CH. 1 FUNDAMENTAL PRINCIPLES OF MECHANICS

Transcription

1 (Solid echanics) Professor Youn, eng Dong CH. 1 FUNDENTL PRINCIPLES OF ECHNICS Ch. 1 Fundamental Principles of echanics 1 / 14

2 (Solid echanics) Professor Youn, eng Dong 1.2 Generalied Procedure u General steps to solve problems in applied mechanics i) Select sstem of interest ii) Postulate characteristics of sstem. This usuall involves idealiation and simplification of the real situation iii) ppl principles of mechanics to the idealied model. Deduce the consequences iv) Compare these predictions with the behavior of the actual sstem. This usuall involves recourse to tests and eperiments v) If satisfactor agreement is not achieved, the foregoing steps must be reconsidered. Ver often progress is made b altering the assumptions regarding characteristics of the sstem, i.e., b constructing a different idealied model of sstem à In this course, steps i), ii), iii) are mainl treated cf. In chapter 1, we appl the principles of mechanics to rigid bod Ch. 1 Fundamental Principles of echanics 2 / 14

3 (Solid echanics) Professor Youn, eng Dong 1.3 Fundamental Principles of echanics u Two different tpes of movement which are important in the mechanics of solids i) Gross overall changes in position with time = motion ii) Local distortions of shapes = deformation à In solid mechanics, we shall consider situations in which there is deformation u nalsis of a mechanical sstem i) Stud of forces applied to a sstem à Focused on the equilibrium of a sstem ii) Stud of motions and deformation in a sstem à Focused on the gross motion and local deformation iii) pplication of laws relating the forces to the motion and deformation 1.4 Concept of Force u Concept i) The development of the idea of force in mechanics has provided us with an effective means for describing a ver comple phsical interaction between bodies in terms of a simple, convenient concept. ii) Newton, in his third law, postulated equal and opposite effectiveness of force on the two interacting sstems Force interactions have two principal effects 1) lter the motion of the sstems 2) Deform or distort the shape of the sstems iii) ver important propert of force is that the superposition of forces satisfies the laws of vector addition iv) Summar 1) Force is a vector interaction which can be characteried b a pair of equal and opposite vectors having the same line of action 2) The magnitude of a force can be established in terms of a standardied eperiment. 3) When two or more forces act simultaneousl, at one point, the effect is the same as if a single force equal to the vector sum of the individual forces were acting. u Units of forces i) SI-unit ~ 1 Newton = 1 kgf m/s ii) United States unit ~ 1 lb ft/s 1 slug= lbm = / Ch. 1 Fundamental Principles of echanics 3 / 14

4 (Solid echanics) Professor Youn, eng Dong 1.5 oment of a force u Definition à The moment or torque of F about the point O is defined as the vector cross product. The resultant vector is perpendicular to the plane composed b F and r. The sense is determined b the right-hand rule. = Ch. 1 Fundamental Principles of echanics 4 / 14

5 (Solid echanics) Professor Youn, eng Dong u agnitude of the = = = h = h à is independent of the position of P along ; u Other epression of the moment (consider two-dimensional structure shown in Fig. 1.7) = = h = ( + ) + = u The magnitude of the moment of F about OQ-ais ( ) = cos = h u Units of the moment à [m N] or [ft lbf] u Eamples 1.1 Determine the moment about the shaft ais OO' due to the force P applied Ch. 1 Fundamental Principles of echanics 5 / 14

6 (Solid echanics) Professor Youn, eng Dong Sol) = = ( ) = [ ] = u Couple à Two equal and parallel forces which have opposite sense = + = ( + ) + = ( + ) + = It is independent of the location of O à The moment of a couple is the same about all points in space agnitude of the couple = h 1.6 Conditions for equilibrium u Equilibrium of a particle à ccording to Newton s law of motion, a particle has no acceleration if the resultant force acting on it is ero. We sa that such a particle is in equilibrium. Ch. 1 Fundamental Principles of echanics 6 / 14

7 (Solid echanics) Professor Youn, eng Dong Necessar and sufficient condition for the equilibrium = = u Equilibrium of particles Eternal Internal 1) Consider an isolated sstem of particles as indicated in Fig We sa that such a sstem is in equilibrium if ever one of its constituent particles is in equilibrium. 2) The forces acting on each particle are of two kinds, eternal and internal. The internal forces represent interactions with other particles in the sstem. 3) For rigid bod ~ necessar and sufficient condition for a perfectl rigid bod to be in equilibrium is that the vector sum of all the eternal forces should be ero and that the sum of the moments of all the eternal forces about an arbitrar point (together with an eternal applied moments) should be ero = = (1.5) = = (1.6) 2) For deformable sstem ~ necessar and sufficient condition for the equilibrium of a deformable sstem is that the sets of eternal forces which act on the sstem and on ever possible subsstem isolated out of the original sstem should all be sets of forces which satisf both (1.5) and (1.6). Two-force member à The line of action of must pass through and the line of action of must pass through. Ch. 1 Fundamental Principles of echanics 7 / 14

8 (Solid echanics) Professor Youn, eng Dong à = Three-force member à The three forces must all lie in the plane C and must all intersect in a common point O The total moment about the intersection of an two of the lines of action should be ero General two-dimensional equilibrium = = (1.7) = (1.8) 1.7 Engineering pplications (Free-bod diagram; Principles of mechanics; Staticall (in)determinate sstem) u The general method of analsis that is followed throughout this book i) Selection of sstem ii) Idealiation of sstem characteristics Table 1.2 Force-transmitting properties of some idealied mechanical elements Ch. 1 Fundamental Principles of echanics 8 / 14

9 (Solid echanics) Professor Youn, eng Dong à frictionless surface à frictional surface à frictionless bearing à frictionless pinned joint à The direction of the force is altered but its magnitude remains constant à n Ideal clamped support iii) These are followed b an analsis based on the principles of mechanics including the following steps: iii-1) Stud of forces and equilibrium requirements à Static conditions (Ch. 3) iii-2) Stud of deformation and conditions of geometric fit iii-3) pplications of force-deformation relations cf. Staticall determinate sstem: Possible to determine all the forces involved b using onl the equilibrium requirements without regard to the deformations Staticall Indeterminate sstem 1.8 Friction u Given pair of surfaces the forces, are proportional to the normal force N = = Ch. 1 Fundamental Principles of echanics 9 / 14

10 (Solid echanics) Professor Youn, eng Dong where : static coefficient of friction : kinetic coefficient of friction. à These coefficients are intrinsic properties of the interface between the materials and, being determined b the materials and and b the state of lubrication or contamination at the interface u Properties of the coefficients of friction i) oth coefficients of friction are nearl independent of the area of the interface. In particular, if bod in Fig were tipped up so that onl an edge or a corner was in contact with, we should still find approimatel the same coefficients of friction. Note that under these circumstances the tangential and normal directions are determined onl b the surface of. ii) oth coefficients are nearl independent of the roughness of the two surfaces, although this is a conclusion which man people find hard to accept. iii) The static coefficient is nearl independent of the time of contact of the surfaces at rest. Similarl, the kinetic coefficient is nearl independent of the relative velocit of the two surfaces. Figure 1.19 shows a schematic representation of tpical static-friction-time and kinetic-friction-velocit plots Hooke s Joint (Reading homework) Ch. 1 Fundamental Principles of echanics 10 / 14

11 (Solid echanics) Professor Youn, eng Dong Comparison an incorrect conclusion of the over-idealied model with a correct solution. u Incorrect solution. Considering the conditions of force balance from Fig.1.33a H - V = H = V = 0 From Fig. 1.33c = (a) lternativel, if we proceed directl with moment equilibrium about the point O, we have i - ( i cosq + k sinq ) - Li V j - L( i cosq + k sinq ) V j = 0 Working out the cross products, V L(1 + cosq ) = - + V Lsinq = The result as above is - sinq V L = 1+ cosq * sinq cosq Incorrect solution. Suppose we were to continue our analsis further b considering the shaft separatel as in Fig The sstem of Fig can t possibl be in equilibrium because there is nothing to balance V in the vertical Ch. 1 Fundamental Principles of echanics 11 / 14

12 (Solid echanics) Professor Youn, eng Dong direction. u Correct solution. To provide for the double-contact tpe of reaction shown in Fig. 1.35, we have four components 1, 2, 3,and 4 at bearing and the four components 1, 2, 3,and 4 at bearing. Ch. 1 Fundamental Principles of echanics 12 / 14

13 (Solid echanics) Professor Youn, eng Dong - From fig. 1-36(b)~ The statements of force and moment balance for the cross (Fig. 1.36b) are å å F = 0 (C = + D (ac + E - ad + F ) i + (E - ae + af ) i + Setting each component separatel equal to ero, + F ) j + (ae (C - af ) j + D ) k = 0 + (-ac + ad ) k = 0 (b) C C = D = -D = -E = E = -F = -F (c) - From fig (a) ~ In Fig. 1.36a, equilibrium of forces parallel to ields F = 0 and equilibrium of moments about the ais ields 2 af =. C C = D = -D = -E = E = -F = -F = 0 = - 2a (d) For completeness, considering the other equilibrium conditions in Fig 1.36a, = 2 = 3 = 4 1 = - From fig (c)~ Using values (d) in Fig. 1.36c 0 - ad = cosq + ac cosq cosq = 0 (e) For completeness, considering the other equilibrium conditions in Fig 1.36c = = - = 0 2 = b sinq Ch. 1 Fundamental Principles of echanics 13 / 14

14 (Solid echanics) Professor Youn, eng Dong To aid visualiation the complete solution is shown in Fig It is important to emphasie that our solution (e) is for the configuration shown if Fig n eact solution for arbitrar angle of orientation f of shaft measured from EF in the direction of twist of 1.32) can be found. The result is (Fig sin f + cos q cos f = (f) cosq When f =0, the result (f) reduces to (e). Ch. 1 Fundamental Principles of echanics 14 / 14