Principle of transmissibility

The principle of transmissibility states that moving a force along its line of action preserves its external mechanical effect on an ideal rigid body. The force vector is unchanged, and its torque about every reference point is unchanged. The statement concerns external rigid-body motion or equilibrium; it does not assert unchanged internal deformation.

For an original proof, a nonzero force $\mathbf{F}$ is applied at $\mathbf{r}$. Move it to $\mathbf{r}+s\hat{\mathbf{F}}$, where $\hat{\mathbf{F}}=\mathbf{F}/\Vert \mathbf{F}\Vert$ and $s$ is a signed distance. The new moment isThe last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.^[ref]

For a numerical example, let $\mathbf{F}=(0,10,0)\mathrm{N}$ act at $(0.30,0.20,0)\mathrm{m}$. Its moment is $(0,0,3)\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80,0)$ preserves that moment because the displacement follows the force direction. Moving it instead to $(0.50,0.20,0)$ gives $(0,0,5)\mathrm{N}\mathrm{m}$. The second move needs an additional compensating couple to preserve the original wrench.

A one-dimensional elastic counterexample shows the rigid-body limitation. Consider a uniform bar along $0\le x\le L$, fixed at the left end, with a rightward axial force $F$ applied to a cross-section at $x=a$. In a small-strain axial model, internal force is $F$ on $0<x<a$ and zero on $a<x<L$. Its right-end displacement iswhere $E$ is Young's modulus and $A$ the section area. Applying the same force at $a=L$ instead gives $FL/(EA)$, although both applied forces share one line of action and give the same external resultant and moment.

With stipulated $F=100\mathrm{N}$, $L=1\mathrm{m}$, $E=200\mathrm{GPa}$ and $A=1\mathrm{m}{\mathrm{m}}^2$, the end displacement is $0.25\mathrm{mm}$ for $a=L/2$ and $0.50\mathrm{mm}$ for $a=L$. The linear stress–strain relation underlying that calculation is described in the primary elasticity section.^[ref]

A proposed comparison therefore measures support reaction and body deformation separately when changing load position along a common line. Equal reactions and moments can coexist with different local stress and displacement. The external free-body diagram may be equivalent while the internal loading model differs.