$3\mathrm{.}1\mathrm{.}1\mathrm{.}1$ CONVERTING DISTRIBUTED LOADS TO POINT LOADS

Unfortunately, both our live and dead loads are distributed loads. We have to convert the distributed loads into point loads at the joints.

To see how to do this, let's looks at the live load acting on member AB $.$ The live load acting on AB is $q\frac{L}{2}($force $q$ per unit length time length $\frac{L}{2}).$ Because the live load is uniformly distributed along AB $,$ we see that half of it can be allocated to joint A and half to joint $B,$ or $q\frac{L}{4}$ at A and $q\frac{L}{4}$ at B $.$

Now lets look at the live load acting on member BC$.$ Again, the total force is $q\frac{L}{2}.$ Half is allocated to each end of BC$,$ or qL$/4$ at $B$ and $q\frac{L}{4}$ at C $.$

The total live load acting on joint A is $q\frac{L}{4}.$ The total live load acting on joint B is $q\frac{L}{2}(q\frac{L}{4}$ from member AB and $q\frac{L}{4}$ from member BC $).$ The total live load acting on joint C is $q\frac{L}{4}$

$L{L}_A=\frac{q*L}{4}$

$L{L}_B=\frac{q*L}{2}$

$L{L}_C=\frac{q*L}{4}$

$L{L}_D=0$

$L{L}_E=0$

We can perform a similar analysis to find the dead load acting at each point. Half the weight of member $AB$ is allocated to joint A$,$ half to Joint B$.$ So on and so forth.

$D{L}_A=\frac{w_{AB}+w_{AE}}{2}$

$D{L}_B=\frac{w_{AB}+w_{BC}+w_{BD}+w_{BE}}{2}$

$D{L}_C=\frac{w_{BC}+w_{CD}}{2}$

$D{L}_D=\frac{w_{BD}+w_{CD}+w_{DE}}{2}$

$D{L}_E=\frac{w_{AE}+w_{BE}+w_{DE}}{2}$

$3\mathrm{.}1\mathrm{.}2$ Overall Equilibrium Equations

Now that the free body diagram is complete, we do our equilibrium equations. This is a $2$D rigid body problem, so there are three degrees of freedom and three equilibrium equations. There are no forces in x $,$ and symmetry tells us that RAy and RBy must be equal.

$R_{Ay}=R_{Cy}=\frac{DL+LL}{2}$

$3\mathrm{.}1\mathrm{.}3$ Jont Level FBDs and Equilibrium Equations$3\mathrm{.}1\mathrm{.}1\mathrm{.}1$ CONVERTING DISTRIBUTED LOADS TO POINT LOADS

Unfortunately, both our live and dead loads are distributed loads. We have to convert the distributed loads into point loads at the joints.

To see how to do this, let's looks at the live load acting on member AB $.$ The live load acting on AB is $q\frac{L}{2}($force $q$ per unit length time length $\frac{L}{2}).$ Because the live load is uniformly distributed along AB $,$ we see that half of it can be allocated to joint A and half to joint $B,$ or $q\frac{L}{4}$ at A and $q\frac{L}{4}$ at B $.$

Now lets look at the live load acting on member BC$.$ Again, the total force is $q\frac{L}{2}.$ Half is allocated to each end of BC$,$ or qL$/4$ at $B$ and $q\frac{L}{4}$ at C $.$

The total live load acting on joint A is $q\frac{L}{4}.$ The total live load acting on joint B is $q\frac{L}{2}(q\frac{L}{4}$ from member AB and $q\frac{L}{4}$ from member BC $).$ The total live load acting on joint C is $q\frac{L}{4}$

$L{L}_A=\frac{q*L}{4}$

$L{L}_B=\frac{q*L}{2}$

$L{L}_C=\frac{q*L}{4}$

$L{L}_D=0$

$L{L}_E=0$

We can perform a similar analysis to find the dead load acting at each point. Half the weight of member $AB$ is allocated to joint A$,$ half to Joint B$.$ So on and so forth.

$D{L}_A=\frac{w_{AB}+w_{AE}}{2}$

$D{L}_B=\frac{w_{AB}+w_{BC}+w_{BD}+w_{BE}}{2}$

$D{L}_C=\frac{w_{BC}+w_{CD}}{2}$

$D{L}_D=\frac{w_{BD}+w_{CD}+w_{DE}}{2}$

$D{L}_E=\frac{w_{AE}+w_{BE}+w_{DE}}{2}$

$3\mathrm{.}1\mathrm{.}2$ Overall Equilibrium Equations

Now that the free body diagram is complete, we do our equilibrium equations. This is a $2$D rigid body problem, so there are three degrees of freedom and three equilibrium equations. There are no forces in x $,$ and symmetry tells us that RAy and RBy must be equal.

$R_{Ay}=R_{Cy}=\frac{DL+LL}{2}$

$3\mathrm{.}1\mathrm{.}3$ Jont Level FBDs and Equilibrium Equations