Steve,
A hacksaw blade is nominally about .032 X .500 inches (= .016 in²). Garden variety low-carbon merchant steel has a yield strength of 36,000 psi. That means that it will not start yielding until a longitudinal (tensile) load of 576 lbs is placed on it. Rc 50 tool steel has a yield strength in the 120,000 psi range. That means that it will not start yielding until a longitudinal load of 1920 lbs is placed on it. A power hacksaw blade is supposed to be tightened to (about) 70% of yield to allow for any "binding" in operation. A hand hacksaw should be somewhat less than that as you are much more likely to "bind" a hand-held blade than a machine blade. Right?
The only screw-tightened hand hacksaw I have left (having changed over to cam-lock type many years ago) has a 5/16-24 thread on it. A very general torque value for a given load is given by: T = kPD (T = torque, k = friction factor, P = load, and D = major diameter of screw) or T = (.2)(576 lbs)(.3125 in) = 36 lb-in (= 3 lb-ft). (Make no mistake, this is a really rough value calculation. The k-factor incorporates friction, thread pitch, and surface finish into one value. The variations from "calculated" can be quite high and this is really only an estimated value!)
I would be highly surprised if your blade did not have a much higher yield strength value than low-carbon merchant steel. It is probably more like 60,000 psi for a cheap "soft back" type blade. However, most people can only exert somewhere in the 10 lb-in range when "torquing" a wingnut by hand. People with very strong hands might be able to hit 25 lb-in of hand torque on a wingnut. (I actually tested this for SAE back in the mid-1970's).
Does this help?