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Chapter 3 Deflections

1. Outline. Deflection diagrams

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Chapter 3 Deflections

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    1. Chapter 3 Deflections

    2. 1 Outline Deflection diagrams & the elastic curve Elastic-beam theory The double integration method Moment-area theorems Conjugate-beam method

    3. 2 3-1 Deflection diagrams & the elastic curve Deflections of structures can come from loads, temperature, fabrication errors or settlement. In designs, deflections must be limited in order to prevent cracking of attached brittle materials. A structure must not vibrate or deflect severely for the comfort of occupants. Deflections at specified points must be determined if one is to analyze statically indeterminate structures.

    4. 3 3-1 Deflection diagrams & the elastic curve In this topic, only linear elastic material response is considered. This means a structure subjected to load will return to its original undeformed position after the load is removed. It is useful to sketch the shape of the structure when it is loaded in order to visualize the computed results & to partially check the results.

    5. 4 3-1 Deflection diagrams & the elastic curve This deflection diagram rep the elastic curve for the points at the centroids of the cross-sectional areas along each of the members. Table 8.1 shows the deflection at some typical joints. If the elastic curve seems difficult to establish, it is suggested that the moment diagram be drawn first. From there, the curve can be constructed.

    6. 5 3-1 Deflection diagrams & the elastic curve

    7. 6 3-1 Deflection diagrams & the elastic curve Consider the beam in Fig 8.3 Due to pin-and-roller support, the displacement at A & D must be zero. Within the region of ve moment, the elastic curve is concave downward. Within the region of +ve moment, the elastic curve is concave upward. There must be an inflection point where the curve changes from concave down to concave up.

    8. 7 3-1 Deflection diagrams & the elastic curve

    9. 8 Example 3.1 Draw the deflected shape of each of the beams shown in Fig. 8.5.

    10. 9 Example 3.1 - solution In Fig 8.5(a), the roller at A allows free rotation with no deflection while the fixed wall at B prevents both rotation and deflection. The deflected shape is shown by the bold line. In Fig. 8.5(b), no rotation or deflection occur at A & B. In Fig. 8.5(c), the couple moment will rotate end A. This will cause deflections at both ends of the beam since no deflection is possible at B & C.

    11. 10 Example 3.1 - solution Notice that segment CD remains undeformed since no internal load acts within. In Fig. 8.5(d), the pin at B allows rotation, so the slope of the deflection curve will suddenly .change at this point while the beam is constrained by its support. In Fig. 8.5(e), the compound beam deflects as shown. The slope changes abruptly on each side of B.

    12. 11 Example 3.1 - solution In Fig. 8.5(f), span BC will deflect concave upwards due to load. Since the beam is continuous, the end spans will deflect concave downwards.

    13. 12 3-2 Elastic Beam theory To derive the DE, we look at an initially straight beam that is elastically deformed by loads applied perpendicular to beams x-axis & lying in x-v plane of symmetry as shown in Fig. 8.7(a). Due to loading, the beam deforms under shear & bending. If beam L >> d, greatest deformation will be caused by bending. When M deforms the element of beam, the angle between the cross sections becomes d? as shown in Fig. 8.7(b).

    14. 13 3-2 Elastic Beam theory

    15. 14 3-2 Elastic Beam theory The arc dx that rep a portion of the elastic curve intersects the neutral axis. The radius of curvature for this arc is defined as the distance, ?, which is measured from centroid of curvature O to dx. Any arc on the element other than dx is subjected to normal strain. The strain in arc ds located at position y from the neutral axis is.

    16. 15 3-2 Elastic Beam theory If the material is homogeneous & behaves in a linear manner, then Hookes law applies The flexure formula also applies

    17. 16 3-2 Elastic Beam theory Combining these equations, we have:

    18. 17 3-2 Elastic Beam theory

    19. 18 3-2 Elastic Beam theory This equations rep a non-linear second DE. V=f(x) gives the exact shape of the elastic curve. The slope of the elastic curve for most structures is very small. Using small deflection theory, we assume dv/dx ~ 0. Equation 3 reduces to

    20. 19 3-2 Elastic Beam theory By assuming dv/dv ~ 0 ? ds in Fig .7(b) is approximately equal to dx This implies that points on the elastic curve will only be displaced vertically and not horizontally.

    21. 20 3-3 The double integration method M = f(x), successive integration of equation 8.4 will yield the beams slope. ? ? tan ? = dv/dx = ? M/EI dx Equation of elastic curve V = f(x) = ? M/EI dx Consider the beam shown in Fig 8.8 The internal moment in regions AB, BC and CD must be written in terms of x1, x2 and x3.

    22. 21 3-3 The double integration method

    23. 22 3-3 The double integration method Once these functions are integrated through the applications of equation 8.4 & the constants determined, the functions will give the slope & deflection for each region of the beam. When applying equation 8.4, it is important to use the proper sign for M as established by the sign convention used in derivation. +ve v is upward, hence, the +ve slope angle, ? will be measured counterclockwise from the x-axis. Reason for this is shown in Fig. 8.9(b).

    24. 23 3-3 The double integration method

    25. 24 3-3 The double integration method The constants of integration are determined by evaluating the functions for slope or displacement at a particular point on the beam where the value of the function is known. These values are called boundary conditions. Consider the beam shown in Fig 8.10 Here x1 and x2 coordinates are valid within the regions AB & BC.

    26. 25 3-3 The double integration method

    27. 26 3-3 The double integration method Once the functions for the slope and deflections are obtained, they must give the same values for slope & deflection at point B. This is so as for the beam to be physically continuous.

    28. 27 Example 3.3 The cantilevered beam shown in Fig. 8.11(a) is subjected to a couple moment Mo at its end. Determine the equation of the elastic curve EI is constant.

    29. 28 Example 3.3 - solution The load tends to deflect the beam as shown in Fig. 8.9(a). By inspection, the internal moment can be represented throughout the beam using a single x coordinate. From the free-body diagram, with M acting in +ve direction as shown in Fig. 8.11(b), we have:

    30. 29 Example 3.3 - solution Applying equation 8.4 & integrating twice yields: Using boundary conditions, dv/dx = 0 at x = 0 & v = 0 at x = 0 then C1 = C2 =0

    31. 30 Example 3.3 - solution Substituting these values into earlier equations, we get: Max slope & displacement occur at A (x = L) for which

    32. 31 Example 3.3 - solution The +ve result for ?A indicates counterclockwise rotation & the +ve result for vA indicates that it is upwards. This agrees with results sketched in Fig. 8.11(a). In order to obtain some idea to the actual magnitude of the slope Consider the beam in Fig. 8.11(a) to: Have a length of 3.6m Support a couple moment of 20kNm Be made of steel having Est = 200GPa

    33. 32 Example 3.3 - solution If this beam were designed w/o as by assuming the allowable normal stress = yield stress = 250kNm/2 A W6 x 9 would be found to be adequate

    34. 33 Example 3.3 - solution Since ?2A = 0.00280(10-6)<<1, this justifies the use of equation 8.4. Also this numerical application is for cantilevered beam, we have obtained larger values for max ? and v than would have been obtained if the beam were supported using pins, rollers or other supports.

    35. 34 Example 3.3 - solution

    36. 35 3-4 Moment-Area theorems To develop the theorems, reference is made to Fig. 8.13(a). If we draw the moment diagram for the beam & then divide it by the flexural rigidity, EI, the M/EI diagram shown in Fig. 8.13(b) results. By equation 8.2,

    37. 36 3-4 Moment-Area theorems d? on either side of the element dx = the lighter shade area under the M/EI diagram. Integrating from point A on the elastic curve to point B, Fig 8.13(c), we have This equation forms the basis for the first moment-area theorem.

    38. 37 3-4 Moment-Area theorems Theorem 1 The change in slope between any 2 points on the elastic curve equals the area of the M/EI diagram between the 2 points. The second moment-area theorem is based on the relative derivation of tangents to the elastic curve. Shown in Fig 8.12(c) is a greatly exaggerated view of the vertical deviation dt of the tangents on each side of the differential element, dx.

    39. 38 3-4 Moment-Area theorems Fig 8.12(c)

    40. 39 3-4 Moment-Area theorems Since slope of elastic curve & its deflection are assumed to be very small, it is satisfactory to approximate the length of each tangent line by x & the arc ds by dt. Using ?s = r ? dt = xd? Using equation 8.2, d? = (M/EI) dx The vertical deviation of the tangent at A with the tangent at B can be found by integration.

    41. 40 3-4 Moment-Area theorems Centroid of an area

    42. 41 3-4 Moment-Area theorems Theorem 1 The vertical deviation of the tangent at a point (A) on the elastic curve with the tangent extended from another point (B) equals the moment of the area under the M/EI diagram between the 2 points (A & B). This moment is computed about point A where the derivation is to be determined.

    43. 42 3-4 Moment-Area theorems Provided the moment of a +ve M/EI area from A to B is computed as in Fig. 8.14(b), it indicates that the tangent at point A is above the tangent to the curve extended from point B as shown in Fig. 8.14(c). -ve areas indicate that the tangent at A is below the tangent extended from B.

    44. 43 3-4 Moment-Area theorems Fig 8.14

    45. 44 3-4 Moment-Area theorems It is important to realize that the moment-area theorems can only be used to determine the angles and deviations between 2 tangents on the beams elastic curve. In general, they do not give a direct solution for the slope or displacement at a point.

    46. 45 Example 3.5 Determine the slope at points B & C of the beam shown in Fig. 8.15(a). Take E = 200GPa I = 360(106)mm4

    47. 46 Example 3.5 - solution This diagram is shown in Fig. 8.15(b). It is easier to solve the problem in terms of EI & substitute the numerical data as a last step. The 10kN load causes the beam to deflect as shown in Fig. 8.15(c). Here the tangent at A is always horizontal. The tangents at B & C are also indicated. By construction, the angle between tan A and tan B (?B/A) is equivalent to ?B.

    48. 47 Example 3.5 - solution Fig 8.15

    49. 48 Example 3.5 - solution Applying Theorem 1, is equal to the area under the M/EI diagram between points A and B.

    50. 49 Example 3.5 - solution Substituting numerical data for E & I The ve sign indicates that the angle is measured clockwise from A as shown in Fig. 8.15(c). In a similar manner, the area under the M/EI diagram between points A and C equals (?C/A).

    51. 50 Example 3.5 - solution

    52. 51 3-5 Conjugate-Beam method The basis for the method comes from the similarity of equations 4.1 and 4.2 to equations 8.2 and 8.4. To show this similarity, we can write these equation as shown

    53. 52 3-5 Conjugate-Beam method Or intergrating

    54. 53 3-5 Conjugate-Beam method Here the shear V compares with the slope , the moment M compares with the displacement v & the external load w compares with the M/EI diagram. To make use of this comparison we will now consider a beam having the same length as the real beam but referred to as the conjugate beam as shown in Fig. 8.22.

    55. 54 3-5 Conjugate-Beam method Fig 8.22

    56. 55 3-5 Conjugate-Beam method The conjugate beam is loaded with the M/EI diagram derived from the load w on the real beam. From the above comparisons, we can state 2 theorems related to the conjugate beam. 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.

    57. 56 3-5 Conjugate-Beam method 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. When drawing the conjugate beam, it is important that the shear & moment developed at the supports of the conjugate beam account for the corresponding slope and displacement of the real beam at its supports.

    58. 57 3-5 Conjugate-Beam method For example, as shown in Table 8.2, a pin or roller support at the end of the real beam provides zero displacement but the beam has a non-zero slope. Consequently from Theorem 1 & 2, the conjugate beam must be supported by a pin or roller since this support has zero moment but has a shear or end reaction. When the real beam is fixed supported, both beam has a free end since at this end there is zero shear & moment.

    59. 58 3-5 Conjugate-Beam method Corresponding real & conjugate beam supports for other cases are listed in the table

    60. 59 Example 3.12 Determine the slope & deflection at point B of the steel beam shown in Fig. 8.24(a). The reactions have been computed. Take E = 200GPa I = 475(106)mm4

    61. 60 Example 3.12 Fig 8.24

    62. 61 Example 3.12 - solution The conjugate beam is shown in Fig. 8.24(b). The supports at A and B correspond to supports A and B on the real beam. The M/EI diagram is ve, so the distributed load acts downward, away from the beam. Since ?B and ?B are to be determined, we must compute VB and MB in the conjugate beam as shown in Fig. 8.24(c).

    63. 62 Example 3.12 - solution

    64. 63 Example 3.12 - solution

    65. 64 Example 3.12 - solution The ve signs indicate the slope of the beam is measured clockwise & the displacement is downward as shown in Fig. 8.24(d).

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