Copyright ©2014 Pearson Education, All Rights Reserved

Copyright ©2014 Pearson Education, All Rights Reserved
Chapter Objectives Navigate between rectilinear co-ordinate system for strain components Determine principal strains and maximum in-plane shear strain Determine the absolute maximum shear strain in 2D and 3D cases Know ways of measuring strains Define stress-strain relationship Predict failure of material Copyright ©2014 Pearson Education, All Rights Reserved

In-class Activities Reading Quiz Applications Equations of plane-strain transformation Principal and maximum in-plane shear strain Mohr’s circle for plane strain Absolute maximum shear strain Measurement of strains Stress-strain relationship Theories of failure Concept Quiz Copyright ©2014 Pearson Education, All Rights Reserved

READING QUIZ 1) Which of the following statement 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 ©2014 Pearson Education, All Rights Reserved

READING QUIZ (cont) 2) Which of the following failure criteria is not suitable for brittle materials? Maximum-normal-stress theory Mohr’s failure criterion Tresca yield criterion None of them Copyright ©2014 Pearson Education, All Rights Reserved

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 ©2014 Pearson Education, All Rights Reserved

EQUATIONS OF PLANE-STRAIN TRANSFORMATION (cont)
Note: Plane-stress case ≠ plane-strain case Copyright ©2014 Pearson Education, All Rights Reserved

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 ©2014 Pearson Education, All Rights Reserved

EXAMPLE 1 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 ©2014 Pearson Education, All Rights Reserved

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 ©2014 Pearson Education, All Rights Reserved

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 ©2014 Pearson Education, All Rights Reserved

EXAMPLE 2 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 ©2014 Pearson Education, All Rights Reserved

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 ©2014 Pearson Education, All Rights Reserved

MOHR’S CIRCLE FOR PLANE STRAIN (cont)
Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 3 The state of plane strain at a point is represented by the components: Determine the principal strains and the orientation of the element. Copyright ©2014 Pearson Education, All Rights Reserved

ABSOLUTE MAXIMUM SHEAR STRAIN
State of strain in 3-dimensional space: Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 The state of plane strain at a point is represented by the components: Determine the maximum in-plane shear strain and the absolute maximum shear strain. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 (cont) Solutions From the strain components, the centre of the circle is on the ε axis at Since , the reference point has coordinates Thus the radius of the circle is Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 (cont) Solutions Computing the in-plane principal strains, we have From the circle, the maximum in-plane shear strain is From the above results, we have Thus the Mohr’s circle is as follow, Copyright ©2014 Pearson Education, All Rights Reserved

MEASUREMENT OF STRAINS BY STRAIN ROSETTES
Ways of arranging 3 electrical-resistance strain gauges In general case (a): Copyright ©2014 Pearson Education, All Rights Reserved

MEASUREMENT OF STRAINS BY STRAIN ROSETTES (cont)
In 45° strain rosette [case (b)], In 60° strain rosette [case (c)], Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 5 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 ©2014 Pearson Education, All Rights Reserved

EXAMPLE 5 (cont) Solutions 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 ©2014 Pearson Education, All Rights Reserved

STRESS-STRAIN RELATIONSHIP
Use the principle of superposition Use Poisson’s ratio, Use Hooke’s Law (as it applies in the uniaxial direction), Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6 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 ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6 (cont) Solutions 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 ©2014 Pearson Education, All Rights Reserved

Maximum-normal-stress theory (for materials having equal strength in tension and compression) Maximum principle stress σ1 in the material reaches a limiting value that is equal to the ultimate normal stress the material can sustain when it is subjected to simple tension. Copyright ©2014 Pearson Education, All Rights Reserved

Mohr’s failure criterion (for materials having different strength in tension and compression Perform 3 tests on the material to obtain the failure envelope Circle A represents compression test results σ1 = σ2 = 0, σ3 = – (σult)c Circle B represents tensile test results, σ1 = (σult)t, σ2 = σ3 = 0 Circle C represents pure torsion test results, reaching the τult. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 7 The solid shaft has a radius of 0.5 cm and is made of steel having a yield stress of σY = 360 MPa. Determine if the loadings cause the shaft to fail according to the maximum-shear-stress theory and the maximum-distortion-energy theory. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 7 (cont) Solutions Since maximum shear stress caused by the torque, we have Principal stresses can also be obtained using the stress-transformation equations, Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 7 (cont) Solutions Since the principal stresses have opposite signs, the absolute maximum shear stress will occur in the plane, Thus, shear failure of the material will occur according to this theory. Using maximum-distortion-energy theory, Using this theory, failure will not occur. Copyright ©2014 Pearson Education, All Rights Reserved

CONCEPT QUIZ 1) Which of the following statement is incorrect? Dilatation is caused only by normal strain, not shear strain. When Poisson’s ratio approaches 0.5, the bulk modulus tends to infinity and the material behaves like incompressible. Von Mises failure criterion is not suitable for ductile material. Mohr’s failure criterion is not suitable for brittle material having different strength in tension and compression. Copyright ©2014 Pearson Education, All Rights Reserved

Copyright ©2014 Pearson Education, All Rights Reserved

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