for short circuit case its simply ohms law
V = 40 V DC
Load = 10.0 Ohm [obviously connected in parallel to supply]
then Current I = 40/10 = 4A
now if an abnormally low resistance path is formed in circuit in parallel to our load (may be because of component failure) then circuit current will be substantially increase like for example, if short circuit path has a resistance of 0.1 Ohm then majority of the current will flow through this low resistance path and current sourced by power supply would be around 40 V /0.1 Ohm = 400 A according to Ohm's law. resulting in damaging power supply, pcb. Assuming designer has not used power supply rated for 16 KW operation for 160W operation.
Other case is of Overload and it has plenty of details like in rush, normal circuit current, I^2 *t for circuit, DC transients etc. for this you can either refer to opti-fuse guide or paper written by Cynthia Cline on fuse selection both are short 6-7 pages and a good read . It specifies various parameters for fuse selection for both AC and DC operation. Otherwise i would suggest select 2 A fuse if selecting local or just next value if selecting any standard part do consider ambient temperature as most fuse wire go off because of I^2*R losses.