Page 1 ( ) ( ) ? = L v c EI dx x M x M 0 ) 1 ( d ? (1) where () x M v d and are the virtual moment resultant and real moment resultant at any section () x M x . Substituting the value of ( ) x M v d and in the above expression, we get () x M ( ) ( ) ? ? + = L L L c EI Mdx EI Mdx 4 / 3 4 / 3 0 0 1 ) 1 ( ? EI ML c 4 3 = ? (2) Vertical deflection at C To evaluate vertical deflection at C , a unit virtual vertical force is applied ac as shown in Fig 5.3d and the bending moment is also shown in the diagram. According to unit load method, C ( ) ( ) ? = L v A EI dx x M x M u 0 ) 1 ( d (3) In the present case, () ? ? ? ? ? ? - - = x L x M v 4 3 d and () 0 M x M + = ? ? ? ? ? ? ? - - = 4 3 0 4 3 L A dx EI M x L u dx x L EI M L ? ? ? ? ? ? ? - - = 4 3 0 4 3 4 3 0 2 2 4 3 L x x L EI M ? ? ? ? ? ? - - = EI ML 32 9 2 - = ( ?) (4) Example 5.2 Find 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. Page 2 ( ) ( ) ? = L v c EI dx x M x M 0 ) 1 ( d ? (1) where () x M v d and are the virtual moment resultant and real moment resultant at any section () x M x . Substituting the value of ( ) x M v d and in the above expression, we get () x M ( ) ( ) ? ? + = L L L c EI Mdx EI Mdx 4 / 3 4 / 3 0 0 1 ) 1 ( ? EI ML c 4 3 = ? (2) Vertical deflection at C To evaluate vertical deflection at C , a unit virtual vertical force is applied ac as shown in Fig 5.3d and the bending moment is also shown in the diagram. According to unit load method, C ( ) ( ) ? = L v A EI dx x M x M u 0 ) 1 ( d (3) In the present case, () ? ? ? ? ? ? - - = x L x M v 4 3 d and () 0 M x M + = ? ? ? ? ? ? ? - - = 4 3 0 4 3 L A dx EI M x L u dx x L EI M L ? ? ? ? ? ? ? - - = 4 3 0 4 3 4 3 0 2 2 4 3 L x x L EI M ? ? ? ? ? ? - - = EI ML 32 9 2 - = ( ?) (4) Example 5.2 Find 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. The reactions and bending moment diagram of the frame due to applied external loading are shown in Fig 5.4b and Fig 5.4c respectively. Since, it is required to calculate 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 are shown in Fig 5.4d. Page 3 ( ) ( ) ? = L v c EI dx x M x M 0 ) 1 ( d ? (1) where () x M v d and are the virtual moment resultant and real moment resultant at any section () x M x . Substituting the value of ( ) x M v d and in the above expression, we get () x M ( ) ( ) ? ? + = L L L c EI Mdx EI Mdx 4 / 3 4 / 3 0 0 1 ) 1 ( ? EI ML c 4 3 = ? (2) Vertical deflection at C To evaluate vertical deflection at C , a unit virtual vertical force is applied ac as shown in Fig 5.3d and the bending moment is also shown in the diagram. According to unit load method, C ( ) ( ) ? = L v A EI dx x M x M u 0 ) 1 ( d (3) In the present case, () ? ? ? ? ? ? - - = x L x M v 4 3 d and () 0 M x M + = ? ? ? ? ? ? ? - - = 4 3 0 4 3 L A dx EI M x L u dx x L EI M L ? ? ? ? ? ? ? - - = 4 3 0 4 3 4 3 0 2 2 4 3 L x x L EI M ? ? ? ? ? ? - - = EI ML 32 9 2 - = ( ?) (4) Example 5.2 Find 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. The reactions and bending moment diagram of the frame due to applied external loading are shown in Fig 5.4b and Fig 5.4c respectively. Since, it is required to calculate 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 are shown in Fig 5.4d. 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 4 ( ) ( ) ? = L v c EI dx x M x M 0 ) 1 ( d ? (1) where () x M v d and are the virtual moment resultant and real moment resultant at any section () x M x . Substituting the value of ( ) x M v d and in the above expression, we get () x M ( ) ( ) ? ? + = L L L c EI Mdx EI Mdx 4 / 3 4 / 3 0 0 1 ) 1 ( ? EI ML c 4 3 = ? (2) Vertical deflection at C To evaluate vertical deflection at C , a unit virtual vertical force is applied ac as shown in Fig 5.3d and the bending moment is also shown in the diagram. According to unit load method, C ( ) ( ) ? = L v A EI dx x M x M u 0 ) 1 ( d (3) In the present case, () ? ? ? ? ? ? - - = x L x M v 4 3 d and () 0 M x M + = ? ? ? ? ? ? ? - - = 4 3 0 4 3 L A dx EI M x L u dx x L EI M L ? ? ? ? ? ? ? - - = 4 3 0 4 3 4 3 0 2 2 4 3 L x x L EI M ? ? ? ? ? ? - - = EI ML 32 9 2 - = ( ?) (4) Example 5.2 Find 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. The reactions and bending moment diagram of the frame due to applied external loading are shown in Fig 5.4b and Fig 5.4c respectively. Since, it is required to calculate 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 are shown in Fig 5.4d. 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 5 ( ) ( ) ? = L v c EI dx x M x M 0 ) 1 ( d ? (1) where () x M v d and are the virtual moment resultant and real moment resultant at any section () x M x . Substituting the value of ( ) x M v d and in the above expression, we get () x M ( ) ( ) ? ? + = L L L c EI Mdx EI Mdx 4 / 3 4 / 3 0 0 1 ) 1 ( ? EI ML c 4 3 = ? (2) Vertical deflection at C To evaluate vertical deflection at C , a unit virtual vertical force is applied ac as shown in Fig 5.3d and the bending moment is also shown in the diagram. According to unit load method, C ( ) ( ) ? = L v A EI dx x M x M u 0 ) 1 ( d (3) In the present case, () ? ? ? ? ? ? - - = x L x M v 4 3 d and () 0 M x M + = ? ? ? ? ? ? ? - - = 4 3 0 4 3 L A dx EI M x L u dx x L EI M L ? ? ? ? ? ? ? - - = 4 3 0 4 3 4 3 0 2 2 4 3 L x x L EI M ? ? ? ? ? ? - - = EI ML 32 9 2 - = ( ?) (4) Example 5.2 Find 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. The reactions and bending moment diagram of the frame due to applied external loading are shown in Fig 5.4b and Fig 5.4c respectively. Since, it is required to calculate 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 are shown in Fig 5.4d. 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 sRead More

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!