9 58. Determine the location of the centroidal ais - of the beam s cross-sectional area. Neglect the size of the corner welds at A and B for the calculation. 15 mm 15 mm B 15 mm 15 mm A = 7.5(15) (15) + 9(15) (15) + 215(p)(5) 2 A C = 1 97 981.5 mm 2 5 mm A = 15(15) + 15(15) + p(5) 2 = 12 353.98 mm 2 = A A = 1 97 981.5 12 353.98 = 154 mm
9 63. Determine the location (, ) of the centroid C of the crosssectional area for the structural member constructed from two equal-sized channels welded together as shown. Assume all corners are square. Neglect the size of the welds..5 in. _ C A = 1.51.521.252 + 11.52152 + 1.51.5219.752 + 1.51.5215.252122 + 11.5214.252 1.5 in. _ = 61.625 in 3 A = 31.51.52 + 11.52 + 1.51.524122 = 13 in 2.5 in. 4 in. 2 in. 4 in. = A A = 61.625 = 4.74 in. 13 A = 1.51.5211.252122 + 11.521.252 + 1.51.521.752 + 11.5215.52 + 1.51.5211.252 = 38.875 in 3 = A A = 38.875 = 2.99 in. 13
9 82. Determine the distance h to which a 1-mm diameter hole must be bored into the base of the cone so that the center of mass of the resulting shape is located at z = 115 mm. The material has a densit of 8 Mg>m 3. z 1 3 p1.1522 1.52A.5 4 B - p1.52 2 1h2A h 2 B 1 3 p1.1522 1.52 - p1.52 2 1h2 =.115 5 mm.4313 -.2875 h =.4688-1.25 h 2 h 2 -.23 h -.3 = h 5 mm C 15 mm _ z Choosing the positive root, h = 323 mm
9 83. Determine the distance z to the centroid of the shape which consists of a cone with a hole of height h = 5 mm bored into its base. z 5 mm z V = 1 3 p (.15)2 A.5B a.5 4 b - p (.5)2 (.5) a.5 2 b = 1.463(1-3 )m 4 h 5 mm C 15 mm z V = 1 3 p (.15)2 (.5) - p(.5) 2 (.5) =.1139 m 3 z = z V V = 1.463 (1-3 ) =.12845 m = 128 mm.1139
9 89. Locate the center of mass z of the assembl. The clinder and the cone are made from materials having densities of 5Mg>m 3 and 9Mg>m 3, respectivel. z Center of mass: The assembl is broken into two composite segments, as shown in Figs. a and b. z = z m m = 5(.4)Cp(.2 2 )(.8)D + 9(.8 +.15)c 1 3 p(.42 )(.6) d 5Cp(.2 2 )(.8)D + 9c 1 3 p(.42 )(.6) d.4 m.2 m.6 m.8 m = 16.6 147.4 =.754 m = 754 mm
9 16. Determine the volume of an ellipsoid formed b revolving the shaded area about the ais using the second theorem of Pappus Guldinus. The area and centroid of the shaded area should first be obtained b using integration. 2 2 1 a 2 b 2 b Area and Centroid: The differential element parallel to the ais is shown shaded in Fig. a. The area of this element is given b Integrating, da = d = a b 2b2-2 d a b a A = da = LA L b 2b2-2 d = pab 4 The centroid can be obtained b appling Eq. 9 4 with =. da = LA da LA b = L c a b 2b2-2 d d pab 4 = 4b 3p Volume: Appling the second theorem of Pappus Guldinus and using the results obtained above, we have V = 2pA = 2pa 4b 3p bapab 4 b = 2 3 pab2
9 19. Determine the volume of the solid formed b revolving the shaded area about the u u ais using the second theorem of Pappus Guldinus. The area and centroid of the area should first be obtained b using integration. u 3 1 9 Area and Centroid: The differential element parallel to the ais is shown shaded in Fig. a. The area of this element is given b da = d = 1 9 3 d 2 ft Integrating, u 1 A = da = LA L 9 3 d = 1 36 4 2 A = 2.25 ft 2 The centroid can be obtained b appling Eq. 9 4 with =. da = LA da LA = L a 1 9 3 db 2.25 = L Volume: Appling the second theorem of Pappus-Guldinus and using the results obtained above, we have 1 9 4 d = 2.25 1 45 5 2 = 2.4 ft 2.25 V = 2p ra = 2p(3-2.4 + 2)(2.25) = 36.8 ft 3
9 114. Determine the surface area of the roof of the structure if it is formed b rotating the parabola about the ais. 16 ( 2 /16) 16 m Centroid: The length of the differential element is dl = 2d 2 + d 2 d = 1 + a d 2 and its centroid is =. Here,. Evaluating the d = - C d b d 8 integrals, we nave 16 m L = L dl = L 16 m Appling Eq. 9 5, we have L 16 m dl = 1 + 2 d = 217.181 m2 L C 64 1 + 2 d = 23.663 m C 64 dl = L L dl = 217.181 23.663 = 9.178 m Surface Area: Appling the theorem of Pappus and Guldinus, Eq. 9 7, with u = 2p, L = 23.663 m, r = = 9.178, we have A = url = 2p(9.178) (23.663) = 1365 m 2
9 123. Determine the magnitude of the resultant hdrostatic force acting on the dam and its location, measured from the top surface of the water. The width of the dam is 8 m; r w = 1. Mg>m 3. 6 m p = 6(1)(1 3 )(9.81) = 58 86 N/m 2 F = 1 2 (6)(8)(58 86) = 1.41(16 )N = 1.41 MN h = 2 3 (6) = 4 m
9 135. Determine the magnitude of the hdrostatic force acting per foot of length on the seawall. g w = 62.4 lb>ft 3. 2 2 8ft A = da = -d = 2 2 d = 2 LA L L 3 3 2-2 -2 w = b g h = 1(62.4)(8) = 499.2 lb>ft F = 5.333(1)(62.4) = 332.8 lb -2 = 5.33 2 2ft F = 1 (499.2)(8) = 1997 lb 2 F R = 2(332.8) 2 + (1997) 2 = 224 lb = 2.2 kip