Module 1. Energy Methods in Structural Analysis. Version 2 CE IIT, Kharagpur

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1 Module 1 Energy Methods in Structural Analysis Version CE IIT, Kharagpur

2 esson 5 Virtual Work Version CE IIT, Kharagpur

3 Instructional Objecties After studying this lesson, the student will be able to: 1. Define Virtual Work.. Differentiate between external and internal irtual work. 3. Sate principle of irtual displacement and principle of irtual forces. 4. Drie an expression of calculating deflections of structure using unit load method. 5. Calculate deflections of a statically determinate structure using unit load method. 6. State unit displacement method. 7. Calculate stiffness coefficients using unit-displacement method. 5.1 Introduction In the preious chapters the concept of strain energy and Castigliano s theorems were discussed. From Castigliano s theorem it follows that for the statically determinate structure; the partial deriatie of strain energy with respect to external force is equal to the displacement in the direction of that load. In this lesson, the principle of irtual work is discussed. As compared to other methods, irtual work methods are the most direct methods for calculating deflections in statically determinate and indeterminate structures. This principle can be applied to both linear and nonlinear structures. The principle of irtual work as applied to deformable structure is an extension of the irtual work for rigid bodies. This may be stated as: if a rigid body is in equilibrium under the action of a F system of forces and if it continues to remain in equilibrium if the body is gien a small (irtual) displacement, then the irtual work done by the F system of forces as it rides along these irtual displacements is zero. 5. Principle of Virtual Work Many problems in structural analysis can be soled by the principle of irtual work. Consider a simply supported beam as shown in Fig.5.1a, which is in equilibrium under the action of real forces F 1, F,..., F n at co-ordinates 1,,..., n respectiely. et u, 1 u,..., u n be the corresponding displacements due to the action of forces F 1, F,..., F n. Also, it produces real internal stresses σ and real internal strains ε inside the beam. Now, let the beam be subjected to second system of forces (which are irtual not real) δ F1, δf,..., δfn in equilibrium as shown in Fig.5.1b. The second system of forces is called irtual as they are imaginary and they are not part of the real loading. This produces a displacement Version CE IIT, Kharagpur

4 configurationδ u, δu,... 1, δun. The irtual loading system produces irtual internal stresses δσ and irtual internal strains δε inside the beam. Now, apply the second system of forces on the beam which has been deformed by first system of forces. Then, the external loads Fi and internal stresses σ do irtual work by moing along δui and δε. The product Fiδu i is known as the external irtual work. It may be noted that the aboe product does not represent the conentional work since each component is caused due to different source i.e. δ ui is not due to Fi. Similarly the product σ δε is the internal irtual work. In the case of deformable body, both external and internal forces do work. Since, the beam is in equilibrium, the external irtual work must be equal to the internal irtual work. Hence, one needs to consider both internal and external irtual work to establish equations of equilibrium. 5.3 Principle of Virtual Displacement A deformable body is in equilibrium if the total external irtual work done by the system of true forces moing through the corresponding irtual displacements of the system i.e. Fiδu i is equal to the total internal irtual work for eery kinematically admissible (consistent with the constraints) irtual displacements. Version CE IIT, Kharagpur

5 That is irtual displacements should be continuous within the structure and also it must satisfy boundary conditions. Fi δui σ δε d (5.1) where σ are the true stresses due to true forces due to irtual displacementsδ. ui Fi and δε are the irtual strains 5.4 Principle of Virtual Forces For a deformable body, the total external complementary work is equal to the total internal complementary work for eery system of irtual forces and stresses that satisfy the equations of equilibrium. δfi ui δσ ε d (5.) where δσ are the irtual stresses due to irtual forces δ Fi and ε are the true strains due to the true displacements u i. As stated earlier, the principle of irtual work may be adantageously used to calculate displacements of structures. In the next section let us see how this can be used to calculate displacements in a beams and frames. In the next lesson, the truss deflections are calculated by the method of irtual work. 5.5 Unit oad Method The principle of irtual force leads to unit load method. It is assumed throughout our discussion that the method of superposition holds good. For the deriation of unit load method, we consider two systems of loads. In this section, the principle of irtual forces and unit load method are discussed in the context of framed structures. Consider a cantileer beam, which is in equilibrium under the action of a first system of forces F 1, F,..., F n causing displacements u, 1 u,..., un as shown in Fig. 5.a. The first system of forces refers to the actual forces acting on the structure. et the stress resultants at any section of the beam due to first system of forces be axial force ( P ), bending moment ( M ) and shearing force (V ). Also the corresponding incremental deformations are axial deformation ( dδ ), flexural deformation ( d θ ) and shearing deformation ( d λ ) respectiely. For a conseratie system the external work done by the applied forces is equal to the internal strain energy stored. Hence, Version CE IIT, Kharagpur

6 n Fu i i P dδ M dθ V dλ + + i 1 P ds EA M ds + + V ds AG (5.3) Now, consider a second system of forces causing irtual displacements δ F1, δf,..., δfn, which are irtual and 1, δu δun respectiely (see Fig. 5.b). et the δ u,..., irtual stress resultants caused by irtual forces be δ P, δm and δv at any cross section of the beam. For this system of forces, we could write n 1 δp ds δm ds δv ds δ Fiδu i + + EA (5.4) AG i 1 where δ P, δm and δv are the irtual axial force, bending moment and shear force respectiely. In the third case, apply the first system of forces on the beam, which has been deformed, by second system of forces δ F1, δf,..., δfn as shown in Fig 5.c. From the principle of superposition, now the deflections will be ( u 1 + δ u1 ), ( u + δu ),...,( u n + δun ) respectiely Version CE IIT, Kharagpur

7 Since the energy is consered we could write, n n n δ δ δ Fu j j + δfjδuj + δfu j j j 1 j 1 j 1 EA AG E 1 1 Pds M ds V ds Pds + A M ds V ds + Pd Md Vd + AG δ Δ+ δ θ + δ λ (5.5) In equation (5.5), the term on the left hand side ( δ F j u j ), represents the work done by irtual forces moing through real displacements. Since irtual forces act Version CE IIT, Kharagpur

8 1 at its full alue, does not appear in the equation. Subtracting equation (5.3) and (5.4) from equation (5.5) we get, n j j j 1 From Module 1, lesson 3, we know that δ F u δp dδ + δm dθ + δv dλ (5.6) Pds Mds Vds d Δ, dθ and d λ. Hence, EA AG n δp Pds δm Mds δvvds δ Fju j + + (5.7) j 1 EA AG 1 Note that does not appear on right side of equation (5.7) as the irtual system resultants act at constant alues during the real displacements. In the present case δ and if we neglect shear forces then we could write equation (5.7) as P n δm Mds δ F u (5.8) j j j 1 If the alue of a particular displacement is required, then choose the corresponding force δ 1 and all other forces j 1,,..., i 1, i + 1,..., n. F i Then the aboe expression may be written as, δ ( ) F j δm Mds (1) ui (5.9) where δ M are the internal irtual moment resultants corresponding to irtual force at i-th co-ordinate, δ 1. The aboe equation may be stated as, F i ( unit irtual load ) unknown true displacement ( )( ) irtual stress resultants real deformations ds. (5.1) The equation (5.9) is known as the unit load method. Here the unit irtual load is applied at a point where the displacement is required to be ealuated. The unit load method is extensiely used in the calculation of deflection of beams, frames and trusses. Theoretically this method can be used to calculate deflections in Version CE IIT, Kharagpur

9 statically determinate and indeterminate structures. Howeer it is extensiely used in ealuation of deflections of statically determinate structures only as the method requires a priori knowledge of internal stress resultants. Example 5.1 A cantileer beam of span is subjected to a tip moment M as shown in Fig 5.3a. 3 Ealuate slope and deflection at a point from left support. Assume of the 4 gien beam to be constant. Slope at C To ealuate slope at C, a irtual unit moment is applied at C as shown in Fig 5.3c. The bending moment diagrams are drawn for tip moment and unit moment applied at C and is shown in fig 5.3b and 5.3c respectiely. et θ c be the rotation at due to moment M applied at tip. According to unit load method, the rotation C at C, θ c is calculated as, M Version CE IIT, Kharagpur

10 where ( x) M ( x) M ( x) δm (1) θ c (1) δ and M x are the irtual moment resultant and real moment resultant at any section expression, we get Vertical deflection at C ( ) x. Substituting the alue of ( x) ( 1) θ c 3 / 4 ( 1) M ( ) + 3 / 4 δ and M x in the aboe M M ( ) 3M θ c () 4 To ealuate ertical deflection at C, a unit irtual ertical force is applied ac C as shown in Fig 5.3d and the bending moment is also shown in the diagram. According to unit load method, (1) u A δm ( x) M ( x) In the present case, ( ) 3 δ M x x 4 M x + M and ( ) (3) Example 5. u A 3 4 M 3 x M M 3 x 3 x x M ( ) (4) 3 Find the horizontal displacement at joint B of the frame ABCD as shown in Fig. 5.4a by unit load method. Assume to be constant for all members. Version CE IIT, Kharagpur

11 The reactions and bending moment diagram of the frame due to applied external loading are shown in Fig 5.4b and Fig 5.4c respectiely. Since, it is required to calculate horizontal deflection at B, apply a unit irtual load at B as shown in Fig. 5.4d. The resulting reactions and bending moment diagrams of the frame are shown in Fig 5.4d. Version CE IIT, Kharagpur

12 Now horizontal deflection at B, B D may be calculated as u B ( x) M ( x) B δm ( 1) uh (1) A C D ( x) M ( x) δm ( x) M ( x) δm ( x) M ( x) + + A δm 5 B.5 ( x)( 5x) (.5 x) 1(.5 x) ( 5x ) (.5 x) C + Hence, u A ( ) () 3 Example 5.3 Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume to be constant for all members. Version CE IIT, Kharagpur

13 Rotation at B Apply unit irtual moment at B as shown in Fig 5.5a. The resulting bending moment diagram is also shown in the same diagram. For the unit load method, the releant equation is, D ( x) M ( x) δm ( 1) θb (1) A wherein, θ B is the actual rotation at B, δ M ( x) is the irtual stress resultant in the D M ( x) frame due to the irtual load and is the actual deformation of the frame A due to real forces. Version CE IIT, Kharagpur

14 ( ) ( ) Now, M x 1. 5 x and δ M ( x).4(. 5 x) Substituting the alues of M ( x ) and δ M ( x) in the equation (1), Rotation at C θ B 4.5 (.5 x) x x x + () 3 3 For ealuating rotation at C by unit load method, apply unit irtual moment at C as shown in Fig 5.5b. Hence, D ( x) M ( x) δm ( 1) θc (3) θ C.5 1 A (.5 x)(.4x) x x 31.5 (4) Unit Displacement Method Consider a cantileer beam, which is in equilibrium under the action of a system of forces F 1, F,..., F n. et u, 1 u,..., un be the corresponding displacements and P, M and V be the stress resultants at section of the beam. Consider a second system of forces (irtual) δ F1, δf,..., δf n causing irtual displacementsδ u1, δu,..., δun. et δ P, δm and δv be the irtual axial force, bending moment and shear force respectiely at any section of the beam. Apply the first system of forces F F,..., on the beam, which has been preiously bent by irtual forces displacements we hae,, 1 F, δf F n δf n δ 1,...,. From the principle of irtual n j 1 F δ u j j V M σ ( x) δm ( x) T δε δ ds (5.11) Version CE IIT, Kharagpur

15 The left hand side of equation (5.11) refers to the external irtual work done by the system of true/real forces moing through the corresponding irtual displacements of the system. The right hand side of equation (5.8) refers to internal irtual work done. The principle of irtual displacement states that the external irtual work of the real forces multiplied by irtual displacement is equal to the real stresses multiplied by irtual strains integrated oer olume. If the alue of a particular force element is required then choose corresponding irtual displacement as unity. et us say, it is required to ealuate F1, then choose δ u 1 1 and δ u i n i,3,...,. From equation (5.11), one could write, () 1 1 M ( δm ) ds F 1 (5.1) where, ( δ M ) is the internal irtual stress resultant for δ u Transposing the aboe equation, we get (δm ) Mds 1 F1 (5.13) The aboe equation is the statement of unit displacement method. The aboe equation is more commonly used in the ealuation of stiffness co-efficient k. Apply real displacements u,...,u in the structure. In that set u 1 n 1and the other all displacements u i ( i 1,3,..., n). For such a case the quantity Fj in equation (5.11) becomes k i.e. force at 1 due to displacement at. Apply irtual displacement δ u 1 1. Now according to unit displacement method, () 1 1 ( δm ) 1M ds k (5.14) Summary In this chapter the concept of irtual work is introduced and the principle of irtual work is discussed. The terms internal irtual work and external irtual work has been explained and releant expressions are also deried. Principle of irtual forces has been stated. It has been shown how the principle of irtual load leads to unit load method. An expression for calculating deflections at any point of a structure (both statically determinate and indeterminate structure) is deried. Few problems hae been soled to show the application of unit load method for calculating deflections in a structure. Version CE IIT, Kharagpur