hw9_sol

# hw9_sol - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—35 Locate the centroid of the shaded area shown in Fig. P5—35. Flg. P5-3S SOLUTION The shaded area can be divided into a rectangle with two circles removed. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5-13. Thus, 1440 -37.71 -113.13 1289.16 M _ Y _ 1289.16 _ . C j Ci y C — 3- — -§TZT§E — 6.00 in. Ans. > X ll [‘1 > >< U 2 X -—_—.._._.—_—_.— ENGINEERING MECHANECS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-36 Locate the centroid of the shaded area shown in P5-359 SOLUTION The shaded area can be divided into three rectangles. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5-13. Thus, Kc; My y51 Mx (mm) mma} [mm] (mmai 40 896,000 140 3,136,000 1 2 120 768,000 240 1,536,000 3 , 200 2,560,000 200 2,560,000 4,224,000 T,232,000 _1 = 4,224 000 = C A ——E§T%ﬁa- 101.5 mm x = _ ;§ _ 7,232,000 _ 3c ' A ‘ 41,500 ‘ 173'8 mm 5-39 Locate the centroid of the shaded area shown in Fig. P5-39. SOLUTION The shaded area can be divided into two squares, a quarter circle, and a triangle with a circle removed. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5-13. Thus, 1 2_ l 2_ . 2 2 — '- 1n. :2- 11 - 33 - 4‘7} = 2 971 in xc2 ' yc2 ‘ 3n 3n are: n(3)2= 28.27 in? My 57 2 _ = M x = -— = TEETH? = 0.431 in. Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-44 Locate the centroid of the shaded area shown in Fig. P5—44. SOLUTION The shaded area can be divided into a square with a circle and a quarter circle removed. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5—13. Thus, = an: n(60)2 = 11,310 mm 2 2: ln(1zo)2 11,310 x _ _ 4(120) _ - 240 3“ - 189.07 mm 4(120) 240 3" ' 189.07 mm Ai XCi My yCi Mx 3 (mmz) (mm) (mma) (mm! (mm ) 1 57,600 120 6,912,000 120 6,912,000 2 —11,310 100 -1,131,000 80 -904,800 3 -11,310 189.07 -2,138,382 189.07 -2,138,382 2 34,980 3,642,618 3,868,818 = 3,642,618 _ 34,980 = 3,868,818 = 34,980 ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-69 Determine the resultant g of the system of distributed loads and locate its line of action with respect to the left support for the beam shown in Fig. P5—69. SOLUTION 200(3) 1.5 ft. 150(3) 4.5 ft. 100(3) 7.5 ft. 2F = A1 + A2 + A3 = 600 + 450 + 300 = 1350 lb = 1350 lb 1 A + A x + A x 1XC‘1 2 C2 3 C3 = 600(1.5) + 4.50(4.5) + 300(7.5) = 5175 ft'lb '"' "'"UL ----- ---. 3.56 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-72 Determine the resultant ﬁ of the system of distributed loads and locate its line of action 25kNm1 with respect to the left support for the beam shown in Fig. 95—72. WE ‘ ‘ ‘ L—nglv‘im Fig. P5-72 SOLUTION 2.5(4) = 10 RN 4 m %(2.5)(4) = 5 MI 2 + gm = 4.667 In SF = A1 + A2 = 10 + = 15.00 kN ¢ + Azxc2= 10(4) + 5(4.667) = 63.335 kN-m 357 The width of the rectangular gate shown in Fig. P5-86 is 2 m. Determine the magnitude of the resultant force ﬂ exerted on the gate 9y the water (0 = 1000 kg/m ) pressure and the location of the center of pressure with respect to the hinge at the top of the gate. SOLUTION p x: pgh 1000(9.807)(3) = 29,421 N/m2 - -l R — V — 2p hw ps max - §<29.421)(3)(2) 88,263 N a 88.3 RN 2 _ 1+3(h)—1+ ...
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## This note was uploaded on 03/26/2008 for the course CVEN 221 taught by Professor - during the Spring '08 term at Texas A&M.

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hw9_sol - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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