Hey all...back for a little help. I am finishing up the front end rebuild and need a wheel hub bearing. Does anyone know the torque setting for the spindle lock nut for a 2000 2500 2WD? I have heard three different things so far...175 lbft, 250 lbft and "it doesn't matter as long as it's tight". Any ideas??

Thanks in advance for any help.

 

250 ft lbs works for me

 

From my 01 FSM page 2-9:
INSTALLATION (1) On models with all-wheel antilock system (ABS), check condition of tone wheel on hub/bearing. If teeth on wheel are damaged, hub/bearing assembly will have to be replaced (tone wheel is not serviced separately). (2) Slide the hub/bearing onto the spindle. (3) Install the new hub/bearing nut and tighten to: • LD 1500: 251N·m (185 ft. lbs.) • HD 2500/3500: 380 N·m (280 ft lbs.)
Driver side was replaced on mine about a year ago. My buddies torque wrench only went to 250 ft lbs. Torqued it that far and then grunted some more.

 

You’re right, 125 ft-lbs does sound like a lot for those relatively small hub bolts, but that spec is pretty common on heavy-duty trucks because the wheel bearings and hub assembly see huge loads, especially on a 3/4-ton. The high torque ensures the 608ZZ bearing hub stays fully seated and doesn’t shift or loosen under braking and cornering forces. If you’re unsure, it’s worth double-checking the service manual by part number or consulting the dealer to confirm the torque spec for your exact bearing hub setup, but in many cases, that higher torque is correct to keep the bearings secure over time.