Consider how much volume of ice - which is only slightly less dense than water, so that won’t help much - and water displacement is needed to support a railcar of any significant weight.
Even ‘back in the day’, a 40,000 lb. gross weight railcar might be 30 to 40 ft. long, or about 1,000 to 1,300 lbs. per foot of length.
Water is about 62.4 lbs. per cubic foot, so to support 1,000 lbs. per foot, about 16.0 cu. ft. would have to be displaced. For example, that means a section of ice about 16 ft. wide (8 ft. on either side of the center of the track, or about 4 ft. beyond the ends of the ties on each side, which intuitively seems about right) would have to sink into the water about 1.0 feet to provide the needed buoyancy uplift to counteract the imposed weight of the car ( or an equivalent, such as 8 ft. wide x 2 feet deep, or 24 ft. wide x 0.67 ft. deep, etc.).
That 1,000 lbs. per ft. is equivalent to a Cooper’s E-10 loading, which is extremely light by later (E-40 to E-60) and modern standards (E-72 to E-80). However, it’s probably not far off from what the "Ice Road Truckers’ mentioned by Charlie above weight - 60,000 kg would be about 132,000 lbs., probably spread out over a little longer wheelbase than a standard US tractor-trailer, too. - maybe as much as 2,000 lbs. per foot. But a 5,000 lb. per ft. load (E-50) - which is roughly what a 175-ton (350,000 lb.) locomotive 70 ft. long, or a 286,000 lb. hopper 57 ft. long, etc. weighs - would have to displace about 5 times as much water, or 80 cu. ft. per longitudinal ft. For a width of 20 ft., that would be about 4.0 ft. down, or for a 30-ft. width, about 2.7 ft. down, either of which is an awful lot. Which is why we don’t see ice bridges for modern railcar loads.