With blocks allowed to repeat and block sizes allowed to differ, you should be able to do lots of things if you do not insist that the number of points be small. I am about to add another answer with a different perspective. The point of this on is that a $(v,b,r,k,\lambda)$ is an $(r,\lambda)$ design and stays so even if some points are deleted. So $(r_i,\lambda_i)$ designs with the same points can be combined to get $(\sum r_i,\sum \lambda_i)$ designs which have somewhat regular block structure.

1) You say that you don't want to just repeat a symmetric design $r$ times because you want at least some disjoint blocks. But you can permute the underlying set. A small example is the Fano plane

013 124 235 346 450 561 602

repeat the blocks three times each but in the third copy swap 0 and 1 to get

013 024 235 346 451 560 612

now the blocks 124 and 560 are disjoint.

I was being lazy there anyway, The 35 triples from a 7-set can be partitioned into 5 groups of 7 with each group a Fano plane, so use three groups to have an honest $(v,b,r,k,\lambda)=(7,21,9,5,3)$ design with distinct blocks.

2) The complements of the lines in a Fano plane gives $(r,\lambda)=(4,2).$ Just taking the pairs gives $(r,\lambda)=(6,1)$ So the anti-lines once and the pairs twice each is $(16,4)$

3) To have a $(v,k,\lambda)$ BIBD is it necessary that $b=\frac{v(v-1)\lambda}{k(k-1)}$ and $r=\frac{(v-1)\lambda}{k-1}$ be integers. That is not sufficient in all cases but for each $k$ it is so long as $v$ is large enough (which is perhaps not all that large, and pretty soon there are (super)exponentially many non-isomorphic ones IIRC). I only know the ancient results that those two conditions *are* sufficient for $\lambda=1,2$ and $k=3,4,5$ with the one exception of $(v,k,\lambda)=(15,5,2).$ First let's work with $k=5.$ So for every $m \gt 1$ there are BIBDs with these $(v,r,k,\lambda)$ values:

$(20m+1,5m,5,1)$ ,$(20m+5,5m+1,5,1)$,$(10m+1,5m,5,2)$ and $(10m+5,5m+2,5,2).$

So we have a $(55,27,5,2)$ design. The projective plane over $GF_7$ is a $(57,8,8,1)$ design. Delete two points to get blocks of sizes $6,7,8$ but still $(r,\lambda)=(8,1).$ Now we can take $p$ copies of the first design and $q$ of the second to get $(r,\lambda)=(27p+8q,2p+q).$ Can we have $(2p+q)^2=27p+8q?$ Yes, if $q=4-2p+\sqrt{11p+16}.$ So $(p,q)=(3,5)$ gives a design with $v=55,r=121,\lambda=11.$ The repeat numbers are low enough to comfortably have distinct blocks.

That was a rather haphazardly constructed example. I just wanted to get away from $\lambda=1.$ I imagine would could do all manner of things with other $v$. I'll stick with $v=55.$ There is not a design with $(v,k,\lambda)=(55,4,1)$ but there is one with $(v,r,k,\lambda)=(55,36,4,2)$ so we are sure that it is not just two $\lambda=1$ designs. We could take 9 copies of this $(r,\lambda)=(36,2)$ design. More fun is one copy of that design along with $8$ copies of the mutilated projective plane to get a total of $(r,\lambda)=(36+8\cdot 8,2+8 \cdot 1)=(100,10).$ Also possible is two copies of the $(r,k,\lambda)=(36,4,2)$, two mutilated projective planes and four copies of the $(r,k,\lambda)=(27,5,2)$ for a total of $(r,\lambda)=(2\cdot36+2\cdot 8+4 \cdot 27,2 \cdot 2+2 \cdot 1+4\cdot2)=(196,14).$

I'll stop there but, since $55$ is of the form $6m+1$ there is a $(v,r,k,\lambda)=(55,27,3,1)$ design and of course there is the $(v,r,k,\lambda)=(55,54,2,1)$ Perusing the Google copy of the CRC handbook of combinatorial designs I see that there are also $(v,r,k,\lambda)=(56,11,11,2),(56,15,12,3))$ which could be mutilated for $v=55.$ So there may be further combinations to consider.

Here are some things I found for $v=56$:

$[0, 3, 1, [10, 100]], [4, 1, 2, [13, 169]], [0, 4, 3, [15, 225]], [6, 1, 9, [24, 576]],$

$ [0, 5, 10, [25, 625]], [0, 5, 15, [30, 900]], [6, 1, 16, [31, 961]], [0, 4, 28, [40, 1600]],$

$ [4, 1, 31, [42, 1764]], [0, 3, 36, [45, 2025]]$

The second item $[4, 1, 2, [13, 169]]$ means that 4 copies of the $(56,11,11,2)$ along with one copy of the $(56,15,12,3)$ and 2 copies of the $(56,55,2,1)$ design give $169=r=\lambda^2.$ There is no choice but to repeat the $k=2$ design however there are (according to the handbook) at least five non-isomorphic $(56,56,11,11,2)$ designs. I did not see that the mutilated projective plane was helpful for $v=56.$