MecMovies 4.0 : M1.4: Load Capacity of Two-Bar Assembly

A trial-and-error approach can be used to determine the maximum allowable load $P$ :

1. Using the allowable stresses given, compute the allowable axial forces for members (1) and (2) from:

$F_{allow} = σ A$

2. Arbitrarily assume that one of the two axial members will control the load-carrying capacity of the structure. For example, assume that member (1) will be stressed to its allowable limit. If this assumption is correct, member (2) should carry less than its allowable stress.

3. Set $F_{1} = F_{allow, 1}$ and solve the two equilibrium equations simultaneously for the unknown forces $F_{2}$ and $P$ .

4. Compare the value of $F_{2}$ computed from the equilibrium equations to its allowable force $F_{allow, 2}$ .

The figure shows a diagonal member labeled 1 and a horizontal member labeled fixed to a column on the right at the points A on top and C below. The diagonal member runs between points A and B and the horizontal member runs between the points B and C. Joint B is at a distance of 4.3 meters to the left of A and 5 meters below A. Joint C is at a distance of 7 meters directly below A. The load at joint B is labeled P and it is directed
downwards.