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# Hw2 - Normal Stress 7 A mechanical test is performed on a...

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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 0-015 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 y-direction, and can be considered rigid relevant to the stifﬁiess of cables BC and BD. Find the displacement of B in the y-direction, 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 y-direction (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 x-direction) 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 =100-9.81=981N. 63c: FBCLBC = 98l-1.2 = EACAAC zoo-10’ 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 _1250-10-12_0.00 a _ 1=,,LB _ 750-1042 B _ _—6=0.018in EBA, 10-10 ~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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