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How to Find Vertical Deflection in Beam Structures

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How to Find Vertical Deflection in Beam Structures How to find vertical deflection in beam structures Only 2D beam structures will be regarded blow. Only vertical deflection will be discussed. Vertical deflection is the vertical distance between a point from the undeformed axis of a structure and the same point which lies on the deformed axis. (see the picture below). The formula and the procedure which are described below allow obtaining the vertical deflection in a given point from a structure. δV The deformed axis of Vertical deflection the structure Finding vertical deflection is a task which contains the following steps: - drawing the Frere Body Diagram (FBD) - determining the support reactions by using the equilibriums - applying the method of the section to determine the internal forces by using the equilibriums (only Mz will be determined) - determining the derivatives of the Mz - applying the formulas for finding deflections The following formula can be used for finding the vertical deflection in 2D loading in bending: MM V zz..dx i L EIF. zi In the above formula: V i - vertical deflection in the i-th point, Mz – internal moment, E – modulus of elasticity (Young’s modulus), Iz – moment of inertia of the cross-section, Mz - derivative of Mz with respect to the concentrated force in the i-th point. Fi The integration is along the length of the beam. In order to find the vertical deflection in a given point there must be a concentrated force in this point. IT IS IMPORTANT TO MAKE A LETTER SOLUTION (not to replace the parameters with their values) UNTIL SUBSTITUTING IN THE FORMULA!!! To demonstrate the above described procedure two examples will be regarded: Example 1 V Find the vertical deflection in point B (B ? ) if F = 18 kN, M = 6 kN.m, L = 3 m. The cross- section is a standard square hollow section 160x160x5 according to EN 10210-2:2006. The material is ductile standard steel S235JR with modulus of elasticity E = 2.1011 Pa. F M C A B L 2L 160x160x5 EN 10210-2:2006 1. FBD F M C x A C Ay B Cy L 2L 2. Determining the support reactions by using the equilibriums FCxx0 0 kN ; FACFACFy0 y y 0 y y FLMFM. MMFLCLCCz 0 . y .3 0 y y = A 3.LLL 3. 3 3. From the second equilibrium FMFM2 AFCAFAy y y y 3 3.LL 3 3. 3. Applying the method of the section to determine the internal forces by using the equilibriums Two sections have to be made in the regarded example F section 1 section 2 M C x A C Ay B Cy L 2L The left part will be taken for section 1. The internal forces are drawn in the point of the section - S1. section 1 Mz1 Nx1 A S1 Ay V x y1 The boundaries of the coordinate x are: 0 xL. Since only Mz will be used in the formula this will be the only internal force to be determined. 22FMFM M0 MAx. 0 MAxM. xMxx zS1 z1 y z 1 y z 1 z 1 3 3.LL 3 3. The right part will be taken for section 2. The internal forces are drawn in the point of the section - S2. section 2 Mz2 Nx2 Cx S2 C V Cy y2 x The boundaries of the coordinate x are: 02xL. Again only MZ will be determined. FMFM M0 MCx. 0 MCxM. xMxx . zS2 z2 y z 2 y z 2 z 2 3 3.LL 3 3. 4. Determining the derivatives of Mz V The formula for B requires the derivative of Mz1 and Mz2 with respect to the force concentrated in point B to be determined. The equation of Mz1 is 2FM M x x z1 3 3.L The derivative of Mz1 with respect to the concentrated in point B force is: MMzz112 x FFB 3 Some comments on how to find derivatives of the internal moment: In this type of problems the derivative of Mz can be found by taking into account with what is the force F 2 multiplied in the expression for Mz , i.e. in the above expression for Mz1 F is multiplied with x 3 22FMM Mz1 2 ( Mxz1 x x .Fx ) and that is why x . 333.L 3.L F 3 The equation of Mz2 is FM M x x z2 3 3.L Following the above explanations the derivative of Mz2 can be found by accounting that F is 1 multiplied with x in the expression for Mz2 which means that 3 MM1 zz22x FFB 3 5. Applying the formula for finding the deflection V Since the structure consists of 2 sections in the formula for B 2 integrals will be used – one for the first section and one for the second one: LLMMMM2 V z1.. z 1dx z 2 z 2 dx B 00EIFEIF..z B z B It has to be minded that the boundaries of each integral coincide with the boundaries of each section. Substituting the expressions for Mz1 and Mz2 and their derivatives with respect to F will give 2FMFM LLx x 2 x x 3 3.LL21 3 3. V ..x dx x dx B 00EIEI.zz 3 . 3 Opening the parenthesis will lead to 42FMFM2 2 2 2 LLx x 2 x x 9 9.LL 9 9. V dx dx B 00EIEI..zz This expression can be divided into 4 integrals 42FMFM2 2 2 2 LLLLx x 22 x x V 9 9.LL 9 9. dx dx dx dx B 0EIEIEIEI....z 0 z 0 z 0 z If all the constants are moved in front of the integrals it will follow 42FMFMLLLL22 V x2 dx x 2 dx x 2 dx x 2 dx B 9..EILEIEILEIz0 9... z 0 9.. z 0 9... z 0 These integrals can be solved by using the following formula L n1L n 1 n 1 n 1 n x L0 L x dx 0 0 n1 n 1 n 1 n 1 that is 3 3 3 3 4FLMLFMLMLM3 2 32LLLL 2 1 3 2 3 2 2 V . . . . 4FF . . . . B 9..39...39..EILEIEILEIEILL 3 9... 3 9.. 3 3 3 3 z z z z z Now the parameters can be replaced with their values from the condition. It has to be minded that the units must be in [N], [m], [Pa]. The moment of inertia of the cross-section Iz can be taken from the tables. For the standard 4 -8 4 shape 160x160x5 according to EN 10210-2:2006 Iz = 1225 cm = 1225.10 m . The loads are F = 18 kN = 18.103 N, M = 6 kN.m = 6.103 N.m 33 1 33 2.6.10 3 3 32.3 6.10 3 2.3 V 4.18.1033 . . 18.10 . . B 11 8 9.2.10 .1225.10 3 3 3 3 3 3 V 113 3 3 3 3 B 648.10 36.10 1296.10 144.10 1836.10 22050.1033 22050.10 V B 0,0833 m The above example shows how to find the vertical deflection V in a point in which there is a concentrated force. The next example is related to cases in which the vertical deflection has to be determined in a point in which there is no concentrated force. Example 2 V Find the vertical deflection in point A ( А ? ) if w = 20 kN/m, L = 2 m. The cross-section is a standard IPE shape – IPE 180. The material is ductile standard steel S235JR with modulus of elasticity E = 2.1011 Pa. w A B L UPE 180 1. FBD In this example there is no concentrated force in point A which makes it impossible to find the derivative of Mz. In such cases a fictitious concentrated force Ff which value is equal to zero is added in the point where needs to be found. The solution continues according to the steps described in the previous example. L/2 Ff = 0 Q=w.L MB A B Bx L By 2. Determining the support reactions by using the equilibriums FBxx0 0 ; Fy0 B y Q F f 0 B y w. L F f . LL M0 M F. L Q . 0 M F. L w .. L zC B f22 C f 3. Applying the method of the section to determine the internal forces by using the equilibriums V One section is needed to determine the internal forces. L/2 Ff = 0 Q=w.L MB A B Bx section 1 L By The left part will be taken for the section. The internal forces are drawn in the point of the section - S1. section 1 x/2 Ff Q’=w.x Mz1 Nx1 A S1 V x 0 y1 xL The boundaries of the coordinate x are: . Since only Mz will be used in the formula this will be the only internal force to be determined. x x x2 M0 MFxQ. '. 0 MFxwx . .. MFxw . zS1 z1 f2 z 1 f 2 z 1 f 2 4. Determining the derivatives of Mz Now the derivative of Mz1 with respect to the force Ff which is concentrated in point A (the V point where A is searched) can be determined. (To find this it has to be accounted with what Ff is multiplied in the expression of Mz1) MMzz11 x FFAf 5. Applying the formula for finding the deflection V The formula for A consists of one integral which boundaries are the same as the boundaries of the section: L MM V zz11. dx A 0 EIF.
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