Projekt za statiko

Projekt pri predmetu statika in kinematika2.P = 1000Nα = 30 ° , β = 60 ° , γ = 90°P F1 F2= =sinγ sin β sinαFF12= 866,02N= 500NFFF= sinα⋅ F = 433,01N1x1= cosα⋅ F = 750,00N1y1= sin β ⋅ F = 433,01N2x2F = cos β ⋅ F = 250N2y22

Projekt pri predmetu statika in kinematika3.Q = 1000Nα = 45°β = 45°γ = 90°Q F1 F2= =sinγ sin β sinαFF12= 707,11N= 707,11NSili v vrveh sta enaki 707,11N.3

Projekt pri predmetu statika in kinematika4.Q = 1000Nα = 30°β = 60°γ = 90°Q DF EF= =sin β sinα sinγDFEF= 577,35N= 1154,70NSila v palici DF je 577,35N, v palici EF pa 1154,70N.4

Projekt pri predmetu statika in kinematika5.Q = 1000Nα = 60°β = 30°γ = 90°Q BC AC= =sinα sin β sinγACBC= 1154,70N= 577,35NSila v palici AC je 1154,70N, v palici BC pa 577,35N.5

Projekt pri predmetu statika in kinematika6.F = 0, 6kNAB = 4mC0 = 2m1D0 = 0,2mN∑iFFF1DM= 10kNi= 0= 1kN⋅ D0DM 1C= =C01CAMN= FA1kN( )A0⋅ F = AB − A0⋅FAA0 ⋅ 1kN = 2, 4kNm − A0 ⋅0, 6kN2, 4kNmA0 = = 1,5m1, 6kN6

Projekt pri predmetu statika in kinematika7.7

Projekt pri predmetu statika in kinematikaPAB= 1, 8kNABBC =3α = β = 30°Q = 20kN∑iMi= 0 AD ⋅ AB −Q ⋅ AB − P ⋅ BC = 0M M ABM ABsinα ⋅ AD ⋅ AB − sinα ⋅Q ⋅ AB − sinαPAB⋅ = 03 PABAD −Q− = 03 PABAD = Q + = 20,6kN3 ADX= cosα⋅ AD = 17,84kN AD = sinα⋅ AD = 10,30kNY∑iXi= 0B − AD = 0 ⇒ B = AD = 17,84kN∑iFX X X XFYi= 0B − P −Q − AD = 0 ⇒ B = P + Q + AD = 32,10kNY AB Y Y AB YR : A = AD = 10,30 kN, A = AD = 10,30kNA X X Y YR : B = 17,84 kN, B = 32,10kNB X Y8

Projekt pri predmetu statika in kinematika8.P1= 200NP = 300NP23∑i= 400NMi= 0P ⋅ 20 + P ⋅ 10 + P ⋅10 − P ⋅ 50 = 0P1 2 3P ⋅ 20 + P ⋅ 10 + P ⋅ 10 4000N + 3000N + 4000N 11000N50 50 501 2 3= = == 220N∑iXi= 0A − P + P = 0X3A = P − P = 400N − 220N = 180NiX∑Yi3= 0A − P − P = 0Y2 1A = P + P = 300N + 200N = 500NYFF2 19

Projekt pri predmetu statika in kinematikaNTM diagram10

Projekt pri predmetu statika in kinematika9.PoznamoP, α, a, b,mbsin β =AFM = 0∑iiN ⋅ AF + m − P ⋅ a = 0DFYYN ⋅sin β ⋅ AF + m − cosα⋅P ⋅ a = 0DFbNDF⋅ ⋅ AF + m − cosα⋅P ⋅ a = 0AFcosα⋅P ⋅ a − mNDF=b11


Projekt pri predmetu statika in kinematika10.G = 50 N, P = 200NBα = 30 ° , γ = 60°A0 = B0 = 2C0∑ii= 0B = G , CD ⋅ cosα= CD = NY B Y CDP ⋅ A0 = CD⋅ C0 + B⋅B0NCDA0GBP ⋅ A0 = ⋅ + ⋅ A0cosα2 cosαN = P ⋅cos −G ⋅ 2 = 246, 41NCDM( α )BP = sinα⋅ P = 100 N, P = cosα⋅ P = 173,20 N, CD = 142,26 N,X Y XCD = 246, 41 N, B = 28,87 N, B = 50N∑iY X YFXi= 0− P + 0 − CD + B = 0 ⇒ 0 = P + CD − B = 213,39N∑iX X X X X X X XFYi= 0− P + 0 − CD − B = 0 ⇒ 0 = P + CD + B = 469,61NY Y Y Y Y Y Y Y13

Projekt pri predmetu statika in kinematika11.QAD= 3kNα = 30°AB = AC = 0, 6AD = 1mN∑iBC= 27,5kNi= 0ADsinα ⋅NBC ⋅ AC − sinα ⋅QAD ⋅ − sinα⋅QM⋅ AD = 02QMMADNBC⋅ AC −QAD⋅= 2 = 15kNAD14

Projekt pri predmetu statika in kinematika12.NB= 400N0B= 30cm0A= 50cmα = 120°∑ii= 0( α )N ⋅ 0B = cos − 90° ⋅N ⋅0ANiBA∑NB⋅ 0B= = 277,18Ncos 90° ⋅0AXi= 0( α − )0 − N = 0 ⇒ 0 = N = 138,59NX AX X AX( α )N = sin − 90° ⋅ N = 138,59N∑iAXYi= 0N + 0 + N = 0 ⇒ 0 = −N − N = −640,04NB Y AY Y B AY( α )N = cos − 90° ⋅ N = 240,04NAYMFFAAA15


Projekt pri predmetu statika in kinematika13.P1= 6, 0kNP = 10,0kN2M1= 30,0kNmq = 1,5 kN m∑iXi= 0− P + A − F = 0∑i1x x xYi= 0− P + A + B − Q + E − P + F = 0∑iFF1Y Y Y Y 2 YFMi= 0A ⋅ X + M + B ⋅ X − Q⋅ X + E ⋅ X − P ⋅ X + F ⋅ X = 0Y 1 1 Y 2 3 Y 4 2 5 Y 617

Projekt pri predmetu statika in kinematikaReakcije1.P11X11Y11= 6, 0kNP = cos 60°⋅ P = 3,0kNP = sin60°⋅ P = 5,20kNM= 30,0kNmq = 1,5 kN m∑iXi= 0− P + A + C = 0 ⇒ A = P − C = 4,64kND1x x X X 1x XXF=−1, 64kN∑iFYi= 0− P + A + B − Q + C = 0 ⇒ A = P − B + Q − C = 13,42kN1Y Y Y Y Y 1Y Y Y∑iMi= 0A ⋅ X + M + B ⋅ X −Q⋅ X + C ⋅ X = 0 ⇒BY 1 1 Y 2 3 Y 4YF−AY=⋅ X − M + Q⋅ X −CY⋅ XX( )( )1 1 3 421Y Y Y 1 1 3 Y 4= =1Y Y 1 1 3 Y 4= =( )22 1− 5,20kN + 1,5 kN m ⋅ 2m + 2,25kN ⋅2m −30,0kNm= +6m− 2m⎛2⎞1, 5 kN m ⋅2m ⋅ ⎜ + 6 m 2, 25kN 8m2⎟ + ⋅⎝ ⎠=−2, 97kN6m− 2mC =−2,25kNY− P − B + Q − C ⋅ X − M + Q⋅ X − C ⋅ X− P + Q − C ⋅ X − M + Q⋅ X −C ⋅ XXX− X=18

Projekt pri predmetu statika in kinematika2.q = 1,5 kN m∑iXi= 0− C + D = 0 ⇒ C = D = −1,64kNDiX∑FX X X X=−1, 64kNFYi= 0−C − Q + D = 0 ⇒ C =− Q + D =−1,5 kN m ⋅ 3m + 2,25kN =−2,25kNY Y Y Y∑iFMi= 03Q⋅X1, 5 kN m ⋅3m ⋅ m−Q⋅ X + D ⋅ X = ⇒ D = = 2 = ,25kN1 Y 210Y2X23m19

Projekt pri predmetu statika in kinematika3.P2= 10,0kNq = 1,5 kN m∑iXi= 0−D − F = 0 ⇒ D =− F =−1,64kNX x X XF = sin30°⋅ F = 1,64kNXF∑iFYi= 0−D − Q + E − P + F = 0 ⇒ E = D + Q + P − F =Y Y 2 Y Y Y 2 Y= 2,25kN + 1,5 kN m ⋅ 1,5m + 10kN − 2,84kN = 11,66kN∑iFMi= 0−Q⋅ X + E ⋅ X − P ⋅ X + F ⋅ X = 01 Y 2 2 3 Y 4( YY )( )−Q⋅ X + D + Q + P − F ⋅ X − P ⋅ X + FFY1 2 2 2 3Q⋅ X − D + Q + P ⋅ X + P ⋅ X1 Y2 2 2 3= =X− X4 2⋅ X = 01, 51, 5 kN m ⋅1,5m ⋅ m − ( 2, 25kN + 1, 5 kN m ⋅ 1,5m + 10, 0kN)⋅1,5m= 2+5m− 1,5m10,0kN ⋅3m 1,6875kNm − 21,75kNm + 30kNm= = 2, 84kN5m − 1,5m 3,5mY4FFY= = 3, 28kNcos30°20


Projekt pri predmetu statika in kinematika14.P1= 12,0kNP = 20,0kN2MM12= 30,0kNm= 24,0kNmq = 2, 0 kN m∑iAiX∑Xi= 0+ P = 0Yi1X= 0A − P + B − Q + D − P + F = 0Y 1Y Y Y 2 Y∑iFFFMi= 0−P ⋅ X − M + B ⋅ X − Q⋅ X + D ⋅ X + M − P ⋅ X + F ⋅ X = 01Y 1 1 Y 2 3 Y 4 2 2 5 Y 622

Projekt pri predmetu statika in kinematikaReakcije1.P1M1= 12,0kN= 30,0kNmq = 2, 0 kN m∑iXi= 0A + P + C = 0 ⇒ A = −P − C = −8,48kNCX 1X X X 1X XX= 0kNP = cos 45°⋅ P = 8, 48kN1X1∑iFFYi= 0A − P + B − Q + C = 0 ⇒ A = P − B + Q − C = −2,22kNY 1Y Y Y Y 1Y Y Y∑iFMi= 0−P ⋅ X − M + B ⋅ X −Q⋅ X + C ⋅ X = 0 ⇒ B =1Y 1 1 Y 2 3 Y 4YP ⋅ X + M + Q⋅ X − C ⋅ XX1Y1 1 3 Y 4= =1Y12⎛2⎞8, 48kN ⋅ 2m + 30, 0kNm + 2, 0 kN m ⋅2m ⋅ ⎜ + 4,5 m 0,1kN 6,5m2⎟ + ⋅=⎝ ⎠=4,5m= 14,80kNP = sin 45°⋅ P = 8, 48kNCY=−0,1kN23

Projekt pri predmetu statika in kinematika2.M2= 24,0kNmq = 2, 0 kN m∑iXi= 0− C + E = 0 ⇒ C = E = 0kNEiX∑X X X X= 0kNYi= 0−C − Q + D + E = 0 ⇒ C = − Q + D + E = −0,1kN∑iY Y Y Y Y YMi= 01 2 Y 3−Q⋅ X1 + DY ⋅ X2 + M2 + EY ⋅ X3= 0 ⇒ DY= =X22,52 kn / m ⋅2,5m ⋅ m − 24kNm + 8,57kN ⋅6m= 2 = 13, 47kN2,5mEYFFF=−8,57kNQ⋅ X − M − E ⋅ X24

Projekt pri predmetu statika in kinematika3.P2∑i= 20,0kNXi= 0− E = 0 ⇒ E = 0∑iXYi= 0−E − P + F = 0 ⇒ E = − P + F = − 20,0kN + 11, 43kN = −8,57kN∑iFFY 2 Y Y 2FMi= 0XP ⋅ X 20,0kN ⋅ 2m−P ⋅ X + F ⋅ X = ⇒ F = = = 1,43kN2 1 Y 22 10Y1X23,5m25


Projekt pri predmetu statika in kinematika15.P1= 8, 0kNP = 14,0kN2M1= 25,0kNmq = 1, 0 kN m∑iXi= 0A + B − P = 0∑iX X 2XYi= 0A − P + B − Q + D − P + F = 0Y 1 Y Y 2Y Y∑iFFFMi= 0−P ⋅ X + M + B ⋅ X −Q⋅ X + D ⋅ X − P ⋅ X + F ⋅ X = 01 1 1 Y 2 3 Y 4 2Y 5 Y 627

Projekt pri predmetu statika in kinematikaReakcije1.P1M1= 8, 0kN= 25,0kNmq = 1, 0 kN m∑iXi= 0A + B + C = 0 ⇒ B = −C − A = −0,55kNCX X X X X XX=−7kNAYAX= A⋅ sin30° = ⋅ sin30° = 7,55kNcos30°F = 0∑iYiA − P + B − Q + C = 0 ⇒ A = P − B + Q − C = 13,08kNY 1 Y Y Y 1 Y Y∑iFFMi= 0−P ⋅ X + M + B ⋅ X −Q⋅ X + C ⋅ X = 0 ⇒ B =1 1 1 Y 2 3 Y 4YP1 ⋅ X1 − M1 + Q⋅ X3− C=X24⎛2,5⎞8,0kN ⋅2m − 25,0kNm + 1,0kN m ⋅2,5m ⋅ ⎜ + 5 m 7,82kN 7,5m2⎟ − ⋅=⎝ ⎠=5m=−10, 41kNCY= 7, 82kNY⋅ X=28

Projekt pri predmetu statika in kinematika2.q = 1, 0 kN m∑iXi= 0− C + E = 0 ⇒ C = E = −7kNEiX∑X X X X=−7kNYi= 0−C − Q + D + E = 0 ⇒ C = − Q + D + E = 7,82kN∑iY Y Y Y Y YMi= 0Q X D X E X 0 D1 Y 3− ⋅1+Y⋅2+Y⋅3= ⇒Y= =X25,51, 0 kN m ⋅5, 5m ⋅ m + 6, 06kN ⋅5, 5m= 2 = 19,38kN2,5mEYFFF=−6,06kNQ⋅ X − E ⋅ X29

Projekt pri predmetu statika in kinematika3.P2∑i= 14,0kNXi= 0−E − P = 0 ⇒ E =− P =−7kNX 2X X 2XP = cos 60°⋅ P = 7kN2X2∑iYi= 0−E − P + F = 0 ⇒ E = − P + F = −6,06kN∑iFFY 2Y Y Y 2Y YFMi= 0P ⋅ X 12,12kN ⋅ 1,5m−P ⋅ X + F ⋅ X = ⇒ F = = ,06kN2Y12Y 1 Y 20Y6X23mP = sin60°⋅ P = 12,12kN2Y230


Projekt pri predmetu statika in kinematika16.P1= 9, 0kNP = 15,0kN2M1= 28,0kNmq = 1, 2 kN / m∑iXi= 0A + P − F = 0∑iX 1X XYi= 0A − Q + C − P + D − P + F = 0Y Y 1Y Y 2 Y∑iFFFMi= 0−Q⋅ X + C ⋅ X − P ⋅ X + D ⋅ X − M − P ⋅ X + F ⋅ X = 01 Y 2 1Y 3 Y 4 1 2 5 Y 632

Projekt pri predmetu statika in kinematikaReakcije1.q = 1, 2 kN m∑iXi= 0A + B = 0 ⇒ A = − B = −14,22kN∑iX X X XYi= 0A − Q + B = 0 ⇒ A = Q − B = 2,4kN∑iFFY Y Y YFMi= 04Q ⋅ X1, 2 kN m ⋅4m ⋅ m−Q ⋅ X + B ⋅ X = ⇒ B = = 2 = ,4kN1 Y 210Y2X24m33

Projekt pri predmetu statika in kinematika2.P1M1= 9, 0kN= 28,0kNmq = 1, 2 kN m∑iXi= 0− B + P + E = 0 ⇒ B = P + E = 14,22kNEXX 1X X X 1X X= 6, 43kNP = cos30°⋅ P = 7,79kN1X1∑iYi= 0−B − Q + C − P + D + E = 0 ⇒ C = B + Q + P − D − E = 3,25kN∑iY Y 1Y Y Y Y Y 1Y Y YMi= 0−Q⋅ X + C ⋅ X − P ⋅ X + D ⋅ X − M + E ⋅ X = 0 ⇒DYFFF1 Y 2 1Y 3 Y 4 1 Y 6Q⋅ X −C ⋅ X=1 Y 2( )( )+ P ⋅ X + M − E ⋅ XX1Y3 1 Y 64Q⋅ X − B + Q + P − D − E ⋅ X + P ⋅ X + M − E ⋅ X1 Y 1Y Y Y 2 1Y 3 1 Y 6= =Q⋅ X − B + Q + P − E ⋅ X + P ⋅ X + M − E ⋅ X1 Y 1Y Y 2 1Y 3 1 Y 6= =44 221,2 kN m ⋅2m ⋅ m − ( 2, 4kN + 1,2 kN m ⋅ 2m + 4,5kN − 8,57kN)⋅2m= 2+6,5m− 2m4,5kN ⋅ 5m + 28, 0kNm − 8,57kN ⋅8,5m6,5m− 2mBEYY= 2, 4kN= 8,57kNP = sin30°⋅ P = 4,5kN1Y1XX− X==−2,52kN34

Projekt pri predmetu statika in kinematika3.P2∑i= 15,0kNXi= 0−E − F = 0 ⇒ E = F = 6,43kNX X X XFYFX= F ⋅ sin 45° = ⋅ sin 45° = 6, 43kNcos 45°F = 0∑iYiE − P + F = 0 ⇒ E = P − F = 8,57kNY 2 Y Y 2 Y∑iFFMi= 0P ⋅ X 15,0kN ⋅ 1,5m−P ⋅ X + F ⋅ X = ⇒ F = = = ,43kN2 1 Y 22 10Y6X23,5m35

Projekt pri predmetu statika in kinematikaNTM diagramć36

Projekt pri predmetu statika in kinematika17.P1= 9, 0kNP = 15,0kNP23MMi12∑= 12,0kN= 24,0kNm= 35,0kNmXi= 0− P + P + D = 0∑i1X 2X XYi= 0A − P + C − P + D + F − P = 0Y 1Y Y 2Y Y Y 3∑iFFFMi= 0−P ⋅ X + C ⋅ X − P ⋅ X + D ⋅ X − M + M + F ⋅ X − P ⋅ X = 01Y 1 Y 2 2Y 3 Y 4 1 2 Y 5 3 637

Projekt pri predmetu statika in kinematikaReakcije1.P1∑i= 9, 0kNXi= 0− P + B = 0 ⇒ B = P = 4,5kN1X X X 1XP = cos 60°⋅ P = 4,5kN1X1∑iYi= 0A − P + B = 0 ⇒ A = P − B = 3,9kNY 1Y Y Y 1Y Y∑iFFFMi= 0P ⋅ X 7,79kN ⋅ 1,5 m−P ⋅ X + B ⋅ X = ⇒ B = = = ,89kN1Y11Y 1 Y 20Y3X23mP = sin60°⋅ P = 7,79kN1Y138

Projekt pri predmetu statika in kinematika2.P2Mi1∑= 15,0kN= 24,0kNmXi= 0− B + P + D + E = 0 ⇒ D = B − P − E = −6,11kNBEXXX 2X X X X X 2X X= 4,5kN= 0kNP = cos 45°⋅ P = 10,61kN2X2∑iYi= 0− B + C − P + D + E = 0 ⇒ C = B + P − D − E = 1,4kN∑iY Y 2Y Y Y Y Y 2Y Y YMi= 0C ⋅ X − P ⋅ X + D ⋅ X − M + E ⋅ X = 0 ⇒DY 1 2Y 2 Y 3 1 Y 4YFFF−CY⋅ X + P ⋅ X + M=X1 2Y2 1( )( )( )3Y 2Y Y Y 1 2Y 2 1 Y 4= =Y 2Y Y 1 2Y 2 1 Y 4= =2Y33 1− 3,89kN + 10,61kn − 6,8kN ⋅ 2,5m + 10,61kN ⋅4,5m= +6,5m− 2,5m24,0kNm −6,8kN ⋅9m= 19,9kN6,5m− 2,5mBEPYY= 3, 89kN=−6,8kN= 10,61kN− EY⋅ X− B + P − D − E ⋅ X + P ⋅ X + M − E ⋅ X− B + P − E ⋅ X + P ⋅ X + M − E ⋅ XXX− X4=39

Projekt pri predmetu statika in kinematika3.P3Mi2∑= 12,0kN= 35,0kNmXi= 0− E = 0kN ⇒ E = 0kN∑iXYi= 0− E + F − P = 0 ⇒ E = F − P = −6,8kN∑iFFY Y 3 Y Y 3FMi= 0M + F ⋅ X − P ⋅ X = 0 ⇒ F =2 Y 1 3 2X− 35,0kNm + 12,0kN ⋅4m= = 5, 2kN2,5mY− M + P ⋅ XX2 3 21=40


Projekt pri predmetu statika in kinematika18.P1= 6, 0kNP = 12,0kNP23M1= 16,0kN= 26,0kNmq = 1, 5 kN m∑iXi= 0A + P + F = 0∑iX 3X XYi= 0A −Q − P + C − P + D − P + F = 0Y 1 Y 2 Y 3Y Y∑iFFFMi= 0−Q⋅ X − P ⋅ X + C ⋅ X − M − P ⋅ X + D ⋅ X − P ⋅ X + F ⋅ X = 01 1 2 Y 3 1 2 4 Y 5 3Y6 Y 742

Projekt pri predmetu statika in kinematikaReakcije1.P1= 6,0kNq = 1, 5 kN m∑iXi= 0A + B = 0 ⇒ B = − A = −6,71kNX X X XAYAX= A⋅ sin 45° = ⋅ sin 45° = 6,71kNcos 45°F = 0∑iYiA −Q − P + B = 0 ⇒ A = Q + P − B = 6,71kNY 1 Y Y 1 Y∑iFFMi= 0Q X P X B X 0 BQ⋅ X + P ⋅ X1 1 2− ⋅1−1⋅2+Y⋅3= ⇒Y= =X34,51,5 kN m ⋅4, 5m ⋅ m + 6, 0kN ⋅2m= 2 = 6, 04kN4,5m43

Projekt pri predmetu statika in kinematika2.P2M1= 12,0kN= 26,0kNmq = 1, 5 kN m∑iXi= 0− B + E = 0 ⇒ E = B = −6,71kNBiX∑X X X X=−6,71kNYi= 0−B − Q + C − P + D + E = 0 ⇒ C = B + Q + P − D − E = 10,83kN∑iY Y 2 Y Y Y Y 2 Y YMi= 0−Q⋅ X + C ⋅ X − M − P ⋅ X + D ⋅ X + E ⋅ X = 0 ⇒DYFFF1 Y 2 1 2 3 Y 4 Y 5Q⋅ X −C ⋅ X + M + P ⋅ X − E ⋅ XX1 Y 2 1 2 3 Y 5= =( )4Q⋅ X1− BY+ Q + P − DY − EY ⋅ X + M + P ⋅ X − EY⋅ X=X( )2 2 1 2 3 51 Y 2 Y 2 1 2 3 Y 5= =44 221,5 kN m ⋅2m ⋅ m − ( 6,04kN + 1,5 kN m ⋅ 2m + 12,0kN + 5,96kN)⋅2m= 2+6m− 2m26,0kNm + 12,0kN ⋅ 3,5m + 5,96kN ⋅8m= 16,17kN6m− 2mB = 6,04kNEYYQ⋅ X − B + Q + P − E ⋅ X + M + P ⋅ X − E ⋅ X=−5, 96kNX− X=44

Projekt pri predmetu statika in kinematika3.P3∑i= 16,0kNXi= 0− E + P + F = 0 ⇒ F = E − P = −14,71kNEXX 3X X X X 3X=−6,71kNP = cos 60°⋅ P = 8,0kN3X3∑iYi= 0−E − P + F = 0 ⇒ E = − P + F = −5,96kN∑iFFY 3Y Y Y 3Y YFMi= 0P ⋅ X 13,86kN ⋅ 2m−P ⋅ X + F ⋅ X = ⇒ F = = = ,92kN3Y13Y 1 Y 20Y7X23,5mP = sin60°⋅ P = 13,86kN3Y345


Projekt pri predmetu statika in kinematika19.P1= 10,0kNP = 16,0kN2M1= 32,0kNmq = 1, 4 kN m∑iFiX∑Xi= 0+ P = 0Yi2X= 0A + C − Q + E − P + F − P = 0Y Y Y 1 Y 2Y∑iFFFMi= 0− M + C ⋅ X −Q⋅ X + E ⋅ X − P ⋅ X + F ⋅ X − P ⋅ X = 01 Y 1 2 Y 3 1 4 Y 5 2Y647

Projekt pri predmetu statika in kinematikaReakcije1.Mii1∑BX∑= 32,0kNmXi= 0Yi= 0= 0A + B = 0 ⇒ A = − B = −8,0kN∑iFFY Y Y YFMi= 0M 32,0kNm− M + B ⋅ X = ⇒ B = = = ,0kN1 Y 110Y8X14m48

Projekt pri predmetu statika in kinematika2.q = 1, 4 kN m∑iXi= 0− B + D = 0 ⇒ D = B = 0kNBiX∑X X X X= 0kNYi= 0− B + C − Q + D = 0 ⇒ C = B + Q − D = 16,15kN∑iFFY Y Y Y Y YFMi= 0Y 1 2Y⋅1− ⋅2+Y⋅3= 0 ⇒Y= =X3C X Q X D X D( ) ( )−C ⋅ X + Q⋅X− B + Q − D ⋅ X + Q⋅ X − B + Q ⋅ X + Q⋅XY Y 1 2 Y1 2= = =X X − X3 3 1⎛2,5⎞− ( 8, 0kN + 1, 4 kN m ⋅ 2,5m)2m + 1, 4 kN m ⋅2,5m⋅ ⎜ + 2 m2⎟=⎝ ⎠4,5m− 2mBY= 8, 0kN=−4, 65kN49

Projekt pri predmetu statika in kinematika3.P1= 10,0kNP = 16,0kN2q = 1, 4 kN m∑iXi= 0− D + F + P = 0 ⇒ F = D − P = −13,86kNDXX X 2X X X 2X= 0kNP = cos30°⋅ P = 13,86kN2X2∑iYi= 0−D − Q + E − P + F − P = 0 ⇒ E = D + Q + P − F + P = −2,56kN∑iY Y 1 Y 2Y Y Y 1 Y 2YMi= 0−Q⋅ X + E ⋅ X − P ⋅ X + F ⋅ X − P ⋅ X = 0 ⇒FYFFF1 Y 2 1 3 Y 4 2Y5Q⋅ X1 − EY⋅ X2 + P1⋅ X + PY⋅ X=X( )( )43 2 5Q⋅ X − D + Q + P − F + P ⋅ X + P ⋅ X + P ⋅ X1 Y 1 Y 2Y 2 1 3 2Y5= =Q⋅ X − D + Q + P + P ⋅ X + P ⋅ X + P ⋅ X1 Y 1 2Y 2 1 3 2Y5= =2Y44 221, 4 kN m ⋅2m ⋅ m − ( − 4, 65kN + 1, 4 kN m ⋅ 2m + 10, 0kN + 8, 0kN ) ⋅2m= 2+5,5m− 2m10,0kN ⋅ 3,5m + 8,0kN ⋅7,5m= 18,71kN5,5m− 2mDPY=−4, 65kNX= sin30°⋅ P2= 8,0kNX− X=50


Projekt pri predmetu statika in kinematika20.P1= 7, 0kNP = 10,0kNP23M1= 14,0kN= 26,0kNmq = 1, 3 kN m∑iAiX∑Xi= 0+ P = 0Yi2X= 0A − P −Q − P + D + E − P = 0Y 1 2Y Y Y 3∑iFFFMi= 0−P ⋅ X − Q⋅ X − P ⋅ X + D ⋅ X + M + E ⋅ X − P ⋅ X = 01 1 2 2Y 3 Y 4 1 Y 5 3 652

Projekt pri predmetu statika in kinematikaReakcije1.P1∑i= 7, 0kNXi= 0A + B = 0 ⇒ A = − B = −5,0kNBiX X X XX∑= 5, 0kNYi= 0A − P + B = 0 ⇒ A = P − B = 3,0kNY 1 Y Y 1 Y∑iFFFMi= 0P ⋅ X 7, 0kN ⋅ 2m−P ⋅ X + B ⋅ X = ⇒ B = = = ,0kN1 1 Y 21 10Y4X23,5m53

Projekt pri predmetu statika in kinematika2.P2= 10,0kNq = 1, 3 kN m∑iXi= 0− B + P + C = 0 ⇒ B = P + C = 5,0kNCXX 2X X X 2X X= 0kNP = cos 60°⋅ P = 5,0kN2X2∑iYi= 0−B −Q − P + C = 0∑iFFY 2Y YFMi= 01 2Y2−Q⋅ X1 − P2 Y⋅ X2 + CY ⋅ X3= 0 ⇒ CY= =X35,51, 3 kN m ⋅5, 5m ⋅ m + 8, 66kN ⋅2m= 2 = 6,72kN5,5mP= sin60°⋅ P = 8,66kN2Y2Q⋅ X + P ⋅ X54

Projekt pri predmetu statika in kinematika3.P3M1= 14,0kN= 26,0kNmq = 1, 3 kN m∑iXi= 0− C = 0 ⇒ C = 0kN∑iXYi= 0−C − Q + D + E − P = 0 ⇒ D = C + Q − E + P = 14,0kN∑iY Y Y 3 Y Y Y 3Mi= 0X−Q⋅ X + D ⋅ X + M + E ⋅ X − P ⋅ X = 0 ⇒EYFFF1 Y 2 1 Y 3 3 4Q⋅ X − D ⋅ X − M + P ⋅ XX1 Y 2 1 3 4= =( )3Q⋅ X − C + Q − E + P ⋅ X − M + P ⋅ X1 YY 3 2 1 3 4= =( )Q⋅ X1 − CY+ Q + P3 ⋅ X2− M=X − XX33 21 3 421,3kN m ⋅2m ⋅ m − ( 6,72kN + 1,3kN m ⋅ 2m + 14,0kN ) ⋅2m −26,0kNm= 2+5m− 2m14,0kN⋅ 7m= 9,32kN5m− 2mC = 6,72kNY+ P ⋅ X=55


Projekt pri predmetu statika in kinematika21.P1= 5, 0kNP = 13,0kN2M1= 30,0kNmq = 2, 0 kN m∑iXi= 0A − P − E = 0∑iX 2X XYi= 0A − Q + C − P − P + E = 0Y Y 1 2Y Y∑iFFFMi= 0−Q⋅ X + C ⋅ X + M − P ⋅ X − P ⋅ X + E ⋅ X = 01 Y 2 1 1 3 2Y4 Y 557

Projekt pri predmetu statika in kinematikaReakcije1.q = 2, 0 kN m∑iXi= 0A + B = 0 ⇒ A = − B = 15,67kNBiX X X XX∑=−15,67kNYi= 0A − Q + B = 0 ⇒ A = Q − B = 5,0kN∑iFFY Y Y YFMi= 05Q⋅X2, 0 kN m ⋅5m ⋅ m−Q⋅ X + B ⋅ X = ⇒ B = = 2 = ,0kN1 Y 210Y5X25m58

Projekt pri predmetu statika in kinematika2.M1= 30,0kNmq = 2, 0 kN m∑iXi= 0− B + D = 0 ⇒ B = D = −15,67kNDiX∑X X X X=−15,67kNYi= 0−B − Q + C + D = 0 ⇒ C = B + Q − D = 29,5kN∑iFFY Y Y Y Y YFMi= 0Q⋅ X −C ⋅ X1 Y 2−Q⋅ X1 + CY ⋅ X2 + M1 + DY ⋅ X3= 0 ⇒ DY= =X3( ) ( )Q⋅ X − B + Q − D ⋅ X − M Q⋅ X − B + Q ⋅ X − M1 Y Y 2 1 1 Y2 1= = =X X − X3 3 22,52, 0 kN m ⋅2,5m⋅ m − ( 5, 0kN + 2, 0 kN m⋅2,5m)⋅2,5m −30, 0kNm=2 =−19,5kN5m− 2,5mBY= 5, 0kN59

Projekt pri predmetu statika in kinematika3.P1= 5, 0kNP = 13,0kN2∑iXi= 0−D − P − E = 0 ⇒ D = −P − E = −15,67kNX 2X X X 2X XEYEX= E ⋅ sin 45° = ⋅ sin 45° = 9,17kNcos 45°P = cos 60°⋅ P = 6,5kN2X2∑iYi= 0−D − P − P + E = 0∑iFFY 1 2Y YFMi= 01 1 2Y2−P1 ⋅ X1 − P2 Y⋅ X2 + EY ⋅ X3= 0 ⇒ EY= =X35, 0kN ⋅ 2m + 11,26 kN ⋅4m=6mP = sin60°⋅ P = 11,26kN2Y2= 9,17kNP ⋅ X + P ⋅ X60


Projekt pri predmetu statika in kinematika22.P1= 18,0kNP = 7, 0kNP23Mi1∑= 12,0kN= 16,0kNmXi= 0A + P − P + E = 0∑iX 1X 2X XYi= 0A − P + B − P − P + E = 0Y 1Y Y 2Y 3 Y∑iFFFMi= 0−P ⋅ X + B ⋅ X − M − P ⋅ X − P ⋅ X + E ⋅ X = 01Y 1 Y 2 1 2Y 3 3 4 Y 562

Projekt pri predmetu statika in kinematikaReakcije1.P1Mi1∑= 18,0kN= 16,0kNmXi= 0A + P + C = 0 ⇒ C = −A − P = −17,13kNX 1X X X X 1XAYAX= sin30°⋅ A = sin30°⋅ = 4, 4kNcos30°P = cos 45°⋅ P = 12,73kN1X1∑iYi= 0A − P + B + C = 0 ⇒ A = P − B − C = −7,63kNY 1Y Y Y Y 1Y Y Y∑iFFFMi= 0P1 Y⋅ X1 + M1−C−P1 Y⋅ X1 + BY ⋅ X2 − M1 + CY ⋅ X3= 0 ⇒ BY=X12,73kN ⋅ 2,5m + 16,0kNm + 11,2kN ⋅7m= = 31,56kN4mC =−11,2kNYP = sin 45°⋅ P = 12,73kN1Y12Y⋅ X3=63

Projekt pri predmetu statika in kinematika2.P2∑i= 7, 0kNXi= 0−C − P + D = 0 ⇒ D = C + P = −13,63kNCXX 2X X X X 2X=−17,13kNP = cos 60°⋅ P = 3,5kN2X2∑iYi= 0−C − P + D = 0 ⇒ C = − P + D = −11,2kNDYY 2Y Y Y 2Y Y=−5,14kNP = sin60°⋅ P = 6,06kN2Y2∑iFFFMi= 0−P ⋅ X + D ⋅ X = 02Y1 Y 264

Projekt pri predmetu statika in kinematika3.P3∑i= 12,0kNXi= 0− D + E = 0 ⇒ E = D = −13,63kNDiX∑X X X X=−13,63kNYi= 0−D − P + E = 0 ⇒ D = − P + E = −5,14kN∑iFFY 3 Y Y 3 YFMi= 0P ⋅ X 12,0kN ⋅ 2m−P ⋅ X + E ⋅ X = ⇒ E = = = ,86kN3 1 Y 23 10Y6X23,5m65

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