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


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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 IntermediateLength 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 AxiallyLoaded Stiffness Total extension or contraction of a uniform bar in tension or compression Spring constant, with k = F/d
7 7 TorsionallyLoaded 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. 33 & 34
11 Example 41 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 #41 Problem 41 Due Monday 17/3/2014
13 13 Beam Deflection Methods 1. Superposition 2. Momentarea method 3. Numerical integration 4. Castigliano energy method 5. Finite element software
14 14 Beam Deflection by Superposition Table A9 Roark s Formulas for Stress & Strain
15 Example 42 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 43 Consider the beam in Fig. 4 4a. Determine the deflection equations using superposition. Fig. 4 4
17 Example 44 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 44 Fig. 4 5
19 19 HW Assignment #42 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. (44), Torsional loading, k = GJ/l from Eq. (47)
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 xsectional shape
24 24 Some Common Strain Energy Formulas Table 4 1: StrainEnergy Correction Factors for Transverse Shear
25 Example 48 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 49 The cantilever of Ex. 4 8 is a carbon steel bar 10 in long with a 1in 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 410 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 411 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 411
34 34 HW Assignment #43 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 412 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 #44 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 414 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 415 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 415 Fig. 4 17
52 52 HW Assignment #45 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. Intermediatelength 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 Pinended column, Other end conditions: apply a constant C for each end condition
57 57 Recommended Values for End Condition Constant Table 42: EndCondition 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 IntermediateLength Columns with Central Loading Intermediatelength columns, (l/k) (l/k) 1, use a parabolic curve between S y and T General form of parabola
63 63 IntermediateLength 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. (448)
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 416 Develop specific Euler equations for the sizes of columns having a. Round cross sections b. Rectangular cross sections
71 Example 417 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 418 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 419 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. 444, the limiting slenderness ratio for strut is
76 Example 420 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 420 Fig. 4 23
78 78 HW Assignment #46 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 A9: 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
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