Virtual Work Work done by a Force (U) U = work done by the component of the force in the direction of the displacement times the displacement or Since same results are obtained irrespective of the direction in which we resolve the vectors Work is a scalar quantity +U Force and Disp in same direction - U Force and Disp in opposite direction Virtual Work Generalized Definition of Work Work done by F during displacement dr Expressing F and dr in terms of their rectangular components Total work done by F from A1 to A2 Virtual Work: Work done by a Force Forces which do no work: • forces applied to fixed points (ds = 0) • forces acting in a dirn normal to the disp (cosα = 0) • reaction at a frictionless pin due to rotation of a body around the pin • reaction at a frictionless surface due to motion of a body along the surface • weight of a body with cg moving horizontally • friction force on a wheel moving without slipping Sum of work done by several forces may be zero: • bodies connected by a frictionless pin • bodies connected by an inextensible cord • internal forces holding together parts of a rigid body Virtual Work Work done by a Couple (U) Small rotation of a rigid body: • translation to A’B’ work done by F during disp AA’ will be equal and opposite to work done by -F during disp BB’ total work done is zero • rotation of A’ about B’ to A” work done by F during disp AA” : U = F.drA/B = Fbdθ Since M = Fb +M M has same sense as θ - M M has opp sense as θ Total word done by a couple during a finite rotation in its plane: Virtual Work Dimensions and Units of Work (Force) x (Distance) Joule (J) = N.m Work done by a force of 1 Newton moving through a distance of 1 m in the direction of the force Dimensions of Work of a Force and Moment of a Force are same though they are entirely different physical quantities. Work is a scalar given by dot product; involves product of a force and distance, both measured along the same line Moment is a vector given by the cross product; involves product of a force and distance measured at right angles to the force Units of Work: Joule Units of Moment: N.m Virtual Displacement Virtual Displacement is not experienced but only assumed to exist so that various possible equilibrium positions may be compared to determine the correct one • Imagine the small virtual displacement of particle (δr) which is acted upon by several forces. • The corresponding virtual work, → → → → → → → → → → U F1 r F2 r F3 r F1 F2 F3 r → → R r Virtual Displacement Equilibrium of a Particle Total virtual work done on the particle due to virtual displacement r: Expressing ∑F in terms of scalar sums and δr in terms of its component virtual displacements in the coordinate directions: The sum is zero since ∑F = 0, which gives ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 Alternative Statement of the equilibrium: U = 0 This condition of zero virtual work for equilibrium is both necessary and sufficient since we can apply it to the three mutually perpendicular directions 3 conditions of equilibrium Virtual Work Principle of Virtual Work: • If a particle is in equilibrium, the total virtual work of forces acting on the particle is zero for any virtual displacement. • If a rigid body is in equilibrium • total virtual work of external forces acting on the body is zero for any virtual displacement of the body • If a system of connected rigid bodies remains connected during the virtual displacement • the work of the external forces need be considered • since work done by internal forces (equal, opposite, and collinear) cancels each other. Example (1) on Virtual Work Principle Equilibrium of a Rigid Body Total virtual work done on the entire rigid body is zero since virtual work done on each Particle of the body in equilibrium is zero. Weight of the body is negligible. Work done by P = -Pa θ Work done by R = +Rb θ Principle of Virtual Work: U = 0: -Pa θ + Rb θ = 0 Pa – Rb = 0 Equation of Moment equilibrium @ O. Nothing gained by using the Principle of Virtual Work for a single rigid body Virtual Work Principle of Virtual Work Virtual Work done by external active forces on an ideal mechanical system in equilibrium is zero for any and all virtual displacements consistent with the constraints U=0 Three types of forces act on interconnected systems made of rigid members Active Forces: Work Done Active Force Diagram Reactive Forces No Work Done Internal Forces No Work Done Virtual Work Major Advantages of the Virtual Work Method - It is not necessary to dismember the systems in order to establish relations between the active forces. Relations between active forces can be determined directly without reference to the reactive forces. The method is particularly useful in determining the position of equilibrium of a system under known loads (This is in contrast to determining the forces acting on a body whose equilibrium position is known – studied earlier). The method requires that internal frictional forces do negligible work during any virtual displacement. If internal friction is appreciable, work done by internal frictional forces must be included in the analysis. Virtual Work Degrees of Freedom (DOF) - Number of independent coordinates needed to specify completely the configuration of system Only one coordinate (displacement or rotation) is needed to establish position of every part of the system Two independent coordinates are needed to establish position of every part of the system U = 0 can be applied to each DOF at a time keeping other DOF constant. ME101 only SDOF systems Virtual Work Systems with Friction - So far, the Principle of virtual work was discussed for “ideal” systems. If significant friction is present in the system (“Real” systems), work done by the external active forces (input work) will be opposed by the work done by the friction forces. During a virtual displacement x: Work done by the kinetic friction force is: -μkNx During rolling of a wheel: the static friction force does no work if the wheel does not slip as it rolls. Virtual Work Mechanical Efficiency (e) - Output work of a machine is always less than the input work because of energy loss due to friction. Output Work e Input Work For simple machines with SDOF & which • operates in uniform manner, mechanical efficiency may be determined using the method of Virtual Work • For the virtual displacement s: Output Work is that necessary to elevate the block = mg s sinθ • Input Work: T s = mg sinθ s + μk mg cosθ s • The efficiency of the inclined plane is: e 1 mgs sin mg sin k cos s 1 k cot As friction decreases, Efficiency approaches unity Work of a Force dU = F dr = work of the force F corresponding to the displacement dr dU = F ds cosα α= 0, dU =+ F ds α= π , dU = − F ds α= π 2 , dU = 0 dU = Wdy Work of a Couple Small displacement of a rigid body: • translation to A’B’ • rotation of B’ about A’ to B” W = F dr1 + F dr1 + dr2 F dr2 = F ds2 = F rdθ M dθ Principle of Virtual Work • Imagine the small virtual displacement of particle which is acted upon by several forces. • The corresponding virtual work, δU = F1 δr + F2 δr + F3 δr = F1 + F2 + F3 δr R δr Principle of Virtual Work: • If a particle is in equilibrium, the total virtual work of forces acting on the particle is zero for any virtual displacement. • If a rigid body is in equilibrium, the total virtual work of external forces acting on the body is zero for any virtual displacement of the body. • If a system of connected rigid bodies remains connected during the virtual displacement, only the work of the external forces need be considered. Work of a Force Forces which do no work: • reaction at a frictionless pin due to rotation of a body around the pin • reaction at a frictionless surface due to motion of a body along the surface • weight of a body with cg moving horizontally • friction force on a wheel moving without slipping Sum of work done by several forces may be zero: • bodies connected by a frictionless pin • bodies connected by an inextensible cord • internal forces holding together parts of a rigid body 10 - 18 Virtual Work for a rigid body Different types of forces • Forces that do work are called active force. • Reactive and internal forces do not do any work. • Virtual displacements are to be given carefully so that the active forces are only the known forces and the forces we are interested in obtaining • Similar to FBD we draw active force diagram (AFD). Degrees of Freedom • DOF in this context is the total number of independent coordinates required to specify the complete location of every member of the structure. • For VW method in this course we will use only 1-DOF systems. 1 DOF 2 DOF To summarize • Principle of virtual work • The virtual work done by external active forces on an ideal mechanical system in equilibrium is zero for any all virtual displacements consistent with the constraints. • Ideal system: – All surfaces, joints etc. are frictionless. – We will deal with ideal system in this course. • Consistent with constraints: – The virtual displacement should be such that they should not do allow the non-active forces to do any work. Why principle of Virtual Work • For complex mechanisms (we will solve some problems) we do not need to dis-member the system. • We obtain the active unknown force in one shot without bothering about the reactive forces. • Such type of analysis will be a stepping stone to VW analysis using deformations when you study Solid Mechanics, Structural Mechanics etc. not to mention powerful Approximate methods like the Finite Element Method. Principle of Virtual Work U e U i u L u A 1 ( P1D1 ) 1 D u dL U o dV 2 P´ = 1 Virtual loadings Apply virtual load P´ first 1•D Su • dL Real displacements u L In a similar manner, u A dL D P1 Then apply real load P1. Virtual loadings 1• Su • dL Real displacements 24 JFTC 25 JFTC IC 442 26 JFTC IC 442 27 JFTC IC 442 28 JFTC IC 442 29 ARMADURAS JFTC IC 442 30 Method of Virtual Work : Truss P1 • External Loading. n1 n3 n7 B 1kN Where: N1 n5 n8 N3 N7 B n9 1 D N N8 5 P2 N9 D nNL AE 1 = external virtual unit load acting on the truss joint in the stated direction of D n = internal virtual normal force in a truss member caused by the external virtual unit load D = external joint displacement caused by the real load on the truss N = internal normal force in a truss member caused by the real loads L = length of a member A = cross-sectional area of a member E = modulus of elasticity of a member 31 • Temperature 1 D n (DT ) L Where: D = external joint displacement caused by the temperature change = coefficient of thermal expansion of member DT = change in temperature of member • Fabrication Errors and Camber 1 D nDL Where: D = external joint displacement caused by the fabrication errors DL = difference in length of the member from its intended size as caused by a fabrication error 32 ARMADURAS- ejemplos JFTC IC 442 33 Sample Problem 10.1 Determine the magnitude of the couple M required to maintain the equilibrium of the mechanism. SOLUTION: • Apply the principle of virtual work U 0 U M U P 0 M PxD xD 3l cos xD 3l sin 0 M P 3l sin M 3Pl sin 10 - 34 • Note that no support reactions were needed to solve the problem, nor was it necessary to take apart the machine at any connection. A clear and accurate FBD is still highly recommended, however. Problem 1 • Assuming frictionless contacts, determine the magnitude of P for equilibrium This problem will be referred to as the Ladder problem Example 8-15 The cross-sectional area of each member of the truss shown in the figure is A = 400 mm2 and E = 200 GPa. (a) Determine the vertical displacement of joint C if a 4-kN force is applied to the truss at C. (b) If no loads act on the truss, what would be the vertical displacement of joint C if member AB were 5 mm too short? (c) If 4 kN force and fabrication error are both accounted, what would be the vertical displacement of joint C. C 4 kN 3m A B 4m 4m 38 SOLUTION Part (a) •Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical 1 kN load is placed at joint C. The n force in each member is calculated using the method of joint. •Real Force N. The N force in each member is calculated using the method of joint. 1 kN C 4 kN C 0 A 0.5 kN 0.667 n (kN) B 2 4 kN A B N(kN) 0.5 kN 1.5 kN 1.5 kN 39 1 kN C 4 kN C A 0.667 B A n (kN) 2 B C N (kN) = A 8 B C L (m) A (1kN )( D Cv ) nNL AE D C 10.67 B nNL (kN2•m) 1 10.67 kN m (10.41 10.41 10.67) kN AE (400 10 6 m 2 )(200 10 6 2 ) m DCv = 0.133 mm, 40 Part (b): The member AB were 5 mm too short 1 kN C A 0.667 n (kN) B 5 mm (1)(D Cv ) n(DL) D Cv (0.667 )(0.005) DCv = -3.33 mm, Part (c): The 4 kN force and fabrication error are both accounted. DCv = 0.133 - 3.33 = -3.20 mm DCv = -3.20 mm, 41 Example 8-16 Determine the vertical displacement of joint C of the steel truss shown. The cross-section area of each member is A = 400 mm2 and E = 200 GPa. F E 4m D A B 4m C 4m 4 kN 4m 4 kN 42 SOLUTION •Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical 1 kN load is placed at joint C. The n force in each member is calculated using the method of joint. •Real Force N. The N force in each member is calculated using the method of joint. 0 A 0.333 0.333 F -0.333 0.333 kN 4m D C 4m -4 F 1 0.667 0.667 B 4m E 4m n (kN) 1 kN 0.667 kN E 4 0 4 4 A 4 B 4m 4 kN 4m 4 D C 4m 4 kN 4m 4 kN 4 kN N(kN) 43 0.333 0.333 F -0.333 B 1 0.667 0.667 E 4 4 4 4 B C n (kN) 1 kN -4 DA 4 C 4 kN N(kN) 4 kN F D = F 5.33 5.33 A E 4 4 B A E 4 C L(m) D 16 10.67 10.67 B 4 4 4 5.33 A F E C nNL(kN2•m) D nNL (1kN )( D Cv ) AE 1 72.4kN m D Cv [15.07 3(5.33) 2(10.67) 16 30.18)] kN AE (400 10 6 m 2 )(200 10 6 2 ) m DCv = 1.23 mm, 44 Example 8-17 Determine the vertical displacement of joint C of the steel truss shown. Due to radiant heating from the wall, members are subjected to a temperature change: member AD is increase +60oC, member DC is increase +40oC and member AC is decrease -20oC.Also member DC is fabricated 2 mm too short and member AC 3 mm too long. Take = 12(10-6) , the cross-section area of each member is A = 400 mm2 and E = 200 GPa. wall C 10 kN D 3m B A 2m 20 kN 45 SOLUTION • Due to loading forces. 1 kN 1 kN 0.667 kN 3m D 0.667 20 kN 1 D 23.33 23.33 kN C 0 20 3m 0 0.667 kN A 2m C 10 kN 20 D 13.33 kN A 2m N (kN) n (kN) D 31.13 60 C D Cv 0 A nNL(kN2•m) B 3 2 B B A 20 kN (1kN )( D Cv ) 0 C 3 0 B 2 L (m) nNL AE 1 (60 31.13 104 .12) (400 )( 200 ) DCv= 2.44 mm, 46 D 1 kN C 0.667 1 D A C +60 0 0 +40 D 2 C 3 3 D 2 B B A C B A DT (oC) n (kN) -2 L (m) A • Due to temperature change. (1kN )( D Cv ) n (DT ) L B Fabrication error (mm) DCv (12 106 )[(1)(60)(3) (0.667)(40)(2) (1.2)(20)(3.61)] = 3.84 mm, • Due to fabrication error. (1kN )(D Cv ) n(DL) DCv (0.667 )(0.002) (1.2)(0.003) = -4.93 mm, • Total displacement . (D Cv )Total 2.44 3.84 4.93 = 1.35 mm, 47 VIGAS JFTC IC 442 48 Principle of Virtual Work • The internal work in transversely loaded beams is taken equal to the strain energy due to bending moment • The virtual force Fi in the ith mass element in ∆=F*iei may be taken equal to the bending moment mij in the ith mass element due to a unit load at coordinate j Principle of Virtual Work (Displacement) • Sometimes referred as the Unit-Load Method • Generally provides of obtaining the displacement and slope at a specific point on structure i.e. beam, frame or truss • In general, the principle states that: Work of Internal Loads PD Work of External Loads uδ Principle of Virtual Work (Displacement) • Consider the structure (or body) to be of arbitrary shape • Suppose it is necessary to determine the displacement D of point A on the body caused by the “real loads” P1, P2 and P3 Principle of Virtual Work (Displacement) • Since no external load acts on the body at A and in the direction of the displacement D, the displacement can be determined by first placing on the body a “virtual” load suc that this force P’ acts in the same direction as D, (see Figure) Principle of Virtual Work (Displacement) • We will choose P’ to have a unit magnitude, P’ =1 • Once the virtual loadings are applied, then the body is subjected to the real loads P1, P2 and P3, (see Figure) • Point A will be displaced an amount D causing the element to deform an amount dL Principle of Virtual Work (Displacement) • As a result, the external virtual force P’ & internal load u “ride along” by D and dL and therefore, perform external virtual work of 1. D on the body and internal virtual work of u.dL on the element Virtual loadings 1.D u.dL Real displacement • By choosing P’ = 1, it can be seen from the solution for D follows directly since D = SudL Principle of Virtual Work (Slope) • A virtual couple moment M’ having a unit magnitude is applied at this point • This couple moment causes a virtual load uθin one of the elements of the body Principle of Virtual Work (Slope) • Assuming that the real loads deform the element an amount dL, the rotation can be found from the virtual-work equation Principle of Virtual Work (Slope) Virtual loadings 1.θ u.dL Real displacement PRINCIPLE OF UNIT LOAD METHOD VIRTUAL LOAD REAL LOAD V X M • The element deform or rotate dθ = (M / EI) dx • The external virtual work done by the unit load is 1. • The internal virtual work done by the moment, m m dθ = m(M/EI) dx L 1 .D 0 Similarly mM EI dx L 1 .θ 0 mM EI by Saffuan Wan Ahmad dx VIGAS- ejemplos JFTC IC 442 60 b y S a f f u a n W a n A h m a d Example 1 Determine the displacement at point B of the steel beam shown in figure. Take E = 200GPa, I = 500x 106 mm4 b y S a f f u a n W a n A h m a d Real Moment, M LHS RHS M M V V b y S a f f u a n W a n A h m a d Real Moment, M RHS M 12 kN/m M x x 0 (clockwise ve) 12x 2 M 0 2 M 6x 2 b y S a f f u a n W a n A h m a d Virtual Moment, m 1 kN A B LHS RHS 1 kN M M V V Virtual Moment, m Considered RHS 0<x<10 1 kN M x 0 (clockwise ve) m 1.x 0 M m 1.x x by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d Virtual-Work Equation 1kN.D L 0 mM dx EI (x)(6x 2 ) dx 0 EI 15103 kN.m 3 EI 10 OR 15103 kN.m 3 DB 200(106 )kN / m 2 (500(106 )mm4 )(1012 m4 / mm4 ) 0.150m 150mm b y S a f f u a n W a n A h m a d Example 2 Determine the displacement at D of the steel beam in figure. Take E = 200GPa, I = 300E6 mm4 b y S a f f u a n W a n A h m a d Real Moment, M M1 M2 M3 Determine the reaction at support… VA M A VB 0 (clockwise ve) 120 30(3) VB (6) 0 VB 35kN F y 0 (upward ve) V A 35 30 0 V A 5kN by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d 0 x3 Member AB : LHS 120 kN.m M1 M x 0 (clockwise ve) M 1 5(x) 120 0 5 kN x Member BC : RHS M 1 120 5x 0 x3 M M2 35 kN x x 0 (clockwise ve) M 2 35(x) 0 M 2 35(x) b y S a f f u a n W a n A h m a d Member CD : RHS 0 x 4.5 M M3 x 0 (clockwise ve) M3 0 x b y S a f f u a n W a n A h m a d Virtual moment, m 1 kN m1 m1 m1 Determine the reaction at support 1 kN VA M A 0 (clockwise ve) 1(10.5) VB (6) 0 VB 1.75kN VB F y 0 (upward ve) V A 1.75 1 0 VA 0.75kN by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d 0 x3 Member AB : LHS m1 M x 0 (clockwise ve) m1 0.75(x) 0 0.75 kN x m1 0.75x 0 x3 Member BC : RHS 1 kN m2 M x 0 (clockwise ve) m2 1.75(x) 1(x 4.5) 0 1.75 kN x m2 0.75(x) 4.5 b y S a f f u a n W a n A h m a d Member CD : RHS 1 kN m3 0 x 4.5 M x 0 (clockwise ve) m3 1.x 0 x m3 x b y S a f f u a n W a n A h m a d Virtual-Work Equation mM dx 1kN.D d 0 EI 3mM 3m M 4.5 m M 2 2 3 3 1 1 dx dx dx 0 0 0 EI EI EI 3 (0.75x)(120 5x) 3 (0.75x 4.5)(35x) dx dx 0 0 EI EI 4.5 (x)(0) dx 0 EI L b y S a f f u a n W a n A h m a d 371 .25 472 .5 0 EI EI EI 3 843 .75 kN .m EI 843 .75 kN .m 3 200 (10 6 )kN / m 2 (300 (10 6 )mm 4 )(10 12 m 4 / mm 4 ) DD DD 0.0141 m 14 .1mm b y S a f f u a n W a n A h m a d Example 3 Determine the slope at A and deflection at C in the beam shown below 9 kN 12 kN/m A B C 3m D 1m 2m b y S a f f u a n W a n A h m a d Solution • Real Generalized coordinates A Load (M) C 3m • 1 unit load • Virtual Load (m ): Slope θ 2 9 kN 12 kN/m B D 1m 2m 1 • 1 unit load • Virtual Load (m ): Deflection Δ Real Load M ? 1. Support reaction, 9 kN 12 kN/m B A C 3m M A 0(clockwise ), 3 RB (6) 9(4) 12(3) 0 2 RB 15kN D 1m 2m Fy 0, R A 12(3) 9 15 0 R A 30kN by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d Real Load (M) : 0 x 3(segment AC) 12kN/m Mx 30kN x M x 0 (clockwise), x2 M x 30x 12 2 M x 30x 6x 2 ..........(i) b y S a f f u a n W a n A h m a d Real Load (M) : 3 x 4 (segment CD) 12kN/m 30kN Mx 3m x M x 0 (clockwise ), M x 30x 12(3)(x 3 ) 2 M x 30x 36x 54 M x 6x 54..........(ii) b y S a f f u a n W a n A h m a d Real Load (M) : 0 x 2 (segment BD) 15 Mx x M x 0 (clockwise ), M x 15x M x 15 x..........(iii) Virtual Load, m for deflection 1 Apply point load P= 1 C D RA RB 3m M A 0 (clockwise ), 1m 2m Fy 0, RB (6) 1(3) 0 R A RB 1 0 1 RB 2 RA by Saffuan Wan Ahmad 1 2 b y S a f f u a n W a n A h m a d Virtual Load (m) : 0 x 3(segment AC) Mx 1/2 x M x 0 (clockwise ), M x 1 x..........(i) 2 1 3m Mx 1/2 x Virtual Load (m) : 3 x 4 (segment CD) M x 0 (clockwise ), M x 0.5x 3..........(ii) by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d Virtual Load (m) : 0 x 2 (segment BD) 1/2 Mx x M x 0 (clockwise ), 1 Mx x 2 1 M x x..........(iii) 2 b y S a f f u a n W a n A h m a d Deflection at C, D D : DD Mm dx EI 1 3 1 2 (30x 6x )(0.5x)dx 0 EI EI 1 2 (15x)(0.5x)dx 0 EI 135.75 EI 4 3 (6x 54)(0.5x 3)dx Virtual Load, m for rotation Apply mθ= 1 1 C D RA RB 3m M A 0 (clockwise ), RB (6) 1 0 RB 1 6 1m 2m Fy 0, RA RB 0 RA by Saffuan Wan Ahmad 1 6 b y S a f f u a n W a n A h m a d Virtual Load (m) : 0 x 3(segment AC) 1 Mx -1/6 x M x 0 (clockwise ), 1 M x 1 x..........(i) 6 b y S a f f u a n W a n A h m a d Virtual Load (m) : 3 x 4 (segment CD) 1 Mx -1/6 x M x 0 (clockwise ), M x 1 1 x..........(ii) 6 b y S a f f u a n W a n A h m a d Virtual Load (m) : 0 x 2 (segment BD) 1/6 Mx x M x 0 (clockwise ), 1 Mx x 6 M x 1 x..........(iii) 6 b y S a f f u a n W a n A h m a d Slope at A, θA : θA 1 EI Mm dx EI 3 0 1 4 x 1 x dx (30x 6x )1 dx (6x 54) 3 EI 6 6 2 1 2 x dx (15x) 0 EI 6 76.75 EI b y S a f f u a n W a n A h m a d Example 4 DeterminetheslopeanddeflectionatBin thebeamshownbelow.Given E=200kN/mm2 5 kN/m IAB = 4x106 mm4 A IBC = 8x106 mm4 D B C 8 kN 0.5m 0.5m 2m Moment equation (deflection): Segment Condition I mm4 m (deflection) M AD 0<x<0.5 4 x 106 0 8x DB 0.5<x<1 4 x 106 0 8x – 2.5 (x – 0.5)2 BC 1<x<3 8 x 106 x–1 8x – 2.5 (x- 0.5)2 by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d Deflection, D B mM dx DB EI 3 (x 1)(2.5x 2 10.5x 0.625) dx 1 6 6 (20010 )(810 ) 1 2.5x 13x 11.125x 0.625x 1600 4 3 2 1 0.012m 4 12mm 3 2 3 Moment equation (slope): Segment Condition I mm4 m (slope) M AD 0<x<0.5 4 x 106 0 8x DB 0.5<x<1 4 x 106 0 8x – 2.5 (x – 0.5)2 BC 1<x<3 8 x 106 –1 8x – 2.5 (x- 0.5)2 by Saffuan Wan Ahmad b y S a f f u a n W a n A h m a d Slope,θB mM dx EI 3 (1)(2.5x 2 10.5x 0.625) dx 6 6 1 (20010 )(810 ) θB 3 2 1 2.5x 10.5x 0.625x 3 2 1600 1 3 19.1 1600 0.0119rad PORTICOS JFTC IC 442 99 Principle of Virtual Work In general, the principle states that: PD Work of Ext loads u Work of Int loads Consider the structure (or body) to be of arbitrary shape Suppose it is necessary to determine the disp D of point A on the body caused by the “real loads” P1, P2 and P3 Principle of Virtual Work It is to be understood that these loads cause no movement of the supports They can strain the material beyond the elastic limit Sincenoexternal load acts onthe body at A and in the direction of D, the disp D, the disp can be determined by first placing on the body a “virtual” load such that this force P’ acts in the same direction as D Principle of Virtual Work We will choose P’ to have a unit magnitude, P’ =1 Once the virtual loadings are applied, then the body is subjected to the real loads P1, P2 and P3, Point A will be displaced an amount D causing the element to deform an amount dL Principle of Virtual Work As a result, the external virtual force P’ & internal load u “ride along” by D and dL & therefore, perform external virtual work of 1. D on the body and internal virtual work of u.dL on the element 1.D u.dL By choosing P’ = 1, it can be seen from the solution for D follows directly since D = SudL A virtual couple moment M’ having a unit magnitude is applied at this point Principle of Virtual Work This couple moment causes a virtual load u in one of the elements of the body Assumingthat the real loads deform the element an amount dL, the rotation can be found from the virtual – work eqn 1. u .dL Method of virtual work: Beams & Frames To compute D a virtual unit load acting in the direction of D is placed on the beam at A The internal virtual moment m is determined by the method of sections at an arbitrary location x from the left support When point A is displaced D, the element dx deforms or rotates d = (M/EI)dx Method of virtual work: Beams & Frames 1.D L 0 mM dx EI external virtual unit load acting on the beam or frame in the direction of D internal virtual moment in the beam or frame, expressed as a function of x & caused by the ext virtual unit load ext disp of the point caused by real loads acting on the beam or frame int moment in the beam or frame, expressed as a function of x & caused by the real loads modulus of elasticity of the material moment of inertia of cross - sectional area, computed about the neutral axis Method of virtual work: Beams & Frames If the tangent rotation or slope angle at a point on the beam’s elastic curve is to be determined, a unit couple moment is applied at the point The corresponding int moment m have to be determined 1. L 0 m M dx EI Method of virtual work: Beams & Frames If concentrated forces or couple moments act on the beam or the distributed load isdiscontinuous,separate x coordinates will have to chosen within regions that have no discontinuity of loading Example 9.4 Determine the disp of point B of the steel beam. Take E= 200GPa and I = 500(106) mm4. Solution Virtual moment m The vertical disp of point B is obtained by placing a virtual unit load of 1kN at B. Using method of sections, the internal moment m is formulated. Real moment M Using the same x coordinate, M is formulated. Solution Virtual work eqn 1kN.D B L 0 mM dx EI 10 0 15(10 3 )kN 2 m 3 1kN.D B EI D B 0.150m 150mm (1x)(6x 2 ) dx EI Method of virtual work: MiMk Table