PLAIN AND CIVIL: EXAMPLE 8.0.1. METHOD OF JOINTS

Determine the force in each member of the truss. State if the forces are tensile or compressive.

Reactions:

$\sum F_{X}=0\rightarrow$

$Rc_{X}=450lb$

$\sum M_{C}=0$

$R_{A}\left ( 8ft \right )-450lb\left ( 4ft \right )=0$

$R_{A}=225lb$

$\sum M_{A}=0$

$Ry_{C}\left ( 8ft \right )-450lb\left ( 4ft \right )=0$

$Ry_{C}=225lb$

Member forces:

With just two equations of equilibrium to be used for concurrent forces. Start with joints containing two members.

Consider Joint A:

$\sum F_{Y}=0\uparrow$ $\sum F_{X}=0\rightarrow$

$Fy_{AD}=Ra$ $F_{AB}=Fx_{AD}$
$\frac{1}{\sqrt{2}}F_{AD}=225lb$ $F_{AB}=\frac{1}{\sqrt{2}}F_{AD}$

$F_{AD}=318.20lb (C)$ $F_{AB}=225lb(T)$

At Joint C:

$\sum F_{Y}=0\uparrow$ $\sum F_{X}=0\rightarrow$

$\frac{1}{\sqrt{2}}F_{CD}=Ry_{C}$ $F_{CB}=Rx_{C}-Fx_{CD}$

$F_{CD}=225lb\sqrt{2}$ $F_{CB}=450lb-\frac{1}{\sqrt{2}}225\sqrt{2}$

${\color{Red} \mathbf{F_{CD}=318.20lb(T)}}$ ${\color{Red} \mathbf{F_{CB}=225lb(T)}}$

At Joint D:

$\sum F_{Y}=0\uparrow$

$\frac{1}{\sqrt{2}}F_{DA}-\frac{1}{\sqrt{2}}F_{DC}-F_{DB}=0$

$\frac{1}{\sqrt{2}}318.20-\frac{1}{\sqrt{2}}318.20-F_{DB}=0$

$F_{DB}=0$

PLAIN AND CIVIL

Pages

Saturday, 9 March 2019

EXAMPLE 8.0.1. METHOD OF JOINTS

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