Displacement Field of a Parallelepiped: 25-Nav-A6 Advanced Strength of Materials (25-Mec-A6), May 2017, Q2

Question 2: Displacement Field of a Parallelepiped (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. A rectangular block with edges $1.0$ m ($x$), $1.5$ m ($y$), $2.0$ m ($z$); the free corner $A=(1.0,1.5,2.0)$ displaces to $A'=(0.9958,1.4980,1.9992)$ under the field $u=c_1xyz,\ v=c_2xyz,\ w=c_3xyz$.

Find. The six strain components at A, the normal strain along AB, and the change in the right angle between AB and AC.

Parallelepiped with corner A at (1.0, 1.5, 2.0) displacing to A′. B is the bottom corner along the front-face diagonal from A; C is the adjacent top corner along the $-x$ edge. (The extraction’s note placing A at the origin is a mis-read: the field $u=c_1xyz$ vanishes at the origin, so the tracked corner must be the far corner.)

Approach. Back out $c_1,c_2,c_3$ from the known displacement of A, differentiate the field for the strain components, then project the small-strain tensor onto the unit vectors of AB and AC.

Recover the constants. The displacement of A is $\boldsymbol{\delta}=A'-A=(-0.0042,-0.0020,-0.0008)$ m and $(xyz)_A=1.0\cdot1.5\cdot2.0=3.0$, so $c_1=-0.0042/3=-1.400\times10^{-3}$, $c_2=-0.6667\times10^{-3}$, $c_3=-0.2667\times10^{-3}$.
Strain–displacement, evaluated at A. $\varepsilon_x=\partial u/\partial x=c_1yz$, $\varepsilon_y=c_2xz$, $\varepsilon_z=c_3xy$, $\gamma_{xy}=c_1xz+c_2yz$, $\gamma_{xz}=c_1xy+c_3yz$, $\gamma_{yz}=c_2xy+c_3xz$. At $A(1.0,1.5,2.0)$: $$\boxed{\varepsilon_x=-4200\,\mu,\ \varepsilon_y=-1333\,\mu,\ \varepsilon_z=-400\,\mu,\ \gamma_{xy}=-4800\,\mu,\ \gamma_{xz}=-2900\,\mu,\ \gamma_{yz}=-1533\,\mu}$$
Direction vectors. $B=(1.0,0,0)$ and $C=(0,1.5,2.0)$, so $\hat n_{AB}=(0,-0.6,-0.8)$ and $\hat n_{AC}=(-1,0,0)$; their dot product is $0$, confirming AB $\perp$ AC.
Normal strain along AB. $\varepsilon_{AB}=\varepsilon_y n_y^2+\varepsilon_z n_z^2+\gamma_{yz}n_yn_z=0.36\varepsilon_y+0.64\varepsilon_z+0.48\gamma_{yz}=\boxed{-1472\,\mu}$.
Shear strain between AB and AC. $\gamma_{AB,AC}=2\,\hat n_{AB}^{\mathsf T}[\varepsilon]\hat n_{AC}=0.6\,\gamma_{xy}+0.8\,\gamma_{xz}=\boxed{-5200\,\mu\ \text{rad}}$ (the initially right angle opens by $5.2\times10^{-3}$ rad).

Check: The field is position-dependent, so the strains are reported at the tracked corner A. AB is the front-face diagonal and AC the top $-x$ edge; both emanate from A and are mutually perpendicular in the undeformed body, so part (c) is well posed.

Final Results
Quantity	Value
ε_x, ε_y, ε_z	−4200, −1333, −400 µ
γ_xy, γ_xz, γ_yz	−4800, −2900, −1533 µ
ε along AB	−1472 µ
γ between AB and AC	−5200 µ rad

Topic: Strain from a prescribed displacement field; strain transformation in 3-D.

Key relations: Small-strain relations $\varepsilon_{ij}=\tfrac12(u_{i,j}+u_{j,i})$; directional strain $\varepsilon_n=\hat n^{\mathsf T}[\varepsilon]\hat n$; angle change $\gamma_{mn}=2\hat m^{\mathsf T}[\varepsilon]\hat n$.

Why this works: Differentiating the displacement field gives the strain tensor everywhere; the single measured corner fixes the three constants. Because normal and shear strains transform as a second-rank tensor, any directional or between-line strain follows by contracting that tensor with the relevant unit vectors—no re-derivation from geometry is needed.

Common pitfall: Forgetting the factor of 2 in the shear-between-lines formula, or evaluating strains at the wrong point because the field is not uniform.

Source: Ugural & Fenster, Advanced Mechanics of Materials and Applied Elasticity, ch. 2 (strain, transformation of strain).

Question 2 of 8: Displacement Field of a Parallelepiped

Question 2: Displacement Field of a Parallelepiped (20 marks)