B.1 Work done by internal and external forces

The work-energy theorem in structural analysis: the sum of the work done by internal and external forces is zero:

$$W_{i}+W_{e}=0$$ (B.1)

where ${W}_{i}$ is the work done by internal forces and ${W}_{e}$ is the work done by external forces.
The Green's functional for a frame element is [KHMW10]

$$\begin{array}{l} \underbrace{ - \int_{a}^{b}N_{x}\hat{\lambd... ... of\hspace*{3pt} external\hspace*{3pt} forces } = 0 \end{array}$$ (B.2)

where we consider two systems (states) of forces associated with respective deformations and displacements [BP13]:

$N_{x}$, $M_{y}$ - are the internal axial force and bending moment of the first load state;

$\hat{\lambda}$, $\hat{\psi}_{y}$ - axial and bending deformations of the second load state;

$N_{x}\vert_{a}^{b},\hspace*{2pt} Q_{y} \vert_{a}^{b},\hspace*{2pt} M_{y} \vert_{a}^{b}$ - axial force, shear force and bending moment of the first load state at boundaries a and b;

$\hat{u}\vert_{a}^{b},\hspace*{2pt}\hat{w}\vert_{a}^{b},\hspace*{2pt}\hat{\varphi}_{y}\vert_{a}^{b}$ - longitudinal and transverse displacements, and the rotation of the cross section of the second load state at boundaries a and b;

${q}_{x}(x),\hspace*{2pt}{q}_{z}(x)$ - distributed loads of the first load state;

$\hat{u}(x),\hspace*{2pt}\hat{w}(x)$ - longitudinal and transverse displacements of the second load state;

$F_{xi},F_{zi}$ - force components of the first load state, applied at point i in the x and z directions, respectively;

$\hat{u}_{i}$, $\hat{w}_{i}$ - longitudinal and transverse displacements of point i of the second load state.

The first and second elements of Eq. (B.2) describe the work $W_{i}$ of internal forces:

$$W_{i}= - \int_{a}^{b}N_{x}\hat{\lambda}dx - \int_{a}^{b}M_{y}\hat{\psi}_{y}{dx}$$ (B.3)

The final four elements of Eq. (B.2) describe the work $W_{f}$ done by active forces:

$$W_{f}=\int _{a}^{b}{q}_{x}(x)\hat{u}{dx}+F_{xi}\hat{u}_{i}+\int_{a}^{b}{q}_{z}(x)\hat{w}{dx}+F_{zi}\hat{w}_{i}$$ (B.4)

The third and fourth elements of Eq. (B.2) describe the work $W_{b}$ done by boundary forces (fixed-end forces and moments at joints^B.1, support reactions):

$$W_{b}=\left[ N_{x}\hat{u}\right]_{a}^{b}+\left[Q_{y}\hat{w}+M_{y}\hat{\varphi}_{y}\right]_{a}^{b}$$ (B.5)

The external work $W_{e}$ can be divided into two parts:

• $W_{f}$ - work done by active forces, e.g. concentrated loads, uniformly distributed loads;
• $W_{b}$ - work done by reaction forces, e.g. support reactions (Fig. 1.10), internal reactions [WP960] (contact forces ^B.2) (Fig. 1.12).

$$W_{e} = W_{b}+W_{f}$$ (B.6)

Applying now Eq. (B.6) to (B.1), we obtain

$$W_{i}+W_{b}+W_{f}=0$$ (B.7)

Equation (B.7) is a shortened form of Eq. (B.2).

With the elastic energy $U_{elastic {\,}{\,} energy}$ and dissipation energy $D$ existing, the work $W_{i}$ done by internal forces is in relation to the internal energy ${U}_{elastic}+D$:

$$\underbrace{ W_{i}}_{internal{\,}{\,}work} = - \underbrace{U}_{elastic {\,}{\,} energy} - \underbrace{D}_{dissipation{\,}{\,} energy} %\Pi^{s} = 0$$ (B.8)

Equations (B.2) and (B.7) are the basic methodical tools of structural analysis.

Figure B.1: Bar member a-b

Example B.1   (conservation of mechanical energy). We consider the bar from Fig. B.1, subjected to the load F. The bar has been split into sections where the normal forces $N_{a}$ and $N_{b}$ exerted on the member at cross-sections a and b are treated as external loads [VrCty] or, to be more precise, as boundary forces.

No loads act on the bar member a-b, and so $W_{f} = 0$ and $W_{i} \neq 0$ in Eq. (B.7). The expression for energy conservation of the bar member a-b is

$$W_{i}+W_{b}=0$$ (B.9)

or

$$\begin{array}{l} \underbrace{ - \int_{a}^{b}N_{x}\hat{\lambd... ...iptsize}\cite{A-W:WP960} \end{scriptsize} } = 0 \end{array}$$ (B.10)