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a)b) Use equations of equilibrium of forces and moments to find the reactions:{:[sum MAz=0],[8in**350 lbf-22 in**Dy=0],[" Dy "=127","27lbf]:}sum Fy=0350 lbf-By-Ay=0Ay=350lbf-127,27lbf=222,73lbfsum MDy=022 in**Az-6in**451 lbf=0Az=123 lbfsum Fz=0451 lbf-Bz-Az=0Dz=451 lbf-123 lbf=328 lbfc) Plane XY:V_(AB)(V(x)=0;V_(0)=Ay)" : "{:[V(x)=Vo+int F(x)dx],[V(x)=Vo+0=Ay=-222","73lbf],[V_(BD)(V(x)=0;V_(0)=Ay+350 lbf):],[V(x)=Vo+int F(x)dx],[V(x)=Vo+0=Ay+350lbf=-222","73lbf+350lbf=127","27lbf],[M_(AB)(Mo=0;V(x)=-222","73;X=8" in ")" : "],[M(x)=Mo+int V(x)dx],[M(x)=0+int-222","73 dx=-222","73 X],[M(8in)=-1781","84 lbf**in],[M_(B0)(Mo=-1781","84;V(x)=127","27;x=14" in "):],[M(x)=Mo+int V(x)dx],[M(x)=-1781","84+int127","27 dx=-1781","84+127","27 X],[MC=M(8in)=-763","68 lbf**" in "],[MD=M(14" in ")=0]:}VMzPlane XZ:V_(AC)(V(x)=0;V_(0)=Az):{:[V(x)=Vo+int F(x)dx],[V(x)=Vo+0=Ay=-123 lbf],[V_(CO)(V(x)=0;V_(0)=Az+451 lbf):],[V(x)=Vo+int F(x)dx],[V(x)=Vo+0=Az+451 lbf=-123 lbf+451 lbf=328 lbf]:}M_(AC)(Mo=0;V(x)=-123;x=16in):M(x)=Mo+int V(x)dxM(x)=0+int-123 dx=-123XMB=M(8in)=-984 lbf**inMC=M(16in)=-1968lbf**inM_(co)(Mo=-196 ... See the full answer