Equivalent Nodal Forces

When deriving the formulation, IVW = EVW, simply we assumed that there is only nodal forces and that’s why EVW = $\delta {𝐪}^{𝖳}𝐏$. Now suppose that there are body forces and surface traction.

Let $𝐛$: body force per unit mass $\overline{𝐭}$: prescribed boundary force or surface traction per unit area. (here I have used a bar over the traction vector to indicate that it is a prescribed one)

Then $$EVW={\int }_{\mathrm{\Omega }}\rho 𝐛\cdot \delta 𝐮dV+{\int }_{{\mathrm{\Gamma }}_t}\overline{𝐭}\cdot \delta 𝐮dS+\delta {𝐪}^{𝖳}𝐏$$ which can be written as $$EVW={\int }_{\mathrm{\Omega }}\rho \delta {𝐮}^{𝖳}𝐛dV+{\int }_{{\mathrm{\Gamma }}_t}\delta {𝐮}^{𝖳}\overline{𝐭}dS+\delta {𝐪}^{𝖳}𝐏$$ However, $$\delta 𝐮=𝐍\delta 𝐪$$ or $$\{u\}=[N]\{\delta q\}$$ Therefore, \begin{aligned} EVW&= \int_\Omega \rho\, \delta \mathbf{q}^{\mathsf T} \mathbf{N}^{\mathsf T} \mathbf{b}\, dV+\int_\Gamma \delta\mathbf{q}^{\mathsf T}\mathbf{N}^{\mathsf T}\bar{\mathbf{t}}\, dS+\delta\mathbf{q}^{\mathsf T}\mathbf{P}\\ &=\delta \mathbf{q}^{\mathsf T} \left(\int_\Omega \rho\, \mathbf{N}^{\mathsf T} \mathbf{b}\, dV+\int_\Gamma \mathbf{N}^{\mathsf T}\bar{\mathbf{t}}\, dS+\mathbf{P}\right) \end{aligned}

Equating EVW = IVW gives us $$𝐊𝐪=𝐅$$ where $$𝐊={\int }_{\mathrm{\Omega }}{𝐁}^{𝖳}𝐄𝐁dV$$ and $$𝐅={\int }_{\mathrm{\Omega }}\rho {𝐍}^{𝖳}𝐛dV+{\int }_{\mathrm{\Gamma }}{𝐍}^{𝖳}\overline{𝐭}dS+𝐏$$

Example: If the beam is loaded as in the following figure,

then $$\overline{t}(x)=-\frac{w}{L}x$$ and the equivalent generalized nodal forces are: $$P_1={\int }_0^L{N}_1(x)\overline{t}(x)dx=-\frac{3wL}{20}$$ $$M_1={\int }_0^L{N}_2(x)\overline{t}(x)dx=-\frac{w{L}^2}{30}$$ $$P_2={\int }_0^L{N}_3(x)\overline{t}(x)dx=-\frac{7wL}{20}$$ $$M_2={\int }_0^L{N}_4(x)\overline{t}(x)dx=\frac{w{L}^2}{20}$$ where \begin{aligned} N_1(x) &= 1 - \frac{3x^2}{L^2} + \frac{2x^3}{L^3} \\ N_2(x) &= x - \frac{2x^2}{L} + \frac{x^3}{L^2} \\ N_3(x) &= \frac{3x^2}{L^2} - \frac{2x^3}{L^3} \\ N_4(x) &= -\frac{x^2}{L} + \frac{x^3}{L^2} \end{aligned}