Question 2.2: An infinitesimal rectangle at a point in a reference state o...
An infinitesimal rectangle at a point in a reference state of a material becomes the parallelogram shown in a deformed state (Figure 2.31). Determine (a) the extensional strain in the dL_1 direction; (b) the extensional strain in the dL_2 direction; and (c) the shear strain corresponding to the dL_1 and dL_2 directions.
Given: Reference and deformed geometries of infinitesimal rectangle.
Find: Normal and shear components of strain.
Assume: Strain definitions are adequate; use of “true” strain integral is unnecessary.
Normal strain in dL_1 direction:
\varepsilon_{_{1}} = \frac{dL_1^{^{′}}-dL_1 }{dL_1} = \frac{1.2 dL_1-dL_1 }{dL_1}=0.2.Normal strain in dL_2 direction:
\varepsilon_{_{2}} = \frac{dL_2^{^{′}} -dL_2 }{dL_2}= \frac{1.3 dL_2-dL_2 }{dL_2}=0.3.Shear strain is the angular deformation, or change in angle between two reference lines. In reference state, the angle between dL_1 and dL_2 is 90 ̊, or π/2. In the deformed state, the angle between dL_1^{^{′}} and dL_2^{^{′}} is 60°, or π/3. The shear strain is thus
\gamma_{_{12}}=\frac{\pi }{6}= 0.524 radians.
Note: If we tried to approximate shear strain by the tangent of this angular deformation instead of using the angle itself, we would get
\gamma =\frac{1.3 dL_2 \sin \pi /6}{dL_2}= 0.650 radians.
This is close, but not that close, to 0.524 radians. The angular change in this problem is not sufficiently small to justify the use of the tangent in place of the angle itself.