Problem 405 | Method of Joints

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Problem 405
Determine the force in each bar of the truss shown in Fig. P-405 caused by lifting the 120 kN load at a constant velocity of 8 m per sec. What change in these forces, if any, results from placing the roller support at D and the hinge support at A?

Solution 405

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The load is lifted at constant velocity (in dynamic equilibrium), thus, the forces involve is similar to forces under static equilibrium.
$\Sigma F_V = 0$

$C_V = 120 + 120(\frac{3}{5})$

$C_V = 192 \, \text{kN}$

$\Sigma F_H = 0$

$C_H = 120(\frac{4}{5})$

$C_H = 96 \, \text{kN}$

By symmetry of vertical forces
$R_A = D_V = \frac{1}{2}(192)$

$R_A = D_V = 96 \, \text{kN}$

$\Sigma F_H = 0$

$D_H = 96 \, \text{kN}$

At Joint A
$\Sigma F_V = 0$

$F_{AB}(\frac{3}{5}) = 96$

$F_{AB} = 160 \, \text{kN}$

$\Sigma F_H = 0$

$F_{AC} = \frac{4}{5}F_{AB} = \frac{4}{5}(160)$

$F_{AC} = 128 \, \text{kN}$

At Joint C
$\Sigma F_V = 0$

$F_{BC} = 192 \, \text{kN}$

$\Sigma F_H = 0$

$F_{CD} + 96 = 128$

$F_{CD} = 32 \, \text{kN}$

At Joint B
$\Sigma F_H = 0$

$F_{BD}(\frac{4}{5}) = 160(\frac{4}{5})$

$F_{BD} = 160 \, \text{kN}$

$\Sigma F_V = 0$

$F_{BD}(\frac{3}{5}) + 160(\frac{3}{5}) = 192$

$160(\frac{3}{5}) + 160(\frac{3}{5}) = 192$

$192 = 192$ Check!

At Joint D
$\Sigma F_V = 0$

$96 = 160(\frac{3}{5})$

$96 = 96$ Check!

$\Sigma F_H = 0$

$32 + 96 = 160(\frac{4}{5})$

$128 = 128$ Check!

Summary
AB = 160 kN compression
AC = 128 kN tension
BC = 192 kN tension
CD = 32 kN tension
BD = 160 kN compression

With the roller support at D and the hinge support at A

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$\Sigma F_V = 0$

$R_A = D_V = \frac{1}{2}(192)$

$R_A = D_V = 96 \, \text{kN}$

$\Sigma F_H = 0$

$D_H = 96 \, \text{kN}$

At Joint A
$\Sigma F_V = 0$

$F_{AB}(\frac{3}{5}) = 96$

$F_{AB} = 160 \, \text{kN}$

$\Sigma F_H = 0$

$F_{AC} + \frac{4}{5}F_{AB} = 96$

$F_{AC} + \frac{4}{5}(160) = 96$

$F_{AC} = -32 \, \text{kN}$

$F_{AC} = 32 \, \text{kN}$ compression

At Joint C
$\Sigma F_V = 0$

$F_{BC} = 192 \, \text{kN}$

$\Sigma F_H = 0$

$F_{CD} = 32 + 96$

$F_{CD} = 128 \, \text{kN}$

At Joint B
$\Sigma F_H = 0$

$F_{BD}(\frac{4}{5}) = 160(\frac{4}{5})$

$F_{BD} = 160 \, \text{kN}$

$\Sigma F_V = 0$

$F_{BD}(\frac{3}{5}) + 160(\frac{3}{5}) = 192$

$160(\frac{3}{5}) + 160(\frac{3}{5}) = 192$

$192 = 192$ Check!

At Joint D
$\Sigma F_V = 0$

$160(\frac{3}{5}) = 96$

$96 = 96$ Check!

$\Sigma F_H = 0$

$160(\frac{4}{5}) = 128$

$128 = 128$ Check!

Summary
AB = 160 kN compression
AC = 32 kN compression
BC = 192 kN tension
CD = 128 kN compression
BD = 160 kN compression

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