際際滷

際際滷Share a Scribd company logo

Distributes force system centroid and moment of inertia

Alemayehu E
Alemayehu E

Distributes force system

Convert to study guideBETA

Transform any presentation into a summarized study guide, highlighting the most important points and key insights.

1 of 66
Download to read offline
Ch. 5: Distributed Forces 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 234
Ch. 5: Distributed Forces 5.0 Introduction 5.0 Introduction 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. 235
Ch. 5: Distributed Forces 5.1 Center of Mass 5.1 Center of Mass Location of the CG and CM are found by the principle of moment CG: point where the resultant gravitational force W acts CM: point at which the total mass is visually concentrated sum of the moments = moment of the sum 236
Ch. 5: Distributed Forces Center of Gravitiy 5.1 Center of Mass Assumption 1) uniform intensity of the earths 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 x y z x y z dW dW = CG r i + j+ k r = i + j+ k r r = 237
Ch. 5: Distributed Forces Center of Mass 5.1 Center of Mass ( ) W mg & dW gdm dm dm dm dV, e.g. x, y, z , dV dxdydz dV dV = = = = = CM CM r r = r r = Assumption 1) uniform intensity of the earths force field over the body 2) parallel field of force due to the gravitational attraction CM = CG if the gravity field is treated as uniform and parallel CM is independent of gravitational effect and hence is a unique property of the body 238
Ch. 5: Distributed Forces Center of Volume 5.1 Center of Mass Assumption 1) constant intensity of the earths force field over the body 2) parallel field of force due to the gravitational attraction CV depends only on the shape of the body, Independent of the material dV dV CV r r = CV = CM if the mass intensity (density) is uniform over the volume 239
Ch. 5: Distributed Forces 5.1 Center of Mass 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 240
Ch. 5: Distributed Forces 5.3 Centroids (Line, Area, Volume) is purely the geometrical (shape) property of the body since any reference to its mass properties has disappeared. 5.3 Centroids Line: dL , not always on the line dL C C r r = assuming the cross sectional area is constant over the length 241
Ch. 5: Distributed Forces dA , not always on the surface dA C C r r = 5.3 Centroids Area: assuming the thickness is constant over the entire area dV , not always inside the volume dV C C r r = dA first moment of area = C r Volume: 242
Ch. 5: Distributed Forces 5.3 Centroids Guidelines in choosing the differential element & setting up the integrals a) Order of element: prefer 1st order diff. element to higher order, so only one integration will be required ( ) ( ) 2 A l y dy dxdy V r y dy dxdydz = = = = 243
Ch. 5: Distributed Forces 5.3 Centroids b) Continuity: choose an element which can be integrated in one continuous operation to cover the whole object ( ) ( ) 2 2 1 x x 1 2 0 x A h x dx h x dx ( ) 2 y 0 A l y dy = h1(x) h2(x) (x2, y2) 244
Ch. 5: Distributed Forces dxdy 5.3 Centroids c) Neglect high order terms: higher order terms may be dropped compared with lower order terms 1 1 A ydx dxdy y dy dx ydx 2 2 note: dxdy dxdy 錚 錚 = + = + = 錚 錚 錚 錚 dxdy 245
Ch. 5: Distributed Forces ( ) O O x , y 5.3 Centroids d) Choice of coordinates: to appropriately describe the boundaries O y O O 0 A x y xdy = O 2 0 1 A r d 2 慮 慮 = 246
Ch. 5: Distributed Forces 5.3 Centroids 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 = = ( ) ( ) 2 C C first moment x dV x z r z dz = = 247
Ch. 5: Distributed Forces 5.3 Centroids P. 5/1 Locate the centroid of the area of a circular sector w.r.t. its vertex 248
Ch. 5: Distributed Forces 5.3 Centroids P. 5/1 o o c o o c r r o o o o o 0 0 by symmetry, Y 0 select differential element to be strip of partial circular ring with r varying from 0 to r r sin XA x dA dA 2 r dr , x r sin 2rsin X 2 r dr 2 r dr , X 3 留 留 留 留 留 留 留 留 留 = 錚 錚 = = = 錚 錚 = = 249
Ch. 5: Distributed Forces 5.3 Centroids P. 5/1 ( ) c c 2 2 by symmetry, Y 0 select differential element to be strip of sector with varying from to 1 2 XA x dA dA r rd , x r cos 2 3 1 2 1 2rsin X r d rcos r d , X 2 3 2 3 留 留 留 留 慮 留 留 慮 慮 留 慮 慮 慮 留 = 錚 錚 錚 錚 = = = 錚 錚 錚 錚 錚 錚 = = 250
Ch. 5: Distributed Forces 5.3 Centroids P. 5/2 Locate the centroid of the area shown in the figure by direct integration 251
Ch. 5: Distributed Forces 5.3 Centroids P. 5/2 ( ) ( ) ( ) c 2 c1 c2 by symmetry, the centroid lies on the line y a x differential area horizontal strip strip of sector with varying from 0 to /2 1 XA x dA dy ad sin , dA ady a ad 2 dA a sin 1/ 2 d 2 x a/2, x a 3 慮 慮 慮 慮 慮 慮 = = 錚 錚 = = = 錚 錚 = = = ( ) ( ) ( ) ( ) / 2 /2 /2 2 2 2 0 0 0 sin 2 1 X a sin 1/ 2 d a/2 a sin d asin a d 3 2 10 3 a 2a X , Y a X 3 4 3 4 慮 慮 慮 慮 慮 慮 慮 錚 錚 = 錚 錚 錚 錚 = = = 慮 xc1 2/3asin慮 252
Ch. 5: Distributed Forces 5.3 Centroids P. 5/3 Determine the coordinates of the centroid of the volume obtained by revolving the shaded area about the z-axis through 90属angle 253
Ch. 5: Distributed Forces 5.3 Centroids P. 5/3 ( ) ( ) 2 2 2 2 2 2 2 c 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 y z a a , z a a y 2 y dV zdy a a y ydy 4 2 x 属 = + = = = = ( ) ( ) ( ) ( ) ( ) ( ) 2 2 c c a a 2 2 2 2 c 0 0 a a 2 2 2 2 2 2 c 0 0 ysin /4 a a y cos /4 2y/ y , z z/2 /4 2 XV x dV X a a y ydy 2y/ a a y ydy 2 2 4 3 X a Y 4 a a y ZV z dV Z a a y ydy a a y ydy 2 2 2 Z a/4 = = = = = 錚 錚 = = 錚 錚 錚 錚 = = 錚 錚 錚 錚 錚 錚 = = 錚 錚 = 4y/(2) z/2 y dy 254
Ch. 5: Distributed Forces 5.3 Centroids P. 5/4 Locate the center of mass G of the steel half ring. 255
Ch. 5: Distributed Forces 5.3 Centroids 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 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 c a a 2 2 2 2 c a a 2 2 from a to a x y a , y a x dV 2 a x dx R x R x sin / 2 2 r R x / 2 2 rV r dV r 2 R x a x dx R x 2 R x a x dx a 4R r 2 R + = = = = = 錚 錚 = = 錚 錚 + = x y x dx R-x 256
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures 5.3 Composite Bodies and Figures 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 = = = 257
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures 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 i ci i i ci m , r , m , m r 258
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/5 Determine the coordinates of the centroid of the shaded area 259
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/5 i Ai xci yci Aixci Aiyci 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 i ci i i ci i A x X 243.6 mm A A y Y 117.7 mm A = = = = 1 3 2 260
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures 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 留 made by the back with the vertical when the container rests in an equilibrium position on a horizontal surface. 261
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/6 i xi yi zi Ai Aixi Aiyi Aizi 1 200 0 100 8e4 16e6 0 8e6 2 200 -150 -300/ 6e4 12e6 -9e6 -18e6 3 0 -150 -200/ 1.125e4 0 -1.6875e6 -2.25e6 4 400 -150 -200/ 1.125e4 4.5e6 -1.6875e6 -2.25e6 262
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/7 i ci i i ci i i ci i A x 16E6 16.5 E6 X 200 mm A 8E4 8.25 E4 A y 12.375 E6 Y 114.62 mm A 8E4 8.25 E4 A z 8E6 22.5E6 Z 42.75 mm A 8E4 8.25 E4 + = = = + = = = + = = = + 1 2 3 4 x y z 263
Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/7 1 150 Y tan 39.6 Z 留 錚 錚 錚 錚 = = 属 錚 錚 錚 錚 留 y z mg Y Z N 264
Ch. 5: Distributed Forces 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 area is the same as the lateral area of a cylinder of length L and radius 5.4 Pappus Theorem A ydL yL 慮 慮 = = y 265
Ch. 5: Distributed Forces Revolved Volume 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 V ydA yA 慮 慮 = = 266
Ch. 5: Distributed Forces 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 267
Ch. 5: Distributed Forces 5.4 Pappus Theorem 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 mm2, and the wheel has a mass of 10 kg. Calculate the distance to the centroid of the half-section. The aluminum has a density of 2.69 Mg/m3. r 268
Ch. 5: Distributed Forces 5.4 Pappus Theorem P. 5/8 10 / 2690 2 7600 6 77.85 mm M V E r V r 錚 錚 = = = 錚 錚 錚 錚 = 269
Ch. 5: Distributed Forces 5.4 Pappus Theorem P. 5/9 Calculate the mass m of concrete required to construct the arched dam shown. Concrete has a density of 2.40 Mg/m3. 270
Ch. 5: Distributed Forces 5.4 Pappus Theorem P. 5/9 ( ) 2 2 find the centroid w.r.t. the rotation axis first 4 70 80 80 40 120 70 120 4 3 80 80 70 4 175.52 m calculate the revolved volume from Pap i i i y A y y A 錚 錚 + + 錚 錚 錚 錚 錚 錚 = = 錚 錚 錚 錚 錚 錚 = [ ] [ ] 2 pus to find the mass / 3 175.52 80 80 70 4 2400 1.126 9 kg 1.126 6 Mg V yA V m V m V E E 慮 錚 錚 = = 錚 錚 錚 錚 = = = = 271
Ch. 5: Distributed Forces 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. 5.5 Fluid Statics dp gdh = pressure increases with depth 272
Ch. 5: Distributed Forces 5.5 Fluid Statics o o o dp gdh With constant, we may integrate and get p p gh absolute pressure, where p is the surface pressure at h 0. If p atmospheric pressure and is used as the reference value, p gh gage pressure Uni = = + = = = = = [ ] [ ] ( ) 3 2 2 2 2 t: Pa kg/m m/s m kg-m/s / m N/m 錚 錚 錚 錚 錚 錚 錚 錚 = = = 錚 錚 錚 錚 錚 錚 錚 錚 273
Ch. 5: Distributed Forces 5.5 Fluid Statics Resultant force due to pressure distribution on surface 1. Flat surface 274
Ch. 5: Distributed Forces 5.5 Fluid Statics Variation of p over the depth is linear (gh) 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 275
Ch. 5: Distributed Forces 5.5 Fluid Statics ( ) ( ) av R dR pdA gh y x y dy volume of pressure distribution or R g hdA ghA p A average altitude h is the depth corresponding to the centroid O of the flat surface area, A, exposed to the pressure. At = = = = = = = av that depth, the pressure gh is the average pressure, p , of the pressure distribution acting on the surface area. Magnitude 276
Ch. 5: Distributed Forces 5.5 Fluid Statics 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 pre = = = = ' ' ' ssured force direction onto the plate piercing point of the line of action R to the plate centroid of the plate area, A ydA if x is constant, Y dA centroidal y-coord of the area A (not A) of the pr = = = essure profile Direction perpendicular to the flat surface Line of action using principle of moment 277
Ch. 5: Distributed Forces 5.5 Fluid Statics Resultant force due to pressure distribution on surface 2. Cylindrical surface with constant width, b dR continuously changes direction 278
Ch. 5: Distributed Forces 5.5 Fluid Statics ( ) ( ) x y x y R volume of pressure distribution ghA h depth corresponding to the centroid O of the curve AB R b pdL b pdy, R b pdL b pdx = = = = = = Magnitude Direction ( ) 1 y x tan R / R 慮 = Line of action y x y x xdR xpdx ydR ypdy X , Y dR pdx dR pdy at centroid of volume of pressure distribution = = = = 279
Ch. 5: Distributed Forces 5.5 Fluid Statics Equilibrium of block of liquid Px, Py = 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 280
Ch. 5: Distributed Forces 5.5 Fluid Statics 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 281
Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 282
Ch. 5: Distributed Forces 5.5 Fluid Statics Buoyancy When o <f and fully immersed, objects weight < buoyancy force object will rise to the equilibrium position (o >f) where the objects weight = 裡buoyancy force F Vg density of the displaced fluid V volume of the displaced fluid = = = 283
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/10 284
Ch. 5: Distributed Forces 5.5 Fluid Statics 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 dA ( ) ( ) y x x x y 2 x 2 2 ring F pdA g hdA ghA F 0 ghA dA 0 h depth of the centroid of circular area 3000 m ghA 1030 9.81 3000 0.75 462.4 MPa A 0.75 0.725 = = = 錚 錚 = = 錚 錚 = = = = = x 285
Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 286
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/11 Assumption: door has negligible weight and volume, i.e., W = 0 and B = 0 ( ) ( ) 1 2 A 1 2 pressure force magnitude volume of pressure distribution and divide into two parts -- rectangle and triangle F 1000 9.81 1.2 1.6 0.8 F 1/ 2 1000 9.81 0.8 1.6 0.8 M 0 P 1.6cos30 F 0.8 F 1.6 2/ = = = 錚 錚 = + + 錚 錚 3 0 P 12566 N = = A R F1 F2 P 1.2 m 1.2+1.6sin30 m 287
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/12 ( ) y pressure referred to atmosphere F 0 900g A 150 1030g A 250 h 800g A 400 0, h 70.4 mm 錚 錚 = + 錚 錚 = = y W BO BW 288
Ch. 5: Distributed Forces 5.5 Fluid Statics 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/m3. 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 angle慮which 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. 289
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/13 W T B W T B buoy is weighted at its base so it remains vertical ( ) ( ) y 2 2 F 0 B Mg T 0 0.3 if h 0.6 m, B Vg 1030 1.8g N 4 T 402.7 N when the cable goes slack, T 0 B decreases h increases 0.3 1030 2.4 h g 90g 0, h 1.164 m 4 錚 錚 = = 錚 錚 = = = = = = = 290
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/13 W T B ( ) 2 O 2 0.3 B Vg 1030 1.5/ cos g 4 M 0 B 0.75tan 90g 1.2sin 0 sin 0, or cos 0.7584 0, 29.44 慮 慮 慮 慮 慮 慮 = = 錚 錚 = = 錚 錚 = = = 賊 属 1.5 m 慮 291
Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 292
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/14 ( ) A R 0.5 1.2 1000 9.81 1.2sin 60 3 M 0 R 0.4 6Csin60 0.6 0, C 2.354 kN = 錚 錚 = + = = 錚 錚 6C F R A h = 1.2sin60 293
Ch. 5: Distributed Forces 5.5 Fluid Statics 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 mO 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 mO, at the center of the hull. 294
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/15 ( ) G O O y O B Vg 1030 15 6 40/ 2 g M 0 Bcos 2.5 Bsin 1 m g 10cos 23sin 0 m 203 Mg F 0 B mg m g 0, m 1651 Mg 慮 慮 慮 慮 = = 錚 錚 = += 錚 錚 = 錚 錚 = = = 錚 錚 mOg mg B 2 3 5 7.5 295
Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 296
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/16 ( ) ( ) 2 st 2 y 36 36 2 y 0 0 36 2 c 0 y ax or x y/a in 1 quadrant contains 27, 36 a 36/ 27 2 A xdy xy 648 m 3 AX x dA x / 2dy, X 10.125 m F volume of the pressure distribution 1/ 2 36 1000 9.81 36 h W weight of the fluid bl = = = = = = = = = = = = = = D ock 1000 9.81 648 h M 0 F 18 W 10.125 W b 0, b 28.125 m = 錚 錚 = + = = 錚 錚 Rx Ry= W F W x 12 m D 297
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/17 A flat plate seals a triangular opening in the vertical wall of a tank of liquid of density . 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. 298
Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/17 ( ) ( ) ( ) ( ) y a O y 0 dR pdA g h y xdy g h y b 1 y/a dy gba M 0 Pa ydR 0, P h a/2 6 = = = = + = + 錚 錚 = + = = + 錚 錚 y dy x x = b(1-y/a) b a O dR P 299

More Related Content

Distributes force system centroid and moment of inertia

  • 1. Ch. 5: Distributed Forces 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 234
  • 2. Ch. 5: Distributed Forces 5.0 Introduction 5.0 Introduction 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. 235
  • 3. Ch. 5: Distributed Forces 5.1 Center of Mass 5.1 Center of Mass Location of the CG and CM are found by the principle of moment CG: point where the resultant gravitational force W acts CM: point at which the total mass is visually concentrated sum of the moments = moment of the sum 236
  • 4. Ch. 5: Distributed Forces Center of Gravitiy 5.1 Center of Mass Assumption 1) uniform intensity of the earths 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 x y z x y z dW dW = CG r i + j+ k r = i + j+ k r r = 237
  • 5. Ch. 5: Distributed Forces Center of Mass 5.1 Center of Mass ( ) W mg & dW gdm dm dm dm dV, e.g. x, y, z , dV dxdydz dV dV = = = = = CM CM r r = r r = Assumption 1) uniform intensity of the earths force field over the body 2) parallel field of force due to the gravitational attraction CM = CG if the gravity field is treated as uniform and parallel CM is independent of gravitational effect and hence is a unique property of the body 238
  • 6. Ch. 5: Distributed Forces Center of Volume 5.1 Center of Mass Assumption 1) constant intensity of the earths force field over the body 2) parallel field of force due to the gravitational attraction CV depends only on the shape of the body, Independent of the material dV dV CV r r = CV = CM if the mass intensity (density) is uniform over the volume 239
  • 7. Ch. 5: Distributed Forces 5.1 Center of Mass 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 240
  • 8. Ch. 5: Distributed Forces 5.3 Centroids (Line, Area, Volume) is purely the geometrical (shape) property of the body since any reference to its mass properties has disappeared. 5.3 Centroids Line: dL , not always on the line dL C C r r = assuming the cross sectional area is constant over the length 241
  • 9. Ch. 5: Distributed Forces dA , not always on the surface dA C C r r = 5.3 Centroids Area: assuming the thickness is constant over the entire area dV , not always inside the volume dV C C r r = dA first moment of area = C r Volume: 242
  • 10. Ch. 5: Distributed Forces 5.3 Centroids Guidelines in choosing the differential element & setting up the integrals a) Order of element: prefer 1st order diff. element to higher order, so only one integration will be required ( ) ( ) 2 A l y dy dxdy V r y dy dxdydz = = = = 243
  • 11. Ch. 5: Distributed Forces 5.3 Centroids b) Continuity: choose an element which can be integrated in one continuous operation to cover the whole object ( ) ( ) 2 2 1 x x 1 2 0 x A h x dx h x dx ( ) 2 y 0 A l y dy = h1(x) h2(x) (x2, y2) 244
  • 12. Ch. 5: Distributed Forces dxdy 5.3 Centroids c) Neglect high order terms: higher order terms may be dropped compared with lower order terms 1 1 A ydx dxdy y dy dx ydx 2 2 note: dxdy dxdy 錚 錚 = + = + = 錚 錚 錚 錚 dxdy 245
  • 13. Ch. 5: Distributed Forces ( ) O O x , y 5.3 Centroids d) Choice of coordinates: to appropriately describe the boundaries O y O O 0 A x y xdy = O 2 0 1 A r d 2 慮 慮 = 246
  • 14. Ch. 5: Distributed Forces 5.3 Centroids 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 = = ( ) ( ) 2 C C first moment x dV x z r z dz = = 247
  • 15. Ch. 5: Distributed Forces 5.3 Centroids P. 5/1 Locate the centroid of the area of a circular sector w.r.t. its vertex 248
  • 16. Ch. 5: Distributed Forces 5.3 Centroids P. 5/1 o o c o o c r r o o o o o 0 0 by symmetry, Y 0 select differential element to be strip of partial circular ring with r varying from 0 to r r sin XA x dA dA 2 r dr , x r sin 2rsin X 2 r dr 2 r dr , X 3 留 留 留 留 留 留 留 留 留 = 錚 錚 = = = 錚 錚 = = 249
  • 17. Ch. 5: Distributed Forces 5.3 Centroids P. 5/1 ( ) c c 2 2 by symmetry, Y 0 select differential element to be strip of sector with varying from to 1 2 XA x dA dA r rd , x r cos 2 3 1 2 1 2rsin X r d rcos r d , X 2 3 2 3 留 留 留 留 慮 留 留 慮 慮 留 慮 慮 慮 留 = 錚 錚 錚 錚 = = = 錚 錚 錚 錚 錚 錚 = = 250
  • 18. Ch. 5: Distributed Forces 5.3 Centroids P. 5/2 Locate the centroid of the area shown in the figure by direct integration 251
  • 19. Ch. 5: Distributed Forces 5.3 Centroids P. 5/2 ( ) ( ) ( ) c 2 c1 c2 by symmetry, the centroid lies on the line y a x differential area horizontal strip strip of sector with varying from 0 to /2 1 XA x dA dy ad sin , dA ady a ad 2 dA a sin 1/ 2 d 2 x a/2, x a 3 慮 慮 慮 慮 慮 慮 = = 錚 錚 = = = 錚 錚 = = = ( ) ( ) ( ) ( ) / 2 /2 /2 2 2 2 0 0 0 sin 2 1 X a sin 1/ 2 d a/2 a sin d asin a d 3 2 10 3 a 2a X , Y a X 3 4 3 4 慮 慮 慮 慮 慮 慮 慮 錚 錚 = 錚 錚 錚 錚 = = = 慮 xc1 2/3asin慮 252
  • 20. Ch. 5: Distributed Forces 5.3 Centroids P. 5/3 Determine the coordinates of the centroid of the volume obtained by revolving the shaded area about the z-axis through 90属angle 253
  • 21. Ch. 5: Distributed Forces 5.3 Centroids P. 5/3 ( ) ( ) 2 2 2 2 2 2 2 c 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 y z a a , z a a y 2 y dV zdy a a y ydy 4 2 x 属 = + = = = = ( ) ( ) ( ) ( ) ( ) ( ) 2 2 c c a a 2 2 2 2 c 0 0 a a 2 2 2 2 2 2 c 0 0 ysin /4 a a y cos /4 2y/ y , z z/2 /4 2 XV x dV X a a y ydy 2y/ a a y ydy 2 2 4 3 X a Y 4 a a y ZV z dV Z a a y ydy a a y ydy 2 2 2 Z a/4 = = = = = 錚 錚 = = 錚 錚 錚 錚 = = 錚 錚 錚 錚 錚 錚 = = 錚 錚 = 4y/(2) z/2 y dy 254
  • 22. Ch. 5: Distributed Forces 5.3 Centroids P. 5/4 Locate the center of mass G of the steel half ring. 255
  • 23. Ch. 5: Distributed Forces 5.3 Centroids 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 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 c a a 2 2 2 2 c a a 2 2 from a to a x y a , y a x dV 2 a x dx R x R x sin / 2 2 r R x / 2 2 rV r dV r 2 R x a x dx R x 2 R x a x dx a 4R r 2 R + = = = = = 錚 錚 = = 錚 錚 + = x y x dx R-x 256
  • 24. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures 5.3 Composite Bodies and Figures 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 = = = 257
  • 25. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures 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 i ci i i ci m , r , m , m r 258
  • 26. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/5 Determine the coordinates of the centroid of the shaded area 259
  • 27. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/5 i Ai xci yci Aixci Aiyci 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 i ci i i ci i A x X 243.6 mm A A y Y 117.7 mm A = = = = 1 3 2 260
  • 28. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures 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 留 made by the back with the vertical when the container rests in an equilibrium position on a horizontal surface. 261
  • 29. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/6 i xi yi zi Ai Aixi Aiyi Aizi 1 200 0 100 8e4 16e6 0 8e6 2 200 -150 -300/ 6e4 12e6 -9e6 -18e6 3 0 -150 -200/ 1.125e4 0 -1.6875e6 -2.25e6 4 400 -150 -200/ 1.125e4 4.5e6 -1.6875e6 -2.25e6 262
  • 30. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/7 i ci i i ci i i ci i A x 16E6 16.5 E6 X 200 mm A 8E4 8.25 E4 A y 12.375 E6 Y 114.62 mm A 8E4 8.25 E4 A z 8E6 22.5E6 Z 42.75 mm A 8E4 8.25 E4 + = = = + = = = + = = = + 1 2 3 4 x y z 263
  • 31. Ch. 5: Distributed Forces 5.3 Composite Bodies and Figures P. 5/7 1 150 Y tan 39.6 Z 留 錚 錚 錚 錚 = = 属 錚 錚 錚 錚 留 y z mg Y Z N 264
  • 32. Ch. 5: Distributed Forces 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 area is the same as the lateral area of a cylinder of length L and radius 5.4 Pappus Theorem A ydL yL 慮 慮 = = y 265
  • 33. Ch. 5: Distributed Forces Revolved Volume 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 V ydA yA 慮 慮 = = 266
  • 34. Ch. 5: Distributed Forces 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 267
  • 35. Ch. 5: Distributed Forces 5.4 Pappus Theorem 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 mm2, and the wheel has a mass of 10 kg. Calculate the distance to the centroid of the half-section. The aluminum has a density of 2.69 Mg/m3. r 268
  • 36. Ch. 5: Distributed Forces 5.4 Pappus Theorem P. 5/8 10 / 2690 2 7600 6 77.85 mm M V E r V r 錚 錚 = = = 錚 錚 錚 錚 = 269
  • 37. Ch. 5: Distributed Forces 5.4 Pappus Theorem P. 5/9 Calculate the mass m of concrete required to construct the arched dam shown. Concrete has a density of 2.40 Mg/m3. 270
  • 38. Ch. 5: Distributed Forces 5.4 Pappus Theorem P. 5/9 ( ) 2 2 find the centroid w.r.t. the rotation axis first 4 70 80 80 40 120 70 120 4 3 80 80 70 4 175.52 m calculate the revolved volume from Pap i i i y A y y A 錚 錚 + + 錚 錚 錚 錚 錚 錚 = = 錚 錚 錚 錚 錚 錚 = [ ] [ ] 2 pus to find the mass / 3 175.52 80 80 70 4 2400 1.126 9 kg 1.126 6 Mg V yA V m V m V E E 慮 錚 錚 = = 錚 錚 錚 錚 = = = = 271
  • 39. Ch. 5: Distributed Forces 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. 5.5 Fluid Statics dp gdh = pressure increases with depth 272
  • 40. Ch. 5: Distributed Forces 5.5 Fluid Statics o o o dp gdh With constant, we may integrate and get p p gh absolute pressure, where p is the surface pressure at h 0. If p atmospheric pressure and is used as the reference value, p gh gage pressure Uni = = + = = = = = [ ] [ ] ( ) 3 2 2 2 2 t: Pa kg/m m/s m kg-m/s / m N/m 錚 錚 錚 錚 錚 錚 錚 錚 = = = 錚 錚 錚 錚 錚 錚 錚 錚 273
  • 41. Ch. 5: Distributed Forces 5.5 Fluid Statics Resultant force due to pressure distribution on surface 1. Flat surface 274
  • 42. Ch. 5: Distributed Forces 5.5 Fluid Statics Variation of p over the depth is linear (gh) 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 275
  • 43. Ch. 5: Distributed Forces 5.5 Fluid Statics ( ) ( ) av R dR pdA gh y x y dy volume of pressure distribution or R g hdA ghA p A average altitude h is the depth corresponding to the centroid O of the flat surface area, A, exposed to the pressure. At = = = = = = = av that depth, the pressure gh is the average pressure, p , of the pressure distribution acting on the surface area. Magnitude 276
  • 44. Ch. 5: Distributed Forces 5.5 Fluid Statics 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 pre = = = = ' ' ' ssured force direction onto the plate piercing point of the line of action R to the plate centroid of the plate area, A ydA if x is constant, Y dA centroidal y-coord of the area A (not A) of the pr = = = essure profile Direction perpendicular to the flat surface Line of action using principle of moment 277
  • 45. Ch. 5: Distributed Forces 5.5 Fluid Statics Resultant force due to pressure distribution on surface 2. Cylindrical surface with constant width, b dR continuously changes direction 278
  • 46. Ch. 5: Distributed Forces 5.5 Fluid Statics ( ) ( ) x y x y R volume of pressure distribution ghA h depth corresponding to the centroid O of the curve AB R b pdL b pdy, R b pdL b pdx = = = = = = Magnitude Direction ( ) 1 y x tan R / R 慮 = Line of action y x y x xdR xpdx ydR ypdy X , Y dR pdx dR pdy at centroid of volume of pressure distribution = = = = 279
  • 47. Ch. 5: Distributed Forces 5.5 Fluid Statics Equilibrium of block of liquid Px, Py = 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 280
  • 48. Ch. 5: Distributed Forces 5.5 Fluid Statics 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 281
  • 49. Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 282
  • 50. Ch. 5: Distributed Forces 5.5 Fluid Statics Buoyancy When o <f and fully immersed, objects weight < buoyancy force object will rise to the equilibrium position (o >f) where the objects weight = 裡buoyancy force F Vg density of the displaced fluid V volume of the displaced fluid = = = 283
  • 51. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/10 284
  • 52. Ch. 5: Distributed Forces 5.5 Fluid Statics 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 dA ( ) ( ) y x x x y 2 x 2 2 ring F pdA g hdA ghA F 0 ghA dA 0 h depth of the centroid of circular area 3000 m ghA 1030 9.81 3000 0.75 462.4 MPa A 0.75 0.725 = = = 錚 錚 = = 錚 錚 = = = = = x 285
  • 53. Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 286
  • 54. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/11 Assumption: door has negligible weight and volume, i.e., W = 0 and B = 0 ( ) ( ) 1 2 A 1 2 pressure force magnitude volume of pressure distribution and divide into two parts -- rectangle and triangle F 1000 9.81 1.2 1.6 0.8 F 1/ 2 1000 9.81 0.8 1.6 0.8 M 0 P 1.6cos30 F 0.8 F 1.6 2/ = = = 錚 錚 = + + 錚 錚 3 0 P 12566 N = = A R F1 F2 P 1.2 m 1.2+1.6sin30 m 287
  • 55. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/12 ( ) y pressure referred to atmosphere F 0 900g A 150 1030g A 250 h 800g A 400 0, h 70.4 mm 錚 錚 = + 錚 錚 = = y W BO BW 288
  • 56. Ch. 5: Distributed Forces 5.5 Fluid Statics 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/m3. 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 angle慮which 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. 289
  • 57. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/13 W T B W T B buoy is weighted at its base so it remains vertical ( ) ( ) y 2 2 F 0 B Mg T 0 0.3 if h 0.6 m, B Vg 1030 1.8g N 4 T 402.7 N when the cable goes slack, T 0 B decreases h increases 0.3 1030 2.4 h g 90g 0, h 1.164 m 4 錚 錚 = = 錚 錚 = = = = = = = 290
  • 58. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/13 W T B ( ) 2 O 2 0.3 B Vg 1030 1.5/ cos g 4 M 0 B 0.75tan 90g 1.2sin 0 sin 0, or cos 0.7584 0, 29.44 慮 慮 慮 慮 慮 慮 = = 錚 錚 = = 錚 錚 = = = 賊 属 1.5 m 慮 291
  • 59. Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 292
  • 60. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/14 ( ) A R 0.5 1.2 1000 9.81 1.2sin 60 3 M 0 R 0.4 6Csin60 0.6 0, C 2.354 kN = 錚 錚 = + = = 錚 錚 6C F R A h = 1.2sin60 293
  • 61. Ch. 5: Distributed Forces 5.5 Fluid Statics 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 mO 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 mO, at the center of the hull. 294
  • 62. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/15 ( ) G O O y O B Vg 1030 15 6 40/ 2 g M 0 Bcos 2.5 Bsin 1 m g 10cos 23sin 0 m 203 Mg F 0 B mg m g 0, m 1651 Mg 慮 慮 慮 慮 = = 錚 錚 = += 錚 錚 = 錚 錚 = = = 錚 錚 mOg mg B 2 3 5 7.5 295
  • 63. Ch. 5: Distributed Forces 5.5 Fluid Statics 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. 296
  • 64. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/16 ( ) ( ) 2 st 2 y 36 36 2 y 0 0 36 2 c 0 y ax or x y/a in 1 quadrant contains 27, 36 a 36/ 27 2 A xdy xy 648 m 3 AX x dA x / 2dy, X 10.125 m F volume of the pressure distribution 1/ 2 36 1000 9.81 36 h W weight of the fluid bl = = = = = = = = = = = = = = D ock 1000 9.81 648 h M 0 F 18 W 10.125 W b 0, b 28.125 m = 錚 錚 = + = = 錚 錚 Rx Ry= W F W x 12 m D 297
  • 65. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/17 A flat plate seals a triangular opening in the vertical wall of a tank of liquid of density . 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. 298
  • 66. Ch. 5: Distributed Forces 5.5 Fluid Statics P. 5/17 ( ) ( ) ( ) ( ) y a O y 0 dR pdA g h y xdy g h y b 1 y/a dy gba M 0 Pa ydR 0, P h a/2 6 = = = = + = + 錚 錚 = + = = + 錚 錚 y dy x x = b(1-y/a) b a O dR P 299