Page 2 Now horizontal deflection at B, may be calculated as B u ( ) ( ) ? = × D A v B H EI dx x M x M u d ) 1 ( (1) () () ( ) ( ) ( ) ( ) ? ? ? + + = D C v C B v B A v EI dx x M x M EI dx x M x M EI dx x M x M d d d ()( ) () ( ) 0 5 . 2 10 5 . 2 2 5 5 . 2 0 5 0 + - - + = ? ? EI dx x x EI dx x x () () ? ? - + = 5 . 2 0 2 5 0 2 5 . 2 20 5 EI dx x EI dx x EI EI EI 3 5 . 937 3 5 . 312 3 625 = + = Hence, EI u A 3 5 . 937 = ( ?) (2) Example 5.3 Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to be constant for all members. Page 3 Now horizontal deflection at B, may be calculated as B u ( ) ( ) ? = × D A v B H EI dx x M x M u d ) 1 ( (1) () () ( ) ( ) ( ) ( ) ? ? ? + + = D C v C B v B A v EI dx x M x M EI dx x M x M EI dx x M x M d d d ()( ) () ( ) 0 5 . 2 10 5 . 2 2 5 5 . 2 0 5 0 + - - + = ? ? EI dx x x EI dx x x () () ? ? - + = 5 . 2 0 2 5 0 2 5 . 2 20 5 EI dx x EI dx x EI EI EI 3 5 . 937 3 5 . 312 3 625 = + = Hence, EI u A 3 5 . 937 = ( ?) (2) Example 5.3 Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to be constant for all members. Rotation at B Apply unit virtual 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 relevant equation is, ( ) ( ) ? = × D A v B EI dx x M x M d ? ) 1 ( (1) wherein, B ? is the actual rotation at B, () v M x d is the virtual stress resultant in the frame due to the virtual load and () D A M x dx EI ? is the actual deformation of the frame due to real forces. Page 4 Now horizontal deflection at B, may be calculated as B u ( ) ( ) ? = × D A v B H EI dx x M x M u d ) 1 ( (1) () () ( ) ( ) ( ) ( ) ? ? ? + + = D C v C B v B A v EI dx x M x M EI dx x M x M EI dx x M x M d d d ()( ) () ( ) 0 5 . 2 10 5 . 2 2 5 5 . 2 0 5 0 + - - + = ? ? EI dx x x EI dx x x () () ? ? - + = 5 . 2 0 2 5 0 2 5 . 2 20 5 EI dx x EI dx x EI EI EI 3 5 . 937 3 5 . 312 3 625 = + = Hence, EI u A 3 5 . 937 = ( ?) (2) Example 5.3 Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to be constant for all members. Rotation at B Apply unit virtual 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 relevant equation is, ( ) ( ) ? = × D A v B EI dx x M x M d ? ) 1 ( (1) wherein, B ? is the actual rotation at B, () v M x d is the virtual stress resultant in the frame due to the virtual load and () D A M x dx EI ? is the actual deformation of the frame due to real forces. Now, and () ( ) x x M - = 5 . 2 10 ( ) ( ) x x M v - = 5 . 2 4 . 0 d Substituting the values of () M x and () v M x d in the equation (1), () ? - = 5 . 2 0 2 5 . 2 4 dx x EI B ? EI x x x EI 3 5 . 62 3 2 5 25 . 6 4 5 . 2 0 3 2 = ? ? ? ? ? ? + - = (2) Rotation at C For evaluating rotation at C by unit load method, apply unit virtual moment at C as shown in Fig 5.5b. Hence, ( ) ( ) ? = × D A v C EI dx x M x M d ? ) 1 ( (3) () ( ) ? - = 5 . 2 0 4 . 0 5 . 2 10 dx EI x x C ? EI x x EI 3 25 . 31 3 2 5 . 2 4 5 . 2 0 3 2 = ? ? ? ? ? ? - = (4) 5.6 Unit Displacement Method Consider a cantilever beam, which is in equilibrium under the action of a system of forces . Let be the corresponding displacements and and be the stress resultants at section of the beam. Consider a second system of forces (virtual) n F F F ,....., , 2 1 n u u u ,....., , 2 1 M P, V n F F F d d d ,....., , 2 1 causing virtual displacements n u u u d d d ,....., , 2 1 . Let v v M P d d , and v V d be the virtual axial force, bending moment and shear force respectively at any section of the beam. Apply the first system of forces on the beam, which has been previously bent by virtual forces n F F F ,....., , 2 1 n F F F d d d ,....., , 2 1 . From the principle of virtual displacements we have, ( ) ( ) ? ? = = n j v j j EI ds x M x M u F 1 d d (5.11) ? = V T v d de s Page 5 Now horizontal deflection at B, may be calculated as B u ( ) ( ) ? = × D A v B H EI dx x M x M u d ) 1 ( (1) () () ( ) ( ) ( ) ( ) ? ? ? + + = D C v C B v B A v EI dx x M x M EI dx x M x M EI dx x M x M d d d ()( ) () ( ) 0 5 . 2 10 5 . 2 2 5 5 . 2 0 5 0 + - - + = ? ? EI dx x x EI dx x x () () ? ? - + = 5 . 2 0 2 5 0 2 5 . 2 20 5 EI dx x EI dx x EI EI EI 3 5 . 937 3 5 . 312 3 625 = + = Hence, EI u A 3 5 . 937 = ( ?) (2) Example 5.3 Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to be constant for all members. Rotation at B Apply unit virtual 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 relevant equation is, ( ) ( ) ? = × D A v B EI dx x M x M d ? ) 1 ( (1) wherein, B ? is the actual rotation at B, () v M x d is the virtual stress resultant in the frame due to the virtual load and () D A M x dx EI ? is the actual deformation of the frame due to real forces. Now, and () ( ) x x M - = 5 . 2 10 ( ) ( ) x x M v - = 5 . 2 4 . 0 d Substituting the values of () M x and () v M x d in the equation (1), () ? - = 5 . 2 0 2 5 . 2 4 dx x EI B ? EI x x x EI 3 5 . 62 3 2 5 25 . 6 4 5 . 2 0 3 2 = ? ? ? ? ? ? + - = (2) Rotation at C For evaluating rotation at C by unit load method, apply unit virtual moment at C as shown in Fig 5.5b. Hence, ( ) ( ) ? = × D A v C EI dx x M x M d ? ) 1 ( (3) () ( ) ? - = 5 . 2 0 4 . 0 5 . 2 10 dx EI x x C ? EI x x EI 3 25 . 31 3 2 5 . 2 4 5 . 2 0 3 2 = ? ? ? ? ? ? - = (4) 5.6 Unit Displacement Method Consider a cantilever beam, which is in equilibrium under the action of a system of forces . Let be the corresponding displacements and and be the stress resultants at section of the beam. Consider a second system of forces (virtual) n F F F ,....., , 2 1 n u u u ,....., , 2 1 M P, V n F F F d d d ,....., , 2 1 causing virtual displacements n u u u d d d ,....., , 2 1 . Let v v M P d d , and v V d be the virtual axial force, bending moment and shear force respectively at any section of the beam. Apply the first system of forces on the beam, which has been previously bent by virtual forces n F F F ,....., , 2 1 n F F F d d d ,....., , 2 1 . From the principle of virtual displacements we have, ( ) ( ) ? ? = = n j v j j EI ds x M x M u F 1 d d (5.11) ? = V T v d de s The left hand side of equation (5.11) refers to the external virtual work done by the system of true/real forces moving through the corresponding virtual displacements of the system. The right hand side of equation (5.8) refers to internal virtual work done. The principle of virtual displacement states that the external virtual work of the real forces multiplied by virtual displacement is equal to the real stresses multiplied by virtual strains integrated over volume. If the value of a particular force element is required then choose corresponding virtual displacement as unity. Let us say, it is required to evaluate , then choose 1 F 1 1 = u d and n i u i ,....., 3 , 2 0 = = d . From equation (5.11), one could write, () ? = EI ds M M F v 1 1 ) ( 1 d (5.12) where, ( 1 v M) d is the internal virtual stress resultant for 1 1 = u d . Transposing the above equation, we get ? = EI Mds M F v 1 1 ) ( d (5.13) The above equation is the statement of unit displacement method. The above equation is more commonly used in the evaluation of stiffness co-efficient . ij k Apply real displacements in the structure. In that set and the other all displacements . For such a case the quantity in equation (5.11) becomes i.e. force at 1 due to displacement at 2. Apply virtual displacement n u u ,....., 1 1 2 = u ) ,......, 3 , 1 ( 0 n i u i = = j F ij k 1 1 = u d . Now according to unit displacement method, () ? = EI ds M M k v 2 1 12 ) ( 1 d (5.14) Summary In this chapter the concept of virtual work is introduced and the principle of virtual work is discussed. The terms internal virtual work and external virtual work has been explained and relevant expressions are also derived. Principle of virtual forces has been stated. It has been shown how the principle of virtual load leads to unit load method. An expression for calculating deflections at any point of a structure (both statically determinate and indeterminate structure) is derived. Few problems have been solved to show the application of unit load method for calculating deflections in a structure.Read More

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