Strain State from a Prescribed Displacement Field: 22-Mec-A7 Advanced Strength of Materials, May 2015, Q4

Question 4: Strain State from a Prescribed Displacement Field (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Corner A at (1.0, 1.5, 2.0) m moves to A′(0.9955, 1.4982, 1.9997) m under $u=c_1xyz$, $v=c_2xyz$, $w=c_3xyz$.

Find. (a) the six strain components at A; (b) the normal strain along AB; (c) the shear strain between the perpendicular lines AB and AC.

Block with corner A (1,1.5,2); AB is the front-face diagonal to B(1,0,0) and AC is the top edge to C(0,1.5,2); AB ⊥ AC.

Approach. Fit the three constants from A’s displacement, differentiate the field for the small-strain components, then use the strain-transformation quadratic forms with the unit vectors of AB and AC.

Constants of the field. The displacement of A is $(-0.0045,-0.0018,-0.0003)$ m and $xyz=3.0$ at A, so $$c_1=-1.5\times10^{-3},\quad c_2=-0.6\times10^{-3},\quad c_3=-0.1\times10^{-3}.$$
(a) Strain components. Differentiating and evaluating at A(1, 1.5, 2), $$\varepsilon_x=c_1yz,\ \varepsilon_y=c_2xz,\ \varepsilon_z=c_3xy,\quad \gamma_{xy}=c_1xz+c_2yz,\ \gamma_{xz}=c_1xy+c_3yz,\ \gamma_{yz}=c_2xy+c_3xz,$$ giving $\varepsilon_x=-4500\mu$, $\varepsilon_y=-1200\mu$, $\varepsilon_z=-150\mu$, $\gamma_{xy}=-4800\mu$, $\gamma_{xz}=-2550\mu$, $\gamma_{yz}=-1100\mu$.
(b) Normal strain along AB. With $\mathbf{n}_{AB}=(0,-0.6,-0.8)$, $$\varepsilon_{AB}=\varepsilon_y n_y^2+\varepsilon_z n_z^2+\gamma_{yz}n_yn_z=\boxed{-1056\ \mu\varepsilon}.$$
(c) Shear strain between AB and AC. With $\mathbf{n}_{AC}=(-1,0,0)$ (and $\mathbf{n}_{AB}\cdot\mathbf{n}_{AC}=0$), $$\gamma_{AB,AC}=2\,\mathbf{n}_{AB}^{\mathsf T}[\varepsilon]\,\mathbf{n}_{AC}=\boxed{-4920\ \mu\varepsilon}.$$ The originally-right angle between AB and AC opens (positive relative rotation would close it), by 4920 microradians.

Final results — Question 4
Quantity	Value (με)
ε_x, ε_y, ε_z	−4500, −1200, −150
γ_xy, γ_xz, γ_yz	−4800, −2550, −1100
Normal strain along AB	−1056
Shear strain AB–AC	−4920

Topic: Strain–displacement relations and 3-D strain transformation.

Key relations: $\varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})$; directional normal strain $\varepsilon_n=\mathbf{n}^{\mathsf T}[\varepsilon]\mathbf{n}$; shear between $\mathbf n_1,\mathbf n_2$ is $2\mathbf n_1^{\mathsf T}[\varepsilon]\mathbf n_2$.

Why this works: Small-strain components are simply the symmetric part of the displacement gradient, evaluated at the point of interest. Once the strain tensor is known, the normal strain in any direction is the quadratic form of the tensor with the unit vector, and the change in angle between two directions is twice the corresponding off-diagonal form—exactly the tensor generalisation of Mohr’s circle.

Common pitfall: using engineering shear $\gamma$ where tensor shear $\gamma/2$ belongs (the transformation uses the tensor matrix), and mislabelling the block corners so AB/AC directions are wrong.

Source: Boresi & Schmidt, Advanced Mechanics of Materials, Ch. 2 (strain); Ugural & Fenster §2.

Question 4 of 8: Strain State from a Prescribed Displacement Field

Question 4: Strain State from a Prescribed Displacement Field (20 marks)