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Problem 3: Given system is shown below:The given system can be converted into an equivalant fore-Moment systom as shown below:Force acting at point CC^(') is F^(')NTorque acting at point ' C ' is T=F xx CD.{:[:.T=F xx0.4=0.4FN*m.],[:.quad T=(0.4F)N*m=(400F)N*mm.]:}Now we have to consider Maximum streses at Points ' A ' and ' O ' due to the given loadings.rarr Point ' A ':At Point A ', Bending Moment due to ' F ' is{:[M_(A)=F xx(CA)=F(0.05+0.3)],[:.quadM_(A)=(0.35F)N*m=(350F)N.mm]:}:. Bending striees on the surbace at point ' A ' due to M_(A)^(') is issigma_(b)|_(A)=(M_(A)*((d_(A))/(2)))/((pi)/(64)(d_(A))^(4))=(32M_(A))/(pid_(A)^(3))we have d_(A)=0.025m=25mm.{:[:.sigma_(b)|_(A)=(32 xx350 F)/(pi(25)^(3)).],[:.quadsigma_(b)|_(A)=(0.2282 F)MPa.]:}Torsional shearstrees on the surbace at Point ' A ' is{:[tau_(A)=(T*((d_(A))/(2)))/((pi)/(32)(d_(A))^(4))=(16 T)/(pid_(A)^(3)).],[:.tau_(A)=(16 xx400(F))/(pi(25)^(3))],[:.tau_(A)=(0.1304F)MPa.]:}Now the Principal Sfreeser at Point ' A ' ore{:[sigma_(1,2)|_(A)=(sigma_(b)|_(A))/(2)+-sqrt(((sigma_(phi)//A)/(2))^(2)+(tau_(A))^(2)).],[sigma_(1,2)|_(A)=(0.2282 F)/(2)+-sqrt(((0.2282 F)/(2))^(2)+(0.1304 F)^(2))],[:.sigma_(1)|_(A)=0.2874F; quadsigma_(2)|_(A)=-0.0592F". "]:}:. Maximum Tervile sfrees at 'A'.sigma_(1)I_(A)=(0.2874F)MPaMaximum comprevive strees ot ' A ' is alsosigma_(1)|_(A)=(0.2874F)MRa(This occurs at the point juct opposite side of the Nentral axis).Point 'O':At Point ' O ' Bending Moment due to ' F ' ... See the full answer