Gina Cavallo
MANE-6940 Studies in Mechanics of Solids
June 2014
1. Explain the meaning and the differences between the shear strain tensor components and
the engineering shear strains.
True Shear Strain is the change in the angle (radians) between 2 material line elements that start out
perpendicular to each other in the undeformed, initial state. The true shear strain is the change this
angle after deformation.
Engineering Shear Strain: The tangent of this angle. It is the length of the maximum deformation
divided by the perpendicular length in the plane of force application.
Shear Strain Tensor Component: Let α =
𝑑𝑢𝑦
𝑑𝑥
and β =
𝑑𝑢𝑥
𝑑𝑦
and ϒ = α + β
The shear strain tensor components can be expressed using ϒ by replacing εxy = ϒxy/2 , εyx =
ϒyx/2, etc.
http://en.wikipedia.org/wiki/Deformation_(mechanics)
2. Explain why the displacement vector components corresponding to real, unique solutions
of linear elastic problems must be such that the equations of compatibility must always be
satisfied.
It is necessary to solve the differential equations of force equilibrium for a 2-D elasticity problem. The
solution must satisfy the boundary conditions. These equations of equilibrium which contain only 3
stress components are not sufficient to determine these components, resulting in a statically
indeterminate problem. Therefore, the elastic deformation must also be considered. The equations of
equilibrium, the boundary conditions, and the compatibility equations provide a system of equations
that are sufficient to find the stress distribution in a 2-D problem.
Class Textbook: S. Timoshenko and J.N. Gooder, Theory of Elasticity, McGraw-Hill Book Company, Inc.
1951
The strain-displacement equations have 6 strain measures, but only 3 displacements, resulting in 6
unknowns for 3 independent variables.
http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/strain.cfm
3.- Consider the problem of simple twist of a long elastic bar of circular cross
section of unit radius. Under these conditions, to a good approximation, the only
stress components are the shear stresses acting on the plane the cross section.
Consider a stress function given by the expression ψ = k (x2 + y2 -1) with k = - G
α /2. Here α is the twist angle per unit length.
a) Plot the function for the case G=1e10 Pa and α= 0.01π .
b) Obtain the laplacian of the stress function.
c) Compute and plot the values of the stresses inside the bar.
See Maple Attachment