Chapter 4, Problem 4.157P

Find the forces in members CD, DH, and HI.

To determine

Find the forces in members $CD,DH$ and $HI$.

Answer to Problem 4.157P

The forces in members $CD,DH$ and $HI$.

  $\underset{\_}{\underset{\_}{F_{CD}=5184lb}},Tension$

  $\underset{\_}{\underset{\_}{F_{DH}=2304lb}},Tension$

  $\underset{\_}{\underset{\_}{F_{HI}=5343.5lb}},compression$

Explanation of Solution

Given information:

Equations of equilibrium,

  ${\displaystyle \sum F_X}=0$

  ${\displaystyle \sum F_Y}=0$

  ${\displaystyle \sum M_O}=0$

Above equations includes all the forces that act on the body, both applied forces and the reactions (forces provide by the supports).

Calculation:

All the external forces acting on truss can be shown as follows.

Consider the vertical equilibrium of the truss.

  ${\displaystyle \sum F_Y}=0$

  $\uparrow +R_{EY}+R_{IY}=1440+1440+720+720$

  $\uparrow +R_{EY}+R_{IY}=\mathrm{4320...........}(1)$

Consider the horizontal equilibrium of the truss.

  ${\displaystyle \sum F_X}=0$

  $\leftarrow R_{EX}+R_{IX}=\mathrm{0...........}(2)$

Take the moment around point $I$ ,

For the equilibrium,

${\displaystyle \sum M_G}=0$

$12\times R_{EX}+8\times 720+16\times 1440+24\times 1440+32\times 720=0$

$R_{EX}=-7200lb$

From equation (2),

$\leftarrow R_{IX}=-R_{EX}$

$R_{IX}=7200lb$

Thus, the external forces acting on truss are as follows.

Then consider the equilibrium of point $E$ ,

  $\to -F_{DE}+R_{EX}=0$

  $F_{DE}=7200lb$ ,Tension

  $\uparrow R_{EY}=0$

From equation (1),

$\uparrow R_{EY}+R_{IY}=\mathrm{4320...........}(1)$

$R_{IY}=4320lb$

Consider the following section.

Take the moment around point $D$.

${\displaystyle \sum M_G}=0$

  $8\times 1440+16\times 1440+24\times 720-10\cos \theta \times F_{HI}=0$

  $8\times 1440+16\times 1440+24\times 720-10\times \frac{4}{\sqrt{17}}\times F_{HI}=0$

  $F_{HI}=1296\sqrt{17}$

  $\underset{\_}{\underset{\_}{F_{HI}=5343.5lb}},Compression$

Consider equilibrium of point $I$.

  $\to F_{DI}\cos \alpha +F_{HI}\cos \theta -R_{IX}=0$

  $F_{DI}\times \frac{2}{\sqrt{13}}+1296\sqrt{17}\times \frac{4}{\sqrt{17}}-7200=0$

  $F_{DI}=1008\sqrt{13}lb,Compression$

Then consider the equilibrium of point $D$ ,

$\to -F_{DI}\cos \alpha +F_{DE}-F_{CD}=0$

$-1008\sqrt{13}\times \frac{2}{\sqrt{13}}+7200-F_{CD}=0$

$\underset{\_}{\underset{\_}{F_{CD}=5184lb}},Tension$

  $\uparrow -720-F_{DH}+F_{DI}\sin \alpha =0$

  $-720-F_{DH}+1008\sqrt{13\times }\frac{3}{\sqrt{13}}=0$

  $\underset{\_}{\underset{\_}{F_{DH}=2304lb}},Tension$

Thus, the forces in members $CD,DH$ and $HI$.

  $\underset{\_}{\underset{\_}{F_{CD}=5184lb}},Tension$

  $\underset{\_}{\underset{\_}{F_{DH}=2304lb}},Tension$

  $\underset{\_}{\underset{\_}{F_{HI}=5343.5lb}},compression$

Conclusion:

The forces in members $CD,DH$ and $HI$.

  $\underset{\_}{\underset{\_}{F_{CD}=5184lb}},Tension$

  $\underset{\_}{\underset{\_}{F_{DH}=2304lb}},Tension$

  $\underset{\_}{\underset{\_}{F_{HI}=5343.5lb}},compression$