3 032 Problem Set 4 Fall 2006 Due Start of lecture 10 20 06 1 Strain can be coupled not only to stress but also to thermal expansion or contraction this is known as thermoelastic coupling The general constitutive equation is i j 1E i j T i j where is the material s thermal expansion coe cient and T is the E kk i j change in temperature For metals is on the order of 10 ppm C Figure 1 Bar of original length L in contact with two rigid walls a Consider the case of a long rod of length L that is in contact with two rigid walls and subject to a temperature increase T Figure 1 We are interested in the lateral strains that is the strains in the 2 and 3 directions Assumption the walls do not limit strain in these directions Is the 2 3 plane in a state of plane stress or plane strain Derive the value of 22 33 as a function of T and the elastic constants b Find the change in volume V of the bar assuming small strains i e neglecting terms of order 2 and smaller Compare this change in volume to the change in volume if the bar were not constrained c What is the change in volume V if the Poisson s ratio of the constrained bar is zero If is 0 5 d What is the nal diameter of a titanium rod of length 10 cm and original diameter 4 mm after a temperature increase of 250 C Use reasonable material properties for titanium Would you expect the rod to buckle due to this temperature increase e Physically what would you do in this experiment to ensure that the assumption in a is met 1 2 Since there are only two independent elastic constants associated with isotropic materials the shear modulus G is related to the Young s modulus E and the Poisson s ratio Your goal is to derive this relationship a Consider a square region in a state of pure shear Figure 2 a We expect the region to deform as shown in Figure 2 b The shear modulus G is de ned as the ratio of shear stress to the angular deformation assuming that 1 By what name is the variable known Figure 2 a Square region under a stress state of pure shear b Deformed shape b Calculate the post deformation distance between points A and B or AB in terms of and Feel free to use the small angle approximations sin and cos 1 c Transform the stress state into the principal stress state by a method of your choosing Plot the magnitudes and directions of the principal stresses on the original square region Figure 2 De ne a new set of axes i and j corresponding to the directions of the principal stresses d Express the strain i j and also the post deformation distance between points A and B AB as a function of the principal stresses By equating your two expressions for AB express G in terms of E and 3 Listed below are force vs strain data from tensile tests on one elemental polycrystalline metal one polycrystalline metallic alloy and one ceramic Force N 0 500 1000 1500 2000 A 0 0 00019 0 00038 0 00050 fracture 2 B C 0 0 0 00017 0 00006 0 00035 0 00013 0 00054 0 00016 0 00072 0 00025 a Graph these data and nd the Young s elastic modulus E of each material assuming an initial length of 16 cm and an initial cross sectional area of 0 4 cm2 for all samples b Which sample A B or C is likely to be the ceramic Why c What is a possible composition of each metallic sample 4 Materials single crystals of metals and ceramics composites and even types of polymers and proteins have symmetry that reduces the number of independent values in fourth rank tensors that modulate second rank tensors You will prove this and consider the number of independent elastic constants in the context of elastic constants for cubic and isotropic materials a i j is the Kronecker delta and 1 i j or 0 i j We can write this as a matrix not a tensor such that i j 1 0 0 0 1 0 0 0 1 where ai j i j and ai j is the direction cosine matrix as always We can use this fact to write down some pretty obvious truths e g i j jl il 1 because of course 11 1 11 0 12 0 13 11 For this reason the matrix i j is also called the substitution matrix Fascinating Use this fact to prove something useful about the sti ness tensor Ci jkl Ci jkl im jn ko lpCmnop 2 b This means that a component of the sti ness and compliance tensors is unchanged in its value if the reference axes are rotated about a center of symmetry of the material For example a cubic materials has symmetry in the 100 010 and 001 directions such that in the contracted two su x notation of the fourth rank tensor Ci jkl C11 C22 C33 C12 C23 C31 C44 C55 C66 3 To reduce from cubic symmetry 3 independent elastic constants to isotropic symmetry 2 independent elastic constants we can consider a rotation of the reference axes in a cubic crystal under uniaxial tensile strain 11 11 all other strains 0 Express the normal stresses i j in terms of i j and Ci jkl and then rewrite this expressing C in the contracted two su x notation c Consider a 45o CCW in plane rotation from this old axis set 1 2 3 to a new axis set 1 2 3 about the 3 axis Express the direction cosine matrix of this transformation ai j 3 d Express the shear strain component 12 in terms of ai j and the old coordinate system strains e Now express the shear stress 12 in terms of the old coordinate system stresses and then in terms of the old axial strains that de ne those stresses f From d and e and the fact that you have proved that certain Ci j have the same value independent of rotation for materials of cubic symmetry equate 12 with 12 in terms of C11 and C12 g Now for an isotropic material you know that shear stress is proportional to engineer ing and tensorial shear strain via the shear modulus Equate 12 with 12 in terms of C44 on this basis h Finally compare f and g to prove that three independent constants in a cubic material reduces to two independent constants for an isotropic material and give the mathemat ical relationship among these three independent elastic constants C11 C12 and C44 for this isotropic approximation 5 Most metals are elastically anisotropic meaning that the measured Young s elastic modulus Ei jk depends on the direction with respect to the crystal structure along which load is applied This can be expressed in terms of the compliance matrix …