Problem 6.3. Principle of stationary potential energy

Determine the reaction forces and the strains in the bars of the structure given in the Figure. Apply the principle of stationary potential energy.

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Derive end deflection, e.

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$e = \frac{3 F h}{13 E A}$

Calculate reaction force in the left hinge, A [kN].

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$A = \frac{8}{13} F$ (↑)

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Steps

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Step 1. Draw the deflected shape. Choose an unknown.

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The deflected shape is shown in the Figure below. Let us choose the displacement, u below force, F to be the unknown. All the other forces, the strains and the displacements can be given as the function of u.

Step 2. Write the potential energy of the structure.

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Eq.(6-28)

π = U + W = \frac{1}{2} N_{1} (2 u) + \frac{1}{2} N_{2} (3 u) - F u = \frac{1}{2} \frac{E A}{h} {(2 u)}^{2} + \frac{1}{2} \frac{E A}{h} {(3 u)}^{2} - F u

Step 3. Determine unknown from the stationarity of the potential energy.

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Eq.(6-43)

$\frac{\partial π}{\partial u} = 0 = \frac{1}{2} \frac{E A}{h} 8 u + \frac{1}{2} \frac{E A}{h} 18 u - F \to u = \frac{F h}{13 E A}$

Step 4. Calculate displacements, strains and reactions.

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The end displacement is

$e = 3 u = \frac{3}{13} \frac{F L}{E A}$

Forces and strains in the tensile rods are

$N_{1} = 2 u \frac{E A}{L} = \frac{2}{13} F, ε_{1} = \frac{2 u}{L} N_{2} = 3 u \frac{E A}{L} = \frac{3}{13} F, ε_{2} = \frac{3 u}{L}$

Vertical reaction of the left hinge is calculated from the equilibrium equation:

$A = N_{1} + N_{2} - F = \frac{8}{13} F$

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Results

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The potential energy of the structure is