PLC Input Buffer/Driver
Your circuit should work assuming you're talking about a ULN2803 or something with an internal base resistor where you show a discrete darlington. 

Noise immunity will be at most some hundreds of mV when low. 

The 10R resistor means the 5V source could be called upon to sink a lot of current if D1 breaks down (quite plausible with a Schottky). Most 5V regulators won't sink any current. 



It's usually worth a bit of effort to reduce the complexity when there are many channels. Consider this approach: 







simulate this circuit &ndash; Schematic created using CircuitLab

If you use a network for the resistors, that's two dual transistors and 3 4-resistor networks for every 4 channels, so 75 parts total. If you use pre-biased dual transistors, then you only need 3 parts for each 4 channels so 45 small parts total for all 60 channels. 

When the switch is closed, the emitter of Q1 is grounded and the transistor is saturated via base current from R2, so the output voltage is less than 0.1V. 

When the switch is open, the emitter and base are close to the same potential and the transistor is off, so the collector is pulled up to 24V by R3.