Copyright © 2011 Pearson Education South Asia Pte Ltd

Name: Copyright © 2011 Pearson Education South Asia Pte Ltd
Uploaded: 2017-07-11T12:27:24+00:00
Duration: PTM31S48
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Description: Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To develop the transformation of stress components from one orientation of the coordinate system to another orientation. To determine the principal stress and maximum in-plane shear stress at a point. To develop the transformation of plane strain components from one orientation of the coordinate system to another orientation. Copyright © 2011 Pearson Education South Asia Pte Ltd

Chapter Objectives (cont)
To determine the principal strain and maximum in-plane shear strain at a point. To develop Mohr’s circle for analyzing stress and strain components. To discuss strain rosettes for strain components. To present the relationships between material properties, such as the elastic modulus, shear modulus, and Poisson’s ratio. Copyright © 2011 Pearson Education South Asia Pte Ltd

In-class Activities Reading Quiz Applications General equations of plane-stress transformation Principal stresses and maximum in-plane shear stress Mohr’s circle for plane stress Absolute maximum shear stress Equations of plane-strain transformation Principal and maximum in-plane shear strain Mohr’s circle for plane strain Measurement of strains Stress-strain relationship Concept Quiz Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ 1) Which of one the following statements is incorrect? The principal stresses represent the maximum and minimum normal stress at the point When the state of stress is represented by the principal stresses, no shear stress will act on the element When the state of stress is represented in terms of the maximum in-plane shear stress, no normal stress will act on the element For the state of stress at a point, the maximum in-plane shear stress usually associated with the average normal stress. Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ 2) Which one of the following statements is incorrect for plane-strain? σz = γyz = γxz = 0 while the plane-strain has 3 components σx, σy and γxy. Always identical to the state of plane stress Identical to the state of plane stress only when the Poisson’s ratio is zero. When the state of strain is represented by the principal strains, no shear strain will act on the element. Copyright © 2011 Pearson Education South Asia Pte Ltd

GENERAL EQUATIONS OF PLANE-STRESS TRANSFORMATION
The state of plane stress at a point is uniquely represented by three components acting on an element that has a specific orientation at the point. Sign Convention: Positive normal stress acts outward from all faces Positive shear stress acts upwards on the right-hand face of the element Copyright © 2011 Pearson Education South Asia Pte Ltd

GENERAL EQUATIONS OF PLANE-STRESS TRANSFORMATION (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd

Sign Convention (continued) Both the x-y and x’-y’ system follow the right-hand rule The orientation of an inclined plane (on which the normal and shear stress components are to be determined) will be defined using the angle θ. The angle θ is measured from the positive x to the positive x’-axis. It is positive if it follows the curl of the right-hand fingers. Copyright © 2011 Pearson Education South Asia Pte Ltd

+ΣFx’ = 0; σx’ ∆A – (τxy ∆A sin θ) cos θ – (σy ∆A sin θ) sin θ – ( τxy ∆A cos θ) sin θ – (σx ∆A cos θ) cos θ = 0 σx’ = σx cos2 θ + σy sin2 θ + τxy (2 sin θ cos θ) +ΣFy’ = 0; τx’y’ ∆A + (τxy ∆A sin θ) sin θ – (σy ∆A sin θ) cos θ – ( τxy ∆A cos θ) cos θ + (σx ∆A cos θ) sin θ = 0 τx’y’ = (σy – σx) sin θ cos θ + τxy (cos2 θ – sin2 θ) σx’ = σx + σy 2 σx – σy cos 2θ + τxy sin 2 θ + τx’y’ = – σx + σy 2 sin 2θ + τxy cos 2 θ σy’ = σx + σy 2 σx – σy cos 2θ – τxy sin 2 θ – Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 The state of plane stress at a point on the surface of the airplane fuselage is represented on the element oriented as shown in Fig. 9–4a. Represent the state of stress at the point on an element that is oriented 30° clockwise from the position shown. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 (cont) Solution Repeat the procedure to obtain the stress on the perpendicular plane b–b. The state of stress at the point can be represented by choosing an element oriented. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 The state of plane stress at a point is represented by the element shown in Fig. 9–7a. Determine the state of stress at the point on another element oriented 30° clockwise from the position shown. Copyright © 2011 Pearson Education South Asia Pte Ltd

IN-PLANE PRINCIPAL STRESS
The principal stresses represent the maximum and minimum normal stress at the point. When the state of stress is represented by the principal stresses, no shear stress will act on the element. Solving this equation leads to θ = θp Copyright © 2011 Pearson Education South Asia Pte Ltd

IN-PLANE PRINCIPAL STRESS (cont)

IN-PLANE PRINCIPAL STRESS (cont)
Solving this equation leads to θ = θp; i.e Copyright © 2011 Pearson Education South Asia Pte Ltd

MAXIMUM IN-PLANE PRINCIPAL STRESS
The state of stress can also be represented in terms of the maximum in-plane shear stress. In this case, an average stress will also act on the element. Solving this equation leads to θ = θs; i.e And there is a normal stress on the plane of maximum in-plane shear stress Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 When the torsional loading T is applied to the bar in Fig. 9–13a, it produces a state of pure shear stress in the material. Determine (a) the maximum in-plane shear stress and the associated average normal stress, and (b) the principal stress. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 4 When the axial loading P is applied to the bar in Fig. 9–14a, it produces a tensile stress in the material. Determine (a) the principal stress and (b) the maximum in-plane shear stress and associated average normal stress. Copyright © 2011 Pearson Education South Asia Pte Ltd

MOHR’S CIRCLE OF PLANE STRESS
Please refer to the website for the animation: Mohr’s Circle A geometrical representation of equations 9.1 and 9.2; i.e. Sign Convention: σ is positive to the right, and τ is positive downward. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 5 Due to the applied loading, the element at point A on the solid shaft in Fig. 9–18a is subjected to the state of stress shown. Determine the principal stresses acting at this point. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 5 (cont) Solution Construction of the Circle From Fig. 9–18a, The center of the circle is at The reference point A(-12,-6) and the center C(-6, 0) are plotted in Fig. 9–18b.The circle is constructed having a radius of Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 5 (cont) Solution Principal Stress The principal stresses are indicated by the coordinates of points B and D. The orientation of the element can be determined by calculating the angle Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 6 (cont) Solution Construction of the Circle We first construct of the circle, The center of the circle C is on the axis at From point C and the A(-20, 60) are plotted, we have Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 7 The state of plane stress at a point is shown on the element in Fig. 9–20a. Represent this state of stress on an element oriented 30°counterclockwise from the position shown. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 7 (cont) Solution Construction of the Circle We first construct of the circle, The center of the circle C is on the axis at From point C and the A(-8, -6) are plotted, we have Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 7 (cont) Solution Stresses on 30° Element From the geometry of the circle, The stress components acting on the adjacent face DE of the element, which is 60° clockwise from the positive x axis, Fig. 9–20c, are represented by the coordinates of point Q on the circle. Copyright © 2011 Pearson Education South Asia Pte Ltd

ABSOLUTE MAXIMUM SHEAR STRESS
State of stress in 3-dimensional space: Copyright © 2011 Pearson Education South Asia Pte Ltd

ABSOLUTE MAXIMUM SHEAR STRESS (cont)
State of stress in 3-dimensional space: Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 8 The point on the surface of the cylindrical pressure vessel in Fig. 9–24a is subjected to the state of plane stress. Determine the absolute maximum shear stress at this point. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 8 (cont) Solution An orientation of an element 45° within this plane yields the state of absolute maximum shear stress and the associated average normal stress, namely, Same result for can be obtained from direct application of Copyright © 2011 Pearson Education South Asia Pte Ltd

EQUATIONS OF PLANE-STRAIN TRANSFORMATION
Please refer to the website for the animation: Strain Transformation EQUATIONS OF PLANE-STRAIN TRANSFORMATION In 3D, the general state of strain at a point is represented by a combination of 3 components of normal strain σx, σy, σz, and 3 components of shear strain γxy, γyz, γxz. In plane-strain cases, σz, γxz and γyz are zero. The state of plane strain at a point is uniquely represented by 3 components (σx, σy and γxy) acting on an element that has a specific orientation at the point. Copyright © 2011 Pearson Education South Asia Pte Ltd

EQUATIONS OF PLANE-STRAIN TRANSFORMATION (cont)
Note: Plane-stress case ≠ plane-strain case Copyright © 2011 Pearson Education South Asia Pte Ltd

Positive normal strain σx and σy cause elongation Positive shear strain γxy causes small angle AOB Both the x-y and x’-y’ system follow the right-hand rule The orientation of an inclined plane (on which the normal and shear strain components are to be determined) will be defined using the angle θ. The angle is measured from the positive x- to positive x’-axis. It is positive if it follows the curl of the right-hand fingers. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 9 A differential element of material at a point is subjected to a state of plane strain which tends to distort the element as shown in Fig. 10–5a. Determine the equivalent strains acting on an element of the material oriented at the point, clockwise 30° from the original position. Copyright © 2011 Pearson Education South Asia Pte Ltd

PRINCIPLE AND MAXIMUM IN-PLANE SHEAR STRAIN
Similar to the deviations for principal stresses and the maximum in-plane shear stress, we have And, Copyright © 2011 Pearson Education South Asia Pte Ltd

PRINCIPLE AND MAXIMUM IN-PLANE SHEAR STRAIN (cont)
When the state of strain is represented by the principal strains, no shear strain will act on the element. The state of strain at a point can also be represented in terms of the maximum in-plane shear strain. In this case an average normal strain will also act on the element. The element representing the maximum in-plane shear strain and its associated average normal strain is 45° from the element representing the principal strains. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 10 A differential element of material at a point is subjected to a state of plane strain defined by which tends to distort the element as shown in Fig. 10–7a. Determine the maximum in-plane shear strain at the point and the associated orientation of the element. Copyright © 2011 Pearson Education South Asia Pte Ltd

MOHR’S CIRCLE FOR PLANE STRAIN
A geometrical representation of Equations 10-5 and 10-6; i.e., Sign convention: ε is positive to the right, and γ/2 is positive downwards. Copyright © 2011 Pearson Education South Asia Pte Ltd

MOHR’S CIRCLE FOR PLANE STRAIN (cont)

EXAMPLE 11 The state of plane strain at a point is represented by the components: Determine the principal strains and the orientation of the element. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 12 The state of strain at point A on the bracket in Fig. 10–17a is measured using the strain rosette shown in Fig. 10–17b. Due to the loadings, the readings from the gauges give Determine the in-plane principal strains at the point and the directions in which they act. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 12 (cont) Solution Measuring the angles counter-clockwise, By substituting the values into the 3 strain-transformation equations, we have Using Mohr’s circle, we have A(60(10-6), 60(10-6)) and center C (153(10-6), 0). Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 13 The copper bar in Fig. 10–24 is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, b = 500 mm, and t = 20 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 13 (cont) Solution From the loading we have The associated normal strains are determined from the generalized Hooke’s law, The new bar length, width, and thickness are therefore Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ 1) Which one of the following statements is untrue? In 2-D state of stress, the orientation of the element representing the maximum in-plane shear stress can be obtained by rotating the element 45° from the element representing the principle stresses. In 3-D state of stress, the orientation of the element representing the absolute maximum shear stress can be obtained by rotating the element 45° about the axis defining the direction of σint. If the in-plane principal stresses are of opposite sign, then the absolute maximum shear stress equals the maximum in-plane stress, that is, τabs max = (σmax – σmin)/2 Same as (c) but the principal stresses are of the same sign. Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

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