Influence Lines for Trusses

Example
For the Pratt truss shown below, draw the influence diagram for members JK, DK, and DE.

Solution
Solve for the reaction at A due to unit load that move across the bottom chord.
$\Sigma M_G = 0$

$18R_A = 1.0(18 - x)$

$R_A = 1 - \dfrac{x}{18}$

Assume forces F_JK, F_DE, and F_DK as tension as shown in the cut section below.

Influence Line for Member JK
For 0 ≤ x ≤ 9 m

$\Sigma M_D = 0$

$3F_{JK} + 9R_A = 1.0(9 - x)$

$3F_{JK} + 9\left( 1 - \dfrac{x}{18} \right) = 9 - x$

$3F_{JK} + 9 - \frac{1}{2}x = 9 - x$

$F_{JK} = -\frac{1}{6}x$ ← straight line

When x = 0, F_JK = 0
When x = 9, F_JK = -1.5

For 9 m ≤ x ≤ 18 m

$\Sigma M_D = 0$

$3F_{JK} + 9R_A = 0$

$3F_{JK} + 9\left( 1 - \dfrac{x}{18} \right) = 0$

$3F_{JK} + 9 - \frac{1}{2}x = 0$

$F_{JK} = \frac{1}{6}x - 3$ ← straight line

When x = 9, F_JK = -1.5
When x = 18, F_JK = 0

Influence Line for Member DE
For 0 ≤ x ≤ 12 m

$\Sigma M_K = 0$

$3F_{DE} + 1.0(12 - x) = 12R_A$

$3F_{DE} + 12 - x = 12\left( 1 - \dfrac{x}{18} \right)$

$3F_{DE} + 12 - x = 12 - \frac{2}{3}x$

$F_{DE} = \frac{1}{9}x$ ← straight line

When x = 0, F_DE = 0
When x = 12, F_DE = 4/3

For 12 m ≤ x ≤ 18 m

$\Sigma M_K = 0$

$3F_{DE} = 12R_A$

$3F_{DE} = 12\left( 1 - \dfrac{x}{18} \right)$

$F_{DE} = 4 - \frac{2}{9}x$ ← straight line

When x = 12, F_DE = 4/3
When x = 18, F_DE = 0

Influence Line for Member DK
For 0 ≤ x ≤ 9 m

$\Sigma F_V = 0$

$F_{DK} \left( \frac{1}{\sqrt{2}} \right) + R_A = 1.0$

$F_{DK} \left( \frac{1}{\sqrt{2}} \right) + \left( 1 - \dfrac{x}{18} \right) = 1$

$F_{DK} = \frac{\sqrt{2}}{18}x$ ← straight line

When x = 0, F_DK = 0
When x = 9, F_DK = sqrt(2) / 2

For 9 m < x ≤ 18 m

$\Sigma F_V = 0$

$F_{DK} \left( \frac{1}{\sqrt{2}} \right) + R_A = 0$

$F_{DK} \left( \frac{1}{\sqrt{2}} \right) + \left( 1 - \dfrac{x}{18} \right) = 0$

$F_{DK} = \frac{\sqrt{2}}{18}x - \sqrt{2}$ ← straight line

When x = 12, F_DK = -sqrt(2) / 3
When x = 18, F_DK = 0

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