Matrix/Vector Calculus I studied civil engineering almost 20 years ago and forget some knowledge of maths. Hope I could get some help here...
Here are some pictures: https://onedrive.live.com/?cid=5c08c07452173dea&id=5C08C07452173DEA!4696
Q1: Picture 1 shows the two ends/nodes (A & B) of a bar move from coordinates $X_Ao$ and $X_Bo$ (i.e. 2 ends) to $X_A$ and $X_B$ and $u_A$ & $u_A$ are displacement vectors at nodes/ends A and B respectively. $x_o$ & $x$ are the length of the bars before and after deformation/nodal movement. Equation (2.13) on Picture 3 shows $\epsilon_G = \frac{1}{l_o^2}(x_o^Tu + 0.5u^Tu)$. I try putting equations (2.11) and (2.12) into $\frac{l^2 - l_o^2}{2 l_o^2}$ and don't have $\epsilon_G = \frac{1}{l_o^2}(x_o^Tu + 0.5u^Tu)$
Q2: Equation (2.15)/sheet 4 shows $\delta\epsilon_G = \frac{1}{l_o^2}(x_O+u)^T\delta u $.
However, from equation (2.13), $\delta\epsilon_G = \frac{1}{l_o^2} [X_o^T\frac{\partial u}{\partial u}\delta u  + 0.5(\frac{\partial u^T}{\partial u}\delta u . u + u^T\frac{\partial u}{\partial u}\delta u)]  $. I believe there must be something wrong but I don't know where....
Q3: Equation (2.53)/sheet 6 includes $\frac{\partial \epsilon}{\partial u_A^T}$. In equation (2.9), $u = u_B - u_A$.So, do I have to use equation (2.10) in doing this differentiation (wrt $u_A$). Also, what is $\frac{\partial u}{\partial u_A^T}$?
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Hi COTO,
Thanks for your answert and help. Would you mind shedding light again on the following questions:
Q1: You mentioned $x^T_0 u = u^T x_0 = x_0⋅u$ because $X_o u$ is a scalar. I don't know whether I learned this property at university. Where can I get more information about it?
Q2: In your answer, you show $\frac{\partial u^T u}{\partial u} = 2 u^T$. Do you assume that $u^T u$ is a scalar and $u^T u = u^T u^T$?
 A: Q1
Note that the symbols ${\mathbf x_{0}}$ and ${\mathbf x}$ represent the initial and final bar vectors, not the bar lengths, which would be $\sqrt{\mathbf x_{0}^{\rm T}\mathbf x_{0}}$ and $\sqrt{\mathbf x^{\rm T}\mathbf x}$, respectively.
Plugging the equations $$l_0^2 = \mathbf x_{0}^{\rm T}\mathbf x_{0}$$ $$l^2 = \left( \mathbf x_{0} + \mathbf u\right)^T\left( \mathbf x_{0} + \mathbf u\right)$$ into $$\varepsilon _G = \frac{l^2 - l_0^2}{2l_0^2}$$ we get
$$\begin{array}{*{20}{l}}
  {{\varepsilon _G}}&{ = \tfrac{1}{{2l_0^2}}\left[ {{l^2} - l_0^2} \right]} \\ 
  {}&{ = \tfrac{1}{{2l_0^2}}\left[ {{{\left( {{{\mathbf{x}}_0} + {\mathbf{u}}} \right)}^T}\left( {{{\mathbf{x}}_0} + {\mathbf{u}}} \right) - {\mathbf{x}}_0^T{{\mathbf{x}}_0}} \right]} \\ 
  {}&{ = \tfrac{1}{{2l_0^2}}\left[ {{\mathbf{x}}_0^T{{\mathbf{x}}_0} + {\mathbf{x}}_0^T{\mathbf{u}} + {{\mathbf{u}}^T}{{\mathbf{x}}_0} + {{\mathbf{u}}^T}{\mathbf{u}} - {\mathbf{x}}_0^T{{\mathbf{x}}_0}} \right]} \\ 
  {}&{ = \tfrac{1}{{2l_0^2}}\left[ {{\mathbf{x}}_0^T{\mathbf{u}} + {{\mathbf{u}}^T}{{\mathbf{x}}_0} + {{\mathbf{u}}^T}{\mathbf{u}}} \right]} \\ 
  {}&{ = \tfrac{1}{{2l_0^2}}\left[ {2{\mathbf{x}}_0^T{\mathbf{u}} + {{\mathbf{u}}^T}{\mathbf{u}}} \right]} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}\left[ {{\mathbf{x}}_0^T{\mathbf{u}} + \tfrac{1}{2}{{\mathbf{u}}^T}{\mathbf{u}}} \right]} 
\end{array}$$
where the second last step follows from the fact that ${\mathbf{x}}_0^T{\mathbf{u}} = {{\mathbf{u}}^T}{{\mathbf{x}}_0} = {{\mathbf{x}}_0} \cdot {\mathbf{u}}$, a scalar.
Q2
Computing $\frac{{\partial {\varepsilon _G}}}{{\partial {\mathbf{u}}}}$ yields
$$\begin{array}{*{20}{l}}
  {\frac{{\partial {\varepsilon _G}}}{{\partial {\mathbf{u}}}}}&{ = \frac{\partial }{{\partial {\mathbf{u}}}}\tfrac{1}{{l_0^2}}\left[ {{\mathbf{x}}_0^T{\mathbf{u}} + \tfrac{1}{2}{{\mathbf{u}}^T}{\mathbf{u}}} \right]} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}\left[ {\frac{\partial }{{\partial {\mathbf{u}}}}{\mathbf{x}}_0^T{\mathbf{u}} + \tfrac{1}{2}\frac{\partial }{{\partial {\mathbf{u}}}}{{\mathbf{u}}^T}{\mathbf{u}}} \right]} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}\left[ {{\mathbf{x}}_0^T + \tfrac{1}{2} \cdot \left( {2{{\mathbf{u}}^T}} \right)} \right]} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}\left[ {{\mathbf{x}}_0^T + {{\mathbf{u}}^T}} \right]} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}{{\left( {{{\mathbf{x}}_0} + {\mathbf{u}}} \right)}^T}} 
\end{array}$$
and thus $$\delta {\varepsilon _G} = \frac{{\partial {\varepsilon _G}}}{{\partial {\mathbf{u}}}}\delta {\mathbf{u}} = \tfrac{1}{{l_0^2}}{\left( {{{\mathbf{x}}_0} + {\mathbf{u}}} \right)^T}\delta {\mathbf{u}} = \tfrac{1}{{l_0^2}}{{\mathbf{x}}^T}\delta {\mathbf{u}}$$
Q3
What the author has done in (2.51) is exploit the fact that $$\delta {\varepsilon _G} = \tfrac{1}{{l_0^2}}{{\mathbf{x}}^T}\delta {\mathbf{u}} = \tfrac{1}{{l_0^2}}\left( {{\mathbf{x}} \cdot \delta {\mathbf{u}}} \right) = \tfrac{1}{{l_0^2}}\delta {{\mathbf{u}}^T}{\mathbf{x}}$$
That is, $\delta {\varepsilon _G}$ is a scalar quantity, equal to its own transpose, hence you can freely transpose the RHS of the equation. More explicitly,
$$\begin{array}{*{20}{l}}
  {\delta {\varepsilon _G}}&{ = {{\left( {\delta {\varepsilon _G}} \right)}^T}} \\ 
  {}&{ = {{\left( {\tfrac{1}{{l_0^2}}{{\mathbf{x}}^T}\delta {\mathbf{u}}} \right)}^T}} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}{{\left( {\delta {\mathbf{u}}} \right)}^T}{{\left( {{{\mathbf{x}}^T}} \right)}^T}} \\ 
  {}&{ = \tfrac{1}{{l_0^2}}\delta {{\mathbf{u}}^T}{\mathbf{x}}} 
\end{array}$$
The reason he does this is because the second sum in (2.50) is expressed in terms of ${\delta {{\mathbf{u}}^T}}$ rather than $\delta {\mathbf{u}}$ as used in (2.15), and he chooses to transpose (2.15) to make the two sums "compatible" (i.e. amenable to sharing terms).
If you took the transpose of the expression $${\tfrac{1}{{2l_0^2}}\left[ {{{\left( {{{\mathbf{x}}_0} + {\mathbf{u}}} \right)}^T}\left( {{{\mathbf{x}}_0} + {\mathbf{u}}} \right) - {\mathbf{x}}_0^T{{\mathbf{x}}_0}} \right]}$$ in the second step of the chain for ${\varepsilon _G}$ shown above in Q1 and carried it through all subsequent steps, you would end up with $${\varepsilon _G} = \tfrac{1}{{l_0^2}}\left[ {{{\mathbf{u}}^T}{{\mathbf{x}}_0} + \tfrac{1}{2}{{\mathbf{u}}^T}{\mathbf{u}}} \right]$$ yielding the derivative $$\frac{{\partial {\varepsilon _G}}}{{\partial {{\mathbf{u}}^T}}} = \tfrac{1}{{l_0^2}}\left[ {{{\mathbf{x}}_0} + {\mathbf{u}}} \right]$$ and thus $$\delta {\varepsilon _G} = \delta {{\mathbf{u}}^T}\frac{{\partial {\varepsilon _G}}}{{\partial {{\mathbf{u}}^T}}} = \tfrac{1}{{l_0^2}}\delta {{\mathbf{u}}^T}\left[ {{{\mathbf{x}}_0} + {\mathbf{u}}} \right] = \tfrac{1}{{l_0^2}}\delta {{\mathbf{u}}^T}{\mathbf{x}}$$ which yields exactly the same as taking the transpose at the last step instead of the first.
