How would the DC equivalent circuit be constructed? I have a vague idea knowing that AC voltage sources are connected to the ground and capacitors are open circuited but what happens to the resistance \$R_B\$? I have seen some solutions where the whole branch is connected to the ground but that doesn't sem right
For the dc equivalent the capacitor becomes an open circuit. Since Rb is in series with the capacitor then no current will flow through either component, and both of them can be removed from the schematic.
What you have left is the four resistors that set the bias point of the amplifier. It seems like you are working backwards here.
The trick is to find the bias (base) voltage of the circuit. For a DC voltage applied without the cap, if the input is exactly at this voltage it will produce no change in the output voltage. The output, of course, will be one diode drop lower than the input.
If you reference your DC input to this bias voltage, for small +/- changes you'll get the appropriate amplified change at the output.
More or less.
Problem is, the base-emitter voltage drop, which contributes to the bias point, will change with temperature. So using this sort of circuit for DC measurements is not a good idea, since the zero point will vary with temperature. It will also vary from transistor to transistor, since transistors vary between lots.
This is why IC op amps are the preferred approach when dealing with DC. They use intrinsically matched differential transistors on the input, and negative feedback to handle the gain. The result is excellent DC performance over temperature.
But even then, it's not perfect.