For a given homogeneous, isotropic, linearly elastic material, E = 15e6 psi and v = 0.3. Solve for the shearmodulus.2.1.1 Homogeneous, isotropic, linearly elastic materialsFor specimens undergoing small deformations, the stress-strain diagram often ex-hibits a linear behavior. Although this is a very crude approximation to the behaviorof actual materials, it is a convenient assumption that is often used for preliminaryevaluation. A linear relationship between stress and strain can be expressed as01 = E €1,(2.1)where the coefficient of proportionality, E, is called Young's modulus or modulus ofelasticity. Since strains are non-dimensional quantities, this coefficient has the sameunits as stress quantities, i.e., Pa. This linear relationship is known as Hooke's law.The elongation of a bar in the direction of the applied stress is accompanied by alateral contraction that is also proportional to the applied stress. The resulting defor-mations for this uniaxial state of stress can therefore be described by the followingstrains(2.2)01, €3Eº1₁where is called Poisson's ratio and is a non-dimensional constant.If a stress component, 72, is applied to the same material, similar deformationswill result€1 =E 1, 2V1€₁=-=02₁ €2= 0₂₁ €3=---1/20 02.E(2.3)Note that the assumption of material isotropy implies identical values of Young'smodulus and Poisson's ratio in eq. (2.2) and (2.3). Similar relationships hold for anapplied stress, 73.

Modulus of lasticity of material, E=15×106 psiPoisson's ratio, ν=0.3

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For a given homogeneous, isotropic, linearly elastic material, E = 15e6 psi and v=0.3. Solve for the shear modulus.

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Concept explainers

Strain energy

The term strain indicates the variation in an object's length, area, and volume when the object is subjected to an external force. The external force may be of normal, shear, moment, torque, etc. The strain in an object depends on the types of external force and cross-section of the object. Mathematically, the evaluation of strain can be obtained with the help of elastic constants.

Question

For a given homogeneous, isotropic, linearly elastic material, E = 15e6 psi and v = 0.3. Solve for the shear
modulus.
2.1.1 Homogeneous, isotropic, linearly elastic materials
For specimens undergoing small deformations, the stress-strain diagram often ex-
hibits a linear behavior. Although this is a very crude approximation to the behavior
of actual materials, it is a convenient assumption that is often used for preliminary
evaluation. A linear relationship between stress and strain can be expressed as
01 = E €1,
(2.1)
where the coefficient of proportionality, E, is called Young's modulus or modulus of
elasticity. Since strains are non-dimensional quantities, this coefficient has the same
units as stress quantities, i.e., Pa. This linear relationship is known as Hooke's law.
The elongation of a bar in the direction of the applied stress is accompanied by a
lateral contraction that is also proportional to the applied stress. The resulting defor-
mations for this uniaxial state of stress can therefore be described by the following
strains
(2.2)
01, €3
Eº1₁
where is called Poisson's ratio and is a non-dimensional constant.
If a stress component, 72, is applied to the same material, similar deformations
will result
€1 =
E 1, 2
V
1
€₁=-=02₁ €2= 0₂₁ €3=-
--1/20 02.
E
(2.3)
Note that the assumption of material isotropy implies identical values of Young's
modulus and Poisson's ratio in eq. (2.2) and (2.3). Similar relationships hold for an
applied stress, 73.

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