Ep 12.FortherectangularplateofProblems 2.109and2.110,determinethetension ineachofthethreecablesknowingthattheweightoftheplate is792N.9450250DB3601307320360450YCDimensions in mmSOLUTIONSee Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and(3) below.SettingP=792Ngives:8Ta+06T1e+T(1)13TD=09.6712TAs0.64TAcT-To +792N=0(2)17917.2.,17 Tas + 0.48TAc -13T4D=0(3)SolvingEquations (1),(2),and (3)byconventionalalgorithmsgivesTAB=510.00NTAB=510NTAc=56.250NTxc=56.2NTAD = 536.25 NTAD=536N
Ep 12
Ep 13. Known: Dimensions A, B, P acting on the BC bar can be translated with X, C,E is smooth contact,thedeadweight of eachbar is not calculated.pinA,B penetrateseach member.Find out: AB bar stress, and explain whether AB bar stress and XFarxyFaxxEATFA会F(b)(c)(a)Bfe(d)Ep 13SOLUTION:Take the whole as the research object, and the force is shown in Figure (a)ZX=0Fx =0b-xpZme(F)=0P(b-x)- F,b=0get FAybPin A was taken as the object of study, and its stress was shown in Figure (b),ZX=0FA +FADG = 0getFADG =0
Ep 13. Known: Dimensions A, B, P acting on the BC bar can be translated with X, C, E is smooth contact, the dead weight of each bar is not calculated, pin A, B penetrates each member. Find out: AB bar stress, and explain whether AB bar stress and X. SOLUTION: Take the whole as the research object, and the force is shown in Figure (a), X = 0 FAx = 0 mE (F) = 0 P(b − x) − FAyb = 0 get P b b x FAy − = Pin A was taken as the object of study, and its stress was shown in Figure (b), X = 0 FAx + FADCx = 0 get FADCx = 0 Ep 13
BC bar was taken as the research object, and its stress was shown in Figure (d),得F=PZm(F)=0F'b-Px=0bADC bar is taken as the research object, and its force is shown in Figure (c),bbaXP-FADG*Zm,(F)=0FADoy*2=0F'Axcy-Fe*2get2bThen pin A is taken as the research object, and there are:ZY=0FAB+FA,+ FAD,=0b-Xp+=p=0That is FAB +FAB=-PgetbbThat is, the AB bar is under the action of pressure P,independentofpositionXExercise: Known: P、a、b, the point E is smoothcontact, and the weight ofeach rod is not countedCalculate:theforceofBCDbaronACEbaratC
BC bar was taken as the research object, and its stress was shown in Figure (d), mB (F) = 0 FC b − Px = 0 得 P b x FC = ADC bar is taken as the research object, and its force is shown in Figure (c), mD (F) = 0 0 2 2 2 − − = b F a F b FADCy ADCx C get P b x FAXCy = Then pin A is taken as the research object, and there are: Y = 0 FAB + FAy + FADCy = 0 That is + = 0 − + P b x P b b x FAB get FAB = −P That is, the AB bar is under the action of pressure P, independent of position X. Exercise: Known: P、a、b, the point E is smooth contact, and the weight of each rod is not counted. Calculate: the force of BCD bar on ACE bar at C
Ep 14.The cuboid's side length is A, and its acting force are F and F,.The actingposition is shown inFigure6.The coupleacting surfacewitha momentof M,is inOBGE plane, while the couple acting surface with a moment of M, is in BCDGplane.Calculate: the projection of each force on the x,y, z axes and the simplified result ofthemoment ofx,y,zaxesandtheforcesystemtoward Opoint↓ 2SOLUTION:5Projection:45V2V2Fi. = 0F.H222-F2,=F,cos45°cos45°FI-F,cos45°sin45°FF2x = -2V2Ep 14F2.=F,sin45°22The moments of the x, y, z axes of each force and its couple are respectivelyV2V2V2m(F)=m,(E):m.(F)=F,aFaFa222J2111m,(F)=m.(E)m.(F)=F2aFaF,aFa-222V2V2Mix=M,cos45°M,Mi, = -M,sin 45°M,M.=022Mzx=0M2v = M,M2: = 0The simplified principal vector principal moment of theforcesystem towards OpointisV21.FR'=ZXF22V21FR'=ZYH2
Ep 14.The cuboid's side length is A, and its acting force are F1 and F2 . The acting position is shown in Figure 6.The couple acting surface with a moment of M1 is in OBGE plane, while the couple acting surface with a moment of M2 is in BCDG plane. Calculate: the projection of each force on the x,y, z axes and the simplified result of the moment of x, y, z axes and the force system toward O point. SOLUTION: Projection: 1 1 2 2 F x = − F 1 1 2 2 F y = F 0 F1z = 2 2 2 2 1 F x = F cos45 cos45 = F 2 2 2 2 1 F y = −F cos45 sin 45 = − F 2 2 2 2 2 F z = F sin 45 = F The moments of the x, y, z axes of each force and its couple are respectively mx F1 F1a 2 2 ( ) = − my F1 F1a 2 2 ( ) = − mz F1 F1a 2 2 ( ) = mx F2 F2 a F2 a 2 2 2 1 ( ) = + m F F a y 2 2 2 1 ( ) = m F F a z 2 2 2 1 ( ) = − 1 1 1 2 2 M x = M cos45 = M 1 1 1 2 2 M y = −M sin 45 = − M M1z = 0 M 2x = 0 M2y = M2 0 M 2z = The simplified principal vector principal moment of the force system towards O point is 1 2 2 1 2 2 Rx = X = − F + F 1 2 2 1 2 2 R y = Y = F − F Ep 14
V2R'=ZZF中2R'=R'i+R'j+R'kV2V2V2M, =Zm(F):MFa+F,a+2222V2V21M, =Zm,(F)=Fa+FaM, +M,2 22V21M,=Zm.(F)=FaF,a22M。=M,i+Mj+M.k2Exercise:The position of three forces F,F,F, isshown in theright figure, F Il x,F, ll y, F, is in theOxzplane,F=100N,F,=100/2N,F,=200N,ifthe principal moment of a simplified vector in the xyplane is in the same square, calculate the coordinatesof point Aand theprincipal vector and principalmoment simplified to point A
2 2 2 Rz = Z = F R R i R j R k x y z = + + 1 2 2 1 2 2 2 2 2 1 2 2 M x = mx (F) = − F a + F a + F a + M 1 2 1 2 2 2 2 1 2 2 M y = my (F) = − F a + F a − M + M M z mz F F1a F2a 2 1 2 2 = ( ) = − M M i M j M k o x y z = + + Exercise: The position of three forces 1 2 3 F ,F ,F is shown in the right figure, F // x 1 ，F // y 3 ，F2 is in the Oxz plane, F1 = 100N ，F2 = 100 2N,F3 = 200N , if the principal moment of a simplified vector in the xy plane is in the same square, calculate the coordinates of point A and the principal vector and principal moment simplified to point A