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ENT151/4 STATICS ENT151 STATICS ZOL BAHRI BIN RAZALI Lecturer School of Mechatronic Engineering Northen Malaysia University College of Engineering (KUKUM) (2004) Reference and textbook : STATICS, By Hibbeler, R.C. (2001) Prentice Hall, Inc. Copyright © 2004 by Zol Bahri Razali. All rights reserved. 1 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 1 GENERAL PRINCIPLES 1.1 Mechanics Can be defined as that the branch of the physical science concerned with the state of rust or motion of the bodies that are subjected to the action of forces. In general, this subject is subdivided into three branches rigid body mechanics and fluid mechanics. Rigid Body Mechanics Divided into two areas: statics and dynamics. Statics Deals with the equilibrium of bodies that are those that are either at rest or more with the constant velocity. Dynamics Is concerned with the accelerated motion of the bodies 1.2 Fundamental Concepts a) 4 basics quantities Length To locate the position of a point in space and there by describe the size of a physical system. Time Is conceived as a succession of events Mass Property manifests it self as a gravitational attraction between two bodies and provided a quantitative measure of the resistance of matter to change in velocity. Force Is considered on a push or poll exerted by one body on another. 2 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS b) Model or idealizations Particle Has a mass but a size that can be neglected. Rigid body Considered on a combination of a large number of particles in which all the particles remain at fixed distance from one another both forever and after applying a load. Concentrated force Represents the effect of a loading which is assumed to act point on the body. c) Newton's Law First Law A particle originally at rest a moving in a straight like with contact velocity, will remain in this state provided the particle is not subjected to an unbalance force F1 F2 ● (equilibrium) F3 Second Law A particle acted upon by an unbalance force F experience an acceleration that has same direction on the force and magnitude that is directly proportional to the force. If F expressed mathematically. F = m.a F ● a Third Law The mutual force of action and reaction between to particles are equal, opposite and collinear. F ●● Force of A on B F Force of B on A 3 (action – reaction) Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS d) Newton's Law of Gravitational Attraction F = G (M1 . M2 / r²) Where : F G M1 , r - Force of gravitation between two particles - Universal constant of gravitation [ G = 66.73 (10‾ ¹²) m³/ (kgs²) ] – mass of each of the two particles – Distance between the two particles Weight, will be the only gravitational force considered in our study of mechanics. W = m.g where, g - accaelaration due to gravity (9.81 m/s2) at sea level. 1.3 SI Units Length (m), time (s), mass (kg), force (N) F = m.a Thus 1 Newton is equal to a force required to give 1 kilogram of mass an acceleration of 1 m/s². N = kg.m/s² 1kg 9.81 N 4 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 1.4 Dimensional Homogeneity The term of any equation want to describe a physical process must be dimensionally homogeneous, that is, each term must be expressed in the same units. e.g.: e.g.: s = vt + ½ at² m = (m/s) (s) + ½ (m/s) (s) m=m where s - (m), v - (m/s), t - (s), a - (m/s) same units – dimensionally homogenous a = 2s/t² - 2v/t m/s² = 2 m/s² - 2 (m/s) s = 2 m/s² - 2m/s² m/s² = m/s² - dimensionally homogenous 5 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 2 FORCE VECTORS 2.1 Scalars & Vectors Scalar - a quality characterized by positive or negative members (mass, volume, length) Vector – a quality that has both a magnitude and a direction. Vector represented by Ā Magnitude epresented by |A|, positive quantity. - Magnitude, length of arrow (A) - Direction, angle between axis and the arrow’s line. 2.2 Vectors Operations 6 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Multiplication & division : Addition of a vector: parallelogram law Substraction of a vector : R=A−B = A + (− B) 7 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Resolution of vector : Adition of force : FR = (F1 + F2) + F3 EXAMPLES Example 2-1 (pp 22) Use cosine law to find R, sine law to find θ. 8 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Note EXAMPLES Example 2-2 (pp 23) Example 2-3 (pp 24) Example 2-4 (pp 25) 2.3 Additional of Coplaner Forces F = Fx + Fy F’ = Fx’ + Fy’ F’ = Fx’ i + Fy’ j 9 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Coplanar force resultant : Resultant : FR = = = = where : F1 + F2 + F3 F1x i + F1y j - F2x i + F2y j + F3x i - F3y j ( F1x - F2x + F3x ) i + ( F1y + F2y - F3y ) j ( FRx ) i + ( FRy ) j FRx = ∑ Fx ; FRy = ∑ Fy Resultant force, FR = √ [(FRx)2 + (FRy) 2 ] Angle θ = tan -1 ( FRy / FRx ) EXAMPLES Example 2-5 (pp 35) Example 2-6 (pp 36) 10 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 2.4 Position Vectors x , y , z coordinates ; A (x , y , z) Position vector, is defined as a fixed vector which locates a point in space relative to another point. 11 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS = = = = rA + rB rB - rA ( xB i + yB j + zB k ) - ( xA i + yA j + zA k ) (xB - xA) i + (yB - yA) j + (zB - zA) k EXAMPLE Example 2- 12 (pp 57) 2.5 Force Vector Directed Along a Line F – force r - position vector u - unit rector = r / r r - radius from origin The force, F directed along the cord AB The position vector, r directed from point A to point B on the cord. This common direction is specified by the unit vector, u = r / r. F = Fu = F ( r / r) where : F is unit of force r is unit of length 12 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 2-13 (pp 59) 2.6 Example 2-14 (pp 60) Dot Product The dot product of vectors A and B, written A . B is defined as the product of the magnitudes of A and B and the cosine of the angle between their tails. A . B = A B Cos θ 0° ≤ θ ≤ 180° Law of operation 1. A . B = B . A 2. a (A . B ) = ( aA ) . B = A . ( aB ) = (A . B )a 3. A . ( B + D ) = ( A . B ) + ( A . D ) Cartesian Vector Formulation i . i = (1) (1) cos 0° = 1 i . j = (1) (1) cos 90° = 0 Therefore : i . i = 1 i . j = 0 j . j = 1 i . k = 0 k . k = 1 k . j = 0 A . B = ( Ax i + Ay j + Az k ) . ( Bx i + By j + Bz k ) = Ax Bx ( i . i ) + Ax By ( i . j ) + Ax Bz ( i . k ) + A y Bx ( j . i ) + A y By ( j . j ) + A y Bz ( j . k ) + Az Bx ( k . i ) + Az By ( k . j ) + Az Bz ( k . k ) 13 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS By dot operation; A . B = Ax Bx + Ay By + Az Bz Thus to determine the dot product of two Cartesian Vectors, multiply their corresponding x , y , z components and sum their products algebraically. The result is a scalar. Application Dot Product 1. Angle formed between two vectors A . B = A B Cos θ If A.B = 0; Hence 2. θ = cos-1 ( A . B / A B ) θ = cos-1 0 = 90° A┴ B The components of a vector parallel and perpendicular to a line. A ╖ = A Cos θ = A.u where u is a unit vector A┴ = A-A╖ since A = A ┴ + A ╖ To determine θ , θ = cos-1 ( A . u / A ) Then A ┴ = A Sin θ or A ┴ = √ ( A2 - A╖2 ) 14 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 2-16 (pp 71) Example 2-17 (pp 72) EXERCISES Exercise 2 - 6 Exercise 2 - 14 Exercise 2 - 21 Exercise 2 - 26 Exercise 2 - 31 Exercise 2 - 55 Exercise 2 - 63 Exercise 2 - 87 Exercise 2 - 88 Exercise 2 - 114 15 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3.1 Condition of the equilibrium The resultant force acting on a particle to be equal to zero. ∑F = 0 ; 3.2 Newton’s Law ∑ F = m.a where a = 0 Free Body Diagram Account for all known and unknown forces, ∑ F which act on the particles, by draw the particles free body diagram. Spring : Linear elastic spring, F = k.s where k – stiffness (spring constant ) s - distance (from unload position) Cable and Pulleys 16 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 3-1 (pp 84) 3.3 Coplaner Force System For equilibrium ∑ F = 0 ∑ Fx i + ∑ Fy j = 0 ∑ Fx = 0 ; ∑ Fy = 0 EXAMPLES Example 3-2 (pp 87) 3.4 Example 3-3 (pp 88) Three Dimensional Force System ∑F = 0 ∑ Fx = 0 ; ∑ Fy = 0 ; ∑ Fz = 0 » ∑ Fx i + ∑ Fy j + ∑ Fz = 0 17 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 3 -5 (pp 100) Example 3 - 7 (pp 102) Example 3 - 6 (pp 101) Example 3 - 8 (pp 103) EXERCISES Exercise 3 - 7 Exercise 3 - 11 Exercise 3 - 17 Exercise 3 - 19 Exercise 3 - 38 Exercise 3 - 41 Exercise 3 - 45 Exercise 3 - 57 18 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 4 FORCE SYSTEM RESULTANTS 4.1 Moment of a force – Scalar formulation Moment - provides a measure of the tendency of the force to cause a body to rotate about the point of axis. This force tends to cause the pipe to turn about the z-axis. Mo = Fx x d Mo - moment of a force / torque d - moment arm Direction of moment - use right hand rule. Resultant moment of a system of coplanar forces MR = ∑F.d 19 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS The moment of any force will be +ve along the +ve z axis and –ve moment along the –ve z axis. EXAMPLES Example 4-1 (pp 116) Example 4-2 (pp 117) Example 4-3 (pp 118) 4.2 Cross product C = AxB; C = AB Sin θ C is specified by right-hand rule. defined the direction of C. (0o ≤ θ ≤ 180o ) AB Sin θ defined the magnitude C and unit vector uc 20 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Law of operation 1. AxB ≠ BxA ; AxB = -BxA 2. a ( A x B ) = ( aA ) x B = A x ( aB ) 3. Ax(B+D) = (AxD)+(AxD) = (AxB)a Cartesian Vector Formulation ixj = k ji x i = - k jxk = i kxj = -i kxi = j i x ki = - j A x B = ( Ay Bz – Az By ) i - ( Ax Bz – Az Bx ) j + ( Ax By – Ay Bx ) k = │ │ │ i j k│ Ax Ay Az │ Bx By Bz │ 21 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 4.3 Moment of a Force – vector formulation Mo = = = = r x F r . F Sin θ F ( r . Sin θ ) F.d where : r - position vector Cartesian vector : Mo = ( ry Fz – rz Fy ) i - ( rx Fz – rz Fx ) j + ( rx Fy – ry Fx ) k Mo = r x F = │ │ │ i rx Fx j ry Fy k rz Fz │ │ │ Resultant moment : MRo = ∑( r x F ) EXAMPLES Example 4-4 (pp 124) Example 4-5 (pp 125) 22 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 4.4 Principle of Moments The moments of a force about a point is equal to the sum of the moments of the force’s components about the point. Mo = r x F1 + r x F2 = r x ( F1 + F2 ) = r x F EXAMPLES Example 4-6 (pp 127) Example 4-7 (pp 128) 4.5 Moment about specified axis To find the components of the moment along a specified axis, Ma = F . da If the line of action of a force F is perpendicular to any specify axis, the magnitud of the moment of F about the axis can be determine by : Ma = F . da 23 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Where : da – the perpendicular / shortest distance between the force line to the axis. By vector analysis : Ma = ua ( r x F ) EXAMPLES Example 4-8 (pp 142) 4.6 Example 4-9 (pp 143) Moment of a couple A couple is defined as two parallel forces that have the same magnitude, have opposite directions and are separated by a perpendicular distance, d. Moment couple = sum of the moments of both couple forces about any arbitrary point. It is simpler to take moments about a point lying on the line of action of one of the forces. e.g At point A: the moment of – F about A is zero, M = r x f M depends only upon the position vector, r directed between the forces. M = F.d 24 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS MR = M1 + M2 = ∑( r x F ) EXAMPLES Example 4-10 (pp 150) Example 4-11 (pp 151) Example 4-12 (pp 152) 4.7 Equivalent system 25 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 4.8 Resultant of a Force and couple system = M1 = r1 x F1 ; = M2 = r2 x F2 ; FR = F1 + F2 MRo = Mc + M1 + M2 = ∑ M c + ∑ Mo FRx = ∑ Fx ; FRy = ∑ Fy EXAMPLES Example 4-14 (pp 164) 4.9 Example 4-15 (pp 165) Reduction of a Force and Couple System If the force and couple moment system is reduce to a resultant system at point P, only the force resultant will have to be applied to the body. 26 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 4-16 (pp 170) Example 4-17 (pp 171) Example 4-18 (pp 172) Example 4-19 (pp 173) 4.10 Reduction of a simple Distributed Loading FR = ∑ F Magnitude : FR = ∫L W (x) dx = ∫L dA = A Magnitude of the resultant force is equal to the total area A under the loading diagram W = W(x) Location : MRo = ∑ Mo ; x = ∫L x W(x) dx ∫L W(x) dx x = ∫L x dA ∫L dA x . FR = ∫L x . W(x) dx The resultant force has a line of action which passed through the centroid C (geometric center) of the area defined by the distributed loading diagram w(x). 27 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 4-20 (pp 182) Example 4-21 (pp 183) Example 44-22 (pp 184) EXERCISES Exercise 4 - 9 Exercise 4 - 14 Exercise 4 - 22 Exercise 4 - 36 Exercise 4 - 53 Exercise 4 - 63 Exercise 4 - 70 Exercise 4 - 82 Exercise 4 - 105 Exercise 4 - 121 Exercise 4 - 145 28 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5.1 Conditions Forces : Fi + fi = 0 ; Moments : ri x (Fi + fi ) = ri x Fi + ri x fi = 0 ∑Mo = ∑ ri x Fi ; ∑Fi + ∑fi = 0 ; ∑Fi = ∑F ; ∑F = 0 ∑Mo = 0 The two equations of equilibrium for a rigid body : ∑F = 0 ∑Mo = 0 5.1 Free Body Diagram (FBD) To draw free body diagram : Refer Table 5.1 below – Supports for rigid body. 5.2 Support Reactions If a support prevents the translation of a body in a given direction, then the force is developed on the body in that direction. If rotation is prevent, a couple moments is exert on the body. Types of support reactions, and forces and couple moments involved : 29 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 5-1 (pp 202) Example 5-2 (pp 203) Example 5-3 (pp 204) Example 5-4 (pp 205) Example 5-5 (pp 206) 30 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 31 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 5.3 Equations of Equilibrium ∑Fx = 0 ∑Fy = 0 ∑Mo = 0 EXAMPLES Example 5-6 (pp 211) Example 5-7 (pp 212) Example 5-8 (pp 213) Example 5-9 (pp 214) Example 5-10 (pp 215) Example 5-11 (pp 216) Example 5-12 (pp 217) 32 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 5.4 Two and three-force members Two force members : Three-force members EXAMPLE : Example 5-13 (pp 220) 33 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 5.5 Equilibrium in Three Dimensions Free body diagram – Support reactions A force is developed by a support that restrict the translation of the attached members whereas a couple moment is develop when rotation of the attached member is prevented. Magnitude of force : F = √ (Fx2 + Fy2 + Fz2) The third direction angle : Cos2 α + Cos2 β + Cos2 γ = 1 [Table 5-2 – Support for Rigid Bodies Subjected to Three Dimensional Force System] 34 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 5-14 (pp 211) 5.6 Example 5-15 (pp 212) Equation of Equilibrium Scalar Equations : ∑F = 0 ∑M = 0 Vector Equations : ∑F = ∑Fx i + ∑Fy j + ∑Fz k =0 ∑M0 = ∑Mx i + ∑My j + ∑Mz k 35 =0 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Where ; ∑Fx = 0 ; ∑Fy = 0 ; ∑Fz = 0 ∑Mx = 0 ; ∑My = 0 ; ∑Mz = 0 5.7 Constraints for a Rigid Body Redundant constraints When the body has a redundant support more supports then are necessary to hold it in equilibrium, it becomes statically indeterminate. Statically indeterminate – there will be more unknown loadings on the body then equations of equilibrium available for their solution. Eg : Two dimensional case – five unknown but only three equation. Three dimensional case – eight unknown but only five equation Improper Constraints Instability of the body because of improper constraining by the supports. 3-Dimensional – if the support reaction all intersec a common axis 2-Dimensional – the axis is perpendicular to the plane of the forces When all the reactive forces are concurrent at the point, the body is improperly constraint. Improper constraining leads to instability occurs when the reactive forces are all parallel. 36 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Proper constraining requires 1. the line of action of the reactive forces does not intersect point on a common axis 2. the reactive force must not all be parallel on one another EXAMPLES Example 5-15 (pp 242) Example 5-16 (pp 243) Example 5-17 (pp 244) Example 5-18 (pp 245) Example 5-19 (pp 246) 37 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXERCISES Exercise 5 - 5 Exercise 5 - 19 Exercise 5 - 22 Exercise 5 - 24 Exercise 5 - 89 Exercise 5 - 95 38 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 6 STRUCTURAL ANALYSIS 6.1 Trusses A structure composed of slender members joined together at their end points. Planar Truss Lie in a single plane and are often used to support roofs and bridges. 39 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Assumption for Design All loading are applied at the joints. The members are joined together by smooth pins. Each truss member acts as a two force member Tension - tend to elongate the member (tensile force) Compression - tend to shorten the member (compressive force) Simple truss To prevent collapse, the form of a truss must be rigid. A simple truss is constructed by starting with a basic triangular element. 40 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 6.2 The method of joints Truss members are all straight two-forces members lying in the same plane, the forces system acting at each joint is coplanar and concurrent. Moment equilibrium is automatically satisfied at the joint or pin. ∑Fx = 0 ; ∑Fy = 0 Analysis should start at a joint at least one known force and at most two unknown. ∑Fx = 0 ; ∑Fy = 0 The member in compression “pushes” on the joint and a member in tension “pulls” on the joint. EXAMPLES Example 6-1 (pp 262) Example 6-2 (pp 263) Example 6-3 (pp 264) Example 6-3 (pp 265) 41 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS * Note : First – get all the equilibrium forces (external) through FBD. Second – do the analysis 6.3 Zero- Force Members Zero force members are used to increase the stability of the truss during construction and provide support of the applied loading is changed. If only two members from a truss joints and no external load or support reaction is applied to the joint, the members must be zero-force member. zero-force member ∑Fx = 0 ; FAB = 0 ∑Fy = 0 ; FAF = 0 If three members from a truss joint for which two of the members are collinear, the third member is a zero-force member provide no–external force or support reaction is applied to the joints. EXAMPLES Example 6-4 (pp 268) 42 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 6.4 Method of Sections To determined the loadings acting within a body. if a body is in equilibrium, then any part of the body is also in equilibrium. Each internal force as external force to the Free Body diagram (FBD) ∑Fx = 0 ; ∑Fy = 0 ; ∑Mo = 0 ; Analysis – select the section passed through not more than three members in which the forces are unknown. First, find the total external forces through FBD. Analysis for the end part to final FBC, FGC, FGF Start with ∑MG = 0 to find FBC 43 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 6-5 (pp 276) Example 6-6 (pp 277) Example 6-7 (pp 278) 6.5 Space Trusses Space truss consists of members jointed together at their ends so form a stable threedimensional structure – e.g tetrahedron (ball and socket joint). Additional member would be redundant. Procedure of analysis : Method of joint / method of section. 44 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Method of joint – if the forces in all the members of the truss must be determined : ∑Fx = 0 ; ∑Fy = 0 ; ∑Fz = 0 ; The analysis start at the joint at least one known and three unknown forces. Use the Cartesian Vector analysis. Method of section – if only a few member forces are to be determined : When an imaginary section is passed through a truss and the truss saparated into two parts, the force system acting on one of the parts must satisfy. ∑Fx = 0 ; ∑Mx = 0 ; ∑Fy = 0 ; ∑My = 0 ; ∑Fz = 0 ; ∑Mz = 0 ; EXAMPLES Example 6-8 (pp 284) 6.6 Frame and Machine Compose of pin – connected multiforces members. Frame are generally stationary and are use to support loads. Machines contain moving parts and are designed to transmit and alter the effect of forces. To analysis Draw free body diagram Show all the forces and couple of moments / known-unknown forces Identify all the two-forces-member Internal force not shown / external forces EXAMPLES Example 6-9 (pp 288) Example 6-10 (pp 289) 45 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Example 6-11 (pp 290) Example 6-12 (pp 291) Example 6-13 (pp 290) Equations of Equilibrium In the x – y plane : ∑Fx = 0 ; ∑Fy = 0 ; ∑Mo = 0 ; Note : Draw FBD of the entire structure The force at the connected parts of the group are internal force Force common to two members with equal magnitude but opposite sense Two-force mebers, equal but opposite collinear forces 46 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS The proper sense of the unknown forces acting A couple moment is a free vector and can act at any point. EXAMPLES Example 6-14 (pp 295) Note : Always identify the two-force members before starting the analysis. EXAMPLES Example 6-15 (pp 296) Example 6-16 (pp 297) Example 6-17 (pp 298) Example 6-18 (pp 299) Example 6-19 (pp 300) Example 6-20 (pp 301) Example 6-21 (pp 302) EXERCISES Exercise 5 - 5 Exercise 5 - 19 Exercise 5 - 22 Exercise 5 - 24 Exercise 5 - 89 Exercise 5 - 95 47 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 7 INTERNAL FORCES 7.1 Internal Forces Developed in Structure Members The internal loadings can be determined by using the method of sections. If the internal loadings acting on the cross section at C are to be determined, then an imaginary sections is paased through the beam, cutting it into two segments. By doing this, the internal loadings at the section become external on the free body diagram. In equilibrium: – each segment is maintained provided rectangular force components Nc, Vc, and Mc. - equal in magnitud but opposite in directions Direct solution – Nc : ∑Fx = 0 ; Vc : ∑Fy = 0 ; 48 Mc : ∑Mo = 0 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS - where V = shear force N = Normal force M = Bending Moment My = torsional / twisting moment All resultant force /moment acts on the center of centroid (c) of the section cross area. Free Body Diagram Frame and machine - multi force members -subjected to internal normal, shear and bending loadings. 49 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Instead, we must dismember the frame and determine the reaction. Note : We cannot apply the three equation of equilibrium to obtain these nine unknown. Analysis Before the member is ‘cut’, determined the all members support reaction. Keep all the forces, moments at exact locations. Draw FBD of the segment that has the least number of loads on it (x,y,z components) Coplaner system only N, V & M Moments should summed at section passing through centre EXAMPLES Example 7-1 (pp 327) Example 7-2 (pp 328) Example 7-3 (pp 329) Example 7-4 (pp 330) Example 7-5 (pp 331) Example 7-6 (pp 332) 7.2 Shear and moment equations and diagram Beams – to support loadings applied perpendicular to their axis. In general, beams are long, straight bars, constant cross-sectional areas. Simply supported beam – is pinned at one end and roller support at the other. Centilever beam – is fixed at one end and free at the other. 50 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS V - internal shear force / shear diagram M - bending moment / bending moment diagram. The variation of V and M as functions of the position x along the beam’s axis can be obtain by using the method of sections. It is necessary to section the beam at an arbitrary distance x from one end rather then at a specified point. If the result are plotted, the graphical variation of V and M as functions of x are termed the shear diagram and bending moment diagram. Because of the internal shear and bending moment functions will be discontinous, the functions must be determined for each segment of the beam located between any two discontinuities of loading. Procedure for analysis Determined all the reactive forces and couple moment acting on the beam, and resolve all the forces into components acting perpendicular and parallel to the beam axis. Specify separate coordinates and having an origin at the beam’s left-end Section the beam perpendicular to the axis. Shear V-summing forces perpendicular to the axis Moment M-slimming moments about section end. Plot the shear diagram (V vs x) and moment diagram (M vs x) EXAMPLES Example 7-7 (pp 342) Example 7-8 (pp 343) 7.3 Relations Betwen Distributed Load, Shear and Moment --- TO BE CONTINUE --51 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXERCISES Exercise 7 - 2 Exercise 7 - 4 Exercise 7 - 10 Exercise 7 - 33 Exercise 7 - 49 Exercise 7 - 59 Exercise 7 - 86 52 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 8 FRICTION 8.1 Characteristic of dry friction Friction A force of resistance acting on a body which prevents slipping of the body relative to a second body or surface with which it is in contact. ∆ F - frictional force ∆ N - normal force For equilibrium; and ∆N W P ∆F Free body diagram ; F - acts tangent to the contacting surface 53 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS - opposite direction of P N - directed upward - balance the block weight (W) Impending motion When P slowly increase, F correspondingly increase until maximum value, Fs - limiting static frictional force. When the value reached, the block unstable equilibrium. Increase P, the block begin to move. Fs directly propotional normal force, Ns. Fs = μs N ; where μs - coefficient of static friction. When the block verge of sliding, N & F create Rs and the angle ø called angle of static friction. Øs = tan -1 (Fs / N) = tan -1 (μs N / N) = tan -1 μs Tabular Value Ms μs - coefficient of static friction – dimensionless - depends on the characteristic of two surfaces. Motion When P > Fs, frictional force drops slingtly to smaller value Fk - kinetic frictional force. P > Fk – block not in equilibrium – begin to slide with increasing speed. 54 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Fk directly propotional to the magnitude of the N. Fk = μk N where μk – coefficient of kinetic friction. = 25 % smaller than μs Øk = tan -1 (Fk / N) = tan -1 (μk N / N) = tan -1 μk Øs ≥ Øk Note : slipping about to occur, 8.2 Fs = μs N ; slipping is occuring, Fk = μk N Problems involving dry friction Equilibrium: The total number of unknowns to be equal to the total number of available equilibrium equations. F ≤ μs N 55 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS From the equations of equilibrium, FA, NA, Fc and Nc are determined. The bar remain in equilibrium provide FA ≤ 0.3 NA , Fc ≤ 0.5 Nc Impending motion at all points : Total number of unknowns = total number of available equilibrium equations + total number of available frictional. If motion is impending, Fs ≤ μs N If the body is shipping, Fk ≤ μk N Impending motion at some point Total number of unknowns < number of available equilibrium equation + the total number of frictional equation. Unknowns : NA, FA, Nc, Fc, Bx, By, P Equations : six equilibrium equation (three of each member) one static frictional equation Conditions : 1. Slipping at A, no slipping at C FA = 0.3 NA , Fc ≤ 0.5 Nc 2. No slipping at A, slipping at C Fc = 0.5 Nc , FA ≤ 0.3 NA ; depend on which P is smaller. Equilibrium Versus Frictional Equestions F (equilibrium force) < μs N 56 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Procedure for analysis Draw free-body diagrams – always show frictional force as unknowns. Determine the number of unknowns. Apply the frictional equations EXAMPLES Example 8-1 (pp 385) Example 8-2 (pp 386) Example 8-3 (pp 387) Example 8-4 (pp 388) Example 8-5 (pp 389) Example 8-5 (pp 390) EXERCISES Exercise 8 - 5 Exercise 8 - 11 Exercise 8 - 39 Exercise 8 - 43 Exercise 8 - 58 57 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9.1 Center of gravity / mass of the particle Center of gravity – G - locates the resultant weight of a system of particles. WR = ∑ W ; Sum of moments of all weights = moments of the resultant weight. ∑ Mall = MR WR = x1 W1 + x2 W2 + x3 W3 = cordinate o gravity (c.g) cordinate => Hence, = ∑xW ∑ W WR = y1 W1 + y2 W2 + y3 W3 = ∑yW ∑ W = ∑zW ∑ W - cordinate of the center of gravity x , y , z - cordinate of the each particles. 58 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Center of mass ; W = m.g = ∑xW ∑ W Hence, = = ∑xm.g ∑ m.g = ∑xm ∑ m ∑ym ∑ m = ∑zm ∑ m 9.2 Center of Gravity / Mass and Centroid of Body Center of gravity (c.g) : Arbitrary particle (x x , y , z ) , weight dW = ∫ x dW ; ∫ dW = If γ - specific weight, dW = γ dv ; ∫ y dW ∫ dW = ∫v x γ dv ∫v γ dv = = ∫ z dW ∫ dW ∫v y γ dv ∫v γ dv = ∫v z γ dv ∫v γ dv 59 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS Center of mass : Density , ρ = mass per unit volume ; γ = ρ.g = ∫v x ρg dv ∫v ρg dv = ∫v y ρg dv ∫v ρg dv = ∫v x ρ dv ∫v ρ dv = ∫v y ρ dv ∫v ρ dv = ∫v z ρg dv ∫v ρg dv = ∫v z ρ dv ∫v ρ dv Centroid : Centroid, C -- the geometric center of an object and depends only the body’s geometric. Volume = ∫v x dv ∫v dv = ∫v y dv ∫v dv = ∫v z dv ∫v dv = ∫A y dA ∫A dA = ∫A z dA ∫A dA = ∫L y d L ∫L d L = ∫L z d L ∫L d L Area = ∫A x dA ∫A dA Line = ∫L x dL ∫L d L Symetry 60 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 9-1 (pp 441) Example 9- 2 (pp 442) Example 9-3 (pp 443) Example 9- 4 (pp 444) Example 9-5 (pp 445) Example 9- 6 (pp 446) Example 9-7 (pp 447) Example 9- 8 (pp 448) 9.3 Composite Bodies Composite body - consists of a series of connected simpler shaped bodies. Devide into its composite parts Provide weight and location of c.g of each parts Determine the c.g for entire body Hence, = ∑xW ∑ W = where - c.g composite body x - c.g each composite part ∑W - total weight of the body ∑yW ∑ W = ∑zW ∑ W When the body has a constant density or specific weight, center of gravity coincide with the centroid of body. EXAMPLES Example 9-9 (pp 461) Example 9- 10 (pp 462) Example 9-11 (pp 463) 61 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXERCISES Exercise 9 - 46 Exercise 9 - 61 Exercise 9 - 70 Exercise 9 - 75 TO BE CONTINUE 62 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS CHAPTER 10 MOMENT OF INERTIA 10.1 Moment of Inertia for an Areas ∫ x 2 dA ; Moment of inertia for areas The moment of inertia of the different planar area dA about x and y axes are : dIx = y2 dA and dIy = x2 dA Hence, Ix = ∫A y2 dA ; Iy = ∫A x 2 dA Second moment of dA about the pole O or z axis, refered to as the polar moment of inertia, dJo = r2 dA ; where r is the perpendicular distance from the pole (z axis) Polar moment of inertia, Jo = ∫A r2 dA = Ix + Iy Where : r2 = x2 + y2 ; Jo, Ix, Iy always positive 10.2 Parallel-axis Theorem for an Area When the moment of inertia for an area is known about an axis passing through its centroid, determine the moment of inertia for an area about a corresponding parallel axis using the Parallel-axis Theorem. Figure below shown, a differential element dA is located at an arbitrary distance y’ from the centroidal x’ axis, whereas the fixed distance between the parallel x and x’ axes is defined as dy . Since the moment of inertia of dA about the x axis is 63 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS dIx = ( y’ + dy ) 2 dA For entire area; Ix = ∫A ( y’ + dy ) 2 dA = ∫A y’2 dA + 2 dy ∫A y’ dA + dy2 ∫A dA The first integral : moment of inertia for an area about centroidal axis Ix’ The second integral : zero, since the x’ axis passed through the area’s centroid The third integral : the total area Therefore ; Ix = ¯Ix’ + A dy2 ; For polar moment of inertia Iy = ¯Iy’ + A dx2 Jo = ¯Jc + Ad2 The three equations state that : The moment of inertia for an area about an axis is equal to the moment of inertia of the area about a parallel axis passing through the area’s centroid plus the product of the area and the square of the perpendicular distance between the axes. 10.2 Radius of Gyration of an Area The radius of gyration of a planar area has units of length and is a quantity that is often used for the design of columns in structural mechanics. Provided the areas and moments of inertia are known, the radii of gyration are: kx = √(Ix / A) ; ky = Where √(Iy / A) ; ko = √(Jo / A) Ix = kx2 A 64 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 10.4 Moment of Inertia for an Areas by Integration Procedure : If a single integration is performed to determine the moment of inertia of an area about an axis, it will be necessary to specify the differential element dA. Most often the element will be rectangular, such that it will have a finite length and differential width. The element should be located so that it intersects the boundary of the area at the arbitrary point (x,y), by two possible ways. Case 1 : The length of the element can be oriented parallel to the axis. Direct application Iy = ∫A x 2 dA can be made since the element has an infinitisimel thickness dx, therefore all parts of the element lie at the same moment-arm distance x from y axis. Case 2 : The length of the element can be oriented perpendicular to the axis. For this case, first , calculate the moment of inertia of the elemen about the horizontal axis passing through the element’s centroid then determine the moment of inertia of the element about the x axis by using the parallel-axis theorem. Integration of this result will yield Ix. EXAMPLES Example 10-1 (pp 501) Example 10-2 (pp 502) Example 10-3 (pp 503) Example 10-4 (pp 504) 65 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS 10.5 Moment of Inertia for Composite Areas A composite area consists of a series of connected “simpler” parts or shape, such as semicircles, rectangulars and triangles. Provide the moment of inertia of aech af these parts is known or can be determined about a common axis, then the moment inertia of the composit area equal the algebric sum of the moments of inertia of all its parts. Procedure : Using a sketch, divide the area into its parts and indicate the perpendicular distance from the centroid of each part to the reference axis. The moment of inertia of aech part should be determined about its centroidal axis, which is parallel to the reference axis. For calculation, use the table given. If the centroidal axis does not coincide with the reference axis, the parallel-axis theorem I = ¯I + A d2 should be used to determine the moment of inertia of the part about the reference axis. Summations: The moment of inertia of the entire area about the reference axis is determine by summing the result of its components parts. If the composite parts has a hole, its moment of inertia is found by substracting the moment of inertia for the hole from the moment of inertia of the entire part including the hole. EXAMPLES Example 10-5 (pp 509) Example 10-6 (pp 510) 10.6 Mass Moment of Inertia The mass moment of inertia of a body is a property that measures the resistance of the body to angular acceleration. Mass moment of inertia as the integral of the second moment about an axis of all the elements of mass dm which compose of the body. For the figure below, the body’s moment of inertia about z axis is I = ∫m r 2 dm 66 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS The moment arm, r is the perpendicular distance from the axis to the arbitrary element dm. If the body consists of material having a variable density, ρ = ρ (x , y , z), the elemental mass dm of the body may be eexpressed in terms of its density and volume as dm = ρ dV subsituting dm into I = ∫m r 2 dm the body’s moment of inertia is then computed using volume elements for integration I = ∫v ρ dV In the special case of ρ being constant I = ρ ∫v dV When the element volume choosen for integration has differential size in all three directions, eg: dV = dx dy dz; the moment inertia of the body must be determine using ‘triple integration’. 67 Zol Bahri Razali School of Mechatronics, KUKUM ENT151/4 STATICS EXAMPLES Example 10-11 (pp 535) Example 10-12 (pp 536) Example 10-13 (pp 539) Example 10-14 (pp 540) EXERCISES Exercise 10 - 2 Exercise 10 - 13 Exercise 10 - 28 Exercise 10 - 27 Exercise 10 - 34 Exercise 10 - 49 68 Zol Bahri Razali School of Mechatronics, KUKUM