© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 1Poisson’s Ratio• For a slender bar subjected to axial loading:0===zyxxEσσσε• The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),0≠=zyεε• Poisson’s ratio is defined asxzxyεεεεν−=−==strain axialstrain lateral© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 2Generalized Hooke’s Law• For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires:1) strain is linearly related to stress2) deformations are smallEEEEEEEEEzyxzzyxyzyxxσνσνσενσσνσενσνσσε+−−=−+−=−−+=• With these restrictions:© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolfDilatation: Bulk Modulus• Relative to the unstressed state, the change in volume is()()()[][]() e)unit volumper in volume (change dilatation 21111111=++−=++=+++−=+++−=zyxzyxzyxzyxEeσσσνεεεεεεεεε• For element subjected to uniform hydrostatic pressure,()()modulusbulk 213213=−=−=−−=ννEkkpEpe• Subjected to uniform pressure, dilatation must be negative, therefore210 <<ν© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 4Shearing Strain• A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shearstrain is quantified in terms of the change in angle between the sides,()xyxyfγτ=• A plot of shear stress vs. shear strain is similar to the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains, zxzxyzyzxyxyGGGγτγτγτ===where G is the modulus of rigidity or shear modulus.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 5Example 2.10A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.SOLUTION:• Determine the average angular deformation or shearing strain of the block.• Use the definition of shearing stress to find the force P.• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 6• Determine the average angular deformation or shearing strain of the block.rad020.0in.2in.04.0tan ==≈xyxyxyγγγ• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.()()psi1800rad020.0psi10903=×==xyxyGγτ• Use the definition of shearing stress to find the force P.()()()lb1036in.5.2in.8psi18003×=== APxyτkips0.36=P© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 7Relation Among E, ν, and G• An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. ()ν+= 12GE• Components of normal and shear strain are related,• If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain.• An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 8Sample Problem 2.5A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses σx= 12 ksi and σz= 20 ksi. For E = 10x106psi and ν = 1/3, determine the change in: a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 9SOLUTION:• Apply the generalized Hooke’s Law to find the three components of normal strain. () ()in./in.10600.1in./in.10067.1in./in.10533.0ksi20310ksi12psi101013336−−−×+=+−−=×−=−+−=×+=⎥⎦⎤⎢⎣⎡−−×=−−+=EEEEEEEEEzyxzzyxyzyxxσνσνσενσσνσενσνσσε• Evaluate the deformation components.()()in.9in./in.10533.03−×+== dxABεδ()()in.9in./in.10600.13−×+== dzDCεδ()()in.75.0in./in.10067.13−×−== tytεδin.108.43−×+=ABδin.104.143−×+=DCδin.10800.03−×−=tδ• Find the change in volume()33333in75.0151510067.1/inin10067.1×××==∆×=++=−−eVVezyxεεε3in187.0+=∆V© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 10Composite Materials• Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polymers embedded in a resin matrix.zzzyyyxxxEEEεσεσεσ===• Normal stresses and strains are related by Hooke’s Law but with directionally dependent moduli of elasticity, xzxzxyxyεενεεν−=−=• Transverse contractions are related by directionally dependent values of Poisson’s ratio, e.g.,• Materials with directionally dependent mechanical properties are anisotropic.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf2 - 11Saint-Venant’s Principle• Loads transmitted through rigid plates result in uniform distribution of stress and strain.• Saint-Venant’s Principle:Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.• Stress and strain distributions become uniform at a relatively short distance from the load application points.• Concentrated