Chapter 2: Deflections of Structures

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Chapter 2: Deflections of Structures Fig. 4.1.

1 Chapter 2: Deflections of Structures Fig (Fig. 2.1.) ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 1

2 (2.1) (4.1) (2.2) Fig.4.2 Fig.2.2 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 2

3 4.2 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 3

4 2.3) 4.3 (2.3) (2.4) (2.5) (2.6) ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 4

5 2.4) 2.4 (2.7) (2.8) 2.5) 4.5 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 5

6 (2.9) (2.10) (2.11) 4.6) 2.6 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 6

7 (2.12) (2.13) (2.14) 2.7) 2.7 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 7

8 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 8

9 Mohr s area-moment method In applications where the slope or deflection of beams or cantilevers is required at only one position, and in particular where loading systems are relatively simple, the Mohr moment-area method provides a rapid solution. Fig ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 9

10 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 10

11 (b) Cantilever with u.d.l.: (c) Simply supported beam with u.d.l. ASTU, Dept. of CEng., Prepared by: Melkamu E Page 11

12 (d) Simply supported beam with central concentrated load Problem 1: ASTU, Dept. of CEng., Prepared by: Melkamu E Page 12

13 P1 P1 ASTU, Dept. of CEng., Prepared by: Melkamu E Page 13

14 Problem 2: P2 P2 P2 ASTU, Dept. of CEng., Prepared by: Melkamu E Page 14

15 Problem 3: P3 Figure P3 Solution: ASTU, Dept. of CEng., Prepared by: Melkamu E Page 15

16 ASTU, Dept. of CEng., Prepared by: Melkamu E Page 16

17 Conjugate Beam Method Conjugate beam method was developed by H. Muller-Breslau in the year This method relies only on the principles of statics, and hence its application will be more familiar. dq/dx = w; di/dx = M/EI On integrating, Q = wdx; d 2 M/dx 2 = w d 2 y/dx 2 = M/EI M = ( wdx) dx i = y = By the above equations, shear Q compares with the slope i, the moment M compares with the displacement y, and the external load w compares with the M/EI. If the beam loaded with w is said to be real beam, and the beam loaded with M/EI is considered to be called as a conjugate beam, then the following two theorems relate the real and conjugate beams. Theorem 1: The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam. Theorem 2: The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam. Conjugate beam supports: When drawing the conjugate beam it is important that the shear and moment developed at the supports of the conjugate beam account for the corresponding slope and displacement of the real beam at its supports, a consequence of theorem 1 and 2. For example, a pin or roller support at the end of the real beam provides zero displacement, but the beam has a non zero slope. Consequently the conjugate beam must be supported by a pin or roller, since this support has zero moment but has a shear or end reaction. The equivalent support conditions of conjugate beam compared to real beam supports is shown in the table below. i has value y = 0 Q has value M = 0 i has value y = 0 Q has value M = 0 i = 0; y = 0 Q = 0; M = 0 Real Beam Conjugate beam ASTU, Dept. of CEng., Prepared by: Melkamu E Page 17

18 i has value y has value Q has value M has value i has value y = 0 Q has value M = 0 i has value y = 0 Q has value M = 0 i has value y has value Q has value M has value Procedure for analysis The following procedure provides a method that may be used to determine the displacement and slope at a point on the elastic curve of a beam using conjugate beam method. Conjugate beam Draw the conjugate beam for the real beam. This beam has the same length as the real beam and has corresponding supports as listed in the above table. In general, if the real support allows a slope, the conjugate support must develop a shear, and if the real support allows a displacement, the conjugate support must develop a moment. The conjugate beam is loaded with the real beam s M/EI diagram. This loading is assumed to be distributed over the conjugate beam and is directed upward when M/EI is positive and downward when M/EI is negative. In other words, the loading always acts away from the beam. Equilibrium ASTU, Dept. of CEng., Prepared by: Melkamu E Page 18

19 Using the equations of equilibrium determine the reactions in the conjugate beam s supports. Section the conjugate beam at the point where the slope i and displacement y of the real beam are to be determined. At the section show the unknown shear Q and moment M acting in their positive sense. Determine the shear and moment using the equations of equilibrium. Q and M equal i and y respectively, for the real beam. In particular, if these values are positive, the slope is counter clockwise and the displacement is upward Example 1 ASTU, Dept. of CEng., Prepared by: Melkamu E Page 19

20 Example 2 Example 2 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 20

21 ENERGY METHOD External Work And Strain Energy Most energy methods are based on the conservation of energy principles, which states that the work done by all the external forces acting on a structure U e, is transformed into internal work or strain energy,u i, which is developed when the structure deforms. Mathematically this can be written as: U e U i. (1) External work- Force. When a force F undergoes a displacement dx in the same direction as the force, the work done is du e F dx.if the total displacement is x, the work becomes x U e F dx 0 (2) If the magnitude of F is gradually increased from zero to some limiting value F=P, the final elongation of the bar becomes. If the material has a linear elastic response, then F (P / )x. Substituting this value of F and integrating from zero to, we get 1 U e P (3) 2 External work- Moment. The work of a moment is defined by the product of the magnitude of the moment M and the angle d through which it rotates, that is, du M d. If the total angle of rotation is radians, the work becomes. (4) U e M d 0 As in the case of force, if the moment is applied gradually to a structure having linear elastic response from zero to M, the work is then 1 U e M. (5) 2 Strain Energy-Axial Force. When an axial force N is applied gradually to the bar, it will strain the material such that the external work done by N will be converted into strain energy, which is stored in the bar. Provided the material is linear elastic, Hooke s law is valid, E, and if the bar has constant cross sectional area A and length L, the normal stress is N / A and the final strain is / L. Consequently, N / A E( / L), and the final deflection is NL (6) AE Substituting into Eq. (3) with P=N, the strain energy in the bar is therefore 2 U N L.. (7) i 2EA Strain Energy- Bending. Consider the beam shown below, which is distorted by the gradually applied loading P and w. These loads create an internal moment M in the beam at a section located a distance x from the left support. The resulting rotation of the differential element dx can be found from the differential equation d (M / EI )dx. Consequently, the strain energy, or work stored in the element, is determined from Eq(5) since the internal moment is gradually developed. Hence, 2 du M dx.. (8) i 2EI ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 21

22 The strain energy for the beam is determined by integrating this result over the beam s entire length L. The result is L M 2 dx.. (9) U i 0 EI P M M w d x dx L dx CASTIGLIANO S THEOREM FOR TRUSES If the displacement of a point is to be determined, the theorem states that it is equal to the first partial derivative of the strain energy in the structure with respect to a force acting at the point and in the direction of displacement. In a similar manner, the slope at a point in a structure is equal to the first partial derivative of the strain energy in the structure with respect to a couple moment acting at the point and in the direction of rotation. Mathematically: U i. (10, a) i P and U i. (10, b) i M 2 The strain energy for a member of a truss is given by Eq (7), U i N L / 2EA. Substituting this equation into Eq (10, a) and omitting the subscript i, we have 2 N L P 2EA In the general case, L, E and A are constant for a given member, and therefore we may write N L N.. (11) P EA Where: = external joint displacement of the truss P = external force applied to the truss joint in the direction of N= internal force in a member caused by both the force P and the loads on the truss L = length of a member A = cross-sectional area of a member E = modulus of elasticity of a member The following procedure provides a method that may be used to determine the displacement of any joint of a truss using Castigliano s theorem. External force P Place a force P on the truss at the join where the desired displacement is to be determined. This force is assumed to have a variable magnitude and should be directed along the line of action of the displacement. ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 22

23 Internal force N Determine the force N in each member caused by both the real loads and the variable force P. Assume tensile forces are positive and compressive forces are negative. Compute the respective partial derivative N / P for each member. After N and N / P have been determined, assign P its numerical value if it has replaced a real force on the truss. Otherwise, set P equal to zero. Castigliano s Theorem Apply Castigliano s theorem to determine the desired displacement. It is important to retain the algebraic signs for corresponding values of N and N / P when substituting these terms into the equation. If the resultant sum N N / P L / AE is positive, is in the same direction as P. If a negative value results, is opposite to P. ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 23

24 CASTIGLIANO S THEOREM FOR BEAMS AND FRAMES The internal bending strain energy for a beam or frame is given by Eq (9) (U i M 2 dx / 2EI ). Substituting this equation into Eq (10, a) and omitting the subscript i, we get L M 2 dx P EI 0 Differentiating prior to integration L will be easier and noting that E and I are constant, we obtain M M dx (12) 0 P EI Where: = external displacement of the point caused by the real loads acting on the beam or frame P = external force applied to the beam or frame in the direction of M = internal moment in the beam or frame, expressed as a function of x and caused by both the force P and the real loads on the beam E = modulus of elasticity of beam material I = moment of inertia of cross-sectional area computed about the neutral axis. If the slope at a point is to be determined, we must find the partial derivative of the internal moment M with respect to L an external couple moment M acting at the point, i.e, M M dx.. (13) M ' EI o ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 24

25 Example E Example ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 25

26 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 26

27 VIRTUAL WORK METHOD The principle of virtual work was developed by John Bernoulli in 1717 and is sometimes referred to as the unit-load method. It provides a general means of obtaining the displacement and slope at a specific point on a structure, be it a beam, frame or truss. TRUSSES This method can be used to determine the displacement of a truss joint when the truss is subjected to an external loading, temperature change, or fabrication errors. External loading: - for the purpose of explanation let us consider the vertical displacement of joint B of the truss shown. P1 P2 B Here a typical element of the truss would be one of its members having a length L. If the applied loadings P1 and P2 cause a linear elastic material response, then this element deforms an amount L NL / AE, where N is the normal or axial force in the member, caused by the loads. Applying the principles of virtual work, the virtual work equation for the truss becomes nnl 1. AE Where: 1= external virtual unit load acting on the truss joint in the direction of n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by the real loads on the truss N = internal normal force in a truss member caused by the real loads on the truss L = length of a member A = cross-sectional area of a member E = modulus of elasticity of a member Here the external virtual unit load creates internal virtual forces n in each of the truss members. The real loads then cause the truss joint to be displaced in the same direction as the virtual unit load, and each member is displaced NL / AE in the same direction as its respective n force. Consequently, the external virtual work 1. equals the internal virtual work or the internal(virtual) strain energy stored in all the truss members, that is nnl / AE. Temperature: In some cases, truss members may change their length due to temperature. If is the coefficient of thermal expansion for a member and T is the change in its temperature, the change in ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 27

28 length of a member is L T L. Hence, we can determine the displacement of a selected truss joint due to this temperature change as follows. 1. = n T L Procedure for analysis The following procedure may be used to determine the displacement of a given joint. Virtual forces n Place the unit load on the truss at the joint where the desired displacement is to be determined. The load should be in the same direction as the specified displacement, e.g. horizontal or vertical. With the unit load so placed, and all the real loads removed from the truss, use the method of joints or the method of sections and calculate the internal n force in each truss member. Assume that tensile forces are positive and compressive forces are negative. Real forces N Use the method of sections or method of joints to determine the N force in each member. These forces are caused only by the real loads acting on the truss. Virtual- work equation Apply the equation of virtual work, to determine the desired displacement. It is important to retain the algebraic sign for each of the corresponding n and N forces when substituting these terms into the equation. If the resultant sum nnl / AE is positive, the displacement is in the same direction as the unit load. If a negative value results, is opposite to the unit load. When applying 1. = n T L, realize that if any of the members undergoes an increase in temperature; T value for T. will be positive, whereas a decrease in temperature results in a negative ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 28

29 Example ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 29

30 Example ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 30

31 BEAMS AND FRAMES The method of virtual work can also be applied to deflection problems involving beams and frames. Strains due to bending are the primary cause of beam or frame deflections. Consider the beam shown below. w A A x x 1 d M x R x V r Here the displacement of point A is to be determined. To compute a virtual unit load acting in the direction of is placed on the beam at A, and the internal virtual moment m is determined by the method of sections at an arbitrary location x from the left support. When the real loads act on the beam, point A is displaced. Provided the material acts in a linear elastic manner due to these loads, the element dx deforms or rotates d ( M EI )dx. Here M is the internal moment at x caused by the real loads. Consequently, the external virtual work done by the unit load is 1. = m(m / EI )dx. Summing the effects of all elements dx along the beam requires an integration, and therefore, L m M 1. dx 0 E I ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 31

32 Where m = internal virtual moment in the beam or frame, expressed as a function of x and caused by the external virtual unit load. = external displacement caused by the real loads In a similar manner, if the tangent rotation or slope angle at a point on the beam s elastic curve is to be determined, a unit couple is applied at the point, and the corresponding internal moments m have to be determined. Since the work of the unit couple is 1., then Example L 0 m M 1. E I dx ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 32

33 Example ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 33

34 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 34

35 Example ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 35

36 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 36

37 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 37

14. *14.8 CASTIGLIANO S THEOREM

14. *14.8 CASTIGLIANO S THEOREM *14.8 CASTIGLIANO S THEOREM Consider a body of arbitrary shape subjected to a series of n forces P 1, P 2, P n. Since external work done by forces is equal to internal strain energy stored in body, by