Well it depends upon your rectifier as to what you actually have, but you will not have a steady voltage of any value. Yes you will have unidirectional current/voltage that may be called DC. A single rectifier will block alternate half cycles so your voltage (assuming a sine wave) will rise from zero to 180 times√2 for a quarter cycle, then fall back to zero for a quarter cycle. Then it will be zero for a half cycle. this will repeat. A bridge rectifier will invert alternate half cycles so the voltage will rise from zero to 180 times√2 for a quarter cycle, then fall back to zero for a quarter cycle. this will repeat. Either way normal meters will display the AVERAGE value of these waveforms, if the frequency is to fast for the meter to track. Neither will read 180.
aac9876, No, you will see a pulsating voltage of either polarity depending on the way the diodes/bridge is wired. A ordinary d'Arsonval galvanometer is an average responding device. When fitted with a rectifier for AC measurement, it is usually calibrated for the RMS of a symmetric sine wave about zero. That means it cannot measure correctly the value of any other waveform directly. Ratch
Assuming full wave rectification and a capacitor it goes to 1.4X the RMS value. RMS is key here, 120VAC becomes 170VDC. This will vary go down as loaded. BTW, these are hazardous voltage, not to be taken lightly.
For a sinusoidal wave ( same as all outlet) the full wave bridge rectified output would be a 100% duty cycle pulsed wave output nearly the same voltage as each 1/2 sine wave input (this considers very little losses from diodes forward bias and circuit resistance). The DC output voltage is the same peak voltage as the AC wave but since this wave is only at peak for a very very short time the active usable working voltage is slightly less. If you want to make the output the exact same peak voltage DC then if you start with a square wave the rectified output voltage peak is the same as the circuit voltage.