Redefining the output of a function

ftn1 = Sin[s]^2 + 3 t^2 + u^5
ftn2[s_, t_, u_] = ftn1
a=ftn2[2, 1, 2]
N[a]

35 + Sin[2]^2

35.8268

If you prefer to used := then

ftn1 = Sin[s]^2 + 3 t^2 + u^5
ftn2[s_, t_, u_] := Evaluate[ftn1]

Your function is defined as,

ftn1 = Sin[s]^2 + 3 t^2 + u^5

3 t^2 + u^5 + Sin[s]^2

If you use ReplaceAll (/.) on ftn1,

ftn1 /. {s -> 2, t -> 1, u -> 2}

35 + Sin[2]^2

You can use N[] to numerically evaluate it,

N[ftn1 /. {s -> 2, t -> 1, u -> 2}]

35.8268

With this knowledge, define ftn2 as ftn2[s1_, t1_, u1_], where s1, t1 and u1 are placeholders for s, t and u respectively. Now we "replace all" variables so that ftn1 is evaluated inside the definition of ftn2 (Techinically, by using :=, fnt1 is not evaluated inside the definition, it is evaluated in the function call of the fnt2 function. We use placeholders so that variables inside ftn2 is clearly separated from the internal variables in ftn1:

ftn2[s1_, t1_, u1_] := ftn1 /. {s -> s1, t -> t1, u -> u1}
ftn2[2, 1, 2]

35 + Sin[2]^2

Suppose now, this is not enough and you need the numerical value, you can use N[] in the definition of the function,

ftn3[s1_, t1_, u1_] := N[ftn1 /. {s -> s1, t -> t1, u -> u1}]
ftn3[2, 1, 2]

35.8268

This should solve your problem!