March 24, Chapter 4. Deflection and Stiffness. Dr. Mohammad Suliman Abuhaiba, PE

Transcription

1 Chapter 4 Deflection and Stiffness 1

2 2 Chapter Outline Spring Rates Tension, Compression, and Torsion Deflection Due to Bending Beam Deflection Methods Beam Deflections by Superposition Strain Energy Castigliano s Theorem Deflection of Curved Members Statically Indeterminate Problems Compression Members General Long Columns with Central Loading Intermediate-Length Columns with Central Loading Columns with Eccentric Loading Struts or Short Compression Members Suddenly Applied Loading

3 3 Force vs Deflection Elasticity:property of a material that enables it to regain its original configuration after deformation Spring: a mechanical element that exerts a force when deformed

4 4 Force vs Deflection Linear spring Nonlinear stiffening spring Nonlinear softening spring Fig. 4 1

5 5 Spring Rate Spring rate For linear springs, k constant is constant, spring

6 6 Axially-Loaded Stiffness Total extension or contraction of a uniform bar in tension or compression Spring constant, with k = F/d

7 7 Torsionally-Loaded Stiffness Angular deflection (radians) of a uniform solid or hollow round bar subjected to a twisting moment T Torsional spring constant for round bar

8 8 Deflection Due to Bending Curvature of beam subjected to bending moment M Curvature of plane curve

9 9 Deflection Due to Bending Slope of beam at any point x along the length If slope is very small, denominator of Eq. 4.9 approaches unity:

10 10 Deflection Due to Bending Recall Eqs. 3-3 & 3-4

11 Example 4-1 For the beam in Fig. 4 2, the bending moment equation, for 0 x l, is Using Eq. (4 12), determine the equations for slope & deflection of the beam, slopes at ends, and max deflection. Fig. 4 2

12 12 HW Assignment #4-1 Problem 4-1 Due Monday 17/3/2014

13 13 Beam Deflection Methods 1. Superposition 2. Moment-area method 3. Numerical integration 4. Castigliano energy method 5. Finite element software

14 14 Beam Deflection by Superposition Table A-9 Roark s Formulas for Stress & Strain

15 Example 4-2 Consider the uniformly loaded beam with a concentrated force as shown in Fig Using superposition, determine the reactions and deflection as a function of x. Fig. 4 3

16 Example 4-3 Consider the beam in Fig. 4 4a. Determine the deflection equations using superposition. Fig. 4 4

17 Example 4-4 Figure 4 5a shows a cantilever beam with an end load. Normally we model this problem by considering the left support as rigid. After testing the rigidity of the wall it was found that the translational stiffness of the wall was k t force per unit vertical deflection, and the rotational stiffness was k r moment per unit angular (radian) deflection (see Fig. 4 5b). Determine the deflection equation for the beam under the load F.

18 Example 4-4 Fig. 4 5

19 19 HW Assignment #4-2 Problems: 4.10, 4.14, 4.41 Due Wednesday 19/3/2014

20 20 Strain Energy External work done on elastic member while deforming it is transformed into strain energy, or potential energy. Strain energy = average force deflection

21 21 Some Common Strain Energy Formulas Axial loading, k = AE/l from Eq. (4-4), Torsional loading, k = GJ/l from Eq. (4-7)

22 22 Some Common Strain Energy Formulas Direct shear loading, Bending loading,

23 23 Some Common Strain Energy Formulas Transverse shear loading, C = modifier dependent on x-sectional shape

24 24 Some Common Strain Energy Formulas Table 4 1: Strain-Energy Correction Factors for Transverse Shear

25 Example 4-8 A cantilever beam with a round cross section has a concentrated load F at the end, as shown in Fig. 4 9a. Find the strain energy in the beam. Fig. 4 9

26 26 Castigliano s Theorem Forces act on elastic systems subject to small displacements Displacement corresponding to any force along its direction = partial derivative of total strain energy wrt force For rotational displacement, in radians,

27 Example 4-9 The cantilever of Ex. 4 8 is a carbon steel bar 10 in long with a 1-in diameter and is loaded by a force F = 100 lbf. a. Find max deflection using Castigliano s theorem, including that due to shear. b. What error is introduced if shear is neglected? Fig. 4 9

28 28 Utilizing a Fictitious Force Apply a fictitious force Q at the point, and in the direction, of the desired deflection. Set up the equation for total strain energy including energy due to Q. Take derivative of total strain energy wrt Q Set Q to zero

29 29 Finding Deflection Without Finding Energy Partial derivative is moved inside the integral. For example, for bending,

30 Common Deflection Equations

31 Example 4-10 Using Castigliano s method, determine the deflections of points A and B due to the force F applied at the end of the step shaft shown in Fig The second area moments for sections AB and BC are I 1 and 2I 1, respectively. Fig. 4 10

32 Example 4-11 For the wire form of diameter d shown in Fig. 4 11a, determine the deflection of point B in the direction of the applied force F (neglect the effect of transverse shear). Fig. 4 11

33 Example 4-11

34 34 HW Assignment #4-3 Problems: 4.67, 4.70 Due Saturday 22/3/2014

35 35 Deflection of Curved Members Four strain energy terms due to: 1. Bending moment M 2. Axial force F q 3. Bending moment due to F q 4. Transverse shear F r

36 36 Deflection of Curved Members Strain energy due to bending moment M r n = radius of neutral axis

37 37 Deflection of Curved Members Strain energy due to axial force F q

38 38 Deflection of Curved Members Strain energy due to bending moment from F q

39 39 Deflection of Curved Members Strain energy due to transverse shear F r

40 40 Deflection of Curved Members Combining four energy terms Deflection by Castigliano s method

41 41 Deflection of Curved Members

42 42 Deflection of Curved Members

43 43 Deflection of Thin Curved Members R/h > 10, eccentricity is small Strain energies approximated with regular energy equations: Rdq dx As R increases, bending component dominates all other terms

44 Example 4-12 The cantilevered hook shown in Fig. 4 13a is formed from a round steel wire with a diameter of 2 mm. The hook dimensions are l = 40 & R = 50 mm. A force P of 1 N is applied at point C. Use Castigliano s theorem to estimate deflection at point D at the tip. Fig. 4 13

45 45 HW Assignment #4-4 Problems: 4.78 Due Monday 24/3/2014

46 46 Statically Indeterminate Problems Redundant supports: extra constraint supports A deflection equation is required for each redundant support.

47 47 Procedure 1 for Statically Indeterminate Problems 1. Choose redundant reactions 2. Write equations of static equilibrium for remaining reactions in terms of applied loads & redundant reactions. 3. Write deflection equations for points at locations of redundant reactions in terms of applied loads and redundant reactions. 4. Solve equilibrium & deflection equations

48 Example 4-14 The indeterminate beam 11 of Appendix Table A 9 is reproduced in Fig Determine the reactions using procedure 1. Fig. 4 16

49 49 Procedure 2 for Statically Indeterminate Problems 1. Write equations of static equilibrium in terms of applied loads & unknown restraint reactions. 2. Write deflection equation in terms of applied loads and unknown restraint reactions. 3. Apply boundary conditions to deflection equation consistent with restraints. 4. Solve the set of equations.

50 Example 4-15 The rods AD & CE shown in Fig. 4 17a each have a diameter of 10 mm. The second area moment of beam ABC is I = 62.5(10 3 ) mm 4. The modulus of elasticity of the material used for the rods and beam is E = 200 GPa. The threads at the ends of the rods are singlethreaded with a pitch of 1.5 mm. The nuts are first snugly fit with bar ABC horizontal. Next the nut at A is tightened one full turn. Determine the resulting tension in each rod and the deflections of points A and C.

51 Example 4-15 Fig. 4 17

52 52 HW Assignment #4-5 Problems: 4.95, 4.97 Due Wednesday 26/3/2014

53 53 Compression Members Column: A member loaded in compression either its length or eccentric loading causes it to experience more than pure compression

54 54 Compression Members Four categories of columns 1. Long columns with central loading 2. Intermediate-length columns with central loading 3. Columns with eccentric loading 4. Struts or short columns with eccentric loading

55 55 Long Columns with Central Loading When P reaches critical load, column becomes unstable & bending develops rapidly

56 56 Euler Column Formula Pin-ended column, Other end conditions: apply a constant C for each end condition

57 57 Recommended Values for End Condition Constant Table 4-2: End-Condition Constants for Euler Columns [to Be Used with Eq. (4 43)]

58 58 Long Columns with Central Loading I = Ak 2, Euler column formula becomes l/k = slenderness ratio, to classify columns according to length categories. P cr /A = critical unit load necessary to place the column in a condition of unstable equilibrium

59 Euler Curve 59 P cr /A vs l/k, with C = 1 gives curve PQR Fig. 4 19

60 60 Long Columns with Central Loading Vulnerability to failure near point Q Buckling is sudden & catastrophic, a conservative approach near Q is desired T is defined such that P cr /A = S y /2, giving

61 61 Condition for Use of Euler Equation (l/k) > (l/k) 1, use Euler equation (l/k) (l/k) 1, use a parabolic curve between S y & T

62 62 Intermediate-Length Columns with Central Loading Intermediate-length columns, (l/k) (l/k) 1, use a parabolic curve between S y and T General form of parabola

63 63 Intermediate-Length Columns with Central Loading If parabola starts at S y, then a = S y

64 64 Columns with Eccentric Loading M = -P(e+y) d 2 y/dx 2 =M/EI Fig. 4 20

65 65 Columns with Eccentric Loading Boundary conditions: y = 0 at x = 0 and at x = l

66 66 Columns with Eccentric Loading At midspan where x = l/2

67 67 Columns with Eccentric Loading Max compressive stress: Substituting M max from Eq. (4-48)

68 68 Columns with Eccentric Loading Using S yc as the maximum value of s c, and solving for P/A, we obtain the secant column formula

69 69 Secant Column Formula ec/k 2 = eccentricity ratio Fig. 4 21

70 Example 4-16 Develop specific Euler equations for the sizes of columns having a. Round cross sections b. Rectangular cross sections

71 Example 4-17 Specify the diameter of a round column 1.5 m long that is to carry a maximum load estimated to be 22 kn. Use a design factor n d = 4 and consider the ends as pinned (rounded). The column material selected has a minimum yield strength of 500 MPa and a modulus of elasticity of 207 GPa.

72 Example 4-18 Repeat Ex for J. B. Johnson columns. (a) For round columns, Eq. (4 46) yields (b) For round columns, Eq. (4 46) yields

73 Example 4-19 Choose a set of dimensions for a rectangular link that is to carry a maximum compressive load of 5000 lbf. The material selected has a minimum yield strength of 75 kpsi and a modulus of elasticity E = 30 Mpsi. Use a design factor of 4 and an end condition constant C = 1 for buckling in the weakest direction, and design for a. a length of 15 in b. a length of 8 in with a minimum thickness of 0.5 in.

74 74 Struts or Short Compression Members Strut: short member loaded in compression If eccentricity exists, max stress is at B with axial compression and bending.

75 75 Struts or Short Compression Members Differs from secant equation in that it assumes small effect of bending deflection If bending deflection is limited to 1% of e, then from Eq. 4-44, the limiting slenderness ratio for strut is

76 Example 4-20 Figure 4 23a shows a workpiece clamped to a milling machine table by a bolt tightened to a tension of 2000 lbf. The clamp contact is offset from the centroidal axis of the strut by a distance e = 0.10 in, as shown in part b of the figure. The strut, or block, is steel, 1 in square and 4 in long, as shown. Determine the maximum compressive stress in the block.

77 Example 4-20 Fig. 4 23

78 78 HW Assignment #4-6 Problems: 4.107, Due Saturday 29/3/2014

79 79 Suddenly Applied Loading Weight falls from distance h and suddenly applies a load to a cantilever beam Find deflection and force applied to beam due to impact

80 80 Suddenly Applied Loading Abstract model considering beam as simple spring Table A-9: beam 1, k= F/y =3EI/l 3 Assume beam to be massless, so no momentum transfer, just energy transfer Loss of potential energy from change of elevation is W(h + d)

81 81 Suddenly Applied Loading Increase in potential energy from compressing spring is kd 2 /2 Conservation of energy W(h + d) = kd 2 /2

82 82 Suddenly Applied Loading Maximum deflection Maximum force