Tuesday, February 11, Chapter 3. Load and Stress Analysis. Dr. Mohammad Suliman Abuhaiba, PE


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1 1 Chapter 3 Load and Stress Analysis
2 2 Chapter Outline Equilibrium & FreeBody Diagrams Shear Force and Bending Moments in Beams Singularity Functions Stress Cartesian Stress Components Mohr s Circle for Plane Stress General 3D Stress Elastic Strain Uniformly Distributed Stresses Normal Stresses for Beams in Bending Shear Stresses for Beams in Bending Torsion Stress Concentration Stresses in Pressurized Cylinders Stresses in Rotating Rings Press and Shrink Fits Temperature Effects Curved Beams in Bending Contact Stresses
3 3 FreeBody Diagram: Example 3 1 Figure 3 1a shows a simplified rendition of a gear reducer where the input & output shafts AB and CD are rotating at constant speeds ω i & ω o, respectively. The input & output torques are T i = 240 lbf.in & T o, respectively. The shafts are supported in the housing by bearings at A, B, C, and D. The pitch radii of gears G 1 and G 2 are r 1 = 0.75 in & r 2 = 1.5 in, respectively. Draw FBD of each member and determine the net reaction forces and moments at all points.
4 4 FBD: Example 31
5 5 FBD: Example 31
6 6 FBD: Example 31
7 7 FBD: Example 31
8 8 Shear Force and Bending Moments in Beams Internal shear force V & bending moment M must ensure equilibrium Fig. 3 2
9 Sign Conventions for Bending and Shear Fig. 3 3
10 10 Distributed Load on Beam Distributed load q(x) = load intensity Units of force per unit length Fig. 3 4
11 11 Relationships between Load, Shear, and Bending
12 12 Singularity Functions A notation useful for integrating across discontinuities Angle brackets indicate special function to determine whether forces and moments are active Table 31
13 13 Singularity Functions Table 3 1
14 14 Singularity Functions Table 3 1
15 Example 32 Derive the loading, shearforce, and bendingmoment relations for the beam of Fig. 3 5a. Fig. 35
16 Example 32
17 Example 33 Figure 3 6a shows the loading diagram for a beam cantilevered at A with a uniform load of 20 lbf/in acting on the portion 3 in x 7 in, and a concentrated ccw moment of 240 lbf.in at x = 10 in. Derive the shearforce and bending moment relations, and the support reactions M 1 and R 1.
18 Example 33
19 Example 33
20 20 Stress Normal stress, s Tangential shear stress, t Tensile Normal stress Compressive Normal stress Units of stress: psi, N/m 2 = Pa
21 21 HomeWork Assignment #31 3, 6 Due Monday 17/2/2014
22 22 Stress element Choosing coordinates which result in zero shear stress will produce principal stresses
23 23 Cartesian Stress Components Shear stress is resolved into perpendicular components 1 st subscript = direction of surface normal 2 nd subscript = direction of shear stress
24 24 Cartesian Stress Components Plane stress occurs = stresses on one surface are zero Fig. 3 8
25 25 PlaneStress Transformation Equations Fig. 3 9
26 26 Principal Stresses for Plane Stress principal directions principal stresses Zero shear stresses at principal surfaces Third principal stress = zero for plane stress
27 27 Extremevalue Shear Stresses for Plane Stress Max shear stresses are found to be on surfaces that are ±45º from principal directions. The two extremevalue shear stresses are
28 28 Maximum Shear Stress Three principal stresses. One is zero for plane stress. Three extremevalue shear stresses. Max shear stress is greatest of these three If principal stresses are ordered so that s 1 >s 2 >s 3, then t max = t 1/3
29 29 Mohr s Circle Diagram A graphical method for visualizing stress state at a point Relation between xy stresses and principal stresses Relationship is a circle with center at C = (s, t) = [(s x + s y )/2, 0] and radius of R sx s 2 y 2 t 2 xy
30 Mohr s Circle Diagram 30 Fig. 3 10
31 Example 34 A stress element has σ x = 80 MPa & τ xy = 50 MPa cw. a. Using Mohr s circle, find the principal stresses & directions, and show these on a stress element correctly aligned wrt xy coordinates. Draw another stress element to show τ 1 & τ 2, find corresponding normal stresses, and label the drawing completely. b. Repeat part a using transformation equations only.
32 Example Principal stress orientation Max shear orientation
33 33 General 3D Stress Fig Principal stresses are found from the roots of the cubic equation
34 34 General 3D Stress Principal stresses are ordered such that s 1 > s 2 > s 3, in which case t max = t 1/3 Fig. 3 12
35 35 HomeWork Assignment # (a, d), 20 Due Saturday 22/2/2014
36 36 Elastic Strain Hooke s law E = modulus of elasticity For axial stress in x direction, n = Poisson s ratio Table A5: values for common materials
37 37 Elastic Strain For a stress element undergoing s x, s y, and s z, simultaneously,
38 38 Elastic Strain Hooke s law for shear: Shear strain g = change in a right angle of a stress element when subjected to pure shear stress. G = shear modulus of elasticity or modulus of rigidity For a linear, isotropic, homogeneous material,
39 39 Uniformly Distributed Stresses Uniformly distributed stress distribution is often assumed for pure tension, pure compression, or pure shear. For tension and compression, For direct shear (no bending present),
40 40 Normal Stresses for Beams in Bending Straight beam in positive bending x axis = neutral axis xz plane = neutral plane Neutral axis is coincident with centroidal axis of the cross section Fig. 3 13
41 41 Normal Stresses for Beams in Bending Bending stress varies linearly with distance from neutral axis, y Fig. 3 14
42 42 Normal Stresses for Beams in Bending Maximum bending stress is where y is greatest. c = magnitude of greatest y Z = I/c = section modulus
43 43 Assumptions for Normal Bending Stress Pure bending Material is isotropic and homogeneous Material obeys Hooke s law Beam is initially straight with constant cross section Beam has axis of symmetry in the plane of bending Proportions are such that failure is by bending rather than crushing, wrinkling, or sidewise buckling Plane cross sections remain plane during bending
44 Example 35 A beam having a T section with the dimensions shown in Fig is subjected to a bending moment of 1600 N m, about the negative z axis, that causes tension at the top surface. Locate the neutral axis and find the maximum tensile and compressive bending stresses. Fig. 3 15
45 45 HomeWork Assignment #33 25, 29, 34.C, 44 Due Monday 24/2/2014
46 46 TwoPlane Bending Consider bending in both xy & xz planes Cross sections with one or two planes of symmetry only For solid circular cross section, the maximum bending stress is
47 Example 36 As shown in Fig. 3 16a, beam OC is loaded in the xy plane by a uniform load of 50 lbf/in, and in the xz plane by a concentrated force of 100 lbf at end C. The beam is 8 in long. a. For the cross section shown determine the max tensile and compressive bending stresses and where they act. b. If the cross section was a solid circular rod of diameter, d = 1.25 in, determine the magnitude of the maximum bending stress.
48 Example 36
49 Example 36
50 Example 36
51 Example 36
52 Example 36 (b) For a solid circular cross section of diameter, d = 1.25 in, the maximum bending stress at end O is given by Eq. (3 28) as
53 53 Shear Stresses for Beams in Bending Fig. 3 17
54 54 Transverse Shear Stress (TSS) Fig TSS is always accompanied with bending stress
55 55 Transverse Shear Stress in a Rectangular Beam
56 56 Maximum Values of TSS Table 3 2
57 57 Maximum Values of TSS Table 3 2
58 58 Significance of TSS Compared to Bending Figure 3 19: Plot of max shear stress for a cantilever beam, combining the effects of bending and TSS. Max shear stress, including bending stress (My/I) and transverse shear stress (VQ/Ib),
59 59 Significance of TSS Compared to Bending Figure 3 19
60 60 Significance of TSS Compared to Bending Critical stress element (largest t max ) will always be either due to: bending, on outer surface (y/c=1), TSS = 0 TSS at neutral axis (y/c=0), bending is zero Transition at some critical value of L/h Valid for any cross section that does not increase in width farther away from the neutral axis. Includes round & rectangular solids, but not I beams and channels
61 Example 37 A beam 12 long is to support a load of 488 lbf acting 3 from the left support, as shown in Fig The beam is an I beam with crosssectional dimensions shown. Points of interest are labeled a, b, c, and d (Fig. 3 20c). At the critical axial location along the beam, find the following information. a. Determine the profile of the distribution of the transverse shear stress, obtaining values at each of the points of interest. b. Determine bending stresses at points of interest. c. Determine max shear stresses at the points of interest, and compare them.
62 Example 37
63 Example 37
64 64 Torsion Angle of twist, in radians, for a solid round bar Fig. 3 21
65 65 Assumptions for Torsion Equations Pure torque Remote from any discontinuities or point of application of torque, Material obeys Hooke s law Adjacent cross sections originally plane & parallel remain plane & parallel Radial lines remain straight: Depends on axisymmetry, so does not hold true for noncircular cross sections Consequently, only applicable for round cross sections
66 66 Torsional Shear in Rectangular Section Shear stress does not vary linearly with radial distance Shear stress is zero at the corners Max shear stress is at the middle of the longest side
67 67 Torsional Shear in Rectangular Section For rectangular bxc bar, where b is longest side
68 68 Power, Speed, and Torque Power equals torque times speed A convenient conversion with speed in rpm H = power, W n = angular velocity, rpm
69 69 Power, Speed, and Torque In U.S. Customary units, with unit conversion built in
70 Example 38 Figure 3 22 shows a crank loaded by a force F = 300 lbf that causes twisting and bending of a 34 diameter shaft fixed to a support at the origin of the reference system. In actuality, the support may be an inertia that we wish to rotate, but for the purposes of a stress analysis we can consider this a statics problem. a. Draw separate FBDs of the shaft AB and the arm BC, and compute the values of all forces, moments, and torques that act. Label the directions of the coordinate axes on these diagrams. b. Compute maxima of the torsional stress and bending stress in the arm BC and indicate where these act. c. Locate a stress element on top surface of shaft at A, and calculate all stress components that act upon this element. d. Determine max normal and shear stresses at A.
71 Example 38 Fig. 3 22
72 Example 38
73 Example 39 The 1.5 diameter solid steel shaft shown in Fig. 3 24a is simply supported at the ends. Two pulleys are keyed to the shaft where pulley B is of diameter 4.0 in and pulley C is of diameter 8.0 in. Considering bending and torsional stresses only, determine the locations and magnitudes of the greatest tensile, compressive, and shear stresses in the shaft.
74 Example 39 Fig. 3 24
75 Example 39
76 Example 39
77 Example 39
78 Example 39
79 79 Closed ThinWalled Tubes t << r t t = constant t is inversely proportional to t Fig. 3 25
80 80 Closed ThinWalled Tubes Total torque T is A m = area enclosed by section median line Solving for shear stress
81 81 Closed ThinWalled Tubes Angular twist (radians) per unit length L m = length of the section median line
82 Example 310 A welded steel tube is 40 in long, has a 1/8 in wall thickness, and a 2.5in by 3.6in rectangular cross section as shown in Fig Assume an allowable shear stress of psi and a shear modulus of 11.5 Mpsi. a. Estimate the allowable torque T. b. Estimate the angle of twist due to the torque.
83 Example 310 Fig. 3 26
84 Example 311 Compare the shear stress on a circular cylindrical tube with an outside diameter of 1 in and an inside diameter of 0.9 in, predicted by Eq. (3 37), to that estimated by Eq. (3 45).
85 85 Open ThinWalled Sections When the median wall line is not closed, the section is said to be an open section Torsional shear stress T = Torque, L = length of median line, c = wall thickness, G = shear modulus, and q 1 = angle of twist per unit length
86 86 Open ThinWalled Sections For small wall thickness, stress and twist can become quite large Example: Compare thin round tube with and without slit Ratio of wall thickness to outside diameter of 0.1 Stress with slit is 12.3 times greater Twist with slit is 61.5 times greater
87 Example 312 A 12 long strip of steel is 1/8 thick and 1 wide, as shown in Fig If the allowable shear stress is psi and the shear modulus is 11.5 Mpsi, find the torque corresponding to the allowable shear stress and the angle of twist, in degrees, a. using Eq. (3 47) b. using Eqs. (3 40) and (3 41)
88 88 HomeWork Assignment #34 49, 51, 52, 57, 64 Due Saturday 1/3/2014
89 89 Stress Concentration Localized increase of stress near discontinuities K t = Theoretical (Geometric) Stress Concentration Factor
90 90 Theoretical Stress Concentration Factor Graphs available for standard configurations A15 and A16 for common examples Many more in Peterson s Stress Concentration Factors Higher Kt at sharper discontinuity radius, and at greater disruption
91 91 Theoretical Stress Concentration Factor
92 92 Theoretical Stress Concentration Factor
93 93 Stress Concentration for Static and Ductile Conditions With static loads and ductile materials Highest stressed fibers yield (cold work) Load is shared with next fibers Cold working is localized Overall part does not see damage unless ultimate strength is exceeded Stress concentration effect is commonly ignored for static loads on ductile materials
94 94 Techniques to Reduce Stress Concentration Increase radius Reduce disruption Allow dead zones to shape flow lines more gradually
95 95 Techniques to Reduce Stress Concentration
96 96 Techniques to Reduce Stress Concentration
97 Example 313 The 2mmthick bar shown in Fig is loaded axially with a constant force of 10 kn. The bar material has been heat treated and quenched to raise its strength, but as a consequence it has lost most of its ductility. It is desired to drill a hole through the center of the 40mm face of the plate to allow a cable to pass through it. A 4mm hole is sufficient for the cable to fit, but an 8mm drill is readily available. Will a crack be more likely to initiate at the larger hole, the smaller hole, or at the fillet?
98 Example 313 Fig. 3 30
99 Example 313 Fig. A 15 1
100 Example 313 Fig. A 15 5
101 101 HomeWork Assignment #35 68, 72, 84 Due Monday 3/3/2014
102 102 Stresses in Pressurized Cylinders Fig Tangential and radial stresses,
103 103 Stresses in Pressurized Cylinders Special case of zero outside pressure, p o = 0
104 104 Stresses in Pressurized Cylinders If ends are closed, then longitudinal stresses also exist
105 105 ThinWalled Vessels Cylindrical pressure vessel with wall thickness 1/10 or less of the radius Radial stress is quite small compared to tangential stress Average tangential stress
106 106 ThinWalled Vessels Maximum tangential stress Longitudinal stress (if ends are closed)
107 Example 314 An aluminumalloy pressure vessel is made of tubing having an outside diameter of 8 in and a wall thickness of ¼ in. a. What pressure can the cylinder carry if the permissible tangential stress is 12 kpsi and the theory for thinwalled vessels is assumed to apply? b. On the basis of the pressure found in part (a), compute the stress components using the theory for thickwalled cylinders.
108 108 Stresses in Rotating Rings Rotating rings: flywheels, blowers, disks, etc. Tangential and radial stresses are similar to thickwalled pressure cylinders, except caused by inertial forces Conditions: Outside radius is large compared with thickness (>10:1) Thickness is constant Stresses are constant over the thickness
109 109 Stresses in Rotating Rings Stresses are
110 110 Press and Shrink Fits Two cylindrical parts are assembled with radial interference d Pressure at interface Fig. 3 33
111 111 Press and Shrink Fits If both cylinders are of the same material
112 112 Press and Shrink Fits Eq. (349) for pressure cylinders applies
113 113 Press and Shrink Fits For the inner member, p o = p and p i = 0 For the outer member, p o = 0 and p i = p
114 114 HomeWork Assignment #36 94, 104, 110 Due Wensday 5/3/2014
115 115 Temperature Effects Normal strain due to expansion from temperature change a = coefficient of thermal expansion
116 116 Temperature Effects Thermal stresses occur when members are constrained to prevent strain during temperature change For a straight bar constrained at ends, temperature increase will create a compressive stress Flat plate constrained at edges
117 Coefficients of Thermal Expansion Table 3 3: Coefficients of Thermal Expansion (Linear Mean Coefficients for the Temperature Range C)
118 118 Curved Beams in Bending In thick curved beams Neutral axis and centroidal axis are not coincident Bending stress does not vary linearly with distance from the neutral axis
119 119 Curved Beams in Bending Fig. 3 34
120 120 Curved Beams in Bending Location of neutral axis Stress distribution
121 121 Curved Beams in Bending Stress at inner and outer surfaces
122 Example 315 Plot the distribution of stresses across section A A of the crane hook shown in Fig. 3 35a. The cross section is rectangular, with b = 0.75 in and h = 4 in, and the load is F = 5000 lbf.
123 Example 315 Fig. 3 35
124 Example 315
125 Example 315
126 Formulas for Sections of Curved Beams (Table 34)
127 Formulas for Sections of Curved Beams (Table 34)
128 Formulas for Sections of Curved Beams (Table 34)
129 129 Alternative Calculations for e Approximation for e, valid for large curvature where e is small with r n r c Substituting Eq. (366) into Eq. (364), with r n y = r, gives
130 Example 316 Consider the circular section in Table 3 4 with r c = 3 in and R = 1 in. Determine e by using the formula from the table and approximately by using Eq. (3 66). Compare the results of the two solutions.
131 131 Contact Stresses Two bodies with curved surfaces pressed together Point or line of contact changes to area contact Stresses developed are 3Dl Called contact stresses or Hertzian stresses Common examples Wheel rolling on rail Mating gear teeth Rolling bearings
132 132 Spherical Contact Stress Two solid spheres of diameters d 1 and d 2 are pressed together with force F Circular area of contact of radius a
133 133 Spherical Contact Stress Pressure distribution is hemispherical Max pressure at the center of contact area Fig. 3 36
134 134 Spherical Contact Stress Maximum stresses on the z axis Principal stresses
135 135 Spherical Contact Stress From Mohr s circle, maximum shear stress is
136 136 Spherical Contact Stress Plot of three principal stress & max shear stress as a function of distance below the contact surface t max peaks below the contact surface Fatigue failure below the surface leads to pitting & spalling For poisson ratio of 0.30, t max = 0.3 p max at depth of z = 0.48a
137 137 Spherical Contact Stress Fig. 3 37
138 138 Cylindrical Contact Stress Two right circular cylinders with length l and diameters d 1 & d 2 Area of contact is a narrow rectangle of width 2b and length l Pressure distribution is elliptical Fig. 3 38
139 139 Cylindrical Contact Stress Halfwidth b Maximum pressure Fig. 3 38
140 140 Cylindrical Contact Stress Maximum stresses on z axis
141 141 Cylindrical Contact Stress Plot of stress components and max shear stress as a function of distance below the contact surface For poisson ratio of 0.30, t max = 0.3 p max at depth of z = 0.786b
142 142 Cylindrical Contact Stress Fig. 3 39
143 143 HomeWork Assignment # , 133, 138 Due Saturday 8/3/2014
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