# 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*, and

_{DE}*F*as tension as shown in the cut section below.

_{DK}

**Influence Line for Member JK**

For 0 ≤

*x*≤ 9 m

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

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

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

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

When *x* = 0, *F _{JK}* = 0

When

*x*= 9,

*F*= -1.5

_{JK}

For 9 m ≤ *x* ≤ 18 m

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

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

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

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

When *x* = 9, *F _{JK}* = -1.5

When

*x*= 18,

*F*= 0

_{JK}

**Influence Line for Member DE**

For 0 ≤

*x*≤ 12 m

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

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

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

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

When *x* = 0, *F _{DE}* = 0

When

*x*= 12,

*F*= 4/3

_{DE}

For 12 m ≤ *x* ≤ 18 m

$3F_{DE} = 12R_A$

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

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

When *x* = 12, *F _{DE}* = 4/3

When

*x*= 18,

*F*= 0

_{DE}

**Influence Line for Member DK**

For 0 ≤

*x*≤ 9 m

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

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

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

When *x* = 0, *F _{DK}* = 0

When

*x*= 9,

*F*= sqrt(2) / 2

_{DK}

For 9 m < *x* ≤ 18 m

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

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

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

When *x* = 12, *F _{DK}* = -sqrt(2) / 3

When

*x*= 18,

*F*= 0

_{DK}