Short answer
Unlike the root of the wing of a fixed wind aircraft, the root of a helicopter blade is either hinged, or made flexible. The hinge has limits on its travel so the blades do not fall to the ground when stationary.
The fact that is it hinged means that the blade does not act as a beam, but as a tie. In operation the axis of the blade is aligned to the resultant of the forces acting on it. Were it not for the need to apply torque to the blade from the hub, you could make it from weighted string with an aerofoil cross section and the first-order physics would be unaffected.
The stiffness of a tie is the ratio of tension to elongation. Rotation has no effect on this ratio. The transverse stiffness of the blade has nothing to do with whether it can support the helicopter. Resistance to flutter, and the ability to apply torque to the blade, require some transverse stiffness. The rotation of the blade increases the effective transverse stiffness, but not the axial stiffness which is what is holding the helicopter up.
Worked example
Consider a Blackhawk, which has a fully articulated rotor.
* 110kg blade
* 300 rpm
* 16m wingspan
* 10,000 kg mass
I originally queried the answer in the other place for logical gaps rather than the conclusion - for example, saying the blades were not strong enough because they droop when on the ground. They do droop, but not significantly more so than the wings of a glider or even a B-52S with full tanks do, especially considering the articulated root.
The force acting horizontally on the rotor due to one blade's rotation is ~480kN. 333
The force acting vertically on the blade ~25kN, or 10,000 kg times g divided between the four blades.
The blade is articulated at the root. If the centre of lift is the centre of mass, then the moments cancel and blade will fly at an angle of arcsin(25/480) = 0.052 rad or 3 degrees.
Measuring off this picture shows the blades are at an angle of around 1/2 arcsin(35/392) or 0.045 rad, so close enough.
So the rotor can work without needing to be rigid at all. The blades can flex and align to the net of the lift and rotational forces, so do not need to be stiff to function. The string on a round-the-pole flying will exert a force upwards on the pole if the plane is above it, if you had two or three of them then you could balance out the horizontal forces and the pole would take off:
https://www.youtube.com/embed/COb9Ws-tVRA
As for the blade, as it is articulated at its root the forces should act along the axis of the blade and not make it any stiffer, but along it there will be uneven forces so it will be fluttering a bit, and (just like the string on the flyer twangs at a higher note as it goes faster) the effective stiffness under that fluttering will be increased by the rotation.
As to whether the blade would break if statically loaded on the ground, 110kg of blade over 7.8m gives 14 kg/m, assuming half of that is a structural tubular spar.
Taking aluminium 7178 as an example, has a density of 2,800 kg/m3 and so 7kg/m gives an area of 0.0025 m2 or 2500mm2. So I will assume the structural spar is an 80mm diameter 10mm wall tube.
Using http://www.tech.plymouth.ac.uk/sme/desnotes/buccalc.htm and http://www.amesweb.info/StructuralBeamDeflection/CantileverBeamStressDeflectionCalculator.aspx with a point load of 25kN half way along exceeds the yield strength by a factor of four or so - 1940 MPa.
The force acting along the blade of 500kN in an area of 2200mm2 gives 230MPa, which is less than half the tensile yield strength of such a spar.
So the original answer was mostly correct, but badly phrased - the important effect is that the rotational forces set up in the blade cause it to act like a tie rather than a beam, rather than meaning it is a more rigid beam. . If similar forces were exerted on the blade as are found in flight, it would not break. If the blade were used as a tie to support the aircraft, it would not break. The is stiffness gained, but that will affect flutter response rather than whether or not it fails. The fact that the blades or wings droop does not reliably indicate whether they can support the weight of their aircraft without being rotated. 333