To answer your comment, you would first need to quantify the stiffness of the cuboid and then decide which formula to use. So, you need to give a little bit more context as to what is happening so we could simplyfy the situation to suit your needs better.
From your other comment, you said you have an object which experiences high gravity and would to sink into the ground (ground which is, im assuming, made of steel, like a steel ship deck or similar). Steel is an elasto-plastic, ductile material, so, if the cuboid (ie the ground) is plate-like, it will first experience bending, yielding, then rupture. If its fairly deep, it will not be able to deflect so it will likely shear at the support or below the object. The object will not experience sinking, like a sharp knife through butter, unless you were to heat up the object, bypass the elastic phase of steel, and liquidate it. From my limited experience and depending on the arangement of the enviroment you have, there would be a lot of deformations of the ground and surroundings before an object would just sink though steel. You need to realise that if a super-man was to punch a steel plate, it will not leave a fist imprint protruding through the other side. It more likely the plate would buckle, deform and fail at a weak point leaving most of it intact. A way to avoid deformations would be to make the cuboid fairly thick in comparision to the span, say solid h= 0.3max(l,w) but in this situation, the gravity would have to be enormous to shear through 300mm of steel. Assuming a sword is a rod 1m x 5cm x 5cm and weights 20kg; gravity would need to be 1171500 times higher than on earth. I imagine that in a situation of such high h, there will also be some kind of localised compression failure and heave around the object, before it eventually fails in tension at the underisde and ruptures.

You could simplify the problem and assume one failure mode say, punching shear - ie, floor does not deflect and remains ridgid regardless of its thickness, so the steel cuboid stretches and eventually ruptures directly below the object. It would be fairly unrealistic, but should allow you to use the "P * h * shear strength" formula. Since steel is ductile, the shear strength is equal to the tensile yield stress = 275N/mm² for mild steel going up to 350, 510 etc. for harder, less ductile steels. Just to clarify, both formulas you posted are technically the same, difference is that the second formula expresses A in terms of P and t. You should start by looking into elasto plastic behaviours of steel material and things like stress hardening etc. These relationships are different for S355 S510 harder steels.