A box barge of 45 m long and 15 m wide floats at a level keel draught of 2 m in sea water, the load being uniformly distributed over the full length. Two masses, each of 30 tonne, are loaded at 10 m from each end & 50 tonne is evenly distributed between them. Sketch shear force diagram & give the maximum shear force.
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Given
L = 45 m
B = 15 m
d = 2 m
To find
1.Shearing force diagram
2.Maximum shear force
Solution
After addition of different masses there will be shearing force present.
So first we have to find the additional buoyancy required, It is given by
Additional buoyancy = (Total masses added / Length) g
= (30+50+30 / 45) g
= 2.444 g
Now consider point A
Shearing force at point A = 0
Shearing force = Area of load diagram on one side of that point
Load = Difference between weight and buoyancy
Now consider point B
Shearing force at left of point B = Load on that point x Length, Load is only buoyancy force
= 2.444 g x 10
= 24.44 g
Shearing force at right of point B = Shearing force at left of B - Shearing force at B
= 24.44 g - 30 g (Because load at point B is 30)
= -5.56 g
Shearing force in Point B to D = (Load / Distance) g (As it is uniformly distributed)
= (50 / 25) g
= 2 g
Shearing force at point C = Shearing force due to buoyancy at C - Shearing force at B - Shearing force at B to D
Shearing force at C = (2.444 g x 22.5) - 30 g - (2 g x 12.5)
= 55 g - 30 g -25 g
= 0
Since the vessel is symmetrically loaded these values will be repeated but in opposite sign.
1)
2) From diagram we can see that maximum shearing force occurs at point B and D
= 24.44 g
= 24.44 x 9.81
= 239.8 kN
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