# Correct axial force distribution between pinned ends of a member?

I have a planar member pinned at both ends with compressive axial loads at 1/3 and 2/3rd position of 14.4 applied between them. Is the following axial force distribution correct? Shouldn't the section E2 carry the 14.4 kN load. Note all material properties are the same.

Tl;DR: yes it is correct (disregarding rounding errors)

assuming points at

• A: ( $$x=0$$)
• B: ( $$x=\frac{1}{3}L$$)
• C: ( $$x=\frac{1}{2}L$$)
• D: ($$x=\frac{2}{3}L$$)
• E: ( $$x=L$$)

It is obvious that there symmetry in the problem. Therefor point C (in the middle will not move at all).

For the half problem (ABC) you can write the equilibrium of forces.

$$F_A + 14.4 +F_C=0$$ $$F_A + F_C=14.4$$

Also about the deformation you can write:

$$\Delta L_{AB} +\Delta L_{BC} =0$$ $$\frac{P_{AB}L_{AB}}{EA} +\frac{P_{BC}L_{BC}}{EA}=0$$ $$P_{A_B}L_{AB} =- P_{BC}L_{BC}$$ $$\frac{P_{AB}}{P_{BC}}=- \frac{L_{BC}}{L_{AB} }=\frac{1/6 L}{1/3 L }=-\frac{1}{2}$$

Because $$F_A = P_{AB}, F_C = -P_{BC}$$, you can obtain that

$$P_{AB}= 4.8 kN, P_{BC}= 9.6 kN,$$

• It might be correct, but it's a statically indeterminate problem - it depends upon the relative stiffness of the supports and the stiffness of the axial member. If the supports have infinite stiffness, the member has a uniform stiffness, and the member is unstressed before the loads are applied, it's correct. If the supports are relatively flexible compared to the axial stiffness of the member it will tend towards higher load in the middle portion. – achrn Feb 26 at 15:01