Problem 411 Cantilever Truss by Method of Joints

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cantilever truss

method of joints

method of sections

change in internal force

Problem 411
Determine the force in members AB, AC, BD, CD, and CE of the cantilever truss shown in Fig. P-411. If the loads were applied at C and E instead of at B, specify which members would have their internal force changed.

Solution 411

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At Joint A
$\Sigma F_V = 0$

$F_{AB}(\frac{2}{\sqrt{13}}) = 100$

$F_{AB} = 180.2776 \, \text{kN}$ tension answer

$\Sigma F_H = 0$

$F_{AC} = \frac{3}{\sqrt{13}}F_{AB}$

$F_{AC} = \frac{3}{\sqrt{13}}(180.2776)$

$F_{AC} = 150 \, \text{kN}$ compression answer

At Joint B
By inspection
$F_{BD} = 180.2776 \, \text{kN}$ tension answer

$F_{BC} = 200 \, \text{kN}$ compression

Length of member DE
$\dfrac{DE}{AE} = \dfrac{FG}{AG}$

$\dfrac{DE}{20} = \dfrac{20}{30}$

$DE = \frac{40}{3} \, \text{m}$

At Joint C
$\Sigma F_V = 0$

$F_{CD}(\frac{4}{5}) = 200$

$F_{CD} = 250 \, \text{kN}$ tension answer

$\Sigma F_H = 0$

$F_{CE} = 150 + \frac{3}{5}F_{CD}$

$F_{CE} = 150 + \frac{3}{5}(250)$

$F_{CE} = 300 \, \text{kN}$ compression answer

The fastest way to solve for BD, CD, and CE is by method of sections. From the figure below:
$\Sigma M_C = 0$

$\frac{3}{\sqrt{13}}F_{BD}(\frac{20}{3}) = 100(10)$

$F_{BD} = 180.2776 \, \text{kN}$ tension Check!

$\Sigma M_A = 0$

$\frac{4}{5}F_{CD}(10) = 200(10)$

$F_{CD} = 250 \, \text{kN}$ tension Check!

$\Sigma M_D = 0$

$F_{CE}(\frac{40}{3}) = 100(20) + 200(10)$

$F_{CE} = 300 \, \text{kN}$ compression Check!

Summary of Answers
AB = 180.2776 kN tension
AC = 150 kN compression
BD = 180.2776 kN tension
CD = 250 kN tension
CE = 300 kN compression

Solving for other members
$\Sigma M_E = 0$

$\frac{3}{\sqrt{13}}F_{DF}(\frac{40}{3}) = 100(20) + 200(10)$

$F_{DF} = 360.5551 \, \text{kN}$ tension

$\Sigma M_A = 0$

$\frac{2}{\sqrt{5}}F_{EF}(20) = 200(20) + 200(10)$

$F_{EF} = 335.4102 \, \text{kN}$ tension

$\Sigma M_F = 0$

$20F_{EG} = 100(30) + 200(20) + 200(10)$

$F_{EG} = 450 \, \text{kN}$ compression

At Joint E
$\Sigma F_V = 0$

$F_{DE} = \frac{2}{\sqrt{5}}(335.4102)$

$F_{DE} = 300 \, \text{kN}$ compression

With 200 kN loads moved from B and D to C and E, respectively

At Joint A
$\Sigma F_V = 0$

$F_{AB}(\frac{2}{\sqrt{13}}) = 100$

$F_{AB} = 180.2776 \, \text{kN}$ tension (no changes)

$\Sigma F_H = 0$

$F_{AC} = \frac{3}{\sqrt{13}}F_{AB}$

$F_{AC} = \frac{3}{\sqrt{13}}(180.2776)$

$F_{AC} = 150 \, \text{kN}$ compression (no changes)

From the figure below
$\Sigma M_C = 0$

$\frac{3}{\sqrt{13}}F_{BD}(\frac{20}{3}) = 100(10)$

$F_{BD} = 180.2776 \, \text{kN}$ tension (no changes)

$\Sigma M_A = 0$

$\frac{4}{5}F_{CD}(10) = 200(10)$

$F_{CD} = 250 \, \text{kN}$ tension (no changes)

$\Sigma M_D = 0$

$F_{CE}(\frac{40}{3}) = 100(20) + 200(10)$

$F_{CE} = 300 \, \text{kN}$ compression (no changes)

At Joint B
By inspection
$F_{BC} = 0$ (changed)

From the figure below
$\Sigma M_E = 0$

$\frac{3}{\sqrt{13}}F_{DF}(\frac{40}{3}) = 100(20) + 200(10)$

$F_{DF} = 360.5551 \, \text{kN}$ tension (no changes)

$\Sigma M_A = 0$

$\frac{2}{\sqrt{5}}F_{EF}(20) = 200(20) + 200(10)$

$F_{EF} = 335.4102 \, \text{kN}$ tension (no changes)

$\Sigma M_F = 0$

$20F_{EG} = 100(30) + 200(20) + 200(10)$

$F_{EG} = 450 \, \text{kN}$ compression (no changes)

At Joint E
$\Sigma F_V = 0$

$F_{DE} + 200 = \frac{2}{\sqrt{5}}(335.4102)$

$F_{DE} = 100 \, \text{kN}$ compression (changed)

Summary of Changes in Internal Forces

BC from 200 kN compression to 0
DE from 300 kN compression to 100 kN compression

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