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Unformatted text preview: I I’I'OI. uaoer I I I (1) (a) (b) (C) =2a 7m 2 _ 2 _
1y — ycsq +115qu ( 10:}
R (a) Find 1,, the moment of inertia about the y
axis, for the shaded region (a square with a
circular hole) in ﬁgure (i). Him: For a circular
region of radius r with its centroid at C,
Ixc = ch = 34—. (20 pts) (b) Find kv, the radius of gyration about the y—axis,
for the shaded region in ﬁgure (i). (10 pts) (c) Find J0, the polar moment of inertia about 0,
for the shaded region in ﬁgure (ii). (15 pts) (d) Find Ly, the product of inertia with respect to
the coordinate axes, for the shaded region shown in ﬁgure (ii). (5 pts)
3
YC—sq : 461(1420) = 21.33a4
4
ch—circ = 7—11;
64 A = AW  Am =15.21a2 +A :2 )=80.37a‘ yc—circ circ circ [I
ky = —’ = 2.30a2
A g n
2 dad =05” 2R3d =7rr—
Ir (r r) [LEJR r r 4 4 2R _ 157:124
4 R (d) I xy = 0 because the shaded region is symmetrical about the x—axis. A cable is strung between supports at A and C, such that the
slope at A is —I.0, as shown. Point B is the lowest point on the
cable. The distributed vertical load qo is uniform per horizontal
projected distance (deﬁned positive in the negative y direction),
and T H is the horizontal component of the cable tension. Let qo
and T H be given, and let the span be w = 4T H/ qo. (a) Find the axial tension in the cable at support A. (10 pts)
(b) Find the elevation h and the axial cable tension at C. (30 pts)
(c) Find the x coordinate at B. (10 pts) (a) Based on the given slope at A, we have TV A = T”. Thus, TIA = ”T; + TV2 = «RT; = x/ETH. (b) We have, I go
x =——x+C
y() TH 1 y(0)=0=>C2=0 => y(x)=§7—;I—x "X
y’(0)=—1=> C1=—1 h=y “(43): qO [4TH)2_[4T”)_4TH
C _ _ _ —
40 2TH ‘10 go ‘10 yé=g£[ﬂ)—l:3
TH q 0
Then the axial tension at C is TC = T; + (3TH)2 = MT”. (c) We have zero slope at B. Thus, T
y’B=—q—'LxB—1=0=>xB=—i TH ‘10 ...
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 Fall '05
 Downing

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