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Unformatted text preview: Normal Stress: 7 A mechanical test is performed on a synthetic rubber material by imposing tensile load
on a pad as the one shown. For a=3 inches, h=l .5 inches, a force P=180 lbs results in 3 0015 inches
vertical elongation of the pad. What is the elasticity modulus of the rubber, E? +—:='—+ 180 L 1.5
E=g=ﬂ=2000psi=2ksi
e 0.01 Bars: 7 To account for approximations in analysis (misused or incorrect fabrications) most structures are
made stronger than suggested by the analysis. This extra margin of strength is described by Margin of Safety = w  1
Maxnmum Stress Yield Stress or Safety factor = _—
Maxtmum Stress The minimum value of margin of safety = 0
The minimum value of safety factor = l Example:
1 0,000
A circular rod is subjected to a tensile force of 10,000 lb. Ifcyﬂd =
45,000 psi and the factor of safety = 5 ﬁnd the required diameter.
D
1 0,000
Solution:
0' “mm“, = 45,000 = 9,000 psi
0",, =5 or A =L=£D2 so D2 = 4F which means D= 4F
0”,, 4 no,” nod,
D = M = 1.189 in
7: X 9,000 Bars: 10 Beam AB is vertical
before the 20KN load is applied
in the ydirection, and can be
considered rigid relevant to the
stifﬁiess of cables BC and BD.
Find the displacement of B in the
ydirection, considering that the
two cables are identical, of
diameter d=2cm and E=2006Pa. Solution
Due to symmetry, the two cables will be subjected to identical
tensile stress. Coordinates of point B (x,y,z) in meters: (0,0,5)
Coordinates of point C: (l,4,0)
Vector BC: (—1,—4,5) Magnitude ofvector BC: ‘/(—1)2 + (—4)2 + 52 = 6.48m . _ . . 1 —4 —5
Unit vector in direction BC: —,—,—
6.48 6.48 6.48 Let Fnc denote the magnitude of the force in cable BC. Then the component of this force in the y
4 direction is _—FBC . Then for equilibrium in the ydirection (since the component of the force in BD X is the same as the force in BC) zilgc =20kN —)F,C =16.2kN. 6. 48
Then, =M=M=o.mlé6m Eac Anc 200 10’ 1t 0022/4
Let E be the point exactly between points C and D. Then, in order to ﬁnd the displacement of B in the y
direction, we need to project 83C onto axis BE. This projection is done by the dot product, i.e. 835 = ﬁncﬁnc ' ﬁBE , where 5 denotes unit vector, and ' denotes dot product.
_ —1 T —4 —. —5 —
 k nBC —— 1 + 6.48 6. 48 6.48
Coordinates of pomt B (x,y,z) in meters: (0,0,5)
Coordinates of point B: (0,4,0)
Vector BE: (0,4,5) Magnitude of vector BE: (0)2 + (—4)2 + 52 = 6.40312!!!
0 —4 —5 J Unit vector in direction BE: , ,
(6.40312 6.40312 6.40312 01' _ —4 —. 5 —
nBE = j + k
6.403 12 6.40312 an =81“;BC in =0.00166[ —4 —4+ —5 —5
6.40312 6—.48+ 6.40312 6.48 —=] 0. 00164m 55— _—= — =0. 002625m
Cos[51. 3°] 0.625 Bars 11 Find the vertical (in xdirection)
displacement of point B imposed by the 100 kg
weight. Cable BC is of diameter d=2cm and
E=2006Pa. Consider beam ABED as being iigid and
pinned at A and D. Solution Taking moments about axis AD, it follows that the
force in cable BC is equal to $6 the weight of lOOkg. Then: Fae =1009.81=981N.
63c: FBCLBC = 98l1.2 =
EACAAC zoo10’ 3140.022 /4 Bars 12: For the stmctute, ﬁnd:
:1) FA
b) F];
c) 81; d) 53. 8' Solution 8
2F,=0—)FAcos(e)=o—>FAcos(E)=o—> FA =1250 lb {L'+f 2E =o—>1=B =13 sin(0) —>F,, = FA 911%)» FB=7501b
8 _ FALA _12501012_0.00 a _ 1=,,LB _ 7501042 B _ _—6=0.018in
EBA, 1010 ~1/2 Bars 13: A bar is elongated by an end load as indicated below. The displacement at point A is measured
to be “A = no . What is the displacement at point B? Solution Constant strain. 8 _. so we have “A = ELA and 11B =£LB. Thus, “A ’LA = ‘13an and so
uB = uoLB /LA ...
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 Spring '08
 FRANTZISKONIS

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