Question 4 of 7: Displacement field of a deformed parallelepiped

Question 4: Displacement field of a deformed parallelepiped

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 displacement field vanishing on every coordinate plane, with one corner $E$ tracked through the deformation.

Find. (a) the strain state (and, with assumed elastic constants, the stress state) at $E$; (b) the normal strain along the front-face diagonal $EA$.

[Figure not reproduced: Parallelepiped as labelled in the exam figure: the axes meet at the hidden corner $O$, so $A=(1.5,0,0)$ lies on the $x$-axis and $E=(1.5,1.0,2.0)$ is the far corner. The dashed red line is the front-face diagonal $EA$ used in part (b). See the official exam paper.]

Approach. The deformed coordinates give the three constants component-by-component; differentiate the field for the strain tensor at $E$; then project that tensor onto the unit vector along $EA$.

1. Constants from the tracked corner. Since $u=c_1xyz$ acts only on $x$ (and likewise for $y,z$), each deformed component isolates one constant. With $x_Ey_Ez_E=3.0$, $$c_1=\frac{1.504-1.5}{3.0}=1.333\times10^{-3},\quad c_2=\frac{1.002-1.0}{3.0}=6.667\times10^{-4},\quad c_3=\frac{1.996-2.0}{3.0}=-1.333\times10^{-3}$$
2. Strain field by differentiation. $$\varepsilon_{xx}=\frac{\partial u}{\partial x}=c_1yz,\quad \varepsilon_{yy}=c_2xz,\quad \varepsilon_{zz}=c_3xy$$ $$\gamma_{xy}=c_1xz+c_2yz,\quad \gamma_{yz}=c_2xy+c_3xz,\quad \gamma_{zx}=c_3yz+c_1xy$$
3. Evaluate at $E=(1.5,1.0,2.0)$ ($yz=2,\ xz=3,\ xy=1.5$): $$\varepsilon_{xx}=2.667\times10^{-3},\quad \varepsilon_{yy}=2.000\times10^{-3},\quad \varepsilon_{zz}=-2.000\times10^{-3}$$ $$\gamma_{xy}=5.333\times10^{-3},\quad \gamma_{yz}=-3.000\times10^{-3},\quad \gamma_{zx}=-6.667\times10^{-4}$$ This is the complete state of strain at $E$ (part a).
4. State of stress at $E$. The problem supplies no elastic constants; using assumed structural steel ($E=200$ GPa, $\nu=0.30$, $\mu=76.9$ GPa, $\lambda=115.4$ GPa) in $\sigma_{ij}=\lambda\,\theta\,\delta_{ij}+2\mu\,\varepsilon_{ij}$ with dilatation $\theta=2.667\times10^{-3}$ gives, illustratively, $$\sigma_{xx}\approx718,\ \sigma_{yy}\approx615,\ \sigma_{zz}\approx0\ \text{MPa};\quad \tau_{xy}\approx410,\ \tau_{yz}\approx-231,\ \tau_{zx}\approx-51\ \text{MPa}$$
5. Normal strain along $EA$ (part b). In the exam figure the axes emanate from the hidden back corner $O$: the $x$-axis runs out through $A$, the $y$-axis through $C$ and the $z$-axis through $G$. Hence $A=(1.5,0,0)$, and $EA$ is the diagonal of the front face $x=1.5$, not a body diagonal. From $E$ to $A$ the vector is $(0,-1,-2)$, of length $\sqrt5$, so $\mathbf{n}=(0,-0.4472,-0.8944)$ with $n_y^2=0.2$, $n_z^2=0.8$, $n_yn_z=0.4$. Projecting the strain tensor, $$\varepsilon_{EA}=\varepsilon_{xx}n_x^2+\varepsilon_{yy}n_y^2+\varepsilon_{zz}n_z^2+\gamma_{xy}n_xn_y+\gamma_{yz}n_yn_z+\gamma_{zx}n_zn_x$$ Every term containing $n_x=0$ drops out, leaving $$\varepsilon_{EA}=0.2(2.000\times10^{-3})+0.8(-2.000\times10^{-3})+0.4(-3.000\times10^{-3})=4.0\times10^{-4}-1.6\times10^{-3}-1.2\times10^{-3}$$ $$\boxed{\varepsilon_{EA}=-2.40\times10^{-3}}$$ The diagonal $EA$ shortens: the $z$-contraction and the negative $\gamma_{yz}$ outweigh the $y$-extension. (Reading $A$ as the origin would make $EA$ the body diagonal, along which this field happens to give exactly zero strain. The figure rules that reading out.)