Download presentation
Presentation is loading. Please wait.
Published bySophie Allison Modified over 6 years ago
1
GOVERNMENT ENGG. COLLEGE B.E Semester III :- Civil Engineering
Dr. S & S. S. GHANDHY, GOVERNMENT ENGG. COLLEGE SURAT B.E Semester III :- Civil Engineering Fluid mechanics Prepared By :- Gajiwala Nirali ( ) Guided By:- Prof. G. M. Savaliya
2
META-CENTRE It is defined as the point about which a body starts oscillating when the body is tilted by a small angle. The meta-centre may also be defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement. Consider a body floating in a liquid. Let the body is in equilibrium and G is the centre of gravity and B the centre of buoyancy. For equilibrium, both the points lie on the normal axis, which is vertical.
3
META-CENTRIC HEIGHT The distance MG, i.e., the distance between the meta-centre of a floating body and the centre of gravity of the body is called meta-centric height.
4
Experimental method to determine metacentric height
Consider a ship floating in water Initially ship is in equillibrium so deck is horizontal. Now weight P is moved transversely through a distance x across deck, so ship tilts through small angle θ and comes to rest in new position of equillibrium. In new position also C.G. and centre of buoyancy will be vertically in line.
5
Experimental method to determine metacentric height
Moment due to movement of P = moment caused due to shifting of G to G₁ P x = W. GG₁ P x = W. GM tanθ So GM = P.x / W. Tanθ If l is the length of plumb line and d is the distance moved by it on the horizontal scale, then tan θ = d/l
6
ANALYTICAL METHOD FOR META-CENTRE HEIGHT
Fig shows the position of a floating body in equilibrium. The location of centre of gravity and centre of buoyancy in this position is at G and B. The floating body is given a small angular displacement in the clockwise direction. The new centre of buoyancy is at 5). The vertical line through B₁ cuts the normal axis at M. Hence M is the meta-centre and CM is meta-centric height.
7
Analytical method for metacentric height
The angular displacement of the body in clockwise direction causes vertical force dFb acting through C.G. of one prism,while the equal and opposite force dFb acting vertically downward through C.G. of another prism. Consider a small elementary strip of thickness dx in portion OBB’ at a distance x from O, towards the right of the axis .
8
Analytical method for metacentric height
Area of the strip dA = height × thickness =(x θ) × dx L is the length of the floating body, then Volume of the strip = area × L = (x θ) × dx × L So weigth of strip = ƍg × volume dFb = ƍg xθL.dx Similarly, weight of strip in OAA’ dFb = ƍg xθL.dx
9
Analytical method for metacentric height
10
Analytical method for metacentric height
11
Analytical method for metacentric height
12
Analytical method for metacentric height
13
Example 1 A rectangular pontoon is 5 m long, 3 m wide and 1.20 m high. The depth of immersion of the pontoon is 0.80 m in sea water. If the centre of gravity is 0.6 m above the bottom of the pontoon, determine the meta-centric height. The density/or sea water = 7025 kg/m3.
14
Example 1 - solution
15
Example 2 A block of wood of specific gravity 0.7 floats in water. Determine the meta-center height of the block if its size is 2 m x 1 m x 0.8 m.
16
Example 2 - solution
17
Example 2 - solution
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.