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 $\mathbf r+s\widehat{\mathbf F}$, where $\widehat{\mathbf F}=\mathbf F/\|\mathbf F\|$ and $s$ is a signed distance. The new moment is
$$
(\mathbf r+s\widehat{\mathbf F})\times\mathbf F
=\mathbf r\times\mathbf F+s\widehat{\mathbf F}\times\mathbf F
=\mathbf r\times\mathbf F.
$$
The last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.<University Physics Volume 1 by William Moebs and others>[§10.6, torque vector and perpendicular lever arm]{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 $(0,0,3)\,\mathrm{N\,m}$. 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 $(0,0,5)\,\mathrm{N\,m}$. The second move needs an additional compensating couple to preserve the original <wrench (mechanics)>.

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 is
$$
u(L)=\int_0^L\frac{N(x)}{EA}\,dx=\frac{Fa}{EA},
$$
where $E$ 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 $A=1\,\mathrm{mm^2}$, 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.<University Physics Volume 1 by William Moebs and others>[§12.3, tensile stress, strain and Young's modulus]{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.


 Principle of transmissibility (source code)