split a dataframe

You could do use pd.cut on the cumulative sum of the B column:

th = 50

# find the cumulative sum of B 
cumsum = df.B.cumsum()

# create the bins with spacing of th (threshold)
bins = list(range(0, cumsum.max() + 1, th))

# group by (split by) the bins
groups = pd.cut(cumsum, bins)

for key, group in df.groupby(groups):
    print(group)
    print()

Output

   A   B
0  0  10
1  1  30

   A   B
2  2  50

   A   B
3  3  20
4  4  10
5  5  30

Here's a method using numba to speed up our for loop:

We check when our limit is reached and we reset the total count and we assign a new group:

from numba import njit

@njit
def cumsum_reset(array, limit):
    total = 0
    counter = 0 
    groups = np.empty(array.shape[0])
    for idx, i in enumerate(array):
        total += i
        if total >= limit or array[idx-1] == limit:
            counter += 1
            groups[idx] = counter
            total = 0
        else:
            groups[idx] = counter
    
    return groups

grps = cumsum_reset(df['B'].to_numpy(), 50)

for _, grp in df.groupby(grps):
    print(grp, '\n')

Output

   A   B
0  0  10
1  1  30 

   A   B
2  2  50 

   A   B
3  3  20
4  4  10
5  5  30

Timings:

# create dataframe of 600k rows
dfbig = pd.concat([df]*100000, ignore_index=True)
dfbig.shape

(600000, 2)

# Erfan
%%timeit
cumsum_reset(dfbig['B'].to_numpy(), 50)

4.25 ms ± 46.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# Daniel Mesejo
def daniel_mesejo(th, column):
    cumsum = column.cumsum()
    bins = list(range(0, cumsum.max() + 1, th))
    groups = pd.cut(cumsum, bins)
    
    return groups

%%timeit
daniel_mesejo(50, dfbig['B'])

10.3 s ± 2.17 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

Conclusion, the numba for loop is 24~ x faster.