User:Eml4500.f08.ateam.boggs.t/Lecture 24

Proving Hookes Law through the Principle of Virtual Work
The equation $$ F = kd $$ can also be written as $$ kd - F = 0 $$ (Equation 3) or which is true for all values of W if $$ W(kd - F) = 0 $$ (Equation 4). This is a weak form of the Principle of Virtual Work. It is important to note that deriving Equation 4 from Equation 3 is a trivial solution, but deriving Equation 3 from Equation 4 is in fact Not trivial.

From the Force-Displacement relationship,


 * $$ F = kd $$

where F is a 6x1 matrix, k is a 6x6 matrix, and d is a 6x1 matrix, the Principle of Virtual Work can be derived as,


 * $$ W \cdot (kd - F) = 0 $$ (Equation 4)

This is true for all W. For this relationship to work W must be multiplied as a dot product, as 0 is not represented as a matrix. It is desired to show that Equation 3 can be derived from Equation 4 for all W.

To do this the first choice is to select W such that W1 = 1 and W2 through W6 = 0. W is originally defined as a 6x1 matrix but needs to be a 1x6 matrix so,



W^{T} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{1} = 0 $$

The second choice is to select 'W such that W1 equals 0, W2 equals 1, and W3 through W6 also equal 0. So,



W^{T} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{2} = 0 $$

This repeats as follows.

Choice 3,



W^{T} = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{3} = 0 $$

Choice 4,



W^{T} = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{4} = 0 $$

Choice 5,



W^{T} = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{5} = 0 $$

Choice 6,



W^{T} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Equation 4 then becomes,



W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,



\sum_{j=1}^6 k_{ij}d_{j} - F_{6} = 0 $$

Hence $$ F = kd $$ (Equation 1).

To account for the boundary conditions of the two-bar truss where,


 * $$ d_{1} = d_{2} = d_{5} = d_{6} = 0 $$

the weighting coefficient must be "kinematically admissible" or cannot violate the boundary conditions. The weighting coefficients are the virtual displacements where,


 * $$ W_1 = W_2 = W_5 = W_6 = 0 $$

For all $$ W_3 $$ and $$ W_4 $$,



\begin{bmatrix} W_3 \\ W_4 \end{bmatrix} \cdot ( \begin{bmatrix}     k_{33} & k_{34} \\      k_{43} & k_{44} \end{bmatrix} \begin{bmatrix}      d_3 \\      d_4 \end{bmatrix} - \begin{bmatrix}      F_3 \\      F_4 \end{bmatrix} ) = 0 $$

Equation 1 can also be found through Equation 3 by adding F to both sides of the equation as shown below,


 * $$ kd - F = 0 $$

Adding F to both sides,


 * $$ kd - F + F = 0 + F $$

Here F cancels out on the left-hand side of the equation leaving,


 * $$ kd = F $$