# 230 Distance from truss member to truss joint

**Problem 230**

For the truss shown in Fig. P-230, compute the perpendicular distance from E and from G to the line BD. Hint: Imagine a force F directed along BD and compute its moment in terms of its components about E and about G. Then equate these results to the definition of moment M = Fd to compute the required perpendicular distances.

**Solution 230**

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d = length of member BD

d

_{x}= 12 ft

d

_{y}= 16 - 12 = 4 ft

$d = \sqrt{12^2 + 4^2}$

$d = 4\sqrt{10}$

Moment about point E

$M_E = 16d_x = 16(12)$

$M_E = 192 \, \text{ ft}^2$

$M_E = d_E \times d$

$192 = d_E(4\sqrt{10})$

$d_E = 15.18 \, \text{ ft}$ *answer*

Moment about point G

$M_G = 16d_x + 12d_y = 16(12) + 12(4)$

$M_G = 240 \, \text{ ft}^2$

$M_G = d_G \times d$

$240 = d_G(4\sqrt{10})$

$d_G = 18.974 \, \text{ ft}$ *answer*

**Checking (by Geometry):**

$\dfrac{x + 12}{12} = \dfrac{3}{1}$

$x = 24 \, \text{ ft}$

$\dfrac{d_E}{x + 24} = \dfrac{1}{\sqrt{10}}$

$d_E = \dfrac{24 + 24}{\sqrt{10}}$

$d_E = 15.18 \, \text{ ft}$ (*okay!*)

$\dfrac{d_G}{x + 36} = \dfrac{1}{\sqrt{10}}$

$d_G = \dfrac{24 + 36}{\sqrt{10}}$

$d_G = 18.974 \, \text{ ft}$ (*okay!*)