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(a)M_(B)=3063.52 l lbfinM_(C)=3194.32lbf fin     :.M_(max)=M_(c)C is the point of maximum moment. (b)sigma_(1)=10.31Kppsisigma_(2)=-10.31Kpsitau_(1)=5.49Kpsi The detailed solution is given below:Given: rarr{:[d_(B)=10in],[d_(C)=sin]:}∣d_("shaft ")=1.5sin#(x-y" planl ")/(y)Put above value in (1), we getput above value in (2), we get(R_(A)|_(Z)=482.5-321.67=160.83lbf:}(a)Bending moment:-{:[" \# In "x" cyplane "],[longrightarrowM_(B)=(R_(A))_(y)xx9=300lbfxx9in=2700lbf"-in "],[rarrM_(C)=(R_(D))_(y)xx9=150lbfxx9in=1350lbf"-in "],[" \# In "x-z" planl "],[rarrM_(B)=[R_(A)|_(Z)xx9=160.83lbfxx" in "=1447.47lbf-in:}],[rarrquadM_(C)=(R_(D))_(Z)xx9=321.67lbfxx9_(in)=2895.03lbf"-in "]:}Resultant B.M{:[rarrM_(B)=sqrt((2700)^(2)+(1447.47)^(2))=3063.52lbf-in],[rarrM_(C)=sqrt((1350)^(2)+(2895.03)^(2 ... See the full answer