Elastic Displacements
6.5 Truss Example 1
To review, using notation similar to that of Hibbeler (?), the general virtual work expression can be given as:
\[1^* \times \Delta = \sum u \times \delta L\]where
$\Delta$ is the desired displacement due to real effects;
$1^*$ is an applied virtual unit force corresponding to the displacement;
$\delta L$ are internal distortions due to the real effects; and
$u$ are internal virtual forces corresponding to the internal distortions and in equilibrium with the virtual unit force.
For trusses, the $u$ terms will be member forces, $n_i$ due to the virtual unit load and the $\delta L$ terms will be real changes in length of the members.
To determine the distortions due to the real applied loads, we have, for each member $i$, its change of length:
\[\delta L_i = \left( {N L \over A E} \right)_i\]where
$N$ is the real normal (axial) force in each member;
$L$ is the length of the member;
$A$ is its cross sectional area, and
$E$ is the modulus of elasticity.
Displacements Due To Loads
Consider the following loaded truss. The bottom portion of the figure shows the member forces corresponding to the 60kN applied load; these are determined by a simple truss analysis (the truss is statically determinate and stable).
All the truss members are steel ($E = 200 000 MPa$) and the cross sectional areas of each are given here (in $mm^2$):
| Member | Area |
|---|---|
| ab | 5000 |
| ac | 4000 |
| bc | 4500 |
| cd | 4600 |
We wish to determine the vertical displacement of joint a due to the applied loads. Therefore we place a unit virtual load at joint a, corresponding to the displacement we desire. The direction that we show for the unit load is the assumed direction of the displacement (i.e., downwards). For the virtual system, the unit load must be the only load acting. The bottom portion of this figure shows the virtual member forces corresponding to the unit load.
The calculation of the displacement is most conveniently done in tabular form. The following figure shows these calculations.
A few notes about the various columns:
(d) this column shows the normal forces in each member due to the real, applied loads. The usual sign convention is that tension is +ive and compression is -ive.
(e) this column shows the normal force in each member due to the application of the unit virtual load. As the virtual load is unitless, so are the values in this column. The sign convention must be the same as for the real forces (tension +ive).
(f) this column need not be computed separately. It is done so here just to show the real distortions due to the real applied loads. These are increases in lengths (or decreases, if -ive) of each of the members due to the real loads.
(g) finally, column (g) is the product of the virtual force in each member times its real change in length. This column shows the contribution toward the final displacement of each member.
The sum of column (g) is then the real vertical displacement of joint a (because that is where the virtual load was placed). It is +ive, so therefore the displacement direction agrees with the direction of the virtual load - downwards.
Therefore, the displacement of joint a is 2.02mm downwards.
Displacements due to temperature change
Suppose that members ac and cd undergo a temperature change of $-30^\circ C$ (i.e., a temperature decrease). What will be the vertical displacement of joint a due only to this temperature change in two of the members?
Now the $\delta L$ term for each member is its change in length due to the temperature change:
\[\delta L_i = \left( \alpha_T L \Delta T \right)_i\]where
$\alpha_T$ is the coefficient of thermal expansion, commonly $12\times10^{-6} mm/mm/{}^\circ C$ for steel,
$L$ is the length of the member, in mm, and
$\Delta T$ is the temperature change, in ${}^\circ C$.
For this problem we can reuse the same virtual forces as in part a) because we want the same displacement.
The calculations are simple enough that we don’t need to use a tabular form - simply compute the contributions of members ac (length = 5000mm) and cd (length = 3000mm):
\[\begin{split} \Delta_a &= \sum u\times\delta L\\ &= 1.6667\times 12\times10^{-6}\times 5000 \times -30 + 1.3333\times 12\times10^{-6}\times 3000 \times -30\\ &= -4.44 \end{split}\]The final displacement is -4.44mm (therefore counter to the direction of the virtual load, therefore upward) due to the contractions of members ac and cd.
Displacements due to ‘Fabrication Errors’
For the same truss, suppose member bc had been fabricated 10mm too long due to a fabrication error. What effect would that have on joint a? In other words, compute the vertical displacement of joint a due to a change in length of member bc of +10mm (increases in length are +ive to match the sign convention established earlier).
Again we can reuse the virtual loads computed in part a) as we are interested in the same displacement of the same joint.
This time, the $\delta L$ term for each member is simply the specified change in length, and only member bc has a $\delta L$ that is non-zero. So:
\[\begin{split} \Delta_a &= \sum u\times\delta L\\ &= -1.0000\times +10mm\\ &= -10 mm \end{split}\]The vertical displacement of joint a due to member bc being 10mm too long is 10mm upward.