# Where should a weight be placed in order to give no bending moment at amidships

#### Statement: A vessel of constant rectangular cross-section is 300 ft long and 35 ft broad, and floats at an even keel draught of 10 ft. in S.W. The weight of the vessel may be taken as being distributed uniformly over its entire length.Where should a weight be placed in order to give no bending moment at amidships?

Solution:
No bending moment at amidships means W = B

Additional moment caused by the weight will be equal to the additional buoyancy caused by it.

Now,

TPI=\frac{300\times35}{420}=25

MCT 1" = \frac{∆BML}{12L} = \frac{BL^2} {420×12} = \frac{35×300^2}{420×12}=625

Let, w= weight placed at a distance aft from amidships
d = parallel sinkage
x’ = trim

Hence,

d=\frac{w}{25} in = \frac{w}{300} ft

x^\prime = \frac{\text{Moment To Change Trim}}{\text{MCT 1}\prime\prime}\times\frac{l}{L}

=\frac{wx\times150}{625\times300} in

=\frac{wx}{625\times2\times12}ft

Now,

W = B

or,\ wx=\frac{d+d+x^\prime}{2}\times\frac{50\times35}{35}\times{75+\frac{150}{6}\times\frac{d+x^\prime-d}{d+x^\prime+d}}

=11250d+7500x\prime

Putting the value of d and x’.
x = 75 ft