PHY.1:5.2 - Reproduce one deformation without claiming the whole field

Consider a straight uniform bar, fixed at the top, with an axial force F pulling down at the lower end. Let L be length, A cross-sectional area, Y Young’s modulus, rho density and g gravity. Use small-strain linear elasticity and uniform material properties. Let x measure height from the bottom. The part below that point contributes weight rho*A*g*x, so the tensile force there is F+rho*A*g*x. Hooke’s law gives local strain:

strain(x) = F/(Y*A) + rho*g*x/Y.

Integrating from 0 to L gives total extension delta and mean strain:

delta = F*L/(Y*A) + rho*g*L^2/(2*Y)
delta/L = F/(Y*A) + rho*g*L/(2*Y).

A geometrically similar bar made of the same material has L'=lambda*L and A'=lambda^2*A. To reproduce the applied-force contribution to strain, use F'=lambda^2*F. Its own weight instead changes by lambda^3. At unchanged g, the self-weight contribution to strain changes by lambda. Merely using a smaller copy with the same material does not reproduce the two load contributions together.

Suppose the original bar hangs under its own weight, with F=0, and the question concerns only total extension relative to length. An added lower-end force on the smaller bar can match that output. Solve the mean-strain equation for the new force:

F' = A'*rho*g*L*(1-lambda)/2.

At quarter scale, F’ is 3/128 of the original bar’s weight. Substitution into the smaller bar’s mean-strain equation gives rho*g*L/(2*Y), the original value. The comparison therefore reproduces the requested normalized extension within this model.

Now ask for the strain at the lower end. The original unloaded bar has zero strain there; the smaller bar with the added force has F'/(Y*A') greater than zero. The output-specific construction cannot answer this new local question. To reproduce the complete strain profile, revisit the distributed loading or the combination rho*g*L/Y. The successful first comparison is retained for total extension.