Odd bean among 90 of unknown direction

A bag holds $90$ beans; all weigh the same except one, which is either a bit heavier or a bit lighter (you do not know which). With a two-pan balance, how many weighings are needed in the worst case to identify the odd bean?

5. Each weighing has three outcomes, and with no known-good reference bean $k$ weighings can handle at most about $(3^k-1)/2$ suspects; $k=4$ covers only $40$, so at least $5$ are required. Five suffice: split into thirds of $30$, then use rotation of sub-groups (e.g. $27+3$) so each later weighing both narrows the suspects and reveals heavy or light.