Ep 15. As is shown in Figure 7(a), threedisjoint and uneven edges along thecuboid act on three equal forces F.Find out: what relationship should edgeA, B and C have so that the forcesystemcanbereducedtoaforce(b)SOLUTION:(a)Ep 15Select A as the simplificationcenter, and establish the coordinate system as shown in Figure (b). The simplificationresult is as follows:R'=Fi+F+FkM,=(Fb-Fc)i-FajWhen R'IM,the final simplified result is a net force, therefore:R'.M,=0F(b-c)-Fa=0that isThus:a=b-cExercise: As shown in the figure on the right, the sidelength of the cube is a, and the forces acting on it areallthreeforcesofsizeF,whicharesimplifiedtopointO. Judge whether the final simplification result is aresultantforce
Ep 15. As is shown in Figure 7(a), three disjoint and uneven edges along the cuboid act on three equal forces F. Find out: what relationship should edge A, B and C have so that the force system can be reduced to a force. SOLUTION: Select A as the simplification center, and establish the coordinate system as shown in Figure (b). The simplification result is as follows: R Fi Fj Fk = + + M Fb Fc i Faj A = ( − ) − When R M A ⊥ , the final simplified result is a net force, therefore: R • M A = 0 that is ( ) 0 2 2 F b − c − F a = Thus: a = b − c Exercise: As shown in the figure on the right, the side length of the cube is a, and the forces acting on it are all three forces of size F, which are simplified to point O. Judge whether the final simplification result is a resultant force. Ep 15 x y z
Ep 16.As shown in Figure 8 (a), the frame is composed of ninerods regardless ofself-weight, AB=BC=CA=3m, BE=CF=AD=4m, ABC forms a horizontal plane, BECF and AD rods are perpendicular, and force P is parallel to rod BC.Calculate:Theinternalforceofeachbar.SOLUTION:Take node Aas theresearch object, and theforce is shown in Figure (b).n,IABthe axial force of AB, AD and AE bar is projected on it to be zero.ZF=0Pcos30-FAccos30°=0FAc = PWe get50°C(b)(a)(c)Ep 16Take ABC as the research object, and the force is shown in Figure. (c),fromZmn2(F)=0Fcpsinα×3cos30°-P×3cos30°=0SPFcDget3Zm3(F)= 0FAesinα×3cos30°-P×3cos30°=02PFAEget3Zmn(F)=0FBrsinα×3cos30°=0FBF = 0getForthe intersection point A, see Figure (b, c)ZFn4 = 0-FAD-FAECOSα=O
Ep 16. As shown in Figure 8 (a), the frame is composed of nine rods regardless of self-weight, AB=BC=CA=3m, BE=CF=AD=4m, ABC forms a horizontal plane, BE, CF and AD rods are perpendicular, and force P is parallel to rod BC. Calculate: The internal force of each bar. SOLUTION: Take node A as the research object, and the force is shown in Figure (b). n1 ⊥ AB , the axial force of AB, AD and AE bar is projected on it to be zero. Fn1 = 0 cos30 − cos30 = 0 P FAC We get FAC = P Take ABC as the research object, and the force is shown in Figure. (c), from mn2 (F) = 0 sin 3cos30 − 3cos30 = 0 FCD P get FCD P 3 5 = mn3 (F) = 0 sin 3cos30 − 3cos30 = 0 FAE P get FAE P 3 5 = mn4 (F) = 0 sin 3cos30 = 0 FBF get = 0 FBF For the intersection point A, see Figure (b, c). Fn4 = 0 − FAD − FAE cos = 0 Ep 16
4PgetFAD3ZFns=0FAB+FAESinα-Pcos60°-FAccos60°=0FAB=0getFor the intersection point B, see Figure (b, c)FromZFm2=0-FBE-FBFCOSα=OGetFBE=0ZFm6=0FBc+FBrSinα+FABCos60°=0GetFBc = 0For the intersection point C, see Figure (b, c)ZFm3 = 0-FcF-Fccosα=04GetPFcF3Exercise: The equilateral triangle ABC, withweight P and side length A, is supported in thehorizontal position by three non-weightlead-bar1,2,3 and three non-weight inclinedstraight bars 4, 5 and 6 at horizontal anglewithball hinge.Aforcecoupleis acted on theV3plate surface, its moment is M Pa.At2point C, a force P parallel to side AB isapplied. To calculate the internal force of eachbar
get FAD P 3 4 = − Fn5 = 0 + sin − cos60 − cos60 = 0 FAB FAE P FAC get = 0 FAB For the intersection point B, see Figure (b, c). From Fn2 = 0 − FBE − FBF cos = 0 Get FBE = 0 Fn6 = 0 + sin + cos60 = 0 FBC FBF FAB Get FBC = 0 For the intersection point C, see Figure (b, c) Fn3 = 0 − FCF − FCD cos = 0 Get FCF P 3 4 = − Exercise: The equilateral triangle ABC, with weight P and side length A, is supported in the horizontal position by three non-weight lead-bar 1, 2, 3 and three non-weight inclined straight bars 4, 5 and 6 at horizontal angle with ball hinge. A force couple is acted on the plate surface, its moment is M Pa 2 3 = . At point C, a force P parallel to side AB is applied. To calculate the internal force of each bar
Ep 17. The weight P of homogeneous rectangular prism ABCDEF is1 500N,ZABE =30°,a couple M is applied in the BCEF plane, M=500Nm, connected by sixweightless rods connected by ball hinge, as shown in Figure 9, a=lm.Calculate theinternalforceofeachrodSOLUTION:Theforceis shown inFigure (b)V2ZY=0-Fcos45°=0V3F.=0Get(b)(a)Ep 17FromZm.(F)=0F,cos45°xa+Msin60°=0F4 =-250V6NGetZma(F)=0FromF,cos45°xa-Msin60°=0Ff=250/6NGet-Fxa-PxaZm2(F)=0From+Mcos60°=02F =-500NGet2aZmm3(F)=0From-(Fa+F,cos45°×a+P×=03F, =-(500+250/3)NGet
Ep 17 Ep 17. The weight P of homogeneous rectangular prism ABCDEF is1 500N, ABE = 30 , a couple M is applied in the BCEF plane, M=500Nm, connected by six weightless rods connected by ball hinge, as shown in Figure 9, a=1m.Calculate the internal force of each rod. SOLUTION: The force is shown in Figure (b) Y = 0 cos45 0 3 2 − 6 = F Get 0 F6 = From mz (F) = 0 4 cos45 + sin 60 = 0 F a M Get F4 = −250 6N From ( ) 0 mn1 F = 5 cos45 − sin 60 = 0 F a M Get F4 = 250 6N From ( ) 0 mn2 F = cos60 0 2 − 3 − + = M a F a P Get F3 = −500N From mn3 (F) = 0 ) 0 3 2 − ( 1 + 5 cos45 + = a Fa F a P Get F5 = −(500 + 250 3)N
-F,xa-F.cos45xa-Fxa-Px号FromZm(F)=0=04F, = (500 250/3)NGetExercise: The reduction gearbox iscomposed of three axes, and the power isinputbyaxisI,M,=697Nm.GearpitchcirclediameterD,=160mm,D,=632mmD,=204mm,pressureAngle is 20°,excludingfriction,wheel,shaftweightCalculate the constrained reaction forces of axes A, B, C and D when rotating atconstant speed
From m (F) = 0 x 0 3 − 2 − 4 cos45 − 3 − = a F a F a F a P Get F5 = −(500 − 250 3)N Exercise: The reduction gearbox is composed of three axes, and the power is input by axis I, M1 = 697Nm . Gear pitch circle diameter D1 = 160mm , D2 = 632mm , D3 = 204mm , pressure Angle is 20 , excluding friction, wheel, shaft weight. Calculate the constrained reaction forces of axes A, B, C and D when rotating at constant speed