The Arduino Uno uses the ATmega328 microcontroller. If you check the DC characteristics (page 313) you'll see that output voltages are specified at 20 mA:
That's the rated current to which you should adhere. So no 40 mA, that's Absolute Maximum Ratings (AMR), and there's a notice about AMR in the datasheet (same page):
Stresses beyond those listed under “Absolute
Maximum Ratings” may cause permanent damage
to the device. This is a stress rating only and
functional operation of the device at these or
other conditions beyond those indicated in the
operational sections of this specification is not
implied. Exposure to absolute maximum rating
conditions for extended periods may affect
(emphasis by me)
You should never operate a device continuously at AMR, stick to Normal Operating Conditions. You'll need an external transistor to sink the 40 mA of the two relays, or use one I/O pin per relay, and switch both outputs simultaneously in software. That way you'll keep the possibility to still operate them individually as well.
Placing the two input circuits in series, like Tony suggests, is a way to reduce required current to that of a single circuit. Two objections:
- Design objection. No details of the parts used are given, but the LED in an optocoupler often has a voltage drop of 1.15 V typical, 1.5 V maximum. The LED will be around 2 V. So we're at 5 V already, and haven't even accounted for the 0.9 V output or tolerance in the 5 V power supply. So this may not work.
- Practical objection. You'll have to remove a LED, replace a resistor, solder a wire and, depending on how you place the resistor, cut a trace. That's not for you yet.
There's a much simpler patch though, which requires just soldering 1 short wire, the red line in following schematic.
If you drive the top input low the optocoupler's transistor will drive both R12 and Q1's, thus activating both relays. The current will normally be sufficient: even at a 20 % CTR the transistor will drive 4 mA for 20 mA LED current, that's 2 mA per Q1. The relay needs 90 mA, so an HFE of only 45 is needed, what almost any general purpose transistor can do. The patch should look like this: