PROBLEM DEFENITIONA Barge of dimension 150m 30m 20m and thickness 5mm is loaded as given in diagram below. Calculate the shear force diagram for this loading.arrangement

Given datadensity of barge material = 7800 kg/m3 density of cargo = 2400 kg/m3

4M2M 1M 1M

4M

LOADED AREA

Assumptions1) Every region in the barge can be treated as a cantilever beam 2) All forces acting upwards(including shear force) are treated as positive and all acting down as negative . 3) The partitions for different regions are dimensionless.

SOLUTION SETWeight of empty barge = volume of barge density of barge material = ( 150 30 20 - 149.95 29.95 19.95) 7800 9.81 = 30.952 MN Weight of empty barge /unit length = 0.2603 MN/metre run

weight of cargo material = 29.95 29.95 (4+2+1+4+2) 2400 9.81 = 274.547 MN/m

Total weight of ship = 274.547 + 30.952 = 305.499 MN

(which is nothing but equal to buoyant force)

Buoyant force acting per unit length = 2.03666 MN/ metre run

Calculation of loading at various regions Divide the different loading parts into five regions as shown below.

REGION 1

REGION 2

REGION 3

REGION 4

REGION 5

Weight of ship(empty)

weight of cargo -2.80648 -1.40323 -0.70162 -2.80648 -1.40323

buoyant force 2.03666 2.03666 2.03666 2.03666 2.03666

All forces per unit metre length(in MN)

Region 1 Region 2 Region 3 Region 4 Region 5

-0.2603 -0.2603 -0.2603 -0.2603 -0.2603

REGION 1load = weight of empty ship +weight of cargo - buoyant force

V = 0.9857x At x= 30m , V = 29.571 MN

wx

x

Vwx x

REGION 2

29.571

V

V = 29.571-0.4224x At x= 30m , V = 16.899MN

wx16.8909

REGION 3

x

V16.899

V = 16.899-1.12633 At x= 30m , V = -16.8909 MN

wx x

REGION 4

V

V = -16.8909+.9857x At x= 30 , V =12.6801 MN 12.6801 REGION 5

wx x

VV = 12.6801-0.4224x At x= 30m , V = 0.0129 MN

Length(m) Shear force(MN)

0 0

30 29.571

60 16.899

90 -16.8909

120 12.6801

150 0.0129

1

4

2

3

5

2

3

5

Effect of shear force on different sections.

Shear force40

30

20

10

Shear force

0 0 -10 30 60 90 120 150

-20

diagram below.