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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. P53S 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. 513. 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
>
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U
2
X —_—.._._.—_—_.— ENGINEERING MECHANECS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 536 Locate the centroid of the shaded area shown
in P5359 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. 513. 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 539 Locate the centroid of the shaded area shown in Fig. P539. 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. 513. 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 544 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 569 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 572 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. P572
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 kNm 357 The width of the rectangular
gate shown in Fig. P586 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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 Spring '08
 

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