NAME&IDDATEMTE119–STATICSPAGEHOMEWORK61MechatronicsEngineeringSOLUTIONS22PROBLEM1-EXERCISE6-73Thecompoundbeamispin-supportedatCandsupportedbyarollerAandB.Thereisahinge(pin)atD.Determinethereactionsatthesupports.Neglectthethicknessofthebeam.SOLUTION1-EXERCISE6-738kipDx(a)306ft4ft2ftDy4kipAyEquationofEquilibrium:fromFBDof(a)
∑MD=04cos3012()+)2(8−Ay)6(=0Ay=.9595kip



Ans∑Fy=0Dy+.9595−4cos30−8=0Dy=.1869kip
∑Fx=0Dx+.9595−4sin30=0Dx=00.2kipNAME&IDDATEMTE119–STATICSPAGEHOMEWORK62MechatronicsEngineeringSOLUTIONS22FromFBDof(b):∑MC=0.186924()+15+12)8.0(−By16()=0By=.8541kip∑Fy=0Cy+.8541−.1869−12)8.0(=0Cy=93.2kip


Ans∑Fx=0Cx−00.2−12)6.0(=0Cx=2.9kipPROBLEM1-EXERCISE6-91DeterminethehorizontalandverticalcomponentsofforcewhichthepinatA,BandCexertonmemberABCoftheframe.NAME&IDDATEMTE119–STATICSPAGEHOMEWORK63MechatronicsEngineeringSOLUTIONS22SOLUTION1-EXERCISE6-91FBD:ExAyEy∑ME=0;-Ay(3.5)++400(2)300(3.5)+300(1.5)=0


AnsAy=657NANSFBDofmemberCD:400NCx1.5m2mDxCyDy∑MD=−0;Cy(3.5)+400(2)=0Cy=229NANSNAME&IDDATEMTE119–STATICSPAGE4HOMEWORK6MechatronicsEngineeringSOLUTIONS22FBDofmemberABC:229N∑MB=0;Cx=0Cx∑Fx=0;FBD=FBE52.5m∑Fy=0;657.1−−2292FBD=0FBD745F=F=368.7N7BDBE75Bx=0←Ans.2.5mFBE5B=(368.7)(2)=429N←Ans.y74PROBLEM1-EXTRAPRACTICE6-76Thethree-hingedarchsupporttheloadsF1=8kNandF2=5kN.DeterminethehorizontalandverticalcomponentsatthepinsupportsAandB.Takeh=2m.NAME&IDDATEMTE119–STATICSPAGEHOMEWORK65MechatronicsEngineeringSOLUTIONS22SOLUTION1-EXTRAPRACTICE6-76MemberAC:∑MA=0−)4(8+Cy)8(+Cx)7(=0∑Fx=08−Ax−Cx=0∑Fy=0−Ay+Cy=0Cy=AyMemberBC:NAME&IDDATEMTE119–STATICSPAGEHOMEWORK66MechatronicsEngineeringSOLUTIONS22∑MB=0)2(5+Cy)6(−Cx)9(=0∑Fx=0Cx−Bx=0∑Fy=0−Cy+By−5=0Solvingtheaboveequationswehave:Ax=.56141=5.61kNAy=.19122=1.91kN


AnsCx=.23859=2.39kNCy=.19122=1.91kNBx=.23859=2.39kNBy=.69122=6.91kNPROBLEM1-EXTRAPR