June : 2016 (CBCS) Body. Load

Transcription

1 Engineering Mecanics st Semester : Common to all rances Note : Max. marks : 6 (i) ttempt an five questions (ii) ll questions carr equal marks. (iii) nswer sould be precise and to te point onl (iv) ssume suitable data if necessar and state tem clearl. Q. (a) Define te term Force and state clearl te effects of force. Wat are te various caracteristics of a force? ns. Force is an external agent wic can cange te state of motion, sape and size of a bod. t means tat, on application of a force, a bod at rest ma start moving, or a moving bod ma come at rest. force can also deform a bod i.e. it can bend, twist or break a bod. Force can give basicall five effects, listed below in table : Table : Effects of forces S. No. Force Diagram Effect. Pull. Pus 3. end 4. Twist 5. Sear Elongation : t ma increase te lengt. Contraction : t ma decrease te lengt. ending : t ma bend te bod. Twisting : t ma twist te bod. Sear : t ma break te bod. From table given above we can conclude tat, a same force can give five tpe of effects, if we onl cange its direction. t simpl means tat force is a direction dependent quantit. Suc tpe of quantities are known as vector quantit. Value of force is given b Newton s second law of motion wic states tat, Te rate of cange of momentum is directl proportional to te impulse of force. F ma S unit : Newton, it is denoted b N. N is a force wic can produce an acceleration of bod of mass kg. kg m kg m N or s s CGS unit : Dne, dne force is te force wic can produce an acceleration of mass gram. Force (Pull) Force Reaction Force gm cm dne s June : 6 (CCS) (Pus) Twist od Load F Force Reaction F Force m/s in a cm/s in a bod of

2 Engineering Mecanics st Semester : Common to all rances n order to define a force completel, te following caracteristics sould be mentioned :. Magnitude of te force : Te magnitude of force is simpl te value of force.. Line of action : t is te imaginar line along wic force is acting. 3. Direction of te force : Te angle troug wic line of action of force is inclined wit te reference axis. 4. Point of application : Te point at wic te force acts is called point of application. F = Magnitude of Force = Direction O Point of application Line of action Reference axis x Q. (b) square CE as sides equal to mm forces of 5 N eac act along and CD and 5 N eac along C and D. Find te moment of te couple. Wic will keep te sstem in equilibrium. ns. Given : 5 N 5 N 5 N mm Moment of couple C and D : M 5. M Moment of couple and CD : 5 Nm(nti Clock Wise) ns. M 5. M 3 Nm(Clock Wise) ns. Total moment of couple wic will bring te sstem in equilibrium : M M M M N mm M Nm or Nm (nti Clock Wise) ns. Q. (a) State and prove parallelogram law of forces. D C ns. Law of parallelogram of forces : f two force acting at a point are represented in magnitude and direction b te adjacent sides of a parallelogram ten te diagonal passing troug teir point of intersection represents te resultant in bot magnitude and direction.

3 3 Engineering Mecanics st Semester : Common to all rances Proof : Let te forces P and Q act at O. Let O and O represent te forces P and Q acting at an angle. Complete te parallelogram OC. Draw CD perpendicular to O. Let CO, OC R denote te magnitude of te resultant and is te direction of resultant. C Q R From DC, cos D C D C cos D Qcos Fig. Now, OD O D O C cos OD P Q cos [ C O Q] lso, DC C sin Q sin Resultant : i.e. Direction of resultant : OC OD DC R ( PQcos ) ( Qsin ) R P Q PQ cos R P Q PQ cos (i) CD Qsin tan OD P Q cos Equation (i) and equation (ii) give te required magnitude and direction of te resultant. Q. (b) wire is fixed at two points and D at same level. Two weigts kn and 5 kn are ns. Given : (ii) suspended at and C respectivel. Wen equilibrium is reaced it is found tat inclination at is O 3 and tat of CD is 6 to vertical. Determine te tension in te segments. C and CD of te rope and also te inclination of C to te vertical. P D D 3 6 C kn 5 kn

4 4 Engineering Mecanics st Semester : Common to all rances FD of : F ( ) 5 (9 ) F C kn ppling lami s teorem at, F F C sin 9 sin F C sin5 sin(5 ) sin( ) (i) FD at C : F sin(9 ) sin( ) (ii) F C (5 ) F CD (9 ) C ppling Lami s teorem at C, F F CD C sin(9 ) sin( ) 5 kn F C F CD From (i) and (iii), we get On solving, 5 sin( ) sin(5 ) 5sin(9 ) sin(5 ) sin(5 ) 5 sin() sin( ) sin(5 ) (iii) (iv) ns. Q.3 (a) State and prove te Varignon s principle of moments. ns. Frenc matematician Varignon s gave te following teorem wic is also known as principle of moments. t states tat : Te algebraic sum of moments of a sstem of coplanar forces about a moment centre is equal to te moment of teir resultant force about te same moment centre.

5 5 Engineering Mecanics st Semester : Common to all rances O d d d R F F x F x F x Fig. : Proof of Varignon s teorem Proof : Consider two forces F and F wose resultant is R as sown in figure. Consider a point O. Perpendicular distance of forces F and F from O are d and d respectivel., and are te direction of F, F and R from x axis. Moment of te force R about O, Rd RO ( cos ) O( Rcos ) Moment of te force F about O, Rd O R x (i) F d F( Ocos ) O( F cos ) F d OF (ii) x Moment of te force F about O, F d F ( Ocos ) OF ( cos ) Fd O Fx (iii) dding equation (ii) and equation (iii), we get ut, FdF d O( F F ) (iv) R F F x x x x x Te sum of te x components of te forces F and F is equal to x components of te resultant R. Terefore, O( R ) O( F F ) x x x So, from equation (i) and equation (iv), we get R x Rd Fd Fd Hence Proved. Q.3 (b) Discuss various tpes of supports and beams wit sketces. ns. (i) Concentrated load : Wen a load is applied at a ver small area, it is called concentrated load or point load. Fig.(a) : Concentrated load

6 6 Engineering Mecanics st Semester : Common to all rances (ii) Distributed load : Wen a load is spread along te span of beam it is called distributed load. Te intensit (load per unit lengt) of distributed load is equal to area of load diagram and it acts at te centroid of te load diagram. Load diagram Fig.(b) : Distributed load Q.4 (a) Wat are te assumptions made, wile finding out te forces in te various members of a framed structure? Discuss te metod of section for te analsis of pin jointed frame. ns.. Te joints of a simple truss are assumed to be connected and frictionless. Te joints terefore cannot resist moments.. Te loads on te truss are applied at te joints onl. 3. Te members of a truss are straigt two fore members wit te forces acting collinear wit te center line of te members. 4. Te weigt of te members are negligibl small unless oterwise mentioned. 5. Te concept is staticall determinate. Q.4 (b) bod consisting of cone and emispere of radius R fixed on te same base rests on a table, te emispere being in contact wit te table. Find te greatest eigt of te cone, so tat te combined bod ma stand uprigt. ns. r s te bod is smmetrical about te vertical axis, its centre of gravit will lie on tis axis. Considering two parts of te bod, viz. emispere and cone. Let bottom of te emispere be te axis reference. Hemispere : coordinate of center of graving of emispere. 5 r. 8 Mass of emispere : ( x ) ( x) C 3 m r 3 O

7 7 Engineering Mecanics st Semester : Common to all rances For cone : coordinate of centre of gravit of cone Mass of cone r 4 m 3 r Distance of centre of gravit of te combined bod from O is, m m m m 5 r r r 3 3 r r 3 3 Condition for stable equilibrium : 3 5 r r r r r r Center of gravit of a bod sould be below te common face C or maximum it coined wit it. Terefore, On solving, 5 r r r 3 3 r r 3 3 3r 3 r r ns. Q.5 (a) Derive an expression for movement of inertia of a triangular section about its centroidal axis ns. parallel to base. Moment of inertia of a triangle wit base widt b and eigt is to be determined about te base. b d b Consider an elemental strip at distance from te base. Let d be te tickness of te strip and d its area. Widt of tis strip is given b : b' b [ similarit of triangle] (i) Moment of inertia of tis strip about d bd '

8 8 Engineering Mecanics st Semester : Common to all rances From equation (i), we get bd o bd 3 4 b b 3 4 o o 3 bd 3 b ns. t is clear tat te centroidal axis will be parallel to base ence from parallel axis teorem, xx Here, is te distance between base and centroidal axis x- x and is equal to b b xx b xx 3 8 Moment of inertia about centroidal axis b b b xx 8 36 ns. Q.5 (b) simpl supported beam of span 6 m is carring a uniforml distributed load of kn/m over a lengt of 3 m from te rigt end. Calculate te support reactions. ns. FD : Calculation for reaction :, F Talking moment about,, M R R 6 (i) R From equation (i),we get R 4.5 kn ns. R R.5 kn ns. kn/m R 3 m 6 m R Q.6 (a) Discuss various basic terms used in dnamics in detail. State general principles in dnamics. ns.. Displacement : t is te sortest distance from te initial to te final position of a point. t is a vector quantit.

9 9 Engineering Mecanics st Semester : Common to all rances Displacement = Final position nitial position Distance Pat taken Fig. : Displacement. Distance travelled : t is te lengt of pat travelled b a particle or bod. t is a scalar quantit. 3. Velocit : Velocit is te rate of displacement of a bod wit respect to time. Velocit Displacement Time interval 4. cceleration : cceleration is te rate of cange of velocit of a bod wit respect to time. cceleration Velocit Time interval 5. verage velocit : t is te average value of te given velocities. verage velocit is displacement over total time. x verage velocit t 6. nstantaneous velocit : t is te velocit at a particular instant of time. t can be obtained from te average velocit b coosing te time interval t and te displacement x. nstantaneous velocit, (or velocit) x dx v limit t t dt Te unit of velocit is m/s. 7. verage acceleration : Let v be te velocit of te particle at an time t. f te velocit becomes ( v v) at a later time ( t t) ten, v verage acceleration t 8. nstantaneous acceleration : t is te acceleration of a particle at a particular instant of time and can be calculated b coosing te time interval t and te velocit v. v dv d x cceleration a limit. t t dt dt cceleration is position if te velocit is increasing. Te unit of acceleration is So, lso, So, dv dx a as v dt dt d x a dt dv dv dx a dt dx dt and as dx dt v dv a v dx Displacement m/s. 9. Uniform motion : particle is said to ave a uniform motion wen its acceleration is zero and its velocit is constant wit respect to time. t also called uniform velocit.

10 Engineering Mecanics st Semester : Common to all rances v v = constant dv a dt t Fig. : Uniform Motion. Uniforml accelerated motion : particle moving wit a constant acceleration (a constant wit respect to time) is said to be in uniforml accelerated motion. v dv a dt constant t Fig. : Uniforml ccelerated Motion Q.6 (b) small steel ball is sot verticall upwards from te top of a building 5 m above te ground ns. wit an initial velocit of 8 m/sec. Find te total time during wic te bod is in motion. s te ball is being sot verticall. Te eigt of building is not important for calculation. Given : nitial velocit ( u) 8 m/s cceleration due to gravit ( g) 9.8 m/s Wen ball is sot up : Using Newton s 3 rd equation v u as [ a g 9.8m/s ] fter reacing at peak point, final velocit is zero, so v S 8 S 9.8 S 6.5 m Time of travel in upward motion S ut at 6.5 8t 9.8 t t.76 s [ a g 9.8 m/ s, u 8 m/ s] Wen te ball is coming down : t will travel a distance S 6.8 wit initial velocit u under te action of gravit. So, using Newton s second equation, S ut at 6.8 t 9.8 t t.85 s Total time taken b ball t t t m g 9.8 m/s 5 m u = a = g = 9.8 t.6 s ns.

11 Engineering Mecanics st Semester : Common to all rances Q.7 Write sort notes on an four of te following : (a) Fundamental laws of mecanics ns.. Newton s law : Newton s laws consist of te law of inertia, te law of motion and te law of action and reaction. Newton s first law (te law of inertia) : Te law states tat, bod will remain at rest or in uniform motion in a straigt line unless it is compelled to cange tis state b forces impressed upon it. Te first law depicts tat if tere is no external effect, an object must be still or moving at a constant velocit, wic leads to a concept of te net force. Newton s second law (te law of motion) : Tis law states tat, bod acted upon b an external unbalanced force will accelerate in proportion to te magnitude of tis force in te direction of applied force. i.e. F a Te second law quantifies te force in terms of acceleration and mass of te object. F ma Newton s tird law (te law of action and reaction) : Tis law states tat, For ever action (or force) tere is an equal and opposite reaction (or force). Te tird law illustrates te existence of te counter force wic is related to normal forces and tension, etc.. Law of conservation of mass : Tis law states tat, Mass can neiter be created nor be destroed troug an psical or cemical process. Matematical expression of tis law is : d ( m ) dt Were, m is mass of a bod. 3. Law of conservation of energ : Tis law states tat, Energ can neiter be created nor be destroed, it can onl transform from one form to anoter. Q.7 (b) ow s Notation ns.. Wen te different sstem of te forces are drawn, ten spaces are formed around it. Tese spaces so formed are named b capital letters,, C, D, E and so on in order.. Tis metod of putting te capital, alpabetic letters on eiter side of te forces in order is called as ow s notation. 3. ow s notation is used to represent or designate or to denote te force in grapical solution of te problem. For example, consider a force of 3 N is acting on a bod. Two spaces are formed around it are named b ow s notation, as sown in figure. To represent tis force in a force diagram or vector diagram, a suitable scale is taken ( cm = N as in figure) and line ab = 3 cm is drawn parallel to te line of action of as sown in figure. Lengt ab sows te magnitude of a force and an arrow ead indicate direction. 3 N b a 3cm Scale cm = N letters, are ow s notation Fig. : Representation of a force b ow s notation

12 Engineering Mecanics st Semester : Common to all rances Q.7 (c) Tpes of loading on beam ns. (i) Concentrated load : Wen a load is applied at a ver small area, it is called concentrated load or point load. (ii) Distributed load : Wen a load is spread along te span of beam it is called distributed load. Te intensit (load per unit lengt) of distributed load is equal to area of load diagram and it acts at te centroid of te load diagram. Load diagram Fig.(a) : Concentrated load Fig.(b) : Distributed load Q.7 (d) Parallel axis teorem ns. Moment of inertia about an axis in te plane of an area is equal to te sum of moment of inertia about a parallel centroidal axis and te product of area and square of te distance between te two parallel axes. Referring to figure, te above teorem means : Were, Moment of inertia about axis. Moment of inertia about centroidal axis GG parallel to. Te area of te plane figure given. Te distance between te axis and te parallel centroidal axis GG. Proof : Consider an elemental parallel strip d at a distance from te centroidal axis as sown in figure, d G Centroid G Ten, Now, d ( ) d d can be written as, ( ) d Fig. d d d Moment of inertia about te centroidal axis d d d n te above term is constant and d is te distance of centroid from te reference axis GG. d Since GG is passing troug te centroid itself is zero and ence te term d is zero. Now, te tird term, Terefore, d d

13 3 Engineering Mecanics st Semester : Common to all rances Q.7 (e) Tpes of motion ns. Tpes of motion : bod ma move in an direction in space. n tis capter, onl motion in a single plane is considered. Tis tpe of motion is called plane motion. Plane motion ma be classified as, (i) Translation : motion is said to be translation, if a straigt line drawn on te moving bod remains parallel to its original position at an time. During translation if te pat traced b a point is a straigt line, it is called rectilinear translation (Fig.(a)) and if te pat is a curve one it is called curvilinear translation (Fig.(b)). Fig.(a) : Rectilinear translation Fig.(b) : Curvilinear translation n te stud of te motion of particles, rectilinear translation and curvilinear translation are usuall referred as linear motion and curvilinear motion. (ii) Rotation : motion is said to be rotation if all particles of a rigid bod move in a concentric circle. C D C Fig. : Rotation General plane motion : Tere are man oter tpes of plane motion, i.e., motion in wic all te particles of te bod move in parallel planes. n plane motion wic is neiter a rotation nor a translation is referred to as a general plane motion. Two examples of general plane motion are sown in Fig.5. D = + Fig.5 : General plane motion

14 4 Engineering Mecanics st Semester : Common to all rances (iv) Rolling : Rolling motion is a combination of bot translation and rotation. Common examples of suc motion are points on weels of moving veicles, a ladder sliding down from its position against wall etc. t is also a kind of general plane motion. v v v v v v Q Q Q v P O + (a) Pure rotation (b) Pure translation (c) Rolling motion Fig.6 : Rolling motion n pure rolling motion te velocit at point of contact between weel and surface is zero. t is also called rolling witout slip. (v) Space motion : t is te motion of bod in 3D space. Example : Motion of satellite. O P v v = v O P v v v