VIRTUAL WORK CHAPTER. Work done by a force. Special cases. If a constant force F acts on a particle and particle is displaced from a point A to

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1 M th VIRTUAL WORK 3 CHAPTER Work doe by a force If a costat force F acts o a particle ad particle is displaced from a poit A to B. Let AB = d, the the work doe by the force F is give by F W = F. d = Fd cos Where is agle betwee F ad d. A d B Special cases Now we discuss some special cases of the above article. (i) Whe = 0 0 Work doe = Fdcos0 0 = Fd So work doe will maximum whe = 0 0 (ii) Whe = 90 0 Work doe = Fdcos90 0 = 0 So work doe will be zero whe applied force is perpedicular to the displacemet. (iii) Whe = Work doe = Fdcos180 0 = Fd Thus, Whe The work doe will positive because cos 0 whe Whe The work doe will egative because cos 0 whe Cotact At: qadri86@yahoo.com

2 Virtual Displacemet & Virtual Work If a set of particles or a body is i equilibrium uder the actio of forces the there is o motio ad cosequetly there is o actual displacemet. Suppose that the set of particles or the body receives a imagiary displacemet, the forces actig thereo beig regarded as costat durig the displacemet. The such a displacemet is called virtual displacemet ad work doe by the forces durig such a displacemet is called virtual work. It may be oted that a virtual displacemet is oly a hypothetical displacemet ivolvig o passage of time ad is quite differet from actual displacemet of a movig body takig place i the course of time. APPLIED FORCES & FORCES OF CONSTRANIT Particles or rigid bodies are geerally subjected to two types of forces: (i) Iteral Forces (ii) Exteral Forces Iteral forces Iteral forces are those forces which the differet part of a system exerts o each other ad such forces obey Newto s 3 rd law of motio. Exteral forces Exteral forces are those forces which are ot due to ay part of a system but which are due to some exteral agecy. Exteral forces are further classified as: (i) Reactive Forces or Forces of Costrait (ii) Active or Applied Forces Reactive Forces or Forces of Costrait Whe a set of particle or a body is made to move alog or rest o a curve or surface, the forces are exerted by such curve or surface are called forces of costrait or reactive forces. Active or Applied Forces Exteral forces which are ot due to ay costrait is called active forces or applied forces. If a particle rests o or moves alog a iclied plae, the reactio of the plae is a reactive force but the weight of the particle is a active force. Ideal or Workless Costraits or workless. If the forces of costrait do o virtual work, the costrait are said to be ideal Now we state ad prove the some importat theorems, kow as priciples of Virtual Work, for a sigle particle, a set of particles, a rigid body ad set of rigid bodies. Cotact At: qadri86@yahoo.com

3 Priciple of Virtual Work for a Sigle Particle 3 STATEMENT: A particle subject to workless costraits, is i equilibrium if ad oly if zero virtual work is doe by the applied forces i ay arbitrary ifiitesimal displacemet cosistet with the costraits. proof: Let the total applied force o the particle be F a ad the total force of costrait is F c. Suppose particle is i equilibrium. The by defiitio of equilibrium F a + F c = 0 (i) Now we have to show that virtual work doe by the applied forces is zero. Let δr be a ifiitesimal displacemet of the particle cosistet with costraits. The by takig dot product of (i) with δr, we get (F a + F c ).δr = δr.0 F a.δr + F c.δr = 0 (ii) Sice the costrait is workless therefore So (ii) becomes F c.δr = 0 F a.δr = 0 Which shows that the virtual work doe by the applied forces is zero. Coversely, suppose that the virtual work doe by the applied forces is zero. i.e. F a.δr = 0 (iii) Now we have to prove that the particle is i equilibrium. i.e. i.e. F a + F c = 0 Sice the costrait is workless therefore F c.δr = 0 Addig (iii) ad (iv), we get (iv) F a.δr + F c.δr = 0 (F a + F c ).δr = 0 F a + F c = 0 δr 0 Which shows that the particle is i equilibrium. This completes the proof. Cotact At: qadri86@yahoo.com

4 Priciple of Virtual Work for a Set of Particles 4 STATEMENT: A set of particles subject to workless costraits, is i equilibrium if ad oly if zero virtual work is doe by the applied forces i ay arbitrary ifiitesimal displacemet cosistet with the costraits. proof: Let the total applied force o the set of particles be F ia ad the total force of costrait is F ic. Suppose the set of particles is i equilibrium. The by defiitio of equilibrium F ia + F ic = 0 (i) Now we have to show that virtual work doe by the applied forces is zero. Let δr i be a ifiitesimal displacemet of the particle cosistet with costraits. The by takig dot product of (i) with δr i, we get F ia + F ic.δr i = δr i.0 F ia.δr i + F ic.δr i = 0 F ia.δr i + F ic.δr i = 0 (ii) Sice the costrait is workless therefore So (ii) becomes F ic.δr i = 0 F ia.δr i = 0 Which shows that the virtual work doe by the applied forces is zero. Coversely, suppose that the virtual work doe by the applied forces is zero. i.e. F ia.δr i = 0 (iii) Now we have to prove that the particle is i equilibrium. i.e. i.e. F ia + F ic = 0 Sice the costrait is workless therefore F ic.δr i = 0 (iv) Cotact At: qadri86@yahoo.com

5 5 Addig (iii) ad (iv), we get F ia.δr i + F ic.δr i = 0 F ia.δr i + F ic.δr i = 0 F ia + F ic.δr i = 0 F ia + F ic = 0 δr i 0 Which shows that the particle is i equilibrium. This completes the proof. Priciple of Virtual Work for a Sigle Rigid Body STATEMENT: A rigid body subject to workless costraits, is i equilibrium if ad oly if zero virtual work is doe by the applied forces ad applied torques i ay arbitrary ifiitesimal displacemet cosistet with the costraits. proof: Note that a system of forces actig o a rigid body ca be reduced to a sigle force R at ay arbitrary poit together with a couple G. Let R R a R c Where R a is sum of applied forces ad R c is sum of forces of costraits. Let G G a G c Where G a is sum of applied torques ad G c is sum of torques of costraits. Suppose the rigid body is i equilibrium. The by defiitio of equilibrium R = 0 (i) ad G = 0 (ii) Now we have to show that the virtual work doe by applied force ad applied torques is zero. Let δr ad δ be a ifiitesimal virtual displacemet ad rotatio of the rigid body. The by takig dot product of (i) with δr ad (ii) with δ, we get R.δr = δr.0 R.δr = 0 (iii) ad G.δ = δ.0 G.δ = 0 (iv) Cotact At: qadri86@yahoo.com

6 6 Addig (iii) ad (iv), we get R.δr + G.δ = 0 (R a R c ).δr + (G a G c ).δ = 0 R a δr R c.δr + G a.δ G c.δ = 0 (v) Sice the costraits are workless therefore So (v) becomes R c.δr = 0 ad G c.δ = 0 R a δr + G a.δ = 0 Which shows that the virtual work doe by the applied forces ad applied torques is zero. Coversely, suppose that the virtual work doe by the applied forces ad applied torques is zero. i.e. R a δr = 0 ad G a.δ = 0 R a δr + G a.δ = 0 (vi) Now we have to show that the rigid body is i equilibrium. i.e. R = 0 ad G = 0 Sice the costraits are workless therefore R c.δr = 0 ad G c.δ = 0 R c.δr + G c.δ = 0 (vii) Addig (vi) ad (vii), we get R a δr + G a.δ + R c.δr + G c.δ = 0 (R a + R c ).δr + (G a + G c ).δ = 0 R.δr + G.δ = 0 Sice δr 0 ad δ 0 Therefore R = 0 ad G = 0 Which shows that the body is i equilibrium. This completes the proof. Priciple of Virtual Work for Set of Rigid Bodies STATEMENT: A set of rigid bodies subject to workless costraits, is i equilibrium if ad oly if zero virtual work is doe by the applied forces ad applied torques i ay arbitrary ifiitesimal displacemet cosistet with the costraits. proof: Cotact At: qadri86@yahoo.com

7 7 Note that a give system of forces actig o i th at ay arbitrary poit i the plae together with a couple G i. rigid body ca be reduced to a sigle force R i Let R i R ia R ic Where R ia is sum of applied forces ad R ic is sum of forces of costraits. Let G i G ia G ic Where G ia is sum of applied torques ad G ic is sum of torques of costraits. Suppose the rigid body is i equilibrium. The by defiitio of equilibrium R i = 0 (i) ad G i = 0 (ii) Now we have to show that the virtual work doe by applied force ad applied torques is zero. Let δr i ad δ i be a ifiitesimal virtual displacemet ad rotatio of the rigid body. The by takig dot product of (i) with δr i ad (ii) with δ i, we get R i.δr = 0 (iii) ad G i.δ = 0 (iv) Addig (iii) ad (iv), we get R i δr G i δ = 0 R ia R ic.δr G ia G ic.δ = 0 R ia.δr R ic.δr G ia.δ G ic.δ = 0 (v) Sice the costraits are workless therefore So (v) becomes R ic.δr 0 ad G ic.δ = 0 R ia.δr G ia.δ = 0 Cotact At: qadri86@yahoo.com

8 8 Which shows that the virtual work doe by the applied forces ad applied torques is zero. Coversely, suppose that the virtual work doe by the applied forces ad applied torques is zero. i.e. R ia.δr R ia.δr 0 ad G ia.δ G ia.δ = 0 Now we have to show that the rigid body is i equilibrium. i.e. R i = 0 ad G i = 0 Sice the costraits are workless therefore R ic.δr 0 R ic.δr ad G ic.δ G ic.δ Addig (vii) ad (vii), we get R ia.δr R ic.δr R ia R ic.δr R i δr R i δr R i δr G i δ δ Sice δr 0 ad δ 0 R i 0 = 0 (vi) = 0 = 0 (vii) G ia.δ G ic.δ G ia G ic.δ = 0 = 0 0 ad δ ad G i = 0 Which shows that the body is i equilibrium. This completes the proof. = 0 = 0 = 0 Cotact At: qadri86@yahoo.com

9 9 Questio 1 A light thi rod 1ft log ca tur i a vertical plae about oe of its poit which is attached to a pivot. If weight of 3lb ad 4lb are suspeded from its eds, it rests i a horizotal positio. Fid the positio of the pivot ad its reactio o the rod. Solutio Let AB be a rod of legth 1ft ad O be pivot ad R be the ormal reactio of pivot o rod. Let δy be the ifiitesimal displacemet vertically upward directio. The R A x O 1 x B 4lb 3lb Equatio of virtual work Rδy 3δy 4δy = 0 (R 7)δy = 0 Sice δy is arbitrary therefore δy 0. So R 7 = 0 R = 7 Let AO = x The BO = 1 x The momets of weights about the pivot are 4x ad 3(1 x). Let δ be the small agular displacemet of the rod about the pivot. The Equatio of virtual work 4xδ 3(1 x)δ = 0 (4x x)δ = 0 (7x 36)δ = 0 Sice δ is arbitrary therefore δ 0. So 7x 36 = 0 x = ft AO = ft ad BO = ft Cotact At: qadri86@yahoo.com

10 10 Questio Four equal heavy uiform rods are freely joited to form a rhombus ABCD, which is freely suspeded from A ad is kept i shape of a square by a iextesible strig coectig A ad C. Show that the tesio i the strig is W where W is the weight of each rod. Solutio A Fixed Level T D G B 4W T C Let W be the weight of each rod. The the weight of the four rod is 4W, which acts at a poit G, where G is the poit of the itersectio diagoals AC ad BD. Let T be tesio i the strig. Let AC = y AG = y The Equatio of virtual work 4W δ(ag) Tδ(AC) = 0 4W δ(y) Tδ(y) = 0 (4W T)δ(y) = 0 4W T = 0 δ(y) 0 4W = T T = W Questio 3 Four equal uiform rods are smoothly joited to form a rhombus ABCD, which is placed i a vertical plae with AC vertical ad A restig o a horizotal plae. The rhombus is kept i shape, with the measure of agle BAC equal to, by a light strig joiig B ad D. Fid the tesio i the strig. Solutio Cotact At: qadri86@yahoo.com

11 11 C D T G T B 4W L Fixed Horizotal Plae A Let W be the weight of each rod. The the weight of the four rod is 4W, which acts at a poit G. Where G is the poit of the itersectio diagoals AC ad BD. Let L be legth of each rod. Let T be tesio i the strig. Give BAC = Equatio of virtual work 4W δ(ag) Tδ(BD) = 0 4W δ(ag) + Tδ(BD) = 0 (i) From Fig. Also AG = Lcos ad BG = Lsi BD = (BG) = Lsi Usig values i (i), we get 4W δ(lcos) + Tδ(Lsi) = 0 4WLsi δ + TLcos δ = 0 (LTcos 4WLsi) δ = 0 Sice δ is a arbitrary therefore δ 0 So LTcos 4WLsi = 0 LTcos = 4WLsi Tcos = Wsi T = Wta Cotact At: qadri86@yahoo.com

12 1 Questio 4 A rhombus ABCD of smoothly joited rods, rests o a smooth table with the rod BC fixed i positio. The middle poits of AD ad DC are coected by a strig which is kept taut by a couple applied to the rod AB. Prove that the tesio of the strig is G ABcos 1 AB C Solutio C E T D H B T F G Fixed Horizotal Plae A Cosider a rhombus ABCD of smoothly joited rods, rests o a smooth table with the rod BC fixed i positio as show i figure. Let T be the tesio i the strig EF where E ad F are the middle poits of the rods DC ad AD respectively. Let L be the legth of each rod ad G be a couple applied o the rod AB. Equatio of virtual work Gδ() Tδ(EF) = 0 (i) From figure. EF = (HF) = L si = Lsi Usig value of EF i (i), we get Gδ() Tδ(Lsi) = 0 Gδ TLcos δ = 0 (G TLcos) δ = 0 Cotact At: qadri86@yahoo.com

13 13 G TLcos = 0 δ 0 T = G Lcos Sice L = AB ad = 1 AB C T = G ABcos 1 AB C Questio 5 Four uiform rods are freely joited at their extremities ad form a parallel-gram ABCD, which is suspeded from the joit A ad is kept i shape by a iextesible strig AC. Prove that the tesio i the strig is equal to half the whole weight. Solutio A Fixed Level B T G W T D C Let W be the weight of the four rods actig at poit G. Where G is the poit of itersectio of the diagoals AC ad BD. Suppose the T be the tesio i the strig AC. Let AC = y The AG = y Equatio of virtual Work Wδ(AG) Tδ(AC) = 0 Wδ(y) Tδ(y) = 0 (W T)δy = 0 W T = 0 δy 0 T = W Cotact At: qadri86@yahoo.com

14 14 Questio 6 A strig of legth a forms the shorter diagoal of a rhombus formed by four uiform rods, each of legth b ad weight W, which are haged together. If oe of the rod is supported i a horizotal positio. Prove that the tesio i the strig is Solutio W b a b 4b a A M B Fixed Horizotal Level T 90 0 G D N C 4W T Sice W is the weight of each rod therefore the weight of four rods is 4W actig at poit G. Where G is the poit of itersectio of the diagoals AC ad BD. Let T be the tesio i the strig. Give AB = BC = CD = DA = b ad AC = a Let ADC = The ADG = ad BDC = Equatio of virtual work 4Wδ(MG) Tδ(AC) = 0 From fig. Also MG = 1 AN = 1 ADsi = 1 bsi AC = AG = ADsi = bsi But AC = a a = bsi si = a b Usig values of MG ad AC i (i), we get 4Wδ 1 bsi Tδ bsi 0 (i) Cotact At: qadri86@yahoo.com

15 15 4Wbcosδ Tbcosδ = 0 (4Wbcos Tbcos)δ = 0 4Wbcos Tbcos = 0 δ 0 4Wbcos = Tbcos T = Wcos cos = W cos si = W 1 si si cos 1 si = W 1 si 1 si W 1 a b = 1 a b W 1 a = 4b 1 a 4b W b a b 4b a Questio 7 = W 4b a b 4b a 4b Six equal rods AB, BC, CD, DE, EF ad FA are each of weight W ad are freely joited at their extremities so as to form a hexago. The rod AB is fixed i a horizotal positio ad the middle poits of AB ad DE are joited by a strig. Prove that its tesio is 3W. Solutio A M B Fixed Horizotal Level T F G C 6W E N D Cotact At: qadri86@yahoo.com

16 16 Let M ad N be the midpoits of the sides AB ad DE of a hexago ABCDEF. Let W be the weight of the each rod. The weight of the six rods is 6W which acts at poit G. Where G is the cetre of the gravity of hexago. Let T be the tesio i the strig MN. Let MN = y The MG = y Equatio of virtual work 6Wδ(MG) Tδ(MN) = 0 6Wδ(y) Tδ(y) = 0 (6W T)δy = 0 6W T = 0 δy 0 T = 3W Questio 8 Six equal uiform rods AB, BC, CD, DE, EF ad FA are each of weight W are freely joited to form a regular hexago. The rod AB is fixed i a horizotal positio ad the shape of the hexago is maitaied by a light rod joiig C ad F. Show that the thrust i this rod is 3W. Solutio A P B Fixed Horizotal Level F G T M N T C E D 6W Sice W be weight of each rod therefore weight of six rod is 6W which acts at the cetre of gravity G of hexago. Let L be the legth of each rod ad be the agle which rod BC makes with horizotal. Let AM ad BN are perpedicular o CF. Let P be the midpoit of the AB. Let T be the thrust i the rod FC. Cotact At: qadri86@yahoo.com

17 Equatio of virtual work 6W δ(pg) + T δ(fc) = 0 From figure. PG = AM = BN = BC si = L si ad CN = Lcos Also FC = FM + MN + CN = CN + AB + CN FM = CN ad MN = AB = CN + AB = L cos + L Usig values of PG ad FC i (i), we get 6W δ(l si) + T δ(l cos + L) = 0 6WL cosδ TLsiδ = 0 (6WL cos TLsi)δ = WL cos TLsi = 0 δ 0 6WL cos = TLsi T = 3Wcot For a regular hexago = 60 0 So T = 3Wcot60 0 = 3W 1 3 3W Questio 9 (i) A hexago ABCDEF, cosistig of six equal heavy rods, of weight W, freely joited together, hags i a vertical plae with AB horizotal ad the frame is kept i the form of regular hexago by a light rod coectig the midpoits of CD ad EF. Show that the thrust i the light rod is 3W. Solutio A M B Fixed Horizotal Level F P G 6W H K T T Q C E N D Cotact At: qadri86@yahoo.com

18 18 Sice W be weight of each rod therefore weight of six rod is 6W which acts at the cetre of gravity G of hexago. Let L be the legth of each rod ad be the agle which rod AB makes with horizotal. Let EH ad DK are perpedicular o PQ where P ad Q are the midpoits of the CD ad EF. Let T be the thrust i the rod PQ. Equatio of virtual work From figure. Also 6W δ(mg) + T δ(pq) = 0 MG = GN = DK = DQ si = L si = L si ad KQ = L cos PQ = PH + HK + KQ = KQ + ED + KQ HK = ED ad KQ = PH = KQ + ED = L cos + L Usig values of PQ ad MG i (i), we get 6W δ(l si) + T δ(l cos + L) = 0 6WL cosδ TL siδ = 0 (6WL cos TL si)δ = 0 6WL cos TL si = 0 δ 0 6WL cos = TL si T = 6Wcot For a regular hexago = 60 0 So T = 6Wcot60 0 = 6W 1 3W 3 Questio 10 (i) A heavy elastic strig whose atural legth is πa, is placed aroud a smooth coe whose axes is vertical ad whose semi-vertical agle has measure α. If W be the weight ad λ the modulus of strig. Prove that it will be i equilibrium whe i the form of a circle of radius Solutio a 1+ W πλ cotα Let the strig be i equilibrium at a depth y below the vertex of the coe. Let the radius of the circle formed by the strig i the equilibrium positio be x. The circumferece of the circle is πx. Cotact At: qadri86@yahoo.com

19 From fig. 19 O taα = x y x = y taα Let T be the tesio i the strig the by Hook s law Chage i legth T = λ Origial Legth πx πa = λ πa = λ x a a Let the strig be displaced dowward by displacemet δy. The equatio of virtual work Wδy Tδ(πx) = 0 Wδy Tδ(πy taα) = 0 Wδy πttaαδy = 0 (W πttaα)δy = 0 W πttaα = 0 δy 0 W = πttaα W = πλ x a taα a Wacotα = πλ x a Wacotα = πxλ πaλ Wacotα + πaλ = πxλ Wacotα + πaλ x πλ a + Wacotα πλ a 1 + Wcotα πλ Questio 11 y α x Two equal particles are coected by two give weightless strigs, which are placed like a ecklace o a smooth coe whose axis is vertical ad whose vertex is uppermost. Show that the tesio i the strig is W π cotα Where W is the weight of each particle ad α the measure of the vertical agle of the coe. Cotact At: qadri86@yahoo.com

20 0 Solutio O y α x Let both particles be i equilibrium at a depth y below the vertex O. Let the radius of the circle formed by the strig i the equilibrium positio be x. The circumferece of the circle is πx. From fig. taα = x y x = y taα Let T be the tesio i the strig. The equatio of virtual work Wδy Tδ(πx) = 0 Wδy Tδ(πy taα) = 0 Wδy πttaαδy = 0 (W πttaα)δy = 0 W πttaα = 0 δy 0 W = πttaα T = W π cotα Questio 1 A weightless tripod, cosistig of three legs of equal legth l, smoothly joited at the vertex, stads o a smooth horizotal plae. A weight W hags from the apex. The tripod is preveted from collapsig by three iextesible strigs each of legth is l, joiig the mid-poits of the legs. Show that the tesio i each strig is Solutio 3 3 W Cotact At: qadri86@yahoo.com

21 1 O G 30 0 A D B E G 30 0 D A C W Sice the legth of each rod is l. Therefore OA = OB = OC = AB = BC = CA = l The vertical lie through O meets the plae of ABC at poit G. Where G is the poit of itersectio of the medias AE ad BD. Let each rod makes a agle with OG. Let T be the tesio i each strig. The Equatio of virtual work W δ(og) 3Tδ(AC) = 0 { The height of O above the horizotal plae = OG } W δ(og) + 3Tδ(AC) = 0 (i) From AOG OG = OA cos = l cos ad AG = OA si = l si AC = AD = AGcos30 = l 3 si = 3 l si 3 l si l AC = l si 1 3 As cos = 1 si = = 3 Cotact At: qadri86@yahoo.com

22 ta 1 Usig value of OG ad AC, we get Wδ lcos +3Tδ 3 l si = 0 Wlsi δ l costδ = 0 Wlsi l cost δ = 0 Wlsi l cost 0 δ 0 Wlsi = 3 3 l cost T 3 3 Wta 3 3 W W Questio 13 Three equal rods each of weight W are freely joited together at oe extremity of each to form a tripod ad rest with their other extremities o a smooth horizotal plae. Each rod is iclied at agle of to the vertical, equilibrium is beig maitaied by three light strigs each joiig two these extremities. Prove that the tesio i each strig is W ta 3 Solutio O B G 30 0 A E D C W Cotact At: qadri86@yahoo.com

23 3 Let the legth of each rod is L ad legth of each strig is x. The OA = OB = OC = L ad AB = BC = CA = x The vertical lie through O meets the plae of ABC at poit G. Where G is the poit of itersectio of the medias AE ad BD. Let each rod makes a agle with OG. Let T be the tesio i each strig. The Equatio of virtual work 3Wδ OG 3Wδ OG 3Tδ(AC) = 0 + 3Tδ(AC) = 0 (i) Where OG is equal the height of midpoit of each rod above the horizotal plae. From AOG OG = OA cos = Lcos ad AG = OA si = Lsi AC = AD = AGcos30 = Lsi 3 = 3L si Usig value of OG ad AC, we get 3Wδ L cos +3Tδ 3L si = 0 3 WLsi δ + 3 3LcosTδ = 0 W si + 3cosT δ = 0 W si + 3cosT 0 δ 0 W si = 3cosT W T 3 ta Questio 14 Six equal uiform rods are freely joited at their extremities to form a tetrahedro. If this tetrahedro is place with oe face o a smooth horizotal table. Prove that the thrust alog a horizotal rod is W 6 Where W is the weight of each rod. Solutio Cotact At: qadri86@yahoo.com

24 4 O B G 30 0 A E D C W Let the legth of each rod is L. Therefore OA = OB = OC = AB = BC = CA = L The vertical lie through O meets the plae of ABC at poit G. Where G is the poit of itersectio of the medias AE ad BD. Let each rod makes a agle with OG. Let T be the thrust i each lower rod. The Equatio of virtual work 3Wδ OG 3Wδ OG 3Tδ(AC) = 0 + 3Tδ(AC) = 0 (i) Where OG is equal the height of midpoit of each rod above the horizotal plae. From AOG OG = OA cos = Lcos ad AG = OA si = Lsi AC = AD = AGcos30 = Lsi 3 = 3L si 3L si = L AC = L 1 si 3 Cotact At: qadri86@yahoo.com

25 5 Now cos = 1 si = ta = 3 1 Usig value of OG ad AC, we get 3Wδ L cos +3Tδ 3L si = 0 3 WLsi δ + 3 3LcosTδ = 0 W si + 3cosT δ = 0 W si + 3cosT 0 δ 0 W si = 3cosT W T 3 ta W 1 3 W 6 Questio 15 Four equal uiform rods, each of weight w, are coected at oe ed of each by meas of a smooth joit, ad other eds rest o a smooth table ad are coected by equal strigs. A weight W is suspeded from the joit. Show that the tesio i each strig is W + w 4 a 4l a Where l is the legth of each rod ad a is the legth of each strig. Solutio Cotact At: qadri86@yahoo.com

26 6 O A D G B W 45 0 C Give that ad OA = OB = OC = OD = l AB = BC = CD = DA = a Let the suspeded weight W makes a agle with each rod. Let T be the tesio i each strig. The equatio of virtual work I OCG 4wδ OG Wδ OG 4Tδ(BC) = 0 (i) Also I ABC CG = OCsi = lsi ad OG = lcos AC = CG = lsi BC = ACcos45 0 = lsi 1 = lsi lsi = a BC = a a si = l Cotact At: qadri86@yahoo.com

27 7 Now cos = 1 si = 1 a l = l a l a ta l a Usig values of OG ad BC i (i), we get 4wδ lcos Wδ lcos 4Tδ( lsi) = 0 wsiδ + Wsiδ 4 Tcosδ = 0 wsi + Wsi 4 Tcos δ = 0 wsi + Wsi 4 Tcos 0 δ 0 wsi + Wsi 4 Tcos T w + W ta 4 w + W ta 4 w + W 4 w + W 4 a l a a 4l a Questio 16 A regular octahedro formed of twelve equal rods, each of weight w, freely joited together is suspeded from oe corer. Show that the thrust i each horizotal rod is Solutio Let legth of each rod be L. The 3 w OA = OB = OC = OD=AB = BC = CD = DA = O A = O B = O C = O D = L Let each upper rod makes a agle with vertical OG. Where G is the poit of itersectio of diagoals AC ad BD. Let T be the thrust i each horizotal rod. Cotact At: qadri86@yahoo.com

28 8 The Equatio of virtual work 1Wδ OG + 4Tδ(BC) = 0 (i) O A D G B W 45 0 C I COG OG = OCcos = Lcos Ad CG = OCsi = Lsi Also AC = CG = Lsi I ABC O BC = ACcos45 0 = Lsi 1 = Lsi Lsi = L BC = L si = 1 = 45 0 Usig values of OG ad BC i (i), we get 1Wδ Lcos + 4Tδ( Lsi) = 0 Cotact At: qadri86@yahoo.com

29 9 3Wsi δ + T cos δ = 0 3Wsi + T cos δ = 0 3Wsi + T cos 0 δ 0 3Wsi = T cos T 3 Wta 3 Wta45 3 W 3 W Questio 17 A uiform rod of legth a rest i equilibrium agaist a smooth vertical wall ad upo a smooth peg at a distace b from the wall. Show that, i the positio of equilibrium, the beam is iclied to the wall at a agle Solutio si 1 b 1 3 a Let AB be the rod of legth a. W is its weight actig dowward from its cetre G. The AB = a ad AG = a Let P be a peg at distace b from the wall. B The NP = b Equatio of virtual work From figure. Wδ(MN) = 0 MN = AM AN = AGcos NPcot (i) M N A b P W G = acos bcot Usig value of MN i (i), we get Wδ( acos bcot) = 0 Cotact At: qadri86@yahoo.com

30 W ( asiδ + bcosec δ) = 0 Wasiδ Wbcosec δ = 0 (Wasi Wbcosec )δ = 0 Wasi Wbcosec = 0 δ 0 Wasi = Wbcosec si cosec = b a 30 si 3 = b a si = b a = si 1 b a %%%%% Ed of The Chapter # 3 %%%%% Cotact At: qadri86@yahoo.com