Problem 6.4. Principle of virtual displacements

 Work done by the external load, W_ext=

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Work done by the internal forces, W_int=

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Step 1.  Assume and draw a virtual displacement system.

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Step 2.  Give the work done by the external load, F.

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Eq.(6-72)

${\mathit{W}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathit{F}\mathit{\delta }\mathit{u}$

Step 3.  Give the internal work of the stresses in the tensile rods.

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Eq.(6-72)

$W_{\mathrm{int}}={\int }_h{\sigma }_1\delta {\epsilon }_{1}dx+{\int }_h{\sigma }_2\delta {\epsilon }_{2}dx={\int }_h\frac{N_1}{A}2\frac{\delta u}{h}dx+{\int }_h\frac{N_2}{A}3\frac{\delta u}{h}dx$

Normal forces in the rods are determined in Problem 6.3

${\mathit{W}}_{\mathbf{i}\mathbf{n}\mathbf{t}}={\int }_h\frac{N_1}{A}2\frac{\delta u}{h}dx+{\int }_h\frac{N_2}{A}3\frac{\delta u}{h}dx=\frac{2F}{13A}2\delta u+\frac{3F}{13A}3\delta u\mathbf{=}\mathit{F}\mathit{\delta }\mathit{u}$

Step 4.  Verify equilibrium with the principle of virtual displacements

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The following virtual displacement system is assumed:

The work done by the external load, F is

Eq.(6-72)

${\mathit{W}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathit{F}\mathit{\delta }\mathit{u}$

The internal work is done by the stresses in the tensile rods:

Eq.(6-72)

$W_{\mathrm{int}}={\int }_h{\sigma }_1\delta {\epsilon }_{1}dx+{\int }_h{\sigma }_2\delta {\epsilon }_{2}dx={\int }_h\frac{N_1}{A}2\frac{\delta u}{h}dx+{\int }_h\frac{N_2}{A}3\frac{\delta u}{h}dx$

Normal forces in the rods are determined in Problem 6.3

${\mathit{W}}_{\mathbf{i}\mathbf{n}\mathbf{t}}={\int }_h\frac{N_1}{A}2\frac{\delta u}{h}dx+{\int }_h\frac{N_2}{A}3\frac{\delta u}{h}dx=\frac{2F}{13A}2\delta u+\frac{3F}{13A}3\delta u\mathbf{=}\mathit{F}\mathit{\delta }\mathit{u}$

We may observe that

 $W_{\mathrm{int}}=W_{ext}=F\delta u$ 

Thus the equilibrium of the structure is verified by the principle of virtual displacements.