Kwame Nkrumah University of Science & Technology ME 161/2 - BASIC MECHANICS PDF

Kwame Nkrumah University of Science & Technology, Kumasi, Ghana ME 161/2 - BASIC MECHANICS D.E.K. Dzebre (PhD) Department of Mechanical Engineering.  Design...

Kwame Nkrumah University of Science & Technology, Kumasi, Ghana ME 161/2 - BASIC MECHANICS D.E.K. Dzebre (PhD) Department of Mechanical Engineering.  Design Image Source: https://cutt.ly/qjCszYh  Selecting from available options to meet desired requirements → designing. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 3  Design Image Source:  Specifying variables to meet desired http://ideasoutdoor.selbermachendeko.com/ wp-content/uploads/2019/06/%E2%88%9A- performance requirements 18-How-to-Build-an-Adirondack-Chair- Plans.jpg → designing. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 5  Why learn Statics of Rigid Bodies? Generic process for designing an engineering structure Too much Deflection/Extension/Yield/Fracture/Failure Choose Internal External SIZE & Structural Start Loads Forces and MATERIAL Performance End Moments for the part Axial Stress Deflections Torsional Strain Yield/Fracture Bending/Flexural Behaviour of Success/Failure Combined materials under (using failure stress criteria) Mechanics of Rigid Bodies/Fluids Mechanics of Materials www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 6  Introduction  The study of the interactions between physical forces and bodies constitutes the field of Engineering Mechanics Statics Mechanics of Rigid Bodies Kinematics (Solid Mechanics) Dynamics Kinetics Theory of Plasticity Mechanics of Theory of Elasticity MECHANICS Deformable Bodies Theories of Failure Fatigue Mechanics of Compressible Fluids Mechanics of Fluids Mechanics of Some Branches of Engineering Incompressible Mechanics (J Antonio, Fluids www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 7  EXPECTED LEARNING OUTCOMES Apply Newton’s laws of motion and other basic theories and laws of Newtonian mechanics to particles and rigid bodies. Analyze 2-D and 3-D equilibrium of system of forces. Sketch free body diagrams for problems and use them to determine resultants and components of forces and moments. Determine centroids and centre of gravity of single and composite bodies. Solve static problems involving dry friction. Perform simple equilibrium analyses on statically determinant structures, and simple machines. Apply the kinematic relationships of position, velocity and, acceleration in solving rectilinear and particle kinematics problems. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 10  Some Recommended Text  Basic Engineering Mechanics, J. Antonio  Engineering Mechanics Statics by W.F. Riley and L.D. Sturges  Engineering Mechanics – Statics and Dynamics, J.L. Meriam and L.G. Kraige.  Engineering Mechanics – Statics and Dynamics, R. C. Hibbler.  Vector Mechanics for Engineers, Beer et al.  Any book on Engineering Mechanics. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 11  Assessment Assignments/Quizzes 15% Mid-Semester Exam 15% End of Semester Exam 70% TOTAL 100% Class Attendance 5 bonus marks www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 12  Outline Some Concepts, Principles and Techniques in Mechanics of Static Rigid Bodies; o Fundamental Principles and Concepts in Newtonian Mechanics. o Forces; Characteristics of Forces, Resultants of systems of forces, Resolving forces into components. o Moment of a Force o Centroids and an introduction to Area Moments of Inertia o Equilibrium Analyses of Particles and Rigid Bodies Some Applications; o Analysis of Structures o Friction and Selected Simple Machines Applications Mechanics of Static of Rigid Bodies; o Particle kinematic relationships o Motion; Rectilinear, Curvilinear, angular (in different coordinate systems) o Kinetics: Force, Mass and acceleration www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 13  SOME FUNDAMENTAL PRINCIPLES & CONCEPTS www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 14  Particles and Bodies  Particle A very small amount of matter which may be assumed to occupy a single point in space. Idealizing bodies as points simplifies problems since body geometry is not considered.  Rigid body A collection of several particles that remain at a fixed distance from each other, even when under the influence of a load. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 15  Some Fundamental Concepts  Space (Length) This is associated with notion of describing a point in terms of co-ordinates measured from a reference point.  Time A measure of the succession of two events or the duration of an event. Of significance in dynamics.  Mass A measure of the inertia of a body, which is its resistance to a change of velocity. The mass of a body affects the gravitational attraction force between it and other bodies.  Force  Space, Time, and Mass are independent of each other. However, Force, is related to the mass of a body and the variation of its velocity with time. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 16  Quantities  Measurements.  Time – second (s)  Mass – kilogram (kg)  Length – metre (m)  Force – Newton (N)  Quantities may be basic (fundamental) or derived.  They may be scalars or vectors.  Derived Units of measurement and Relations (formulae) are often in terms of fundamental quantities. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 17  Quantities and Dimensions Quantity Dimension Common SI Units Length L m, cm, mm Time T s Mass M kg Area L2 m2, cm2, mm2 Force ML/T2 or MLT-2 N Linear velocity L/T or LT-1 m/s or ms-1 Linear acceleration L/T2 or LT-2 m/s2 or ms-2 Angular velocity 1/T or T-1 rad/s Angular acceleration 1/T2 or T-2 rad/s2 Moment of a force ML2/T2 or ML2T-2 N.m or N-m Pressure, Stress M/LT2 or ML-1T-2 Pa, kPa, MPa Work and Energy ML2/T2 or ML2T-2 J, kJ Power ML2/T3 or ML2T-3 W, kW Momentum and linear impulse ML/Tor MLT-1 N.s or N-s www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 18  Prefixes Multiplication Factor Name of prefix Symbol of prefix Example 1012 tera T 1.23 TJ = 1 230 000 000 000 J 109 giga G 4.53 GPa = 4 530 000 000 Pa 106 mega M 7.68 MW = 7 680 000 W 103 kilo k 5.46 kg = 5 460 g 10-2 centi c 3.34 cm = 0.0334 m 10-3 milli m 395 mm = 0.395 m 10-6 micro μ 65 μm = 0.000 065 m 10-9 nano n 34 nm = 0.000 000 034 m www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 19  Some Principles/Laws: The Principle of Dimensional Homogenity  When combining quantities via addition/subtraction, the quantities being combined must be dimensionally homogenous.  For example, if the equation 2= + (in which x, v and t displacement, velocity and time respectively) is dimensionally homogenous, the dimensions of A and B can be determined as; = = = ∴ The equation can be re−written as = + ⇒ = , = www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 20  Newtons Laws of Motion  First Law A body will maintain it state of motion (remain at rest or continue to move in a straight line) unless the resultant force on it is not zero.  Second Law A body under the influence of a force experiences a proportionate acceleration in the direction of that force.  Third Law Action and Reaction are equal and opposite. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 21  Newton’s Law of Universal Gravitation  Two particles are attracted to each other by a force defined mathematically as; Mm F G 2 r On earth, F  mg GM g  2 r g varies from place to place on earth. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 22  Some other Laws and Principles  Principle of Transmissibility  Principle of Moments/Varigon’s Theorem  Parallel Axis Theorem  Principle of Virtual Work  Principle of Potential Energy  Principle of Work and Energy www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 23  FORCES Characteristics and Resultants of Forces www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 24  Forces www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 25  Forces: Some Characteristics  Forces are vector quantities. 15 cm 1 cm = 10 N 150 N 68o 68o  Forces considered equal if they have the same magnitude and direction.  Forces are equivalent if they produce the same resultant effect on a rigid body. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 26  Systems of Forces F3 F2 F1 F2 F3 F4 F1 T4 Non-Concurrent Concurrent Coplanar Force System Coplanar Force System T1 T3 F4 F2 T2 F1 F F3 Concurrent Spatial Force System F Parallel spatial Force System Collinear Force System A system of forces can be replaced with an equivalent system of forces or a Resultant Force (one equivalent force). www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 27  How do we find the Resultant of a System of Forces? www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 28  The Resultant of a Systems of Forces There are several approaches: Graphical approach –Parallelogram, Triangle, and Polygon rules of vector addition. Force Triangle with Sine and Cosine rules. Summation of Rectangular/Perpendicular Force components www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 29  Resultants of Forces: Graphical Approach The parallelogram law of vector The triangle law of vector addition. addition B B A A www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 30  Resultants of Forces: Graphical Approach 1N = 1mm 150 323.55 150 19o 19o 323.55 200 200 Triangle Law www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 31  Resultants of Forces: Graphical Approach The Polygon law of vector addition A 60 N C B 40 N 50 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 32  Resultants of Forces: Force Triangle Approach  The sine and cosine laws: a b c Law of Sines   sin  sin  sin   b a a 2  b 2  c 2  2bc cos   Law of Cosines b 2  c 2  a 2  2ca cos   c c 2  a 2  b 2  2 ab cos  www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 33  Resultants of Forces: Force Triangle Approach Example 150 N 68o 200 N 23o From the Cosine Rule, R 2  200 2  150 2 - 2(200)(150 )cos135 o R  323.9 N. R 150 135o 68o From the Sine Law 323.9 N 150 N   , θ  19.11 o 200 sin 135 o sin θ o 23o www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 34  Resultants of Forces: Force Triangle Approach 60 N 40 N 50 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 35  Resultants of Forces Example Determine the magnitude of the resultant force acting on the screw eye and its direction measured clockwise from the x-axis. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 36  Resultants of Forces Example Determine the magnitude of the resultant force acting on the hook. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 37  Resultants of Forces: Force Triangle Approach 60 N 40 N 50 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 38  Resultants of Forces: Force Components Approach 1. This approach requires the forces to be decomposed/resolved into Rectangular/Cartesian/perpendicular components (along principal directions). 2. Like components (along the same direction) are then summed to get the resultant components. 3. Magnitude and direction of the resultant force can be obtained through appropriate Trigonometry techniques. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 39  Resultants of Forces: Force Components Approach  Resolve each force into rectangular components and sum like components to get the components of the resultant. y  R   Fx   Fy Pyj    P R  ( Px  Qx  S x )i  ( Py  Qy  S y ) j Pxi Sx i Qx i x S Q Syj Qyj 2 2 The magnitude of the resultant force is given by; R  F     F  x y The direction w.r.t any axis can be determine using an appropriate technique. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 40  Resultants of Forces: Resolving Forces into Components y F v c a  Fy  b  x Fx h      F Fx  Fy  Fx i  Fy j Pythagoras Theorem Directional Vectors    F  Fx  Fy     bi  a j   F cos   F sin  F  F  F    c  = F sin   F cos     bi  aj Note :  F  F   c  c  a b    cos  ,  sin  , c  b2  a2  Fx i  Fy j c c www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 41  Resultants of Forces; Alternate method for resolving Forces www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 42  Resultants of Forces Example y Force Fx Fy 200 N 200 cos 23 N 200 sin 23 N x 150 N 150 cos 68 N 150 sin 68 N ∑ 240.292 N 217.224 N 150 N 2 2 68o 200 N F   F    F  x y 23o  240.292 2  217.224 2  323.924 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 43  Resultants of Forces Example Determine the magnitude of the resultant force acting on the screw eye and its direction measured clockwise from the x-axis. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 44  Resultants of Forces Example Determine the magnitude of the resultant force acting on the hook. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 45  Resultants of Forces Example 2 2 F  F    F  x y Find the resultant of the forces shown. So,   Fx  60 cos 25 + 40 cos 60  50sin 50  36.076 N  F y  60 sin 25  40 sin 60  50 cos 50  41.423 N Or, Force Fx (+→) Fy(+↑) 60 N 60 N 60 cos 25 60 sin 25 40 N 40 cos 60 - 40 sin 60 50 N - 50 cos 40 - 50 sin 40 ∑ 36.076 N - 41.523 N Or, Force Fx (+→) Fy(+↑) 60 N 60 cos 25 60 sin 25 40 N 40 N 40 cos 300 40 sin 300 50 N 50 N 50 cos 220 50 sin 220 ∑ 36.076 N - 41.523 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 46  Resultants of Forces Example Find the resultant of the forces shown. 2 2 60 N F   F    F  x y F  36.0762  (41.423)2  54.930 N 40 N 50 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 47  Resultants of Forces; Alternate method for resolving forces. Example Find the resultant of the forces shown y 800 800 N 600 x 424 N 400 N 900 Dimensions are in mm 560 480 www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 48  Resultants of Forces; Alternate method for resolving forces. Example - Solution     xi  y j  F  F  F   y  x2  y2    800       800 i  600 j  800i 600 j   F1  800    800.  800.  640i  480 j  800 2  600 2  1000 1000 F1=800 N 600      560i  900 j    F2  424    224i  360 j 2 2  560  900    F3=400 N x   480i  900 j  F2=424 N F3  400  2 2   480  900  900    Resultant, F   F i   F j   2  2 560 480 The magnitude of the Resultant, F   F    i   j F  www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 49  Resultants of Forces Example Determine the magnitude of the resultant force acting on the screw eye and its direction measured counter-clockwise from the x-axis. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 50  Resultants of Forces Example Determine the magnitude of the resultant force acting on the screw eye and its direction measured counter-clockwise from the x-axis. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 51  Resultants of Forces Example Cables AB and AD help support pole AC. Knowing that the tension is 120 N in AB and 40 N in AD, determine the magnitude of the resultant of the forces exerted by the cables at A. 10 cm 8 cm 6 cm www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 52  Resultants of Forces Example Two men are trying to roll the boulder by applying the forces as shown. Determine the magnitude and direction of the force that is equivalent to the forces the two men are applying. 300 N 150 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 57  Resultants of Forces Example The angle Ɵ = 50°. determine the length of the line representing vector rAC. (Hint: all three lines lie in the same plane) rAC = 181 mm www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 60  Resultants of Forces Example The forces FA = 40 N, FB = 50 N, and FC = 40 N act on the screw pin as illustrated in the Figure. α = 50° and β = 80°. Determine the magnitude of the resultant of the three forces on the eye of the screw pin, assuming they are coplanar. R = 83 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 61  Non-Planar/Spatial/3-D Forces. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 62  y Forces in Space O x z     F  Fx  Fy  Fz  F  F cos  x  F cos  y  F cos  z b y F z x    O   ai  bj  ck  c F  F  F   2 2 2  a b c      Fi  Fj  Fk a www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 63  Forces in Space Example A rectangular plate is supported by three cables as shown. Knowing that the tension in cables AC, AB and AD are 60 N, 80 N and 90 N respectively, determine the components of the forces being exerted at C, B and D. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 64  Forces in Space For components of force at D, FDA; Soln DA  250i  480 j  360k    250i  480 j  360k  DA   1   2502  4802  3602 2         250i  480 j  360k  FDA  90 1   2502  4802  3602 2      y' x' For components of force at B, FBA z' BA  320i  480 j  360k BA  www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 65  Forces in Space Example A transmission tower is held by three guy wired anchored by bolts B, C and D. If the tension in wire AD is 315 N, determine the components of the force exerted by the wire on the bolt at D... Dimensions are in dm (1 dm = 0.1 m = 10 cm) www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 66  Forces in Space Example A rectangular plate is supported by three cables as shown. Knowing that the tension in cables AC, AB and AD are 60 N, 80 N and 90 N respectively, determine the magnitude of a force that the three cables are exerting at A. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 67  Moment of Forces Scalar and vector approaches of calculating Moments Principles of Moments Equivalent Force-Moment Systems www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 68  Moment of a Force Forces have the tendency to cause two types motions in rigid bodies; translational and rotational motions. A measure of the tendency of a force to rotate a body is referred to the Moment of the force. A moment may occur about a point; the Moment Centre. F MO    A Mo  r  F r θ rsin  d O d M o  (r sin  ) F  dF www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 69  Moment of a Force d must be perpendicular to the Force’s line of action, so, the Force is treated as a sliding vector, due to the Principle of Transmissibility in rigid body mechanics. F F’.. A. A. If F and F’ have the same magnitude, Moment of F about A = Moment of F’ about A www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 70  Moment of a Force  Scalar Approach Mo  d  F  Only the magnitude of the moment is calculated using only the magnitudes of the force and the moment arm, d, defined as the perpendicular distance between the line of action of the force and the moment centre.  Often used when the moment, d can easily be determined. The sense of the moment is determined by inspection.  Vector Approach  The position vector for the point of application of the force is multiplied by the components of the force to get the components of the Resultant moment. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 71  Moment of a Force Example A force of 800 N acts on a bracket as shown. Determine the moment of the force about B. 800 N M B  0.2 m  800  160 Nm www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 72  Example Moment of a Force A force of 800 N acts on a bracket as shown. Determine the moment of the force about B. 800 N M B  0.16m  800  128 Nm www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 73  Moment of a Force Example A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O. A Determine: a) moment about O of the 100-N force, 100 N b) horizontal force at A which creates the same moment, 60o O www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 74  Moment of a Force Solution FBD Moment about O is equal to the product of the force and the A perpendicular distance between the line of action of the force and O. M O  dF 100 N d   24 m  cos 60   12 m. M O  12 m. 100 N   60o O d Mo www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 75  Moment of a Force Solution FBD Horizontal force at A that produces the same moment, A F d   24 m  sin 60  20.8 m M O  dF d 1200 Nm.   20.8 m  F 1200 Nm F 20.8 m. 60o  O Mo www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 76  The Principle of Moments (Varignon’s Theorem) The moment of a force about a given point, is equal and equivalent to the sum of the moments of an equivalent system of forces in the same plane, about the same point. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 77  The Principle of Moments (Varignon’s Theorem) Determine the magnitude and sense of the moment due to the 100 N force shown below about A. 1 cm  42  42  MA    100  4 100 cos 45  2    www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 78  The Principle of Moments (Varignon’s Theorem) Determine the moment of the 800N force about B using the principle of moments,. 800sin60 800cos60 M B  (0.16m  800 cos 60)  (0.2m  800sin 60)  202.56 Nm www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 79  Moment of a Force: Vector Formulation  The position vector for the point of application of the force (from the moment centre) are multiplied by the force (expressed in vector components form) to get moment components. z Fzk A (x,y,z) c  Direction determination follows the right hand rule; Fyj r A O Fxi thumb must point towards +ve direction of axis. b y a C x  Multiplication of the position and force vectors may be done in one of two ways;  Matrix approach  Expansion and simplification www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 83  Moment of a Force: Vector Formulation  Vectors are expressed in components, arranged in a matrix form, and the determinant of the matrix taken.         If r  ai  bj  ck and F  Fx i  Fy j  Fz k Expressing as a matrix, z    Fzk i j k    A (x,y,z) MO  r  F  a b c c Fyj Fx Fy Fz r A O Fxi Taking the determinant of the matrix, b     y M O   bFz  cFy  i   aFz  cFx  j   aFy  bFx  k a    C  M xi  M y j  M z k x www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 84  Moment of a Force: Vector Formulation  Alternatively, a sort of expansion and simplification is done.   MO  r  F           ai  bj  ck  Fx i  Fy j  Fz k                      ai  bj  ck  Fx i  ai  bj  ck  Fy j  ai  bj  ck  Fz k           But   z i  i  0 j  i  k k  i  j Fzk i  j  k j  j  0 k  j  i i j A (x,y,z) ik   j jk  i k k  0  c Fyj A Therefore, k r       O Fxi M O  bFx k  cFx j  aFy k  cFy i  aFz j  bFz i b y    a   bFz  cFy  i   aFz  cFx  j   aFy  bFx  k    C  M xi  M y j  M z k x www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 85  Moment of a Force: Vector Formulation Directions of component moments www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 86  Moment of a Force: Vector Formulation Determine the magnitude and sense of the moment due to the 100 N force shown below about A.  4  3 F   200i   200 j 5 5  160i  120 j   r  rAB  4i  6 j y i j k x      z M A  r  F  rAB  F   4 6 0  160 120 0  480k lb.in www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 87  Moment of a Force: Vector Formulation  A 1000-N force is applied to a beam cross section as show below. Determine the moment of the force about O, and the perpendicular distance from point B to the line of the action of the force. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 88  Moment of a Force: Vector Formulation Determine the moment of the force F, about point C. z z 2m 2m B B F = 500 N A F = 500 N 3m A y 3m y rCA rCA 4m C 4m C x x www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 89  Moment of a Force: Vector Formulation Determine the moment of the force F, about point C.    M C  rCA  F   rCA  2 mi z     2i  4 j  3k  F  F    500 N    22  4 2  32      2m     (185.53 N) i   371.06 N  j   278.29 N  k B F = 500 N    A i j k  3m M C  2 0 0 y rCA 185.53 371.06 278.29 4m C  556.58 j  742.12k x M C  (556.58) 2  (742.12) 2  927.64 Nm M 927.64 Nm The perpendicular distance between F and point C =   1.86 m F 500 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 90  Moment of a Force: Vector Formulation The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. y x z www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 91  Moment of a Force: Vector Formulation    M A  rAC  F    rAC   0.3 m  i   0.08 m  j         0.3 m  i   0.24 m  j   0.32 m  k  F  F    200 N     0.5 m        120 N  i   96 N  j  128 N  k    rAC i j k  M A  0.3 0 0.08  120 96  128     M A  7.68 N  m  i  28.8 N  m  j  28.8 N  m k www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 92  Moment of a Force: Vector Formulation The turn buckle is tightened until the tension is cable AB is 2.4 kN. Determine the moment about point O of the cable force acting on point A and the magnitude of this moment. -2.74i+4.39j+2.19k kN.m www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 93  Equivalent Force-Moment Systems www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 94  Equivalent Force-Moment Systems A system of forces act on a body, can be reduced to a force-couple system. The force-couple system comprises a resultant force (evaluated with the particle idealization at a desired point) and a resultant moment about that desired point. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 97  Equivalent Force-Moment Systems Example For the machine part shown in the Figure below, shift the line of application of the applied load of 150 kN acting at point A to point B. A 150 kN 80 mm 40 mm B 80 mm 40 mm C www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 98  Equivalent Force-Moment Systems Solution M B  150(0.08 m  0.04 m)  -18 kNm A A 150 kN 80 mm 80 mm 40 mm 40 mm 150 kN B 80 mm B 80 mm 40 mm 40 mm MB = 18 kNm C C www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 99  Equivalent Force-Moment Systems Example Shift the line of application of the 350-N force shown below to point B. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 100  Equivalent Force-Moment Systems Example For the beam, reduce the system of forces shown to (a) an equivalent force- couple system at A, (b) an equivalent force couple system at B. (Ignore the support reactions) 150 N 600 N 100 N 250 N A B 1.6 m 1.2 m 2m www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 101  Equivalent Force-Moment Systems Solution (@ B) The Resultant force will be 150 N 600 N 100 N 250 N   R  F  150 N   600 N   100 N   250 N    600 N  A B The Resultant Moment  2m M B   r  F  3.2 m  (250 N  0m)  100 N  2m   600 N  3.2m   150 N  4.8m  4.8 m  1000 Nm 600 N Equivalent system = 1000 Nm www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 102  Equivalent Force-Moment Systems Example Reduce the system of forces and couple acting on the arm to an equivalent force couple system at O. Take M to be 15 kNmm and ignore support reactions. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 104  Equivalent Force-Moment Systems Soln. FBD Equivalent system 695.78 N 60o 15 kNmm 320 N 76.7o 30o 30o O O 133 kNmm 400 N www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 105  Equivalent Force-Moment Systems Example Four tugboats are used to bring an ocean liner to its pier. Each tugboat exerts a 5000-lb force in the direction shown. Determine the equivalent force-couple system at the foremast O. Also determine the angle the resultant force makes with the horizontal as well as the direction of rotation of the moment. R = 13.33 klb (@ 47.3° CW from +ve x axis) MRO = 1035 klb.ft www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 106  Outline Some Concepts, Principles and Techniques in Mechanics of Static Rigid Bodies; o Fundamental Principles and Concepts in Newtonian Mechanics. o Forces; Characteristics of Forces, Resultants of systems of forces, Resolving forces into components. o Moment of a Force o Centroids and an introduction to Area Moments of Inertia o Equilibrium Analyses of Particles and Rigid Bodies Some Applications; o Analysis of Structures o Friction and Selected Simple Machines Introduction to Dynamics o Particle kinematic relationships o Motion; Rectilinear, Curvilinear, angular (in different coordinate systems) o Kinetics: Force, Mass and acceleration www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 121  Static Equilibrium Of Particles And Rigid Bodies Static Equilibrium Procedure for analyzing static equilibrium problems Free Body Diagrams www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 122  Static Equilibrium  A particle or body is said to be in equilibrium if the resultant force and moment acting on it is zero. In other words, the sum of forces (or moments) must be equal to zero.  R   F 0  For particles;   Fx  0 F y 0 F z 0  R   F 0  For bodies;   Fx  0  F y 0 F z 0 M   M 0  Mx  0 M y 0 M z 0 www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 123  Solving Static Equilibrium Problems Involves three main steps; Sketch a free body diagram for the problem Sum up forces and moments to obtain the equations of equilibrium for the problem. Solve the equations and interpret your results. www.knust.edu.gh DEK Dzebre/2022/ ME162 (Basic Mechanics) / ME164 (Statics of Solid Mechanics) 124  Sketching Free Body Diagrams Select the extent of the body that is of interest, detach it from the ground and all other bodies and supports, and (basically) sketch the outline of the detached (free) body. Indicate force and moment reactions that are exerted on the “free-body” by the ground and other supports that keep it in equilibrium (Newton’s 3rd