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Engineering Mechanics

Engineering Mechanics . د. عبد الرزاق طوقان. Introduction. Mechanics is a physical science concerned with the behavior of bodies subjected to forces. To study mechanics we need: Input: decompose problem into components know degree of components

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Engineering Mechanics

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  1. Engineering Mechanics د. عبد الرزاق طوقان

  2. Introduction Mechanics is a physical science concerned with the behavior of bodies subjected to forces. To study mechanics we need: Input: decompose problem into components know degree of components draw a mathematical model Processing: define laws governing mechanics provide details needed to solve problem design methodology to solve problem (manual, computer) Output:deliversolution in a nice way develop capability to solve problems (intuition) enhance intuition to values (dean)

  3. Decomposition • Bodies are • Rigid • Flexible • Fluids • Forces are • Static • Dynamics • Kinematics • Kinetics

  4. Degree of components (hierarchy of creation)

  5. Define laws governing mechanics (versus theories) • Laws: • constitutive relationships • equilibrium equations • compatibility equations • Theories: • Based on: • Assumptions based on: • Available knowledge is constrained with: • True dynamics is

  6. Details needed: definitions • Space:is the geometric region occupied by bodies • A particle:is a body of negligible dimensions • Rigid body:a body whose relative movement between its parts is negligible • Mass: • Quantity of matter in a body (or measure of the inertia of a body (resistance to change of motion)). • Force:the action of one body on another • Time:is the measure of the succession of events

  7. Details needed to solve problem:System of Units Base units are units of mass, length and time (1ft = 0.3048 m, 1lb = 4.448 N. 1slug = 1lb s²/ft = 14.59 kg) • *Derived unit 1 N = (1 kg)(1 m/ s²) • 1 Newton is the force required to give a mass of 1kg an acceleration of 1m/ s² • The weight of 1 kg Mass is: W = mg = (1kg)(9.81m/ s²) = 9.81 N

  8. Details needed to solve problem:System of Units: SI Prefixes When a numerical quantity is either very large or very small, the units used may be modified by using prefix

  9. Design methodology for solving problems • Input • Problem to be solved • Physics of problem • Mathematical model • Processing • Propose theory • Formulate equations • Solve equations • Output • 1. Verify laws • 2. Build engineering sense • 3. Start a new cycle

  10. develop capability to solve problems Only by working many problems can you truly understand the basic principles and how to apply them (muscles cannot be built by reading hundreds of books without practice!!!)

  11. End of introduction Let Learning Continue

  12. CHAPTER 12 RECTILINEAR KINEMATICS: 1D MOTION Objectives: -find the kinematic quantities of a particle traveling along a straight path.

  13. APPLICATIONS The motion of large objects, such as rockets …, can often be analyzed as if they were particles. Why?

  14. Mechanism definition • Motion of rigid bodies are • Translational: • Rotational • General • Definition of a particle

  15. POSITION AND DISPLACEMENT Define rectilinear motion The position relative to the origin, O, is defined by r, or s, units? The displacement is change in position. Vector : r = r’ - r Scalar :  s = s’ - s The totaldistance, sT, is …

  16. VELOCITY Velocity is... It is a vector quantity, its magnitude is called speed, units m/s The average velocity is vavg = r/t The instantaneous velocity is v = dr/dt Speed is … v = ds/dt Average speed is (vsp)avg = sT/  t

  17. ACCELERATION Acceleration is …, its units are … Vector form: a = dv/dt Scalar form: a = dv/dt = d2s/dt2 We can also express: a ds = v dv

  18. Position: Velocity: v t v s s t ò ò ò ò ò ò = = = dv a dt or v dv a ds ds v dt v o v s s o o o o o Thus • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds •Integrate acceleration for velocity and position. • so and vo are the initial position and velocity at t = 0.

  19. v t ò ò = = + dv a dt v v a t yields c o c v o o s t ò ò = = + + ds v dt s s v t (1/2)a t yields o o c 2 s o o v s ò ò = = + v dv a ds v (v ) 2a (s - s ) yields 2 2 c o c o v s o o CONSTANTACCELERATION (particular case) Constant acceleration(e.g. gravity ac =g= -9.81 m/s2 ), then

  20. EXAMPLE Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2. Find: The distance the motorcycle travels before it stops.

  21. 1) Determine the velocity. a = dv / dt => dv = a dt => v – vo = -3t2 => v = -3t2 + vo v t ò ò = - dv ( 6 t ) dt v o o 3) Calculate distance using so = 0: v = ds / dt => ds = v dt => => s – so = -t3 + vot => s – 0 = -(3)3 + (27)(3) => s = 54 m s t ò ò = - + ds ( 3 t v ) dt 2 o s o o EXAMPLE (solution) 2) Find time to stop (v = 0). Use vo = 27 m/s. 0 = -3t2 + 27 => t = 3 s

  22. GROUP PROBLEM SOLVING Given: Ball A is released from rest at a height of 12m at the same time that ball B is thrown upward, 1.5m from the ground. The balls pass one another at a height of 6m Find: The speed at which ball B was thrown upward.

  23. GROUP PROBLEM SOLVING (solution) 1) Time required for ball A to drop to 6m. t = 1.1s 2) Ball B: sBo = 1.5m, a=…, s=… (vB)o = 9.5m/s

  24. End of 12.1-12.2 Let Learning Continue

  25. ERRATIC MOTION Objectives: Determine position, velocity, and acceleration using graphs.

  26. APPLICATION Having a v-s graph, can we determine a at s = 300m? How?

  27. GRAPHING -Better to handle complex motions difficult to describe with formulas. -Graphs provide a visual description of motion. -Graphs are true meaning of dynamics

  28. S-T GRAPH V-t: Find slope of s-t (v = ds/dt) at various points.

  29. V-T GRAPH a-t: Find slope of v-t (a = dv/dt) at various points.

  30. A-T GRAPH v-t is the area under the a-t curve. We need initial velocity of the particle.

  31. s2 ½ (v1² – vo²) = ò a ds s1 A-S GRAPH Area under a-s curve= one half difference in the squares of the speed (recall ∫a ds = ∫ v dv ). We need initial velocity

  32. V-S GRAPH Knowing velocity v and the slope (dv/ds) at a point, acceleration is. a = v (dv/ds)

  33. EXAMPLE Given: v-t graph for a train moving between two stations Find: a-t graph and s-t graph

  34. a(m/s2) 4 3 t(s) -4 3 EXAMPLE (Solution:) a0-30 = dv/dt = 40/30 = 4/3 m/s2 a30-90 = 0 a90-120 = -4/3 m/s2

  35. EXAMPLE (continued)

  36. GROUP PROBLEMSOLVING Given: The v-t graph shown Find: The a-t graph, average speed, and distance traveled for the 30 s interval

  37. GROUP PROBLEM SOLVING a(m/s²) t(s)

  38. GROUPPROBLEM SOLVING (continued) Ds0-10 = ò v dt = = 400/3 m Ds10-30 = ò v dt = = s0-30 = = 1133.3 m vavg(0-30) = total distance / time =

  39. End of 12.3 Let Learning Continue

  40. CURVILINEAR MOTION: RECTANGULAR COMPONENTS: 2D MOTION Objectives: a) Describe the motion of a particle traveling along a curved path. b) Relate kinematic quantities in terms of the rectangular components of the vectors.

  41. Methodology • Understand previous 1D • Perform analogical solutions between 1D and 2D models • Build up experience with 2D models • 2D Models: • Rectangular: 3 distances • Cylindrical: 2 distances + 1 angle • Spherical: 1 distance + 2 angles

  42. APPLICATIONS Motion of a plane can be tracked with radar relative to a point recorded as a function of time. How to determine v and a? A car travels down a fixed, helical path at a constant speed. How to determine v and a? To design the track, is it important to predict a?

  43. POSITIONAND DISPLACEMENT Define a curvilinear motion? A particle moves along a curve defined by the path function, s. The position is designated by r = r(t). the displacement is Δr = r’ - r

  44. VELOCITY Average velocity is: vavg = Δr/ Δt . Instantaneous velocity is: v = dr/dt . v is always tangent to the path What is speed v? Δ s → Δr as t→0, then v = ds/dt.

  45. ACCELERATION Average acceleration is: aavg = Dv/Dt = (v’- v)/Dt Instantaneous acceleration is: a = dv/dt = d2r/dt2 A hodograph (also named velocity diagram) is a plot of the locus of points defining the arrowhead of velocity vectors. It is used in physics, astronomy and fluid mechanics. Acceleration is tangent to the hodograph. Is it tangent to the path function?

  46. RECTANGULAR COMPONENTS: POSITION The position can be defined as r = x i + y j + z k where x = x(t), y = y(t), and z = z(t) . Magnitude is: r = (x2 + y2 + z2)0.5 Direction is defined by the unit vector: ur = (1/r)r

  47. Velocity vector is : v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt Since i, j, k are constants then? RECTANGULAR COMPONENTS: VELOCITY Magnitude is v = [(vx)2 + (vy)2 + (vz)2]0.5 Direction is tangent to the path.

  48. Acceleration vector is: a = dv/dt = d2r/dt2 = axi + ayj + azk where ax = = = dvx /dt, ay = = = dvy /dt, az = = = dvz /dt • • •• •• vx vy x y • •• vz z RECTANG. COMPONENTS: ACCELERATION Magnitude is a = [(ax)2 + (ay)2 + (az)2 ]0.5 What about direction of a?

  49. EXAMPLE Given: The motion of two particles (A and B) is rA = [3t i + 9t(2 – t) j] m rB = [3(t2 –2t +2) i + 3(t – 2) j] m Find: The point at which the particles collide and their speeds just before the collision.

  50. EXAMPLE (Solution:) - Collision: rA = rB, so xA = xB and yA = yB . - x-components: 3t = 3(t2 – 2t + 2) t2 – 3t + 2 = 0 => t = 2 or 1 s - y-components: 9t(2 – t) = 3(t – 2) 3t2 – 5t – 2 = 0 => t = 2 or – 1/3 s - So, t = 2s. Substituting yields: xA = xB = 6 m, yA = yB = 0

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