234 5.0 Outline 234 Introduction 235 Center of Mass 236 Centroids (Line, Area, Volume) 241 Composite Bodies and Figures 257 Theorem of Pappus 265 Fluid Statics 272 5.0 Outline
5.0 Introduction Ch. 5: Distributed Forces 235 When forces are applied over a region whose dimensions are not negligible compared with other pertinent dimensions, we must account for the actual manner in which the force is distributed by summing up the effects of the distributed force over the entire region. For this purpose, we need to know the intensity of the force at any location and we will use the integration to find their total effects. 5.0 Introduction
236 5.1 Center of Mass CG: point where the resultant gravitational force W acts CM: point at which the total mass is visually concentrated Location of the CG and CM are found by the principle of moment sum of the moments = moment of the sum 5.1 Center of Mass
237 Center of Gravitiy Assumption 1) uniform intensity of the earth s force field over the body 2) parallel field of force due to the gravitational attraction the moment of the resultant gravitational force W about any axis = the sum of the moments about the same axis of the gravitational force dw acting on all particles of the body r = xi+ yj+ zk r = xi+ yj+ zk r = CG rdw dw 5.1 Center of Mass
Center of Mass Ch. 5: Distributed Forces 238 Assumption 1) uniform intensity of the earth s force field over the body 2) parallel field of force due to the gravitational attraction W = mg & dw = gdm r = CM rdm dm ( ) dm = ρdv, e.g. ρ = ρ x, y, z, dv = dxdydz r = CM rρdv ρdv CM is independent of gravitational effect and hence is a unique property of the body CM = CG if the gravity field is treated as uniform and parallel 5.1 Center of Mass
Center of Volume Ch. 5: Distributed Forces 239 Assumption 1) constant intensity of the earth s force field over the body 2) parallel field of force due to the gravitational attraction r = CV rdv dv CV depends only on the shape of the body, Independent of the material CV = CM if the mass intensity (density) is uniform over the volume 5.1 Center of Mass
240 Look for symmetry to simplify the calculation rectangular body rectangular coordinates circular body polar coordinates For a homogeneous body, CM will lie on a line or plane of symmetry 5.1 Center of Mass
241 5.3 Centroids (Line, Area, Volume) is purely the geometrical (shape) property of the body since any reference to its mass properties has disappeared. Line: r = C r C dl, not always on the line dl assuming the cross sectional area is constant over the length 5.3 Centroids
Area: r = C r C Ch. 5: Distributed Forces da, not always on the surface da assuming the thickness is constant over the entire area 242 r C da = first moment of area Volume: r = C r C dv, not always inside the volume dv 5.3 Centroids
Guidelines in choosing the differential element & setting up the integrals a) Order of element: prefer 1 st order diff. element to higher order, so only one integration will be required 243 ( ) A = l y dy = dxdy ( ) = π = 2 V r y dy dxdydz 5.3 Centroids
244 b) Continuity: choose an element which can be integrated in one continuous operation to cover the whole object (x 2, y 2 ) h 1 (x) h 2 (x) x x 2 2 1( ) 2( ) A = ( ) A = h x dx h x dx 0 x 1 y 2 0 l y dy 5.3 Centroids
245 c) Neglect high order terms: higher order terms may be dropped compared with lower order terms dxdy dxdy 1 1 A = ydx + dxdy = y + dy dx = ydx 2 2 note: dxdy dxdy 5.3 Centroids
246 d) Choice of coordinates: to appropriately describe the boundaries ( x, y ) O O y O A x y xdy = O O 0 A θ O = 0 1 2 2 rd θ 5.3 Centroids
247 e) Centroidal coordinate of the element: use the coordinate of the centroid of the element for the moment arm in setting up the moment of the differential element C C ( ) ( ) first moment = x da = x y l y dy C C 2 ( ) π ( ) first moment = x dv = x z r z dz 5.3 Centroids
248 P. 5/1 Locate the centroid of the area of a circular sector w.r.t. its vertex 5.3 Centroids
249 P. 5/1 by symmetry, Y = 0 select differential element to be strip of partial circular ring with r varying from 0 to r o XA xcda = da = 2αrodr o, xc = r r o α o o α o o α 0 0 X 2 r dr ro sinα α r sinα 2rsinα = 2 r dr, X = 3α 5.3 Centroids
250 P. 5/1 by symmetry, Y = 0 select differential element to be strip of sector with θ varying from α to α 1 2 XA xcda da r ( rd θ), xc r cos = = = θ 2 3 α α 2 2 X rd rcos rd, X α 1 2 1 2rsinα θ = 2 θ θ = 3 2 3α α 5.3 Centroids
251 P. 5/2 Locate the centroid of the area shown in the figure by direct integration 5.3 Centroids
252 P. 5/2 θ by symmetry, the centroid lies on the line y = a x differential area = horizontal strip strip of sector with θ varying from 0 to π/2 1 XA xcda = dy = ad sin, da = ady a ad 2 = θ ( θ ) 2 da a sin 1/ 2 d 2 xc1 = a/2, xc2 = asinθ 3 ( ) ( π) ( θ) θ ( θ) π / 2 π/2 π/2 2 2 2 1 2 X a sinθ 1/ 2 dθ = a/2 a sinθdθ asinθ a dθ 3 2 0 0 0 ( π ) ( π) 2a 10 3 a X =, Y= a X= 34 34 2/3asinθ x c1 5.3 Centroids
253 P. 5/3 Determine the coordinates of the centroid of the volume obtained by revolving the shaded area about the z-axis through 90 angle 5.3 Centroids
254 P. 5/3 by symmetry, the centroid lies on the vertical plane making 45 with the x-axis X = Y differential volume as shown in the figure with y varying from 0 to a ( ) 2 2 2 2 2 y + z a = a, z = a a y c 2 2 ( ) 2πy π dv = zdy = a a y ydy 4 2 x ( π ) ysin /4 a a y = cos ( π/4) = 2y/ π = y c, zc = z/2 = π /4 2 2 2 a π ( ) ( ) 2 2 π 2 2 a π c 0 0 XV = x dv X a a y ydy = 2y/ a a y ydy 2 2 4 3 X= a = Y π 4 2 2 a a y π 2 2 ( ) ( ) a a 2 2 π c 0 0 ZV z dv Z a a y ydy a a y ydy = = 2 2 2 Z = a/4 z/2 4y/(π 2) dy y 5.3 Centroids
255 P. 5/4 Locate the center of mass G of the steel half ring. 5.3 Centroids
256 P. 5/4 by symmetry, the centroid lies on the intersection line of the zero horizontal plane and the middle vertical plane, only r needed to be specified differential volume as shown in the figure with x varying from a to a x + y = a, y= a x 2 2 2 2 2 ( ) 2 2 π ( ) dv = 2 a x dx R x ( ) ( π ) R x sin / 2 2 rc = = R x π /2 π ( ) a a 2 2 2 2 2 c π ( ) ( ) π ( ) π a a rv r dv r 2 R x a x dx R x 2 R x a x dx = = y 2 2 a + 4R r = 2π R x x R-x dx 5.3 Centroids
5.3 Composite Bodies and Figures 257 Discrete version in finding the centroid CM may be easily known for subparts of the object mx mx mz X = Y = Z= m m m 5.3 Composite Bodies and Figures
258 Depending on the geometry and the density of the object, mass m may be replaced by l, A, or V. Addition vs. Subtraction Table consultation Systematic tabulation of m i, r ci, m i, mr i ci 5.3 Composite Bodies and Figures
259 P. 5/5 Determine the coordinates of the centroid of the shaded area 5.3 Composite Bodies and Figures
260 P. 5/5 i A i x ci y ci A i x ci A i y ci 1 0.1 0.2 0.125 0.02 0.0125 2 0.01875 0.45 0.083 8.4375e-3 1.5625e-3 3-0.0113 0.2 0.125-2.262e-3-1.4137e-3 1 3 2 X Y Ax i ci = = A i ci = = i Ay A i 243.6 mm 117.7 mm 5.3 Composite Bodies and Figures
261 P. 5/6 A cylindrical container with an extended rectangular back and semicircular ends is all fabricated from the same sheet-metal stock. Calculate the angle α made by the back with the vertical when the container rests in an equilibrium position on a horizontal surface. 5.3 Composite Bodies and Figures
262 P. 5/6 i x i y i z i A i A i x i A i y i A i z i 1 200 0 100 8e4 16e6 0 8e6 2 200-150 -300/π 6πe4 12πe6-9πe6-18e6 3 0-150 -200/π 1.125πe4 0-1.6875πe6-2.25e6 4 400-150 -200/π 1.125πe4 4.5πe6-1.6875πe6-2.25e6 5.3 Composite Bodies and Figures
263 P. 5/7 z 1 x y 2 4 3 X Y Z Ax 16E6 + 16.5π E6 A 8E4 + 8.25π E4 i ci = = = i Ay 12.375π E6 A 8E4 + 8.25π E4 i ci = = = i Az 8E6 22.5E6 A 8E4 + 8.25π E4 i ci = = = i 200 mm 114.62 mm 42.75 mm 5.3 Composite Bodies and Figures
264 P. 5/7 mg z Y α Z 150 Y 1 α = tan = 39.6 Z y N 5.3 Composite Bodies and Figures
265 5.4 Theorem of Pappus is about the surface or the volume created by revolving the planar curve or the planar area about the nonintersecting line in its plane. Revolved Surface A = θ ydl = θyl area is the same as the lateral area of a cylinder of length L and radius y 5.4 Pappus Theorem
266 Revolved Volume V = θ yda = θya volume is obtained by multiplying the planar area by the circumference of the circular path of the centroid about the revolving axis 5.4 Pappus Theorem
267 Usage 1. determining the area or volume of revolution 2. find the centroid of planar curve or planar area when the corresponding area or volume are known 5.4 Pappus Theorem
268 P. 5/8 A hand-operated control wheel made of aluminum has the proportions shown in the cross-sectional view. The area of the total section shown is 15,200 mm 2, and the wheel has a mass of 10 kg. Calculate the distance r to the centroid of the half-section. The aluminum has a density of 2.69 Mg/m 3. 5.4 Pappus Theorem
269 P. 5/8 M ρ = V = 10 / 2690 = 2π 7600E 6 r V r = 77.85 mm 5.4 Pappus Theorem
270 P. 5/9 Calculate the mass m of concrete required to construct the arched dam shown. Concrete has a density of 2.40 Mg/m 3. 5.4 Pappus Theorem
271 P. 5/9 find the centroid w.r.t. the rotation axis first π 2 4 70 80 80 ( 40 + 120) 70 + 120 ya i i 4 3π y y = = A π i 2 80 80 70 4 = 175.52 m calculate the revolved volume from Pappus to find the mass π V = ya V = π / 3 175.52 80 80 70 4 [ θ ] [ ρ ] m= V m= 2400 V = 1.126E9 kg = 1.126E6 Mg 2 5.4 Pappus Theorem
272 5.5 Fluid Statics consider the equilibrium of bodies subjected to forces due to fluid pressure. Fluid force on a surface normal compressive force shear force ( = 0, static) In equilibrium, fluid pressure at any given point is the same in any direction. The pressure is a function of the vertical dimension. dp = ρgdh pressure increases with depth 5.5 Fluid Statics
273 dp = ρgdh With ρ constant, we may integrate and get p = p + ρgh = absolute pressure, where p is the surface pressure at h = 0. o o If p o = atmospheric pressure and is used as the reference value, p = ρgh = gage pressure [ ] [ ] ( ) Unit: Pa = kg/m 3 m/s 2 m kg-m/s 2 / m 2 = = N/m 2 5.5 Fluid Statics
274 Resultant force due to pressure distribution on surface 1. Flat surface 5.5 Fluid Statics
275 Variation of p over the depth is linear (ρgh) section of the pressure distribution is trapezoidal Pressure distribution 12345678 An element of plate area over which p acts is da = xdy An increment of the resultant force is dr = pda 5.5 Fluid Statics
276 Magnitude ( ) ( ) R = dr = pda = ρgh y x y dy = volume of pressure distribution or R = ρg hda = ρgha = p A average altitude h is the depth corresponding to the centroid O of the flat surface area, A, exposed to the pressure. At that depth, the pressure ρgh is the average pressure, p, of the pressure distribution acting on the surface area. av av 5.5 Fluid Statics
277 Direction perpendicular to the flat surface Line of action using principle of moment ypxdy RY = ydr, Y = pxdy = centroidal (C) y-coord of the volume of pressure distribution through which the resultant actually passes center of pressure, P = projection of centroidal volume C along the pressured force direction onto the plate = piercing point of the line of action R to the plate centroid of the plate area, A if x is constant, Y = = yda da ' ' ' centroidal y-coord of the area A (not A) of the pr essure profile 5.5 Fluid Statics
278 Resultant force due to pressure distribution on surface 2. Cylindrical surface with constant width, b dr continuously changes direction 5.5 Fluid Statics
279 Magnitude R = volume of pressure distribution = ρgha h = depth corresponding to the centroid O of the curve AB x ( ) ( ) R = b pdl = b pdy, R = b pdl = b pdx Direction θ = x y y tan 1 Line of action ( R / R ) y x y x, Y y x xdr xpdx ydr ypdy X = = = = dr pdx dr pdy at centroid of volume of pressure distribution 5.5 Fluid Statics
280 Equilibrium of block of liquid P x, P y = resultants of the pressures W = weight of the liquid block passing through centroid of area ABC (constant width b) R = reaction of the surface on the block of liquid from equilibrium equations 5.5 Fluid Statics
281 Buoyancy reaction force the fluid exerted on the boundary of the cavity Resultant of the pressure forces over the surface = weight of the fluid Resultant must pass through the C.M. of the fluid lump 5.5 Fluid Statics
282 Buoyancy Replace the fluid lump by an object of the same dimension, the surface forces acting on the body will be the same as those acting on the fluid lump. Resultant force (buoyancy) exerted on the surface of an object immersed in a fluid = weight of the fluid displaced Its line of action passes through the C.M. of the displaced fluid. 5.5 Fluid Statics
283 Buoyancy F = ρvg ρ = density of the displaced fluid V = volume of the displaced fluid When ρ o <ρ f and fully immersed, object s weight < buoyancy force object will rise to the equilibrium position (ρ o >ρ f ) where the object s weight = Σbuoyancy force 5.5 Fluid Statics
284 P. 5/10 5.5 Fluid Statics
285 P. 5/10 Stress in the shell is induced by the compressive force from the surrounding sea water Write FBD of the vehicle sectioned in hemisphere and display only force in horizontal direction F σda x ( ) ρ ( ) ρ y F = pda = g hda = gha x Fx = 0 ρgha y σda = 0 h = depth of the centroid of circular area = 3000 m σ ρgha 1030 9.81 3000 π 0.75 2 x = = = 2 2 Aring π 0.75 π 0.725 x 462.4 MPa 5.5 Fluid Statics
286 P. 5/11 The cross section of a fresh-water tank with a slanted bottom is shown. A rectangular door 1.6 x 0.8 m in the bottom of the tank is hinged at A and is opened against the pressure of the water by the cable under a tension P as shown. Calculate P. 5.5 Fluid Statics
287 P. 5/11 Assumption: door has negligible weight and volume, i.e., W = 0 and B = 0 F 2 P F 1 A 1.2 m R 1.2+1.6sin30 m pressure force magnitude = volume of pressure distribution and divide into two parts -- rectangle and triangle 1 ( ) ( ) F = 1000 9.81 1.2 1.6 0.8 F = 1/ 2 1000 9.81 0.8 1.6 0.8 P 2 MA = 0 P 1.6cos30 + F1 0.8 + F2 1.6 2 / 3= 0 = 12566 N 5.5 Fluid Statics
288 P. 5/12 y W B O B W pressure referred to atmosphere ( ) Fy = 0 900g A 150 + 1030g A 250 h 800g A 400 = 0, h = 70.4 mm 5.5 Fluid Statics
289 P. 5/13 A channel-marker buoy consists of a 2.4 m hollow steel cylinder 300 mm in diameter with a mass of 90 kg and anchored to the bottom with a cable as shown. If h = 0.6 m at high tide, calculate the tension T in the cable. Also find the value of h when the cable goes slack as the tide drops. The density of sea water is 1030 kg/m 3. Assume the buoy is weighted at its base so that it remains vertical. If the C.M. of the buoy is in the geometric center of the cylinder, calculate the angleθwhich would be made by the buoy axis with the vertical when the water surface is 1.5 m above the lower end of the cylinder. Neglect the diameter compared with the length when locating the center of buoyancy. 5.5 Fluid Statics
290 P. 5/13 B W W T T B buoy is weighted at its base so it remains vertical Fy = 0 B Mg T = 0 2 π 0.3 if h = 0.6 m, B = ρvg = 1030 1.8g N 4 T = 402.7 N ( ) when the cable goes slack, T = 0 B decreases h increases 2 π 0.3 1030 ( 2.4 h) g 90g = 0, h = 1.164 m 4 5.5 Fluid Statics
291 P. 5/13 W θ 1.5 m B T 2 π 0.3 B = ρvg = 1030 ( 1.5/ cosθ) g 4 MO = 0 B 0.75tanθ 90g 1.2sinθ = 0 θ = θ = 2 sin 0, or cos 0.7584 θ = 0, ± 29.44 5.5 Fluid Statics
292 P. 5/14 A fresh water channel 3 m wide is blocked at its end by a rectangular barrier, shown in section ABD. Supporting struts BC are spaced every 0.6 m along the 3 m width. Determine the compression C in each strut. Neglect the weights of the members. 5.5 Fluid Statics
293 P. 5/14 R A h = 1.2sin60 F 6C ( ) R = 0.5 1.2 1000 9.81 1.2sin 60 3 MA = 0 R 0.4 + 6Csin60 0.6 = 0, C = 2.354 kn 5.5 Fluid Statics
294 P. 5/15 The barge crane of rectangular proportions has a 6x15 m cross section over its entire length of 40m. If the maximum permissible submergence and list in sea water are represented by the position shown, determine the corresponding maximum safe mass m O that the barge can handle at 10 m extended position of the boom. Also find the total displacement m in the metric tons of the unloaded barge. The distribution of machinery and ballast places the CG of the barges, minus the mass m O, at the center of the hull. 5.5 Fluid Statics
295 P. 5/15 mg B = ρ Vg = 1030 15 6 40 / 2 g m O 3 m O g 7.5 2 5 B ( ) MG = 0 Bcosθ 2.5 Bsinθ 1 mog 10cosθ + 23sinθ = 0 = 203 Mg Fy = 0 B mg mog = 0, m = 1651 Mg 5.5 Fluid Statics
296 P. 5/16 The fresh water side of a concrete dam has the shape of a vertical parabola with vertex at A. Determine the position b of the base point B through which acts the resultant force of the water against the dam face C. 5.5 Fluid Statics
P. 5/16 Ch. 5: Distributed Forces W x F 12 m R y = W 297 36 0 ( ) ( ) y = ax or x = y/a in 1 quadrant contains 27, 36 a = 36 / 27 2 st 2 y= 36 2 A = xdy = xy = 648 m 3 y= 0 36 2 AX = xcda = x / 2dy, X = 10.125 m 0 F = volume of the pressure distribution = 1/ 2 36 1000 9.81 36 h W = weight of the fluid block = 1000 9.81 648 h 2 D MD = 0 F 18 W 10.125 + W b = 0, b = 28.125 m R x 5.5 Fluid Statics
298 P. 5/17 A flat plate seals a triangular opening in the vertical wall of a tank of liquid of density ρ. The plate is hinged about the upper edge O of the triangle. Determine the force P required to hold the gate in a closed position against the pressure of the liquid. 5.5 Fluid Statics
P. 5/17 Ch. 5: Distributed Forces O b y x dr dy a 299 P x = b(1-y/a) ( ) ρ ( ) ( ) dr = pda = ρg h + y xdy = g h + y b 1 y/a dy y= a ρgba MO = 0 Pa + ydr = 0, P = h + a/2 6 y= 0 ( ) 5.5 Fluid Statics