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Unformatted text preview: Review 3 ’7120. Express the shear and moment acting in the
pipe as a function of y, where 0 s y s 4 ft. +T):F,=o; 164y—V=0 V={16—4y}lb Am (+2M=o; M+4y(§) +40—16y=0 M: {—zyz + 16y—40}1b.ft Ans 7121. Determine the normal force, shear force, and ma
ment at points B and C of the beam. 7:5 kN )i! v' 6kN Free body Diagram: The Suppun reactions need not be computed fnr
his case. lnlenml Forces: Applying the equations of equilibrium [0 segment
DC [FED (n)], we have 32520;  Nc=0 Ans +TEF,.=0; Vc—3.00—6=0 vc=9.oom Ans
{+2Mc=o; —Mc3.00(l.5)—6(3)—40=0
Mc=—62.5 kN'm Ans Applying III: equations of equilibrium to segment DB [FBD (b)}.
we have —> SE. = 0; N, = 0 Ans
+7 2F, :0; v, — 10.0—7.5—4.oo—6=o
v5 = 27.5 kN Ans
(+21% = o; —M,, — 10.005) — 75(5)
4.00m — 6(9) — 40 = 0
My 2 —184.5 kNm Ans l(3)=3.0kN 6kN
)5 m 1.5m ﬁkN 2 40mm 2.5m 25m 2m 2m
(b) 7172. A chain is suspended between points at the same
elevation and spaced 3 distance of 60 ft apart. If it has a
weight of 0.5 mm and the sag is 3 ft, determine the
maximum tension in the chain. dr
_1
2 {1+}12(w0ds)1} Performing the integration yields : F _ 1
f{smh ‘[F(o.5s+ c1)]+cl} [1] H From Eq. 713 At —0 dy—o 11 C—0
5—, d—x— ence 1 dy 0.53
.__. = tang = _ [2]
dx 5: Applying boundary conditions at x = 0; .r = 0 to Eq.[1] and using the result C1 = 0
yields C = 0. Hence )3}, . 0.5
s = —~ smh —x [3]
0.5 F”
Substituting Eq.[3] into [2] yields :
dy .nh 0.5.1: [4]
_.. = s, __
dx F” Performing the integration F 0.5
y = ﬁcosh[FHx)+ C3 F
Applying boundary conditions at x = 0; y = 0 yields C3 = 731% . Therefore _F,, has 1
y"0.5 cos th— F 0.5
Atx=30ft; y=3ft 3=—” cosh —(30) —1
.5 F"
By trial and error F” = 75.251b At 1 = 30 ft; 6: 9m“. From Eq.[4] dy 0.5(30) tanGmu = — = ' 9 a, =1L346°
dx x=30ft S! ( 75.25 J "'
F 75.25 I“, = H = 76.7113. Ans cos 6,”, = cos_11.346° 7123. Draw the shear and moment diagrams for the beam. 954. The gravity wall is made of concrete. Determine
the location (2 y) of the center of gravity G for the wall. 222A = 1.8(3.6)(0.4) + 2.1(3)<3) — 3.4G)(3)(0.6) — 1.2(%)(1.3)(3)
= 15.192 m3 2}}! = 0.2(3.6)(O.4) + 1.9(3)(3) — 1.4G)(3)(0.6)  2.4G)(1.8)(3) : 9648 m3 24 = 3.6(0.4) + 3(3) — game) — gum) = 6.84m E = 244 = 15192 = 2,2211: Ans
IA 634  2' . yzﬂ =2ﬂ§ =1,41m Ans VON) M m
"5" 7 s
MUN'M}
IS'b x0")
m
'59 l 0.4 111 *960. The wooden table is made from a square board
having a weight of 15 lb. Each of the legs weighs 2 lb and
is 3 ft long, Determine how high its center of gravity is
from the ﬂoonAlso. what is the angle, mEasured from the horizontal, through which its lop surface can he tilted on
two of its legs before it begins to overturn? Neglect the thickness of each leg. i = 25W = _._——15(3) + 4mm” = 2.48 ft Ans
EW 15 + 4(2)
e — tan—Ki) : 38.9” Ans
_ 2.43
1025. The polar moment of inertia of the area is .7C = .V' .V 28 in4 about the z axis passing through the centroid C.
The moment of inertia about the x axis is 17 in‘. and the moment of inertia about the y’ axis is 56 in4. Determine the areaA.
Jc = zsm‘ = I, +1, ] 28=i7+1, [ .C X
[y =iiin“ :
I=7+Adz
53in;
56:11+A(3)2 ,
y y A = 5.00 m2 Ans . .
1049. Determine the moments of inertia of the triangular area about the x' and y’ axes. which pass
through the centroid C of the area. L :3%(a)h3+31—6(b—a)h3 =39»? Ans
__ 2,24 _ §a[;(h)(a)]+[a+ b+“][%h(b—a)] _ b+_a
X_ 2A _ %(h)(a)+%h(ba) 3 _
a _.'
1; #ij+%'w(%§a)2+%h(b~a)3+%h(ba>(a+‘?9¥)Z i—— »——J ll ;—6hb(b1—ab+a2) Ans ...
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 Fall '08
 Troxel

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