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LEARNING OUTCOME Use the principle of stress and deformation analysis in stress analyses problem. CHAPTER 2 STRESS AND DEFORMATION ANALYSIS 1 2 CONTENT CONTENT Representing stress on a stress element. Torsion in members having noncircular cross Mohr’s circle for plane stress Mohr’ sections. Direct stresses: tension and compression. Vertical shearing stress. Deformation under direct axial loading. Stress due to bending. Direct shear stress. Flexural center for beams. Torsional shear stress. Combined normal stresses: superposition principle. Torsional deformation. Stress concentrations factors. Types of loading and stress ratio 3 4 1 REPRESENTING STRESS ON A STRESS ELEMENT General case of combined stress Positive shear stresses tend to rotate the element in a clockwise direction Consider a small stress element which combined Negative shear stresses tend to rotate the element in a stresses act. counterclockwise direction Stresses can be normal stresses and/or shear stresses. Consider only two dimensional stresses, i.e. in direction. the x and y direction. 5 6 Principal stresses Combination of applied stresses Max shear stress (normal and shear) that produces maximum normal stress. ⎛ σ x −σ y ⎞ 2 Max/min principal stress τ max = ⎜ ⎜ ⎟ + τ xy 2 ⎟ ⎝ 2 ⎠ 2 σ x +σ y ⎛ σ x −σ y ⎞ σ1 = + ⎜ ⎜ 2 ⎟ + τ xy 2 ⎟ Angle for maximum shear stress element 2 ⎝ ⎠ φσ = arctan[− (σ x − σ y ) 2τ xy ] 1 2 σ x +σ y ⎛σ x −σ y ⎞ 2 σ2 = ⎜ 2 ⎟ + τ xy − ⎜ ⎟ 2 2 ⎝ ⎠ Average normal stress Angle of principal stress element σ avg = (σ x + σ y ) 2 x- Measured from the positive x-axis (shown in figure). Positive sign calls for a clockwise rotation of element and vise versa. φσ = arctan[2τ xy (σ x − σ y )] 1 2 7 8 2 MOHR’S CIRCLE FOR PLANE STRESS Summary Mohr’ The Mohr’s circle enables the following computation with relative ease and better accuracy: Max/min principal stresses and their directions. Max shear stress and the orientation of the planes. Value of normal stresses that act on the planes where the max shear stresses act. Values of the normal and shear stresses that act on an element with any orientation. 9 10 Mohr’ Drawing the Mohr’s circle Perform stress analysis and determine the magnitudes and directions of the normal and shear stresses acting at the point of interest, using the convention: Tensile stresses are positive; compressive negative. Shear stresses that tend to rotate element clockwise are positive; otherwise negative. (σ Set up the normal and shear stress (σ, τ) plane and locate the point on it. Connect the points. The line that cuts the σ-axis is the center of the circle, O, and radius of the circle, R: Locating the points: Mohr’ The Mohr’s circle σ x −σ y R= + τ xy 2 11 12 3 Case when the both principle stresses have the same sign using MDESIGN 13 14 The true maximum shear stress on the element will not be found if the two principal stresses are of the same sign. sign. Hoop stress, σ x = pD 2t Where; p = internal pressure in cylinder D = diameter of the cylinder Longtitudinal stress, σ y = pD 4t t = thickness of the cylinder wall 3- Use 3-D stress element g N in us SIG E D M 15 16 4 Mohr’s circle special stress conditions Pure torsional shear Pure uniaxial tension Pure uniaxial compression Uniaxial tension combined with torsional shear 17 18 DIRECT STRESSES: TENSION AND COMPRESSION Stress: Internal resistance offered by a unit area normal stress, σ = force F area = A [N m2 ] of material to an external load Perpendicular to element Conditions: Compressive stresses: Crushing action. Negative • load- load-carry member must be straight by convention • line of action of the load must pass through the centroid of cross section of the member Tensile: Pulling action. Positive by convention. • member must be of uniform cross section • material must be homogeneous and isotropic • member must be short in the case of compression members 19 20 5 DEFORMATION UNDER DIRECT SHEAR STRESS DIRECT AXIAL LOADING FL σL Occurs when the applied force tends to cut through the δ= = [m] member as scissors. Ex: tendency for a key to be sheared EA E off at the section between the shaft and the hub of a machine element when transmitting torque(see next Where: slide). δ = total deformation of the member carrying the axial Apply force is assumed to be uniformly distributed across the cross section. F = direct axial load L = original load length of the member τ= shearing force F area in shear = As [N m2 ] E = modulus of elasticity of the material A = cross-sectional area of the member cross- σ = direct/normal stress 21 22 TORSIONAL SHEAR STRESS A torque will twist a member, causing a shear stress in the member τ max = Tc T = J Zp [N m2 ] A general shear stress formula: Tr τ= J 23 24 6 Where: The distribution of stress is not uniform across the cross section T = torque c = radius of shaft to its outside surface J (Appendix) = polar moment of inertia (Appendix) r = radial distance from the center of the shaft to the point of interest Zp = polar section modulus (Appendix) (Appendix) 25 26 TORSIONAL DEFORMATION TORSION IN MEMBERS HAVING NONCIRCULAR θ= TL [deg] CROSS-SECTIONS GJ CIRCULAR NONCIRCULAR Where: SECTION SECTION T = torque τ max = Tc T = J Zp τ max = T Q [N m2 ] L = length of the shaft over which the angle of twist is being computed θ= TL θ= TL [rad ] GJ GK G = modulus of elasticity of the shaft material in shear J = polar moment of inertia (Appendix) (Appendix) 27 28 7 VERTICAL SHEAR STRESS Beam carrying transverse loads experience shearing TORSION IN (V forces (V) which cause shearing stress: MEMBERS HAVING τ= VQ It [N ] m 2 ; Q = Ap y [m ] 3 NONCIRCULAR CROSS- CROSS- SECTIONS 29 30 Where: V = shearing force In the analysis of beams, it is usual to compute the variation in shearing force across the entire length of Q = first moment the beam and to draw the shearing force diagram. I = moment of inertia Vertical shear stress = Horizontal shear stress, because t = thickness of the section any element of material subjected to a shear stress on Ap = area of the section above the place where the one face must have a shear stress of the same shearing force is to be computed magnitude on the adjacent face for the element to be in equilibrium. y = distance from the neutral axis of the section to the centroid of the area Ap 31 32 8 STRESS DUE TO BENDING A beam is a member that carries load transverse to its axis. Where: Such loads produce bending moments in the beam, which result in the development of bending stress. M = magnitude of bending moment at the section Bending stress are normal stresses, that is, either tensile or c = distance from the neutral axis to the outermost compressive. fiber of the beam cross section The maximum bending stress in a beam cross section will occur in the part farthers from the neutral axis of the section. I = moment of inertia At that point, the flexure formula gives the stress: flexure formula; σ = Mc I [N m2 ] 33 34 The flexure formula was developed subject to the following conditions: - Beam must pure bending. No shearing stress and axial loads. - Beam must not twist or be subjected to torsional load. Hooke’ - Material of beam must obey Hooke’s law - Modulus of elasticity of the material must be the same in both tension and compression. - Beam is initially straight and has constant cross section. - No part of the beam shape fails because of buckling or wrinkling. 35 36 9 FLEXURAL CENTER FOR For design, it is convenient to define the term section BEAMS modulus, S: To ensure symmetrical bending i.e. no tendency to twist under loading, action of load pass through the line of S=I c [m3 ] symmetry: The flexure formula then becomes: σ =M S [ N m2 ] Then, in design, it is usual to define a design stress, σd ,and with the bending moment known, then: S = M σd [m3 ] 37 38 COMBINED NORMAL STRESSES: SUPERPOSITION PRINCIPLE If there is no vertical axis symmetry: load- When the same cross section of a load-carrying member is subjected to both a direct tensile and compressive stress and a stress due to bending, the resulting normal stress can be computed by the method of superposition: σ =± Mc F I ± A [N m2 ] 39 40 10 COMBINED NORMAL EXAMPLE 3.8 STRESSES: SUPERPOSITION Shigley’s text book PRINCIPLE 41 42 STRESS CONCENTRATIONS FACTORS Example: Any geometric discontinuities will cause the actual maximum stress to be higher than the calculated value Diagram shows a round bar subjected to an axial force, F. Stress concentration factor, Kt= factor by which the Compute maximum stress. actual maximum stress exceeds the nominal stress Given Kt = 1.60 (Appendix), F = 9800N (σnom,τnom) predicted by calculations. That is: σ max = K tσ nom ; τ max = K tτ nom [ N m2 ] σnom = F/A = ( 9800N)/ [π( 10mm)2/4] = 124.8 MPa σmax = Kt σnom = (1.60)(124.8) = 199.6 MPa 43 44 11 TYPES OF LOADING AND STRESS RATIO From figure below, the highest stress occurs in the fillet Mean stress, σ m = (σ max + σ min ) 2 Alternating stress, σ a = (σ max − σ min ) 2 Stress ratio, R = σ min σ max Stress ratio, A = σ a σ m 45 46 Static Repeated and Reversed Load is applied slowly without shock and is held at constant load- Reversed: when a load-carrying component is subjected to value. certain level of tensile load followed by a same level of R = 1.0 because σmax = σmin compressive load. Repeated: when loading is repeated many thousand times. Also known as fatigue loading. 47 48 12 Fluctuating load non- Alternating loading with non-zero mean. Example: R.R Moore Fatigue test device 49 50 Fluctuating load Shock or Impact loading one- A special case of fluctuating stress is repeated, one-direction stress. Loads applied suddenly and rapidly. E.g. Hammer blow, Rock crushing. Random Loading Varying loads that are not regular in their amplitude. 51 52 13 CONCLUSION NORMAL STRESS DIRECT STRESS: STRESS DUE TENSION & THANK YOU TO BENDING COMPRESSION SHEAR STRESS SUPERPOSITION DIRECT VERTICAL PRINCIPLE SHEAR STRESS SHEARING STRESS TORSIONAL SHEAR STRESS 53 54 14

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posted: | 3/25/2011 |

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