From the setup of two trusses as shown in this illustration, how can I solve for the axial reaction forces in the trusses, and in the points A and B (Ax, Ay, Bx, By)? It is assumed that the joints exert no torque/bending moment, and the lengths don't really matter, only the angles alpha and beta.

I've drawn two slightly different setups, while the top one reflects the true setup, and the lower one might be easier to solve:


I'm primarily interested in solving this by hand and calculator, and I'm a little rusty on constraint equations and Lagrange multipliers (though that might be overkill).

  • $\begingroup$ All You need is parallelogramm of forces, analyze the angles and apply law of cosines. $\endgroup$
    – Georg
    Commented Nov 28, 2011 at 13:14
  • 3
    $\begingroup$ The first setup drawn is not statically indeterminate. If you had a truss member vertically from A to B, then it would be. The second setup drawn is not actually constrained and will collapse at the slightest imbalance. Use the joint method and consider the point at which the force is applied. You know the forces in the two members are in the direction of the members. You've got two equations (x and y) and two unknowns - the magnitudes. $\endgroup$
    – Doresoom
    Commented Nov 28, 2011 at 20:43
  • $\begingroup$ Is force $F$ known? Are $\alpha$ and $\beta$ known? $\endgroup$
    – 71GA
    Commented Feb 28, 2013 at 2:42

1 Answer 1


Like @Doresoom said, the truss member here are "Two Force Members" carrying only tangential forces. Thus the sum of the forces on the triangle tip (Point ?C?) should yield the answer.

$$ T_A \cos \alpha + T_B \cos\beta = 0 \\ T_A \sin \alpha - T_B \sin\beta - F = 0 $$

where $T_A$ and $T_B$ are the tensions on the two members.

Next you construct the strain energy

$$ U = \frac{T_A^2}{2 A E} + \frac{T_B^2}{2 A E} $$ where $A$ is the cross-sectional area and $E$ is the (Young's) Modulus of elasticity.

and solve for the deflection $\delta$ along $F$ as

$$ \delta = \frac{\partial U}{\partial F} = \frac{\cos^2 \alpha + \cos^2 \beta}{\sin^2 (\alpha+\beta)} \frac{F}{A E} $$

  • $\begingroup$ What's A? Cross sectional Area? $\endgroup$ Commented Jun 28, 2013 at 13:41
  • $\begingroup$ Yes $A$ is cross sectional area, and $E$ is the modulus of elasticity. $\endgroup$ Commented Jun 28, 2013 at 13:43

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