Ep 3. Known: P, 1, R, Each bar and pulley shall be excluded; Find the constraintreactionatfixedendAFaFM0(b)(c)Ep 3SOLUTION:TaketheCDrod(includingthepulley)astheresearchobject,anditsforceisshowninFigure(b),fromZm,(F)=0FcB×21+F(1+R)-PR= 0PWe getFcB =2Take rod AB and its force diagram is shown in Figure (c), fromZX=0FA+FBC+F=0PWe getFx =2FromZY=0FA,=0Zm,(F)=0MA-F(1+R)-FBc×21=0We getM =PRExercise:Known:g=500N/m,P=2000N,M=500Nm,a=lm,Not countingtheweight of eachpoleCalculate:theforceofABbaronCDbaratBE
Ep 3. Known: P, l, R, Each bar and pulley shall be excluded; Find the constraint reaction at fixed end A. SOLUTION: Take the CD rod (including the pulley) as the research object, and its force is shown in Figure (b), from mD (F) = 0 FCB 2l + FT (l + R) − PR = 0 We get 2 P FCB = − Take rod AB and its force diagram is shown in Figure (c), from X = 0 FAx + FBC + FT = 0 We get 2 P FAx = − From Y = 0 FAy = 0 mA (F) = 0 MA − FT (l + R) − FBC 2l = 0 We get M A = PR Exercise: Known: q = 500N / m,P = 2000N,M = 500Nm,a =1m, Not counting the weight of each pole. Calculate: the force of AB bar on CD bar at B. Ep 3
Ep4.Known:P,a, M=Pa,Notcounting the weight of eachpoleFind the constraint reaction at support ACand D.BMDSOLUTION:Firstly, BC bar is taken as the object ofstudy, and its force diagram is shown inFigure(c),fromZmc(F)=0-F×2a+Pa-M=0We getFby = 0CB7M公FFAFAxa20T(b)(a)FrFeFcFeNFcFtyMpCBMDFix(d)(c)Ep 4Finally, take the CD rod as the research object, and the force is shown in Figure (d),PZX=0-Fc+FDx=0getFpx=2ZY=0Fd, = PFdy-Fo, =0get
Ep 4. Known: P ,a,M = Pa ，Not counting the weight of each pole. Find the constraint reaction at support A and D. SOLUTION: Firstly, BC bar is taken as the object of study, and its force diagram is shown in Figure (c), from mC (F) = 0 − FBy 2a + Pa − M = 0 We get FBy = 0 取折杆 AB 为研究对象，受力如图（b）所示，由 Y = 0 FAy − P − FBy = 0 得 FAy = P mB (F) = 0 − FAy 2a + Pa + FAx 2a = 0 得 2 P FAx = X = 0 FAx + FBx = 0 得 2 P FBx = − 取 BC 为研究对象，受力图（c），由 X = 0 − FBx + FCx = 0 得 2 P FCx = − Y = 0 FBy + FCy − P = 0 得 FCy = P Finally, take the CD rod as the research object, and the force is shown in Figure (d), X = 0 − FCx + FDx = 0 get 2 P FDx = − Y = 0 FDy − FCy = 0 get FDy = P Ep 4
Zmp(F)=0M,+2aFα+2aF,=0get Mp =-PaExercise:Known:AB=2BC=2CD=2m,q=2000N/m,XDM=500Nm, Each rod is homogeneous and itsweight per unit length is 500N/m.Calculate: Constraint reaction at A.WM29
mD (F) = 0 MD + 2aFCx + 2aFcy = 0 get M Pa D = − Exercise: Known: AB=2BC=2CD=2m，q=2000N/m， M=500Nm, Each rod is homogeneous and its weight per unit length is 500N/m. Calculate: Constraint reaction at A
Ep 5.Determine thex andycomponents ofeach of theforces shown.U28mm84mm96 mm50N80mm29NO51N90 mm-48mm-SOLUTIONo.Compute the following distances:OA=(84)2+(80)2SON=116 mmOB= (28)2 +(96)2.51N=100mmOC=/(48)2 +(90)2=102mm84F,=+(29 N)F,=+21.0N29-NForce:11680F, = +(29 N)-F, =+20.0 N11628F=-(50 N)F,=-14.00N50-NForce:10096F, =+(50 N),F, =+48.0 N10048F=+(51N),F,=+24.0N51-NForce:10290F, = (51 N)F, =-45.0 N102
Ep 5
Ep 6.Knowing that α =35°, determine the resultant of the threeforces shown.aa30%100N200N150NSOLUTIONF,=+(100N)c0s35°=+81.915N100-N Force:F, = -(100 N)sin35°= -57.358 N150-NForce:F,=+(150N)cos65°=+63.393NF, = (150 N)sin 65°= 135.946 NF,=(200N)cos35°=163.830N200-NForce:F, = (200 N)sin35°= 114.715 NForcex Comp. (N)yComp. (N)100N+81.91557.358150N+63.393135.946200N163.830114.715R, =18.522R, =308.02R=Ri+R,jR,=-18.522j=(18.522 N)i + (308.02 N)jRtanα=R.308.0218.522Ry= -308.02jα=86.5590K308.02 NR=R=309N786.6°sin 86.559
Ep 6