1. 2.6 FORCE COMPONENTS ON CURVED SURFACES When the elemental forces p δA vary in direction, as in the case of a curved surface, they must be added as vector quantities  their components in three mutually perpendicular directions are added as scalars, and then the three components are added vectorially. With two horizontal components at right angles and with the vertical component - all easily computed for a curved surface - the resultant can be determined. The lines of action of the components also are readily determined.

2. Horizontal Component of Force on a Curved Surface The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface. The vertical plane of projection is normal to the direction of the component. Fig. 2.16: the surface represents any three-dimensional surface, and δA an element of its area, its normal making the angle θ with the negative x direction. Then Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane  force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection. To find the horizontal component at right angles to the x direction, the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined.

3. Figure 2.16 Horizontal component of force on a curved surface Figure 2.17 Projections of area elements on opposite sides of a body

4. When looking for the horizontal component of pressure force on a closed body, the projection of the curved surface on a vertical plane is always zero, since on opposite sides of the body the area-element projections have opposite signs (Fig. 2.17). Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C. If the element of area of the body cut by the prism at B is δAB and at C is δAC, then and similarly for all other area elements To find the line of action of a horizontal component of force on a curved surface, the resultant of the parallel force system composed of the force components from each area element is required. This is exactly the resultant of the force on the projected area, since the two force systems have an identical distribution of elemental horizontal force components. Hence, the pressure center is located on the projected area by the methods of Sec. 2.5.

5. Example 2.9 The equation of an ellipsoid of revolution submerged in water is x 2 /4 + y 2 /4 + z 2 /9 = 1. The center of the body is located 2 m below the free surface. Find the horizontal force components acting on the curved surface that is located in the first octant. Consider the xz plane to be horizontal and y to be positive upward. Projection of the surface on the yz plane has an area of Its centroid is located m below the free surface 

6. Vertical Component of Force on a Curved Surface The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface In Fig.2.18 an area element is shown with the force p δA acting normal to it. Let θ be the angle the normal to the area element makes with the vertical. Then the vertical component of force acting on the area element is p cos θ δA, and the vertical component of force on the curved surface is given by (2.6.2) p replaced by its equivalent γh; cos θ δA is the projection of δA on a horizontal plane  Eq. (2.6.2): (2.5.3-4) in which δ ϑ is the volume of the prism of height h and base cos θ δA, or the volume of liquid vertically above the area element

7. Figure 2.18 Vertical component of force on a curved surface Figure 2.19 Liquid with equivalent free surface

8. Fig. 2.19: the liquid is below the curved surface and the pressure magnitude is known at some point (e.g., O), an imaginary or equivalent free surface s-s can be constructed p/γ above O, so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point. The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface. In constructing an imaginary free surface, the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface; otherwise, the pressure distribution over the surface will not be correctly represented. With an imaginary liquid above a surface, the pressure at a point on the curved surface is equal on both sides, but the elemental force components in the vertical direction are opposite in sign  the direction of the vertical force component is reversed when an imaginary fluid is above the surface. In some cases a confined liquid may be above the curved surface, and an imaginary liquid must be added (or subtracted) to determine the free surface.

9. The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force. With the axis at O (Fig.2.18), in which is the distance from 0 to the line of action Since Fv = γ ϑ the distance to the centroid of the volume  the line of action of the vertical force passes through the centroid of the volume, real or imaginary, that extends above the curved surface up to the real or imaginary free surface

10. Example 2.10 A cylindrical barrier (Fig. 2.20) holds water as shown. The contact between cylinder and wall is smooth. Considering a 1-m length of cylinder, determine (a) its gravity force and (b) the force exerted against the wall. (a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water. (The imaginary free surface for CD is at elevation A.) The vertical force on BCD is The vertical force on AB is Hence, the gravity force per metre of length is (b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD. The horizontal components of force on BC and CD cancel; the projection of BCD on a vertical plane is zero , since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa.

11. Figure 2.20 Semifloating body

12. Tensile Stress in a Pipe and Spherical Shell Fig. 2.21: a circular pipe under the action of an internal pressure is in tension around its periphery; assuming that no longitudinal stress occurs, the walls are in tension Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart). If one-half of this ring is taken as a free body, the tensions per unit length at top and bottom are respectively T 1 and T 2 The horizontal component of force acts through the pressure center of the projected area and is 2pr, in which p is the pressure at the centerline and r is the internal pipe radius. For high pressures the pressure center may be taken at the pipe center; then T 1 = T 2, and T is the tensile force per unit length. For wall thickness e, the tensile stress in the pipe wall is

13. Figure 2.21 Tensile stress in pipe

14. Example 2.11 A 100 mm-ID steel pipe has a 6 mm wall thickness. For an allowable tensile stress of 70 MPa, what is the maximum pressure? From Eq. (2.6.6)

15. 2.7 BUOYANT FORCE Buoyant force: the resultant force exerted on a body by a static fluid in which it is submerged or floating  Always acts vertically upward (there can be no horizontal component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero) The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side Figure 2.22

16. Figure 2.22 Buoyant force on floating and submerged bodies

17. Fig. 2.22: the upward force on the bottom is equal to the gravity force of liquid, real or imaginary, which is vertically above the surface ABC, indicated by the gravity force of liquid within ABCEFA. The downward force on the upper surface equals the gravity force of liquid ADCEFA. The difference between the two forces is a force, vertically upward, due to the gravity force of fluid ABCD that is displaced by the solid. In equation form F B is buoyant force, V is the volume of fluid displaced, and γ is the unit gravity force of fluid The same formula holds for floating bodies when V is taken as the volume of liquid displaced

18. Fig.2.23: the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is δV is the volume of the prism. Integrating over the complete body gives γ is considered constant throughout the volume To find the line of action of the buoyant force, moments are taken about a convenient axis O and are equated to the moment of the resultant, thus, i is the distance from the axis to the line of action. This equation yields the distance to the centroid of the volume;  the buoyant force acts through the centroid of the displaced volume of fluid; this holds for both submerged and floating bodies. The centroid of the displaced volume of fluid is called the center of buoyancy.

19. Figure 2.23 Vertical force components on element of body

20. Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force, volume, unit gravity force, and relative density. Figure 2.24: two free-body diagrams for the same object suspended and gravity force determined in two fluids, F 1 and F 2 ; γ 1 and γ 2 are the unit gravity forces of the fluids. W and V, the gravity force and volume of the object, are to be found. The equations of equilibrium are written and solved:

21. Figure 2.24 Free body diagrams for body suspended in a fluid

22. A hydrometer uses the principle of buoyant force to determine relative densities of liquids Figure 2.25: a hydrometer in two liquids with a stem of prismatic cross section a Considering the liquid on the left to be distilled water (unit relative density S = 1.00), the hydrometer floats in equilibrium when V 0 is the volume submerged, γ is the unit gravity force of water, and W is the gravity force of hydrometer The position of the liquid surface is marked 1.00 on the stem to indicate unit relative density S. When the hydrometer is floated in another 1iquid, the equation of equilibrium becomes where ΔV = aΔh. Solving for Δh with Eqs. (2.7.2) and (2.7.3)

23. Figure 2.25 Hydrometer in water and in liquid of relative density

24. Example 2.12 A piece of ore having a gravity force of 1.5 N in air is found to have a gravity force 1.1 N when submerged in water. What is its volume, in cubic centimetres, and what is its relative density? The buoyant force due to air may be neglected. From Fig. 2.24

25. 2.8 STABILITY OF FLOATING AND SUBMERGED BODIES A body floating in a static liquid has vertical stability. A small upward displacement decreases the volume of liquid displaced  an unbalanced downward force which tends to return the body to its original position. Similarly, a small downward displacement results in a greater buoyant force, which causes an unbalanced upward force. A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position. A body has rotational stability when a restoring couple is set up by any small angular displacement.

26. Methods for determining rotational stability are developed in the following discussion A body may float in  stable equilibrium  unstable equilibrium (any small angular displacement sets up a couple that tends to increase the angular displacement)  neutral equilibrium (any small angular displacement sets up no couple whatever) Figure 2.26: three cases of equilibrium a. a light piece of wood with a metal mass at its bottom is stable b. when the metal mass is at the top, the body is in equilibrium but any slight angular displacement causes it to assume the position in a c. a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation; i.e., no couple results from an angular displacement

27. Figure 2.26 Examples of stable, unstable, and neutral equilibrium

28. A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig. 2.27a) When the object is rotated counterclockwise, the buoyant force and gravity force produce a couple in the clockwise direction (Fig. 2.27b) Figure 2.27 Rotationally stable submerged body

29. Normally, when a body is too heavy to float, it submerges and goes down until it rests on the bottom. Although the unit gravity force of a liquid increases slightly with depth, the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body Example: a ship is sure to go to the bottom once it is completely submerged, owing to compression of air trapped in its various parts

30. Determination of Rotational Stability of Floating Objects Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig. 2.26a). Certain floating objects, however, are in stable equilibrium when their center of gravity is above the center of buoyancy. Figure 2.28a: a cross section of a body with all other parallel cross sections identical. The center of buoyancy is always at the centroid of the displaced volume, which is at the centroid of the cross-sectional area below liquid surface in this case.

31. Figure 2.28 Stability of a prismatic body

32.  when the body is tipped (Fig. 2.28b), the center of buoyancy is at the centroid B' of the trapezoid ABCD, the buoyant force acts upward through B', and the gravity force acts downward through G, the center of gravity of the body When the vertical through B' intersects the original centerline above C, as at M, a restoring couple is produced and the body is in stable equilibrium The intersection of the buoyant force and the centerline is called the metacenter (M) When M is above G, the body is stable; when below G, it is unstable; and when at G, it is in neutral equilibrium The distance MG is called the metacentric height and is a direct measure of the stability of the body. The restoring couple is in which θ is the angular displacement and W the gravity force of the body

33. Example 2.13 In Fig. 2.28 a scow 6 m wide and 20 m long has a gross mass of 200 Mg. Its center of gravity is 30 cm above the water surface. Find the metacentric height and restoring couple when Δy = 30 cm. The depth of submergence h in the water is The centroid in the tipped position is located with moments about AB and BC, By similar triangles AEO and B'PM,

34. G is 1.97 m from the bottom; hence The scow is stable, since is positive; the righting moment is #

35. Nonprismatic Cross Sections For a floating object of variable cross section (e.g., a ship) (Fig. 2.39a), a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ The horizontal shift in center of buoyancy r (Fig. 2.29b) is determined by the change in buoyant forces due to the wedge being submerged, which causes an upward force on the left, and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right. The force system, consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges, must have as resultant the equal buoyant force at B'. With moments about B to determine the shirt r,

36. Figure 2.29 Stability relations in a body of variable cross section

37. The amount of the couple can be determined with moments about O, the centerline of the body at the liquid surface For an element of area δA on the horizontal section through the body at the liquid surface, an element of volume of the wedge is xθ δA. The buoyant force due to this element is γxθ δA, and its moment about O is γx2θ δA, in which θ is the small angle of tip in radians. By integrating over the complete original horizontal area at the liquid surface, the couple is determined to be I is the moment of inertia of the area about the axis y-y (Fig.2.29a) Substitution into the above equation produces V is the total volume of liquid displaced Since θ is very small and

38. Example 2.14 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig. 2.30. Its center of buoyancy is 2.0 m below the water surface, and its center of gravity is 0.5 m below the water surface. Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis). GB = 2 – 0.5 = 1.5 m For rolling For pitching #

39. Figure 2.30 Horizontal cross section of a ship at the waterline