Interesting question. We may start from the definition of the Beta function:
$$B(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1}\ \text{d}t$$
and rewrite it when $m\to 2m$:
$$B(2m, n) = \int_0^1 t^{2m-1}(1-t)^{n-1}\ \text{d}t$$
$$B(2m, n) = \int_0^1 t^{2m-1} \frac{(1-t)^n}{1-t}\ \text{d}t$$
Now, since the range of integration is $[0, 1]$, we are allowed to make use of the geometric series
$$\frac{1}{1-t} = \sum_{k = 0}^{+\infty} t^k$$
Hence
$$B(2m, n) = \int_0^1 t^{2m-1} (1-t)^n \sum_{k = 0}^{+\infty} t^k\ \text{d}t = \sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1} (1-t)^n t^k\ \text{d}t$$
And easily write:
$$\sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1+k} (1-t)^n\ \text{d}t$$
Calling now
$$2m-1+k = a ~~~~~~~~~~~ n = b$$
We notice that the integral is well known:
$$ \int_0^1 t^a (1-t)^b\ \text{d}t \equiv B(a+1, b+1)$$
Then we end up with the partial result (re-expanding $a$ and $b$):
$$B(2m, n) = \sum_{k = 0}^{+\infty} B(2m+k, n+1)$$
That series does exist and it does converge to a known result:
$$\sum_{k = 0}^{+\infty} B(2m+k, n+1) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
What you end up with is a sort of recursive relation for the Beta function:
$$B(2m, n) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
BUT
The above expression can be simplified!
$$\frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)} \equiv \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
What we obtained is actually nothing than what we would have obtained by simply substituting at the beginning $m\to 2m$ in the Gamma function / Beta function definition.
$$B(2m, n) = \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
This really suggest that such a particular duplication formula for the Beta function may not exist at all, since all you need is the Gamma function and ITS duplication formula, through which you can evaluate $\Gamma(2m)$.
Seems like that this is the only "duplication formula" for the beta function.
(Also, I found nothing on reviews or literature).
Best Answer
A way to solve the Beta-function without applying clever identities will be provided in this answer.
We want to evaluate $$\operatorname{B}[3/4,5/4] = \int\limits_{0}^{1}t^{3/4-1}(1-t)^{5/4-1}\,\mathrm{d}t=\int\limits_{0}^{1}\frac{\sqrt[4]{1-t}}{\sqrt[4]{t}}\,\mathrm{d}t.$$ Substituting $u=\frac{\sqrt[4]{t}}{\sqrt[4]{1-t}}$ will give us $$4\int\limits_{0}^{\infty}\frac{u^2}{(u^4+1)^2}.$$ From now on there are many ways to proceed. I believe you want to get to the $\arctan$ integral somehow. Splitting up the denomitor, Ostrogradski method, complex analysis are the first ones that come to my mind.
Edit: I would do it via complex analysis. You can add a semi-circle and show that the integral over it goes to zero. Then you have to calculate the residues.