Published on

this is useful for structural students

Published in: Design
  • Be the first to comment

  • Be the first to like this

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide


  1. 1. Module 1Energy Methods inStructural Analysis Version 2 CE IIT, Kharagpur
  2. 2. Lesson 5Virtual Work Version 2 CE IIT, Kharagpur
  3. 3. Instructional ObjectivesAfter studying this lesson, the student will be able to:1. Define Virtual Work.2. Differentiate between external and internal virtual work.3. Sate principle of virtual displacement and principle of virtual forces.4. Drive an expression of calculating deflections of structure using unit loadmethod.5. Calculate deflections of a statically determinate structure using unit loadmethod.6. State unit displacement method.7. Calculate stiffness coefficients using unit-displacement method.5.1 IntroductionIn the previous chapters the concept of strain energy and Castigliano’s theoremswere discussed. From Castigliano’s theorem it follows that for the staticallydeterminate structure; the partial derivative of strain energy with respect toexternal force is equal to the displacement in the direction of that load. In thislesson, the principle of virtual work is discussed. As compared to other methods,virtual work methods are the most direct methods for calculating deflections instatically determinate and indeterminate structures. This principle can be appliedto both linear and nonlinear structures. The principle of virtual work as applied todeformable structure is an extension of the virtual work for rigid bodies. This maybe stated as: if a rigid body is in equilibrium under the action of a F − system offorces and if it continues to remain in equilibrium if the body is given a small(virtual) displacement, then the virtual work done by the F − system of forces as ‘itrides’ along these virtual displacements is zero.5.2 Principle of Virtual WorkMany problems in structural analysis can be solved by the principle of virtual work.Consider a simply supported beam as shown in Fig.5.1a, which is in equilibriumunder the action of real forces F1 , F2 ,......., Fn at co-ordinates 1,2,....., n respectively.Let u1 , u 2 ,......, u n be the corresponding displacements due to the action offorces F1 , F2 ,......., Fn . Also, it produces real internal stresses σ ij and real internalstrains ε ij inside the beam. Now, let the beam be subjected to second system offorces (which are virtual not real) δF1 , δF2 ,......, δFn in equilibrium as shown inFig.5.1b. The second system of forces is called virtual as they are imaginary andthey are not part of the real loading. This produces a displacement Version 2 CE IIT, Kharagpur
  4. 4. configuration δu1 , δu 2 ,........., δu n . The virtual loading system produces virtual internalstresses δσ ij and virtual internal strains δε ij inside the beam. Now, apply thesecond system of forces on the beam which has been deformed by first system offorces. Then, the external loads Fi and internal stresses σ ij do virtual work bymoving along δ ui and δε ij . The product ∑ F δu i i is known as the external virtualwork. It may be noted that the above product does not represent the conventionalwork since each component is caused due to different source i.e. δu i is not dueto Fi . Similarly the product ∑σ δεij ij is the internal virtual work. In the case ofdeformable body, both external and internal forces do work. Since, the beam is inequilibrium, the external virtual work must be equal to the internal virtual work.Hence, one needs to consider both internal and external virtual work to establishequations of equilibrium.5.3 Principle of Virtual DisplacementA deformable body is in equilibrium if the total external virtual work done by thesystem of true forces moving through the corresponding virtual displacements ofthe system i.e. ∑ Fi δu i is equal to the total internal virtual work for everykinematically admissible (consistent with the constraints) virtual displacements. Version 2 CE IIT, Kharagpur
  5. 5. That is virtual displacements should be continuous within the structure and also itmust satisfy boundary conditions. ∑ F δu = ∫ σ i i ij δε ij dv (5.1)where σ ij are the true stresses due to true forces Fi and δε ij are the virtual strainsdue to virtual displacements δu i .5.4 Principle of Virtual ForcesFor a deformable body, the total external complementary work is equal to the totalinternal complementary work for every system of virtual forces and stresses thatsatisfy the equations of equilibrium. ∑ δF u = ∫ δσ i i ij ε ij dv (5.2)where δσ ij are the virtual stresses due to virtual forces δFi and ε ij are the truestrains due to the true displacements u i .As stated earlier, the principle of virtual work may be advantageously used tocalculate displacements of structures. In the next section let us see how this canbe used to calculate displacements in a beams and frames. In the next lesson, thetruss deflections are calculated by the method of virtual work.5.5 Unit Load MethodThe principle of virtual force leads to unit load method. It is assumed throughoutour discussion that the method of superposition holds good. For the derivation ofunit load method, we consider two systems of loads. In this section, the principle ofvirtual forces and unit load method are discussed in the context of framedstructures. Consider a cantilever beam, which is in equilibrium under the action ofa first system of forces F1 , F2 ,....., Fn causing displacements u1 , u2 ,....., un as shown inFig. 5.2a. The first system of forces refers to the actual forces acting on thestructure. Let the stress resultants at any section of the beam due to first system offorces be axial force ( P ), bending moment ( M ) and shearing force ( V ). Also thecorresponding incremental deformations are axial deformation ( dΔ ), flexuraldeformation ( dθ ) and shearing deformation ( dλ ) respectively.For a conservative system the external work done by the applied forces is equal tothe internal strain energy stored. Hence, Version 2 CE IIT, Kharagpur
  6. 6. 1 n 1 1 1 2 ∑ Fu i =1 i i = 2 ∫P dΔ + 2 ∫M dθ + 2∫ V dλ P 2 ds M 2 ds V 2 ds L L L =∫ +∫ +∫ (5.3) 0 2 EA 0 2 EI 0 2 AGNow, consider a second system of forces δF1 , δF2 ,....., δFn , which are virtual andcausing virtual displacements δu1 , δu2 ,....., δun respectively (see Fig. 5.2b). Let thevirtual stress resultants caused by virtual forces be δPv , δM v and δVv at any crosssection of the beam. For this system of forces, we could write 1 n δP ds δM v ds L 2 δVv ds L 2 L 2 ∑ δFiδui = ∫ 2vEA + ∫ 2 EI + ∫ 2 AG 2 i =1 (5.4) 0 0 0where δPv , δM v and δVv are the virtual axial force, bending moment and shear forcerespectively. In the third case, apply the first system of forces on the beam, whichhas been deformed, by second system of forces δF1 , δF2 ,....., δFn as shown in Fig5.2c. From the principle of superposition, now the deflections will be(u1 + δu1 ), (u2 + δu2 ),......, (un + δun ) respectively Version 2 CE IIT, Kharagpur
  7. 7. Since the energy is conserved we could write, 1 n 1 n n L δ Pv 2 dsL δ M v 2 dsL δ Vv 2 dsL P 2 ds ∑ Fj u j + 2 ∑ δ Fjδ u j + ∑ δ Fj u j = ∫ 2 EA + ∫ 2 EI + ∫ 2 AG + ∫ 2 EA + 2 j =1 j =1 j =1 0 0 0 0 (5.5) M 2 ds V 2 ds L L L L L ∫ 2 EI 0 2 AG + ∫ δ Pv d Δ + ∫ δ M v dθ + ∫ δ Vv d λ 0 +∫ 0 0 0In equation (5.5), the term on the left hand side (∑ δF u ), represents the work j jdone by virtual forces moving through real displacements. Since virtual forces act Version 2 CE IIT, Kharagpur
  8. 8. ⎛1⎞at its full value, ⎜ ⎟ does not appear in the equation. Subtracting equation (5.3) ⎝2⎠and (5.4) from equation (5.5) we get, n L L L ∑ δFju j = ∫ δPvdΔ + ∫ δM vdθ + ∫ δVvdλ j =1 (5.6) 0 0 0From Module 1, lesson 3, we know that Pds Mds Vds dΔ = , dθ = and dλ = . Hence, EA EI AG n L δPv Pds L δM v Mds L δVvVds ∑ δF u = ∫ j =1 j j EA +∫ EI +∫ AG (5.7) 0 0 0 ⎛1⎞Note that ⎜ ⎟ does not appear on right side of equation (5.7) as the virtual system ⎝2⎠resultants act at constant values during the real displacements. In the presentcase δPv = 0 and if we neglect shear forces then we could write equation (5.7) as n L δM v Mds ∑ δFju j = ∫ j =1 EI (5.8) 0If the value of a particular displacement is required, then choose thecorresponding force δFi = 1 and all other forces δF j = 0 ( j = 1,2,...., i − 1, i + 1,...., n ) .Then the above expression may be written as, L δM v Mds (1)ui = ∫ (5.9) 0 EIwhere δM v are the internal virtual moment resultants corresponding to virtual forceat i-th co-ordinate, δFi = 1 . The above equation may be stated as, (unit virtual load ) unknown true displacement (5.10) = ∫ ( virtual stress resultants )( real deformations ) ds.The equation (5.9) is known as the unit load method. Here the unit virtual load isapplied at a point where the displacement is required to be evaluated. The unitload method is extensively used in the calculation of deflection of beams, framesand trusses. Theoretically this method can be used to calculate deflections in Version 2 CE IIT, Kharagpur
  9. 9. statically determinate and indeterminate structures. However it is extensively usedin evaluation of deflections of statically determinate structures only as the methodrequires a priori knowledge of internal stress resultants.Example 5.1A cantilever beam of span L is subjected to a tip moment M 0 as shown in Fig 5.3a. ⎛ 3L ⎞Evaluate slope and deflection at a point ⎜ ⎟ from left support. Assume EI of the ⎝ 4 ⎠given beam to be constant.Slope at CTo evaluate slope at C , a virtual unit moment is applied at C as shown in Fig 5.3c.The bending moment diagrams are drawn for tip moment M 0 and unit momentapplied at C and is shown in fig 5.3b and 5.3c respectively. Let θc be the rotationat C due to moment M 0 applied at tip. According to unit load method, the rotationat C , θc is calculated as, Version 2 CE IIT, Kharagpur
  10. 10. L δM v (x )M ( x )dx (1)θ c = ∫ (1) 0 EIwhere δM v (x ) and M (x ) are the virtual moment resultant and real momentresultant at any section x . Substituting the value of δM v (x ) and M (x ) in the aboveexpression, we get 3L / 4 (1)Mdx + L (0)Mdx (1)θ c = ∫ 0 EI ∫ 3L / 4 EI 3ML θc = (2) 4 EIVertical deflection at CTo evaluate vertical deflection at C , a unit virtual vertical force is applied ac C asshown in Fig 5.3d and the bending moment is also shown in the diagram.According to unit load method, L δM v ( x )M ( x )dx (1)u A = ∫ (3) 0 EI ⎛ 3L ⎞In the present case, δM v (x ) = −⎜ − x⎟ ⎝ 4 ⎠ and M (x ) = + M 0 3L ⎛ 3L ⎞ 4−⎜ − x ⎟M uA = ∫ ⎝ 4 ⎠ dx 0 EI 3L M ⎛ 3L ⎞ =− EI ∫ 0 4 ⎜ ⎝ 4 − x ⎟dx ⎠ 3L M ⎡ 3L x2 ⎤ 4 =− ⎢ x− ⎥ EI ⎣ 4 2 ⎦0 9 ML2 =− (↑ ) (4) 32 EIExample 5.2Find the horizontal displacement at joint B of the frame ABCD as shown in Fig.5.4a by unit load method. Assume EI to be constant for all members. Version 2 CE IIT, Kharagpur
  11. 11. The reactions and bending moment diagram of the frame due to applied externalloading are shown in Fig 5.4b and Fig 5.4c respectively. Since, it is required tocalculate horizontal deflection at B, apply a unit virtual load at B as shown in Fig.5.4d. The resulting reactions and bending moment diagrams of the frame areshown in Fig 5.4d. Version 2 CE IIT, Kharagpur
  12. 12. Now horizontal deflection at B, u B may be calculated as D δM v (x )M (x )dx (1) × u = ∫ B H (1) A EI B δM v (x )M (x )dx C δM v (x )M (x )dx D δM v ( x )M ( x )dx =∫ +∫ +∫ A EI B EI C EI =∫ 5 (x )(5 x )dx + 2.5 2(2.5 − x )10(2.5 − x )dx + 0 0 EI ∫ 0 EI =∫ 5 (5x )dx + 2 20(2.5 − x ) dx 2.5 2 0 EI ∫ 0 EI 625 312.5 937.5 = + = 3EI 3EI 3EI 937.5Hence, uA = ( →) (2) 3EIExample 5.3Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI tobe constant for all members. Version 2 CE IIT, Kharagpur
  13. 13. Rotation at BApply unit virtual moment at B as shown in Fig 5.5a. The resulting bendingmoment diagram is also shown in the same diagram. For the unit load method, therelevant equation is, D δM v ( x )M ( x )dx (1) × θ B = ∫ (1) A EIwherein, θ B is the actual rotation at B, δ M v ( x) is the virtual stress resultant in the D M ( x)frame due to the virtual load and ∫ dx is the actual deformation of the frame A EIdue to real forces. Version 2 CE IIT, Kharagpur
  14. 14. Now, M (x ) = 10(2.5 − x ) and δM v ( x ) = 0.4(2.5 − x )Substituting the values of M ( x) and δ M v ( x) in the equation (1), 2.5 4 θB = ∫ (2.5 − x ) dx 2 EI 0 2.5 4 ⎡ 5 x 2 x3 ⎤ 62.5 = ⎢6.25 x − + ⎥ = (2) EI ⎣ 2 3 ⎦0 3EIRotation at CFor evaluating rotation at C by unit load method, apply unit virtual moment at C asshown in Fig 5.5b. Hence, D δM v ( x )M (x )dx (1) × θC = ∫ (3) A EI 10(2.5 − x )(0.4 x ) 2.5 θC = ∫ 0 EI dx 2.5 4 ⎡ 2.5 x 2 x3 ⎤ 31.25 = ⎢ − ⎥ = (4) EI ⎣ 2 3 ⎦0 3EI5.6 Unit Displacement MethodConsider a cantilever beam, which is in equilibrium under the action of a system offorces F1 , F2 ,....., Fn . Let u1 , u2 ,....., un be the corresponding displacements and P, M and V be the stress resultants at section of the beam. Consider a secondsystem of forces (virtual) δF1 , δF2 ,....., δFn causing virtualdisplacements δu1 , δu2 ,....., δun . Let δPv , δM v and δVv be the virtual axial force,bending moment and shear force respectively at any section of the beam.Apply the first system of forces F1 , F2 ,....., Fn on the beam, which has beenpreviously bent by virtual forces δF1 , δF2 ,....., δFn . From the principle of virtualdisplacements we have, n M ( x )δM v ( x )ds ∑ F δu = ∫ j =1 j j EI = ∫ σ T δε δv (5.11) V Version 2 CE IIT, Kharagpur
  15. 15. The left hand side of equation (5.11) refers to the external virtual work done by thesystem of true/real forces moving through the corresponding virtual displacementsof the system. The right hand side of equation (5.8) refers to internal virtual workdone. The principle of virtual displacement states that the external virtual work ofthe real forces multiplied by virtual displacement is equal to the real stressesmultiplied by virtual strains integrated over volume. If the value of a particular forceelement is required then choose corresponding virtual displacement as unity. Letus say, it is required to evaluate F1 , then choose δu1 = 1 and δui = 0 i = 2,3,....., n .From equation (5.11), one could write, (1) F1 = ∫ M (δM v )1 ds (5.12) EIwhere, (δM v )1 is the internal virtual stress resultant for δu1 = 1 . Transposing theabove equation, we get (δM v )1 Mds F1 = ∫ (5.13) EIThe above equation is the statement of unit displacement method. The aboveequation is more commonly used in the evaluation of stiffness co-efficient kij .Apply real displacements u1 ,....., un in the structure. In that set u2 = 1 and the otherall displacements ui = 0 (i = 1,3,......, n) . For such a case the quantity F j inequation (5.11) becomes kij i.e. force at 1 due to displacement at 2. Apply virtualdisplacement δu1 = 1 . Now according to unit displacement method, (1) k12 = ∫ (δM v )1 M 2ds (5.14) EISummaryIn this chapter the concept of virtual work is introduced and the principle of virtualwork is discussed. The terms internal virtual work and external virtual work hasbeen explained and relevant expressions are also derived. Principle of virtualforces has been stated. It has been shown how the principle of virtual load leads tounit load method. An expression for calculating deflections at any point of astructure (both statically determinate and indeterminate structure) is derived. Fewproblems have been solved to show the application of unit load method forcalculating deflections in a structure. Version 2 CE IIT, Kharagpur