Mechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection


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1 Mechanics of Materials II Chapter III A review of the fundamental formulation of stress, strain, and deflection
2 Outline Introduction Assumtions and limitations Axial loading Torsion of circular shafts Beam in bending Bending of symmetric beams in two planes Thinwalled pressure vessels Superposition Statically indeterminate problems Stress and strain transformations* Buckling instability of columns
3 Introduction The basic topics from mechanics of materials I include Direct axial load Shear load Torsion in circular shafts Transverse loading of long, straight, narrow beam The purpose of this chapter is to provide a concise review of the fundamental formulation (stress, strain, and deflection)
4 Assumptions and limitations Homogeneous, Isotropic and linear strainstress relations Cross section are exact, and constant or gradually varying in the normal direction. The point of load or support connection are at suffer large distance for interested point The applied load/ or support connection are perfectly positioned geometrically Load are static and applied very gradually No initial stress effect or residual stresses
5 Axial loading Axial stresses P x A (a) elongation of the prismatic bar (b) freebody diagram
6 Axial loading Axial strains and deflection Hooke s law 1D x x E x y z E x y z P x AE P x AE
7 Important equations = d/l (Geometry of deformation) = P/A (Equilibrium Condition) = E (Material behavior) d PL/AE (Deformation) The product (AE) is called axial rigidity
8 SaintVenant s principle
9 STRESS CONCENTRATIONS max K t nom
10 STRESS CONCENTRATION CHARTS
11
12 Stepped bar with multiple loadings
13 Flexibility versus stiffness Flexibility, f Stiffness, k
14 Nonuniform bars
15 Example 3.21: The limitation of homogeneous materials
16 Example 3.22: Deformation due to weight of the rod
17 Example 3.23: the displacement of two cables
18 Example of a stepped bar  Section plane at the changeofload point  determine the internal force in each section  determine the deformation in each section
19 Example: Displacement of a threemember device Given: The rigid bar BC is supported by the 5/8 in diameter (rod AB) and ½ in (rod CD) which under load P = 12 kip at point E. Each of the rods is mad of aluminum alloy6061 with E=10x10 6 psi Find: The deflection at point E
20 Example: Displacement of a threemember device
21 Quiz: Axial loading As shown in Figure below, a rigid rod ABC is suspended by two wires. The wire AD is made of steel have crosssectional area As = 0.14 in 2 and elastic modulus Es = 30x10 6 psi. For the aluminumalloy wire CE, crosssection area and elastic modulus Aa =0.28 in 2 and Ea = 10x10 6 psi, respectively. Determine the displacement of point B caused by the load P = 6 kips
22 Torsion of circular shafts Assumptions The plane cross sections perpendicular to the axis of the bar remain plane after the application of a torque: points in a given plane remain in that plane after twisting. Furthermore, expansion or contraction of a cross section does not occur, nor does a shortening or lengthening of the bar. Thus all normal strains are zero. The material is homogeneous and isotropic.
23 The maximum shear strain at the outer radius c is given by: Shear strain at any arbitrary radius r is given by:
24 Torsion formula for circular shafts (C. A. Coulomb, in about 1775
25 Important equations In SI units if T is in Nm, c is in meters and J is in m 4, then t is in Pa. In English units if T is in inlb/sec, c is in inch and J is in inch 4, then t is in psi. J is called the polar moment of inertia of the entire cross section of a bar. For solid circular and hollow cross sections, J is given by the formulas (on the next slide):
26 Important formulas for J J solid = 0.5pc 4, where c is the radius of the bar In terms of the bar diameter, J = pd 4 J hollow = 0.5p(c 4 b 4 ), where c and b are the inner and outer radii of the bar In terms of the diameters, J = p(d o4 d i4 )
27 Important formulas for J For thinwalled circular members (i.e., r/t 10), an approximate formula for J is given by: J = 2pr 3 t, where r is the mean radius and t is the thickness of the tube, respectively.
28 Example3.31: A transmitted torque of step shaft
29 Example of an application
30 Coupling bolts
31 Angle of twist
32 Equation for angle of twist Geometry of deformation: g max = cf/l Equilibrium condition: t max = Tc/J Material behavior: g max = t max /G
33 Equation for angle of twist Combining the previous equations yield: f TL/GJ Torsional spring stiffness:
34 Stepped bar with multiple loads
35 Torsion failures (a) Brittle material (b) Ductile material
36 Torsion tests
37 Design of circular shafts
38 In SI units: In English units:
39 Case study
40 Beam in bending Shear Force (SF) & Bending Moment (BM) Equations and Diagram Isolation internal transverse planar surface Singularity functions Bending stresses Transverse shear stresses Bending strain and deflections Bending of symmetric beams in two planes
41 Types of beams Simple beam Cantilever beam Overhang beam
42 Sign convention In this text book.. For some other text book..
43 In this text book..
44 For some other text book..
45
46 Singularity functions Forming a single equation for describe any discontinuous functions.. F ( x) x a n n Where n is any integer For n 0 For n 0 n when x Fn ( x) x a 0 when x n n ( x a) when x a Fn ( x) x a 0 when x a a a
47
48
49 Example D.21: loadintensity equation q( x) R x 0 w x 0 w x a M x b P x c R x L A C D E q( x) RA x w x w x a MC x b PD x c RE x L
50 Integration rules x n a dx n 1 x a c for n 1 n 1 n 1 x a c for n 0 0 Example 2 1 x a dx x a c x a dx x a c 2
51 Example 3.41: SFD and BMD of Overhang beam
52 Normal stresses Distribution of bending stress in a beam. In this formula, S is called the section modulus (S = I/c)
53 Doubly symmetric crosssectional shapes.
54 Example 3.42: Simply supported beam
55 Example: Single overhang beam with distributed load and Tcross section
56 Twoplane bending problem for practice Calculate the diameter of the shaft based on maximum bending stress
57 Stress concentration in bending
58
59
60 Transverse shear stresses
61
62 The shear formula Q is the first moment of area given by
63 Shear Stress Distribution in Rectangular Beams
64 Example 3.43: Shear stress distribution in solid rectangular cross section beam
65 SOLID CIRCULAR SECTION BEAMS
66 HOLLOW CIRCULAR SECTION BEAMS
67 Comparison of Shear and Bending Stresses
68 If L = 10h ( long beam), then this ratio is only 1/20, which means that t max is only 5% of max
69 Bending strain and deflection x Mz y EI z My z y z EI z g xy 2(1 ) VQ y E I b z
70 Bending strain and deflection Positive loads and internal force resultants.
71 BernoulliEuler Law
72
73 Boundary conditions
74 Method of integration Find: (a) Determine the equation of the elastic curve using the doubleintegration approach. (b) Derive the equation of the elastic curve using the multipleintegration approach. (c) Obtain the maximum deflection and the slopes.
75 The bending moment is given by: Double integration method:
76 Integrating twice in x gives: Applying the boundary conditions we obtain: C 2 = 0 and C 1 =  w L 3 /24
77 Slope and deflection are then given by:
78 Largest displacements and largest slopes:
79 Example 3.45: Simple supports with intermediate load Find: Deflection as function of x of the centroidal axis  Discontinuous equations  singularity functions
80 Bending of symmetric beam in two planes M z y x It Mz y I z x Mz y I z M M M 2 2 z y z I I I z y z
81 Example 3.51: Pulley shaft system
82 Thinwalled pressure vessels If the ratio of the wall thickness (t) to the inner radius (r) is equal or less than about 1/10 (or r/t 10), the vessel is classified as thinwalled In fact, in thinwalled vessels, there is often no distinction made between the inside and outside radii because they are nearly equal.
83 Membrane equation f 2 2 F 0: 2 tr f sin 2 tr sin p r r f 0 n f f f p f r r t f
84 Cylindrical pressure vessel Tangential stress: Axial (longitudinal) stress: z
85 Spherical pressure vessels Tangential stress due to internal pressure:
86 Example : Pressure capacity of a cylindrical vessel A cylinder of thickness t = 3 mm and diameter d = 1.5 m is made of steel with yield strength 240 MPA Find: the internal pressure p that can be carried by the vessel based on a safety factor of n 2 on yielding s y 240 / n all y s MPa For circumferential p 6 t all (0.003) r kpa all For axial or longitudinal p 2 t 960 r kpa So the gage pressure may not exceed 480 kpa
87 Superposition Works well for beams with small deflections that follow linear Hooke s law For the special case when a = b = L/2:
88 Example: Slope and Deflection of a beam with an overhang
89 Example 3.71: Superposition
90 Example 3.72: Deflection equation (overhang beam)
91 Example 3.73: State of stress (Transverse load+torque)
92 Statically Indeterminate Problems Step 1 : Solve for all possible unknown reactions Step 2 : Subtract the number for remaining Eq n from the number of remaining unknown Ex. (n = 32) = 1 Step 3 : Eliminate n of unknown reactions (redundant unknown) Step 4 : Considering redundant unknown as applied force  the deflection and/or rotation at the n point Step 5 : Using the method of superposition Step 6 : Solve the simultaneous Eq n Step 7 : Substitute the results of the Step 6 into Step 1
93 Example 3.82: Deflection equation (overhang beam)
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