Engineering Services Examination Analysis of system of forces, Friction, Centroid and Center of Gravity, Dynamics THEORY WITH SOLVED PROBLEMS (3 XX)

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1 SYLLUS Graduate ptitude Test in ngineering Free-bd diagrams and equilibrium; Trusses and Frames; Virtual Wrk; Kinematics and namics f particles; and f rigid bdies in plane mtin; Impulse and Mmentum (linear and angular) and energ frmulatins; llisins. ngineering Services aminatin nalsis f sstem f frces, Frictin, entrid and enter f Gravit, namics TL OF ONTNT THORY WITH SOLV PROLMS (3 XX) SR. NO. HPTR PG NO. 1. INTROUTION 3. TRUSSS N FRMS 1 3. MSS MOMNT OF INRTI 4 4. FRITION WORK, NRGY, MOMNTUM N OLLISION KINMTIS N YNMIS 57 PRVIOUS GT QUSTIONS (70 11) PRTIS PROLMS (113-19)

2 nalsis f Previus GT Papers am Year 1 Mark Marks 3 Marks 5 Marks Ttal Ques. Ques. Ques. Ques. Marks Set Set Set Set Set Set Set Set Set Set

3 UNIT 1 INTROUTION ngineering Mechanics deals with the stud f rigid bdies under the actin f frces. Mechanics is divided int tw fields, first is statics and secnd is dnamics. Statics cncern the things which dn t mve under the actin f frce and deals with the bd which is the under the frce equilibrium. namics (kinetics) is cncerned abut frces and mtin caused b frces. In dnamics r kinetics we are cncerned abut the frces r cause f mtin. While kinematics is the stud f mtin withut thinking f what cause that mtin. Rigid bd bd is said t be rigid when it des nt defrm under the actin f eternal frces. Hwever it is just an assumptin t simplif the prblem. QUIVLNT FOR SYSTM Frce Frce is defined as an eternal agent which changes r tends t change the state f rest r f unifrm mtin f a bd. Frce is a vectr quantit. haracteristics f a frce The quantities b which a frce is full represented are called characteristics f the frce. The varius characteristics f a frce are (i) (ii) (iii) (iv) Magnitude. The quantit f a frce is called its magnitude. irectin. The line alng which the frce acts is called directin r line f actin f the frce. It is represented b (a) Indicating the angle that the line f actin f the frce makes with the fied r reference aes such as OX, OY r with an ther hrizntal r vertical line. (b) Indicating the angle that the line f actin f the frce makes with the gegraphical aes such as Nrth, Suth, ast r West. Sense. The wa in which the frce acts alng its line f actin is knwn as sense r nature f the frce, i.e. push r pull. Pint f pplicatin. The pint at r thrugh which the frce acts n a bd is called its pint f applicatin. SPTIL FOR SYSTMS Spatial frce sstems are f the fllwing categries: (a) ncurrent sstem: It cnsists f frces which intersect at a pint. (b) Parallel Sstem: It cnsists f frces which d nt intersect. (c) Nn-cncurrent, nn-parallel Sstem: It cnsists f the frces which are nt cncurrent and nt parallel. RSULTNT OF SPTIL FOR SYSTM (i) Resultant f a spatial frce sstem is a frce R and a cuple, where R= F, the vectr sum f all frces f the sstem, and M, the vectr sum f mments f all the frces f the sstem.

4 The value f R is independent f the chice f the reference pint, but the value f depends n the reference pint. Fr an frce sstem, the reference pint is s selected that the vectr representing the cuple is parallel t R. This special cmbinatin is called a wrench r screw. The cllinear cuple F is the prjectin f cuple upn frce F,.nˆ nˆ. F F F If cuple at an pint O is knwn, then cuple at an ther pint can be determined b equatins; M M r R where, R resultant frce, and r psitive vectr f R (ii) The resultant f spatial frces sstem can als determined b appling the fllwing derived scalar equatin. (a) ncurrent Sstem: The resultant R ma be (i) single frce thrugh the cncurrence, r (ii) Zer z lgebraicall, R F F F F F Fz with directin csines, cs,cs and cs z R R R where F, F, Fz are algebraic sums f the, and z cmpnents f the frces f the sstem respectivel.,, ngles which the resultant frce R make with the, and z aes z respectivel. (b) Parallel Sstem: The resultant R ma be (i) single frce R parallel t the sstem (ii) cuple, r (iii) Zer lgebraicall, R F, R M, Rz M where F algebraic sum f the frces f the sstem. perpendicular distance frm the z plane t the resultant. z perpendicular distance frm the plane t the resultant. z M, M algebraic sums f mments f the frces f the sstem abut and z aes respectivel. If F 0, then resultant cuple M M M with tan M where angle which the vectr representing the resultant cuple makes with the -ais. (c) Nn-cncurrent, Nn-parallel Sstem: set f, and z aes is placed with their rigin at the reference pint. Replace each frce f the given sstem b z

5 (i) n equal parallel frce but acting thrugh an chsen rigin, (ii) cuple acting in the plane cntaining given frce and rigin. Magnitude f the resultant f the cncurrent sstem at the rigin z R F F F With directin csines, F F F cs,cs and cs z R R R Magnitude f the resultant cuple, z M M M With directin csines, M M Mz cs,cs and cs z where M, M, M lgebraic sums f the mments f the frces f the z sstem abut the, and z aes respectivel.,, ngles which the vectr representing cuple makes with, and z z aes respectivel. 1. The resultant f tw frces P and Q is R. If ne f the frces is reversed in directin, ns. the resultant is S. Then fr the identit R S P Q z t hld gd, a) Frces are cllinear b) Frces act at right angles t each ther c) Frces are inclined at 60 t each ther d) Frces can have an angle f inclinatin between them (d) planatin : R P Q PQcs S P Q P Q cs P Q PQcs dding we get, R S P Q Hence fr a given identit t hld gd, the frces can have an angle f inclinatin between them. QUILIRIUM OF FORS n sstem f frces which keeps the bd at rest is said t be in equilibrium, r when the cnditin f the bd is unaffected even thugh a number f frces acted upn it, it is said t be in equilibrium. Laws f quilibrium There are tw laws f equilibrium depending upn the frces r the cuple acting n a rigid bd. 1. Frce law f equilibrium. Mment law f equilibrium

6 Frce Law f quilibrium: Fr an sstem f frces keeping a bd in equilibrium, the algebraic sum f frces, in an directin is zer, i.e. F 0 This law is applicable t all sstems f cplanar frces and includes a. Tw frce principle: If a bd in equilibrium is acted upn b tw frces, the must be equal, ppsite and cllinear. b. Three frce principle: If a bd in equilibrium is acted upn b three frces, then the resultant f an tw frces must be equal, ppsite and cllinear with the third frce. c. Fur frce principle: If a bd in equilibrium is acted upn b fur frces, then the resultant f an tw frces must be equal, ppsite and cllinear with the resultant f ther tw frces. Mment Law f quilibrium: Fr an sstem f frces keeping a bd in equilibrium, the algebraic sum f the mments f all the frces abut an pint in their plane is zer, i.e. M0 r Fd 0 This law is applicable nl t cplanar, nn-cncurrent (parallel r nn-parallel) frce sstem. nditins f quilibrium f an Sstem f planar Frces These cnditins can be divided int the fllwing tw categries: 1. naltical cnditins f equilibrium. Graphical cnditins f equilibrium 1. naltical nditins f quilibrium a. Fr cplanar cncurrent frces: (i) The algebraic sum f cmpnents f all the frces alng -ais, in their plane is zer, i.e. F 0 (ii) The algebraic sum f cmpnents f all the frces alng -ais, in their plane is zer, i.e. F 0 Thus, if an number f frces acting n a bd are in equilibrium, the algebraic sum f their cmpnents in tw directins at right angles t each ther, must be equal t zer. LMI S THORM If cplanar frces acting n a pint in a bd keep it in equilibrium, then each frce is prprtinal t the sine f the angle between the ther tw frces. Thus, if three frces P, Q and R, acting n a pint O in a bd, are in equilibrium as shwn in the figure, then accrding t Lami s Therem ;

7 P : Q : R ::sin :sin :sin P Q R r sin sin sin P d Q Fr sstem t be in equilibrium under the actin f cplanar nn-current frces, it must satisf the abve tw cnditins f equilibrium, viz. F 0 and F 0, and a third cnditin viz. the algebraic sum f mments f all the frces abut an pint in their plane must be zer, i.e. M 0. Fr cplanar cncurrent frces cnditin M 0 is nt necessar t be satisfied. The cnditin F 0 and F 0 ensure that the sstem des nt reduce t a single resultant frce and M 0 ensures that it des nt reduce t a cuple.. n electric light fiture weighing 150 N hangs frm a pint b tw hinges and. The tensin is string wuld be apprimatel a. 78 N b. 86 N c. 99 N d. 98 N 45 0 R 60 0 ns. (a) Frm the gemetr f the sstem W sin 75 T 0 0 sin150 sin150 T W sin T T W = 150N N string m lng is tied t the ends f a unifrm rd that weight 60 N and is 1.6 m lng. The string passes ver a nail s that the rd hangs verticall. The tensin in the string will be a. 4 N b. 30 N c. 4 N d. 50 N ns. (d)

8 1m; m sin 0.8 and cs 0.6 Since three frces W, T and T are in equilibrium, W T sin sin 180 W T sin cs sin W 60 T 50N cs 0.6 T T 0.8 m 0.8 m W = 60 N T (180 - ) W 4. mass f 1 kg. is attached t the middle f a rpe, which is being pulled frm bth ends in the ppsite directins. Taking g 10 m/sec, the minimum pull required t cmpletel straighten the rpe will be a. 5 N b. 0 N c. 5 N d. ns. (d) Lami s therem If P,Q and R are the three vectrs acting at a pint, and their resulatant is zer (i.e. the sstem is in equilibrium) P Q R then sin sin sin T cmpletel straighten the rpe, and 90 sin 90 1 P Q R sin unifrm beam hinged at, is kept hrizntal b supprting and setting a 40 kn weight with the help f a string tied at and passing ver a smth peg as shwn in the figure. The bar weights 0 kn. t the supprts and as well as the tensin in the string. (i) Tensin in the string will be a. 40 kn b. 5 kn c. 0 kn d. 15 kn (ii) Reactin at the supprt a will be 10i 10j kn a. b. 10i 0j kn

9 c. 0i d. 0i 0j kn 10j kn ns. (iii) Reactin at the supprt will be 0i 40j kn planatins: (i) M 0 40 T k 0 3k 6i T cs i sin j m T 6T sin 0 3 Taking sin 0.6, 5 T 5kN m F 0; Tcs kN 5 F 0; T sin 40 T 0 0 a. b. 0i c. 40i d. 40i (i) b, (ii) c, (iii) a 0kN 0j kn 40j kn 0j kn 4m OUPL N MOMNT Mment It is a measure f the turning effect f a frce. Onl ne frce is taken int accunt. It is dependent n pint f cnsideratin. uple When tw equal & ppsite frces, but with separate line f actin, are present in a frce sstem, it is called a cuple. It is independent n pint f cnsideratin. f f mment = f uple = f P f ONPT OF FR-OY IGRM

10 free-bd diagram cnsists f a diagrammatic representatin f a single bd r a subsstem f bdies islated frm its surrundings but shwn under the actin f frces and mments due t eternal actins. nsider, fr eample, a bk ling flat n a table. The bk eerts its weight n the table and the table eerts its wn weight as well as transmits the weight f the bk n the grund. free-bd diagram fr the bk alne wuld cnsist f its weight W acting thrugh the centre f gravit and the reactin eerted n the bk b the table tp as shwn in the figure belw. nsider, as anther eample, tw clinders placed in a V-grve. Free-bd diagrams f the tw bdies islated frm the V-grve as well as f each bd separatel are shwn in Figs. (d), (e) and (f) respectivel. k n Table-tp (a) W R/ (b) W ( c) R 1 W 1 W 1 W 1 R 1 W R 1 1 R R R 1 R 1 R R (d) free-bd diagram ma be drawn fr an single member f a sstem, an subsstem f the sstem r the entire sstem irrespective f whether the sstem is in equilibrium: at rest, in unifrm mtin r in a dnamic state f mtin. (f) (i) (ii) (iii) Uses f Free d iagrams in Statics The prblems invlving equilibrium f bdies under an sstem f frces can be simplified b drawing free bd diagram f each bd separatel. ll equatins f equilibrium can be applied t each free bd diagram. The unknwn frces fr equilibrium f each bd can be btained ver easil. LMRT S PRINIPL The equatin f mtin f particle P ; F ma r F ma 0 Which means that the resultant f the eternal frces ( F) and the frce (-ma) is zer. The frce (-ma) is called inertia frce. The inertia f a bd can be defined as the resistance t the change in the cnditin f rest r f unifrm mtin f a bd. The magnitude f the inertia frce is equal t the prduct f the mass and acceleratin f the particle and it acts in a directin ppsite t the directin f acceleratin f the particle. The equatin in the frm

11 F ma 0 r F ( ma) 0 r in the cmpnent frm F ( ma ) 0 & F ( ma ) 0 are called the equatins f dnamic equilibrium f the particle. S, t write the equatin f dnamic equilibrium f a particle add a frictin frce equal t the inertia frce t the frces acting n the particle and equate the sum (resultant) t zer. This cncept is knwn as lembert s principle. 6. Tw blcks with masses M and m are in cntact with each ther and are resting n a hrizntal frictinless flr. When hrizntal frce is applied t the heavier, the blcks accelerate t the right. The frce between the tw blcks is M m F / m ns. a. b. MF / m c. mf / M mf / M d. m (d) planatin: Frm free bd diagram F N Ma... i r N ma... ii Frm equatins (i) and (ii), we get mf N m M 7. frce f 50 N acts n a bd f mass m=100 kg. Find the acceleratin f the bd. ns. Let the acceleratin f the bd in the directin f the frce be a. The equatin f mtin f the bd can be written as F ma a a =.5 m/s The equatin f equilibrium: ppl a frictin frce equal t ma t the bd, in a directin ppsite t the directin f acceleratin f the bd. F ma a 0 a =.5 m/s F M m

12 UNIT TRUSSS N FRMS STRUTURS structure ma cnsist f a truss r a frame pin-cnnected r rigidl secured truss is an assemblage f slender bars fastened tgether at their ends b smth blts r ball- and scket jints acting as hinges. truss, b definitin, is a cnnected structure. The bar members, therefre, act as tw-frce members which can either be in tensin r in cmpressin: there can be n transverse frce in a member f a truss. frame structure, n the ther hand, cnsists f members which ma be subjected t a transverse lad in additin t the aial lad. simple structure is thus a pin cnnected frame r truss. truss cnsists f slenderbar members which can carr n transverse lads. It fllws that the lading in a truss must be at the jints nl. truss cnsisting f members which lie in plane and are laded in the same plane is called plane truss. If a truss is made r nn-cplanar members, it is referred t as space truss. Similarl, a frame ma be a plane frame r a space frame depending upn its structure. Trusses are classified as just-rigid, ver-rigid and nn-rigid mechanism. If the members are allwed an relative mvement then it is called a rigid truss. just-rigid truss is that which, n the remval f an single member, becmes nnrigid. n ver-rigid truss is the ne that has redundant members which ma be remved t render the truss just-rigid. The number f jints j in a truss is related t the number f members m as rigid if m and j are related therwise but a truss ma nt be just-rigid even if m and j are related as befre. F F 1 F F (a) Nn-Rigid Truss- Mechanism (a) Just-Rigid Truss (c) Over-Rigid Truss Tpes f Trusses ccrding t the law, the number f members must be (j - 3) fr a truss-rigid but if the number is different frm (j - 3) the fllwing ma happen m j 3 the truss cannt be just-rigid; a parts f which must be under and a part ma be just-rigid r ver-rigid as shwn in the figure. m j3 the truss cannt be just-rigid; parts f which must be ver-rigid. NLYSIS OF SIMPL PLN TRUSS ssumptins (i) ach truss is cmpsed f rigid members ling in ne plane. (ii) Weight f the members are neglected because the are small in cmparisn with the lads.

13 (iii) Frces are transmitted frm ne member t anther thrugh smth pins fitted in the members. Methds f nalsis f a framed Structure (i) naltical methd (ii) Graphical methd naltical methd: This methd includes a) Methd f Jints: In this methd, the free bd diagram f each jint is separatel analzed t btain the magnitude f stresses in the truss members. The unknwn frces are then determined b equilibrium equatins viz. V 0 and H 0,i.e. sum f all the vertical frces as well as the hrizntal frces is equated t zer. The analsis als ields nature f stresses (whether cmpressin r tensin). The principle f triangle f frces, r reslutin f frces is applied fr analsis. Lgic is applied s that islated jint is in equilibrium under the actin f eternal frces and internal stresses (called frces in the member). are shuld be taken t select a jint which des nt have mre than tw unknwn frces. i b a e f b) Methd f Sectin: This methd is cnvenient, when the frces in the few members f a truss is required t be fund ut. Take such a sectin that the eternal frces n ne side f the sectin ield the frces in the members cut b the sectin. Use f w s ntatin has been fund cnvenient in the analsis f frames. In this methd, either left r right side f the sectin is cnsidered. While taking mments at a pint, the apprpriate sign f the mment shuld be used. e.g. cnsidering left hand side f the sectin X-X, the frces ae, ab and hi shuld be used fr cmputatin and nt gf, dc, and hd. Nte: nl that jint can be slved either analticall r graphicall which has nt mre than tw knwn frces. W h c d g P Q S R 1 R R w s Ntatin. w s ntatins are used t name the different members f a truss. ver members f truss is dented b tw capital letters, placed in the spaces n either side f the member f the space diagram as shwn in the figure. Therefre the lad W will be named as PQ, the reactin R 1 as RP and reactin R as QR. The members f the frame will be named as PS, QS and RS respectivel.

14 P Q Graphical methd: This methd is based n a) Law f triangle f frces b) Law f plgn f frces c) w s ntatin d) Link r funicular plgn: The funicular r link plgn is used fr the graphical determinatin f supprt reactins. nsider a hrizntal string supprted at and. If a frce PQ applied at pint f the string the string will distrt in the shape shwn b dtted lines. This distrted shape f the string is called funicular string r link plgn. 1. Using the methd f jints, find the aial frces in all the members f a truss with the lading shwn in Fig. belw 000 N 4000 N R m 3 m Slutin In general, the reactin at a hinge can have tw cmpnents acting in the hrizntal and vertical directins. s there is n hrizntal eternal frce acting n the truss, s the hrizntal cmpnent f reactin at is zer. Taking mments abut, M 0: R 6 0 R 3500 N F 0: R R R 500 N efre cnsidering the equilibrium f the jints, mark b inspectin, the directins f aial frces in all the members as shwn in Fig N 4000 N 60 R F R R R N Jint Let us begin with the jint at which there are nl tw unknwn frces. We cannt begin with the jint because, there are three unknwn frces acting at the jint. F

15 nsider the free-bd diagram f the jint. quatins f equilibrium can be written as F 0: F F cs i The magnitudes f the frces F and F are bth cming ut be psitive, therefre, the assumed directin f the frces are crrect. Jint F 0: R cs 60 F 0... iii F F 0: R F sin ii R 500 F sin F 887 N() ns. Using (i) F F cs F 0: R R sin iv R 3500 F sin F 4041 N ns. Frm (iii) F F 1443 N T ns. F 4041 cs F 60 R Jint 3500 N F 00.5 N (T) ns. Jint F 887 N (knwn) F 0: F cs60 F cs60 F 0... v F 0: F sin 60 F sin vi F F 577 N(T) ns. Jint F 173 N F 4041 N (knwn) F 0 : F F cs60 F cs vii Or F cs 60 F cs 60 F Or F 0.5 F 577 N ns. Frm (v) F F cs 60 F cs 60 F F 173 N ns. F (887 N) F F 000 N Jint 4000 N F F F

16 Jint. Find the frce in the member F f the truss laded and supprted as shwn in Fig. belw. Y 45 m n P p X 1m Slutin Taking mment abut. M 0: P 0.5 P 0.5 R 0 3 R P 4 F 0: R Y P 0 3 Y P P 4 Y 1.5P F 0: P X 0 F X P T find the frce in the member F let us cnsider the sectin thrugh the line mn. Three members are cut, but as these are cncurrent, we cannt slve fr the three unknwns. If we take a sectin thrugh the line pq then fur members are cut. S, first pass a sectin thrugh rs cutting the members and F. Write the equatins f equilibrium F 0: F F cs i F 3 F 0: F sin 45 P 0 4 3P F 4sin 45 3 F P 3 1 Frm (i) FF F cs 45 P 3 FF PT 4 45 q 1m s P r R 1m

17 F F F r F F P F F F 3 R 1m 1m P 1.5 P 3 4 P 4 nsider the equilibrium f Fig. the 9.1 right (b) hand prtin f the Fig. truss 9. as btained b the sectin pq Taking mments abut F, MF 0: F 1 P1 P 0 4 P 1 F 0.354P() sin 45 1 cs 45 Taking mments abut M 0: P1 P FF 1 FF P Substituting FF P, FF T 4 Taking mments abut, M 0: P P FF 1 FF P Substituting, FF 3 P P 1 P FF 0 4 P FF T ns. 3. Find the aial frces in the members, G, F, G, GF and G f the truss supprted and laded as shwn. Use the methd f jints. P F 45 F F P 1m

18 W l l l F Y X G 60 Slutin: Let us first evaluate the angle which ma be needed while reslving the frces. FG Frm triangle FG, tan 30 F FG Ftan30 FG 3 R Frm triangle FG, tan tan 1 1 FG R 3 3 FG tan F Supprt Reactins: nsider the equilibrium f the entire truss as a free-bd. Taking mments abut. M 0: W R 0 3 W R 0 cs30 W R 3 F 0: Wsin30 X 0 lng X W F 0: Y R Wcs30 0 (Nrmal t ) Y Wcs30 R W 3 W Y 3 Jint 3 W

19 G F 0: F cs 60 F 0 F 0: F sin 60 R 0... i R W F sin W F ns. 3 F G R 60 w 3 Jint F 30 Substituting fr F in (i) W1 FG F cs 60 3 W FG T ns. 6 Jint F 0: F F cs 0... ii F G F 0: F sin F 0 G F FG sin W 1 sin cs F G G G GF F 0: FGF sin 60 FG sin F 0: F F F cs 79.1 F cs iii FG sin w Or FGF sin F 0.5 W ns. GF G F 0.441, W T ns. G Substituting fr F in (ii) G F F cs W F G F 0.89W ns. Slving (iii) F 0.5W T ns. F F Jint F G F 60 F GF G Jint G F G 30 F G 4. cantilever truss is laded and supprted as shwn in Fig. belw. Find the value f lads P which wuld prduce an aial frce f magnitude 3 kn in the member.

20 3m m P 3m P m n F 1.5m 3m Slutin: In this case we need nt determine the supprt reactins. The frce in the member can be determined using the methd f sectins. Pass a sectin mn cutting the members, and F, nsider the equilibrium f the right hand prtin f the truss as shwn in Fig. belw. 3m F P 3m P F m F 1.5m F Taking mment abut, M 0: F P 1.5 P F F 6P 3P s the frce in the member is 3 kn, F 3 kn 3P P 1 kn ns. 5. n X-frame is laded and supprted as shwn in Fig. belw. Find the hrizntal and vertical cmpnents f the reactin at and. W w X Y / / Slutin: ntire Frame nsider the equilibrium f the entire frame as a free-bd. F 0: X W X 0 X X W... i F 0: Y Y W 0... ii Taking mments abut, Y X

21 M 0: Y a W a tan 30 0 Y W tan 30 Y W Frm (ii) Y W Y W W Y W Y X w w X Member X Y 30 / / F 0: X W X 0... iii F 0: Y Y 0 Y Y 0.43 W Taking mments abut, a a M 0: Y W a tan 30 X tan30 0 X tan30 W tan30 Y W W X.733 W Frm (iii) X W X W.733 W X W. Frm (i) X X W X W W X.733, X W, X.733 W Y 0.43 W, Y W ns. Y 30 Y X 6. Frame is laded and supprted as shwn in the Fig. belw. Find the frce in the bar. kn 1 kn m 1m 1m 1m Y

22 Slutin ntire Frame nsider the entire frame as a free-bd. The unknwn reactins invlved are Y and Y. F 0: Y Y 1 0 Y Y 3 Taking mments abut. M 0: Y Y 1.5 kn Y 3 Y 31.5 Y 1.75 kn ismember Member. It is a tw-frce member (tw frces acting at the ends are equal in magnitude, ppsite in sense and are cllinear), therefre, fr equilibriums kn Y X X 1 kn Y X Y X Member F 0: X X 0 r X X F 0: Y Y 0 Y Y 1.75 Y 0.5 kn Taking mments abut, M 0: 1 Y X tan X X 1.99 kn Frm (i) and (ii) X X X 1.99 kn Tw frce member Frce in the tie bar 1.99 kn (Tensin) ns. We need nt cnsider the equilibrium f the bar as the desired frces have been determined. ata fr Q.7 and Q.8 are given belw. Slve the prblems and chse crrect answers. Fr the fllwing truss is shwn in figure. X Y X

23 10 m 10 m 15 m 7. Find the frce transmitted b member. a. 5 N b. 10 N c. 0 N d. 5 N 8. Find the frce transmitted b F member. a. 000 N b N c N d. Nne f these F 10 m 10 m 5000 N 000 N Slutin 0. (c), 1 (a) planatin: Taking mment abut, R R Y Y 3000N R R 7000 R 4000 N Taking mment abut R 15 R R 0 15 Taking mment abut. we have R 0 N Taking mment abut R X R Y 10 F N 000 N R R R R F F 15 R 10 F FF N mpressive F F F F F R 000 N R

24 UNIT 3 MSS MOMNT OF INRTI The mass mment f inertia is ne measure f the distributin f the mass f an bject relative t a given ais. r Where 0-0 is the ais arund which ne is evaluating the mass mment f inertia, and r is the perpendicular distance between the mass and ais 0-0. s can be seen frm the abve equatin, the mass mment f inertia has the units f mass times length squared. The mass mment f inertia shuld nt be cnfused with the area mment f inertia which has units f length t the pwer fur. Mass mments f inertia naturall appear in the equatins f mtin, and prvide infrmatin n hw difficult (hw much inertia there is) it is t rtate the particle arund given ais. Mass mment f inertia fr a rigid bd: When calculating the mass mment f inertia fr a rigid bd, ne thinks f the bd as a sum f particles, each having a mass f dm. Integratin is used t sum the mment f inertia f each dm t get the mass mment f inertia f bd. The equatin fr the mass mment f inertia f the rigid bd is m r dm I r dm r dm m The integratin ver mass can be integratin ver vlume, area, r length. Fr a full three dimensinal bd using the densit ' ' ne can relate t element f mass t the element f vlume. In this case the densit has units f mass per length cubed and the relatin is given as l r dv v Radius f gratin: Smetime in place f the mass mment f inertia the radius f gratin k is prvided. The mass mment f inertia can be calculated frm k using the relatin l mk Where m is the ttal mass f the bd. One can interpret the radius f gratin as the distance frm the ais that ne culd put a single particle f mass m equal t the mass f the rigid bd and have this particle have the same mass mment f inertia as the riginal bd. Parallel-ais therem: The mment f inertia arund an ais can be calculated frm the mment f inertia arund parallel ais which passes thrugh the center f mass. The equatin t calculate this is called the parallel ais therem and is given as m l l md

25 d m M r r Where d is the distance between the riginal ais and the ais passing the center f mass, m is the ttal mass f the bd, and T is the mment f inertia arund the ais passing thrugh the centre f mass. Mass Mment f Inertia f Hmgenus dies 1. The mass mments f inertia f the bject in terms f the z crdinate sstem are: dm ais m ais m z ais zz l l z dm l l z dm l l dm z m m m l dm, l z dm l z dm z m z z. 1 l z z Slender ar l. 1 l ais 0, l ais l z ais m 3 l l l 0 z z 1 l 'ais 0, l 'ais l z'ais m 1 l l l 0 ' 'z' z'' 3.

26 R z 1 l 'ais l 'ais mr 4 1 lz'ais mr, l l l 0 '' 'z' z'' Thin ircular Plate 4. 1 b b h 1 h f z z Thin Rectangular Plate 1 l ais mh, l ais mb, lz ais mb h l mbh, lz lz l'ais mh 1 1 l'ais mb 1 1 lz'ais mb h, 1 l l l 0 '' 'z' z'' 5. b a z c

27 Vlume abc 1 l'ais ma b 1 1 l'ais ma c 1 1 z'ais l m b c,l'' l'z' lz'' l Vlume R, z,z 1 1 l ais l ais m R, lz ais mr, l l l 0 z z 1 1 l 'ais l 'ais m R, lz'ais mr, l l l 0 '' 'z' z'' R l ircular linder 7. R 3 h 4 h Z,Z 1 Vlume R h l ais l ais m h R ' lz ais mr, 10 ircular ne

28 l l l 0. z z 3 3 l 'ais l 'ais m h R l z'ais mr, 10 l l l 0 '' 'z' z'' Vlume R 3 l 'ais l 'ais l z'ais mr, 5 z R Sphere 1. Find the centrid f the crss-sectinal area f a Z-sectin as shwn in Fig. belw G 10 cm 3 0 cm F 3 H.5 cm F.5 cm cm O X 0 cm Slutin Since the figure has n ais f smmetr s an cnvenient aes can be chsen as the aes f

29 reference. hsing ne f the aes alng the bttm edge and the ther alng the vertical edge G, c Figure rea (cm ) Rectangle Rectangle FF Rectangle FGH c c cm. ns c c crdinate f the centrid (cm) /, / c, c 7.69 cm. ns crdinate f the centrid (cm) 1 5/ / / / 1.5. square hle is punched ut f a circular lamina as shwn. The diagnal f the square which is punched ut is equal t the radius f circle. Find the centrid f the remaining lamina. Slutin Let the radius f the circle be a. a s the diagnal f the square is a, side f the square 0.095a c 0 1 Fig hsing the reference aes as shwn in Fig. 4.17, it is seen that figure is smmetrical abut -ais therefre, c 0. a

30 ifferent areas and crdinates f their centrid are tabulated belw: Figure ircle f radius a Square f side a -crdinate f -crdinate f rea the centrid the centrid 1 a a a. a a 0 c a a 3 a (0) a 4 a a ve rea and ve crdiante c a a c c a a ns.