Soln = for chaing bolting

An=(b-ndo)8

=(210-4 X 18)8

An=1104 mm$^{2}$

For zig zag bolting

1) along path AB & FG (2 holes) net area will be

An=$(b-nda)t+(\frac{\sum p^{2}}{4g})$

=(210 X 2 X 18)8+0

=1392 mm$^{2}$

2) Along path cnf (three holes)

An=(h-nda) X t+$\frac{\sum p^{2}}{4g}$

=(210-3 X18)X8+0

An=$1248mm^{2}$

3) Along path ACDE (four holes and staggered path

$An=(b-nda)X t+(\sum \frac{p^{2}}{4g}$

=$[(210-4X18)+(\frac{45^{2}}{4 X50})]\times8$

$An=1185 mm^{2}$

4) Along path FCDE(prove holes & one staggered path)

$An=[(b-ndo+\frac{\sum p^{2}}{4g}]\times t$

$[(210-4 \times 18)+\frac{40^{2}}{4 \times 50}]\times 8$

=$1168mm^{2}$

5) Along path FCB(three holes two staggered path

An=$[(210-3 \times 18)+(\frac{40^{2}}{4\times 50}+\frac{45^{2}}{4 \times 50})]\times 8$

=$1393mm^{2}$

min net are of zigzag bolting is $1168 mm^{2}$ along the path FCDE