rule of calculus we can then find the relation Time Derivatives:- the above equation needs the time between r and  , and r and . derivatives of r , r ,  , and  in order to determine the r and 𝜃 components of v and a. Sept. 2024  Kinematics of Particles Curvilinear Motion: Cylindrical Solution:- Components Polar coordinate will be used since the angular motion of the arm is known. Work example: The amusement park ride consists of a chair that is Velocity and Acceleration:- rotating in a horizontal circular path of radius r such that It is first necessary to specify the first and second time the arm OB has an angular velocity  and angular derivatives of r and . Since r is constant, we have acceleration . Determine the radial and transverse components of velocity and acceleration of the car. r =r r =0 r =0 Neglect its size in the calculation. Thus , vr = r = 0 v = r ar = r − r 2 = − r 2 a = r + 2r = r Sept. 2024  Kinematics of Particles Dependent Motion Analysis of Similarly the acceleration is : Two Particles Some particle motions are interdependent. Example 2: Two position coordinate length minus lCD and lT remain constant, SA and SB measure the segments the constant length of the chord that changes in length, we have Since l and h are constant during the motion, the two time derivatives yield Hence when B moves downward (+SB), A moves to the left (-SA) with twice the motion. Sept. 2024