hokhani said:

TL;DR: Does a measurement affect the expectation value?

If we do this measurement over this ensemble again, we expect to get to the last result

I do not think so in general. Some counter examples.
Photons disappear on the screen leaving dots in double slit experiment. We cannot repeat the experiment on the same photons.
Measurement of electron momentum by scattering ends to scattered states which does not hold the property of the original state.
In another thread of you, it was mentioned that in loose
position observation center of dispersing Gaussian packet does inertial motion in general so <x> changes with time.(the immediate second measurement would give the same value with the first one)

Measurement is performed by interaction between the object system and measuring apparatus. Some measurement process destroy or disturb the object system.

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hokhani said:

TL;DR: Does a measurement affect the expectation value?

So, it seems that consequent measurements disturb the time dependency of ⟨A⟩. Doesn't it

Quantum Zeno effect might be of your interest.

hokhani said:

On the same ensemble after the first measurement.

Others have discussed some practical issues with doing this experiment but if we ignore those, we can analyze this experiment based on the pure mathematical formalism. At least it's relatively easy to do in standard QM, ignoring QFT effects of particle annihilation, creation etc.

From that perspective, we have to ask a critical question, which is, how long after the first measurement on any given system is the second measurement made?

If the second measurement is made immediately after the first, then you will get the same answer as your first measurement. This is just the projection/collapse postulate (process 1 according to Von Neumann). If the second measurement takes some time (and remember, in QM typical time scales can be quite short) then the wave function will have already evolved via the Schroedinger equation (assuming ##A## doesn't commute with the Hamiltonian) and you will have to solve that for the new distribution.