4: Method of Virtual Work (1): Principle of Virtual Displacements
\[\def\kN{~\text{kN}} \def\m{~\text{m}} \def\kNm{~\text{kN}\!\cdot\!\text{m}} \def\kNpm{~\text{kN/m}}\]4.1 Introduction
4.1.1 Principle of Virtual Displacements
The main principle is offered here without proof:
Some proofs are available here:
- CIVE 3203 Lecture, 2019
- Learning the Virtual Work Method in Statics. Note that the first portion of this paper has a good discussion of small virtual rigid body displacements.
4.1.2 Procedure
The above principle can be used in a procedure to determine structural actions (reactions, internal forces) by the method of virtual displacements.
- Identify the single action you wish to determine (e.g., a vertical reaction, a bending moment at a point, or a shear force at a point).
- Identify the constraint that corresponds to that action (e.g, against vertical displacement, against relative rotation from one side of the point to the other, or against relative transverse displacement from one side of the point to the other).
- Relax the constraint and impose an arbitrary small displacement in its place.
- Maintain all other constraints and draw a consistent deflected shape. In particular, all other support constraints are honoured, and no additional curvature is allowed in structural members.
- Assign one of the displacements an arbitrary value, and express all other displacements in the deflected shape in terms of that.
- Write a virtual work equation in terms of the unknown action and known displacements and forces and set it to zero.
- Solve for the unknown action.
This procedure, with the use of properly displaced structures, ensures that we can write equations that contain only one unknown action.
Some notes:
- You may use anything, including a symbolic term, as the value you assign to the particular displacement. Whatever you use will cancel out of the final equation. A particularly convenient value is ‘1’.
- The resulting equation is really an equilibrium equation. This is just a different (and sometimes easier) way to develop it.
4.1.3 Frame Reactions
The procedure will be illustrated by determining all four reactions in the following frame (note that by inspection it is stable and statically determinate). Each reaction will be determined seperately and independently. For each, the virtual deflected shape will be drawn below the structure and key displacements will be shown.
Vertical Reaction at a
The relevant constraint is against vertical displacement at joint $a$. That constraint is relaxed and a unit displacement assigned in its place. Joint $a$ is free to rotate and move horizontally as required. Frame portion $b c d$ is constrained from any displacement by its rigidity and by the fixed support at $d$. The pin at $b$ allows relative rotation between $ab$ and $bcd$. The final deflected shape then is a rotation of $ab$ clockwise to accommodate the unit upward displacement of $a$. The deflected shape drawing shows the end displacement (assigned a value of ‘1’) and also the displacement at the center of gravity of the portion of the distributed load above $ab$.
Writing the virtual work expression:
\[\begin{split} &V_a \times 1 - 16\kNpm \times 12\m \times 0.5 = 0\\ &V_a = 96\kN\\ \end{split}\]Vertical Reaction at d
The constraint against vertical displacement at $d$ is relaxed and a unit upward displacement is assigned in its place (upward, because the drawing assumes that $V_d$ is upward). Constraints still prevent horizontal displacement and rotation at $d$ and therefore frame portion $bcd$ must displace straight vertically with no rotation nor horizontal translation. The pin at $b$ allows relative rotation of $ab$ to $bcd$, and point $a$ is constrained against vertical displacement (but not against rotation). The final deflected shape then shows a counter-clockwise rotation of $ab$ and a vertical translation of $bcd$.
Virtual work (noting that all virtual displacements are unitless):
\[\begin{split} &V_d \times 1 - 16\kNpm\times 8\m \times 1 - 16\kNpm\times12\m\times0.5 = 0\\ &V_d = 224\kN\\ \end{split}\]Horizontal Reaction at d
The constraint against horizontal displacement at $d$ is relaxed and a unit displacement assigned in its place. The frame drawing shows that $H_d$ is assumed acting rightwards and so the displacement is assigned in the same direction.
The fixed support at $d$ constrains the frame portion $bcd$ from rotating and from displacing vertically, so that portion can only move horizontally a unit amount everywhere (especially note joint $c$, the point of application of the $30\kN$ load).
The pin at $b$ prevents differential horizontal displacements af $ab$ and $bcd$ so $ab$ must displace a unit amount rightward as well. That is allowed by the roller support at $a$.
Note that vertical displacements of the distributed load are zero everywhere, so that load does not contribute to the horizontal reaction.
Virtual work:
\[\begin{split} &H_d\times 1 + 30\kN\times 1 = 0\\ &H_d = -30\kN~~~~~~~~(\therefore\leftarrow)\\ \end{split}\]Note that the computed value turned out negative. That only means that we assummed incorrect directions on the drawings and that the reaction does act toward the left, instead of toward the right as assumed.
Moment Reaction at d
The drawing shows an assumed clockwise direction for the moment reaction, $M_d$ at $d$. Therefore, the constraint against rotation at $d$ is relaxed and a unit clockwise rotation at point $d$ is assigned. Because $bcd$ is rigid, that means that point $c$ must move horizontally to the right. Point $b$ will move horizontally the same amount as point $c$, but it will also move vertically (the total movement is perpendicular to a radius drawn from $d$ to $b$, as the centre of rotation is point $d$).
The pin at $b$ drags the portion $ab$ rightward by the same amount (allowed by the roller at $a$). Member $ab$ rotates counter-clockwise in order not to violate the constraint against relative transverse displacement at $b$, while honouring the constraint against vertical displacement at $a$.
Deflection amounts are shown. Although the rotation is unitless, the member lengths are not. Therefore the displacements on the deflected shape all have units.
Virtual work:
\[\begin{split} &M_d\times1 - 16\kNpm\times12\m\times4\m - 16\kNpm\times8\m\times4\m + 30\kN\times8\m = 0\\ &M_d = 1040\kNm\\ \end{split}\]4.1.4 Frame Internal Forces
The following sections show the virtual displaced shape for a number of internal quantities. On each drawing is shown the assigned displacement, other key displacements, and displacements at the centroids of the distributed load, above each portion.
Note that consistent displacements for the three types of internal actions are as follows:
-
For moment, a unit rotation of one side of the point relative to the other. Normal beam sign convention would have the left portion rotating counter-clockwise and the rigth portion clockwise (relatively). Relative transverse and longitudinal displacements are prevented.
-
For shear, a unit relative displacement of one side of the point relative to the other. This displacement is transverse (perpendicular to the axis of the member). Unless the bening moments are forced to be zero (by a pin), there may be no relative rotations. Normal beam sign convention will have the left portion moving dow relative to the right portion.
-
For axial force, a unit relative displacement in the longitudinal direction. If tension is assumed +ive, the displacement is an overlap.
General Comments on Stability
The frame used in the following examples is shown here:
Many of the following examples rely on one particular consequence of the stability of this frame, so we will explain it once here and rely on this explanation many times below.
The pin supports at b and g prevent horizontal and vertical displacements at those points, but allow rotation. The internal pin at d allows relative rotation, only, between parts abd and deg.
Now, if part deg rotates about point g, points d and e will have horizontal components of displacement. But that is prevented by support b and member bd. We will often say something like “point d cannot move” in the explanations below. That is what we mean.
Moment at b
As the portion to the right of b cannot rotate, all of the rotation must occur in the left portion ab. It is assigned a unit counter-clockwise rotation. The centre of ab is 1m away, so that point displaces downwards $1\times1\m = 1\m$.
Virtual work:
\[\begin{split} &M_b\times1 + 12\kNpm\times2\m\times1\m = 0\\ &M_b = -24\kNm\\ \end{split}\]Moment at c
The beam is given a unit rotation at point c. Point d cannot move vertically, so the section cd must rotate about d to allow the upward displacement of point c. Section abc rotates about b to maintain the vertical displacement constraint at b.
To compute the displacements, extend the straight line abc to over point d. As the angle in the triangle above and to the right of c is 1 and the horizontal distance cd is 2m, the vertical distance from d to the extended line is $1\times2\m=2\m$. Similar triangles give the remaining displacements.
Virtual work:
\[\begin{split} &M_c \times 1 + 12\kNpm\times2\m\times\frac{2}{3}\m - 12\kNpm\times 1\m\times\frac{1}{3}\m - 12\kNpm\times 2\m\times\frac{1}{3}\m = 0\\ &M_c = 4\kNm\\ \end{split}\]Moment at e
Here we compute the bending moment just to the left of joint e.
As member eg cannot rotate (see above), all rotation must occur in ed, which is given the normal unit rotation, counter-clockwise as that is the direction of +ive moment on that side by the beam sign convention. See the following figure showing an expanded view of joint e for clarity:
The vertical displacement of d is given as $1\times2\m=2\m$. Use similar triangles to get the other displacements.
Virtual work:
\[\begin{split} &M_{ed}\times 1 - 12\kNpm\times2\m\times\frac{2}{3}\m + 12\kNpm\times3\m\times1\m = 0\\ &M_{ed} = -20\kNm\\ \end{split}\]Moment at f
A unit rotation is imposed at point f, causing that point to move perpendicularly to the line eg. In order for that to happen, def must rotate about a point on that same line (ie., on a line through e and g).
At the same time, point d must move perpendicularly
to the line through e, d and b (constrained by db).
The only way that can happen is if def rotates about a point on the line
through bde. Point e is the only point
that statisfies both rotation criteria, therefore def rotates about e.
As the distance ef is 2m, the distance from e to the extension of rotated fg is $1\times2\m=2\m$. By similar triangles, the displacement of f is $2\m\times3/5=6/5\m$. As de is the same length as ef, and def rotates rigidly, the downward displacement of d is the same as the displacement of f.
Virtual work:
\[\begin{split} &M_{f}\times 1 - 12\kNpm\times2\m\times\frac{2}{5}\m + 12\kNpm\times3\m\times\frac{3}{5}\m = 0\\ &M_{f} = -12\kNm\\ \end{split}\]Shear to the left of b
Here we are determining shear force at a point of interest just to the left of support b. For the displaced shape for shear forces, the rotations must almost always be kept equal (see ‘Shear at d’ below for the only exception). The member bd to the right of the point of interest is prevented from translating or rotating (see general comments above). Therefore, all of the translation must occur on the left side of the point of interest, and the rotation must be kept to zero (to match the right side).
Virtual work:
\[\begin{split} &V_{ba}\times 1 + 12\kNpm\times2\m\times1 = 0\\ &V_{ba} = -24\kN\\ \end{split}\]Shear to the right of b
Now we determine shear force at a point of interest just to the right of support b. The left side of that point is prevented from moving vertically by support b (but it can rotate). Therefore all transverse displacement must occur to the right of the point of interest. Point d still cannot move (see above), so member bd rotates upward about d. The slope to the left must be made equal to the slope to the right, leading to the displacements shown.
Virtual work:
\[\begin{split} &V_{bd}\times 1 - 12\kNpm\times2\m\times\frac13 - 12\kNpm\times3\m\times\frac12 = 0\\ &V_{bd} = 26\kN\\ \end{split}\]Shear at c
The left side of point c moves downward relative to the right side, with the total displacement at that point a unit amount. The right side moves upward. In order to keep the slopes equal ($\theta$ is equal on both sides) portions abc and cd rotate the same amount.
In order to compute the displacements, extend rotated cd to over the support at b. As both portions have rotated the same amount, the deflected shapes are parallel, so the distance from point b to the extension of cd is 1. That allows us to compute the angle $\theta = 1/3\m = 1/3 \m^{-1}$. All other displacements can be calculated from the horizontal distances and $\theta$.
Virtual work:
\[\begin{split} &V_{c}\times 1 - 12\kNpm\times2\m\times\frac13 + 12\kNpm\times1\m\times\frac16 - 12\kNpm\times2\m\times\frac13 = 0\\ &V_{c} = 14\kN\\ \end{split}\]Shear at d
Because there are no internal bending moments at d (it is an internal pin), the slopes of the deflected shape do not have to be equal on each side of d (this is the single exception to the requirement for equal slopes). As d attached to deg cannot move, all of the transverse displacement must occur on the left side (in abd).
Virtual work:
\[\begin{split} &V_{d}\times 1 - 12\kNpm\times2\m\times\frac13 + 12\kNpm\times3\m\times\frac12 = 0\\ &V_{d} = -10\kN\\ \end{split}\]Shear to the left of e
Here our point of interest is just to the left of the joint at e.
For reasons explained above, the portion of joint e attached to eg cannot have any displacement or rotation. Therefore all displacement must be on the left side of our point of interest, and the rotation at e must be zero. Relative rotation is allowed at the pin at d, and so the displaced shape is as shown.
Virtual work:
\[\begin{split} &V_{ed}\times 1 - 12\kNpm\times2\m\times\frac13 + 12\kNpm\times3\m\times\frac12 = 0\\ &V_{ed} = -10\kN\\ \end{split}\]Note that the 60Kn load is not included because it is just to the right of the point at which we are finding the shear, and does not appear on any displaced portion of the virtual displacement diagram.
Also note that as there are no transverse loads between d and just to the left of e, the shear forces at these two locations will be equal.
Shear at the top of eg
Now the point of interest is just below joint e, at the top of member eg. As we are interested in shear force, the directions of interest are perpendicular to eg. Point e on member de cannot move in that direction, so all translation must occur in eg just below our point of interest, resulting in a rotation of eg about g. de above the point of interest must rotate the same amount to keep the slopes equal. The amount of rotation is easy to determine as the length of eg is 5m, thus $\theta = 1/5\m^{-1}$.
Virtual work:
\[\begin{split} &V_{eg}\times 1 + 12\kNpm\times2\m\times\frac{2}{15} - 12\kNpm\times3\m\times\frac{1}{5} = 0\\ &V_{eg} = 4\kN\\ \end{split}\]Shear at f
The shear force at f is determined very similar to that at c, above – the only complexity being the directions are perpendicular to the sloped member, eg.
Virtual work:
\[\begin{split} &V_{f}\times 1 + 12\kNpm\times2\m\times\frac{2}{15} - 12\kNpm\times3\m\times\frac{1}{5} = 0\\ &V_{f} = 4\kN\\ \end{split}\]As there are no transverse loads between e and f, the shear force at f must be the same as at ef (above).
Axial force at c
Assuming tension forces +ive, the virtual displacement imposed is a unit overlap in member bd at c. As c cannot move rightward, both points d and e must move leftward a unit amount to accommodate this. The only way this can happen is if deg rotates an amount $\theta$ about point $g$.
From the relationship that horizontal displacement = vertical distance times angle, and the vertical distance is 4m, then $\theta = 1/4\m^{-1}$. The horizontal distance from g to d is 5m, so its vertical displacement is $5\theta = 5/4$.
Virtual work:
\[\begin{split} &N_{c}\times 1 - 12\kNpm\times2\m\times\frac{5}{12} + 12\kNpm\times3\m\times\frac{5}{8} + 60\kN\times\frac{3}{4} - 30\kN\times1= 0\\ &N_{c} = -27.5\kN\\ \end{split}\]Axial force at f
A unit overlap at point f must be accommodated by displacement in the portion above, def, as the portion below f is constrained by support g. Point e, therefore, has to move a unit amount in the direction eg. Also, point e must move straight vertically, as it is constrained from moving horizontally (by the support at b and member bde). Therefore, the displacement in the direction eg is a component of the vertical displacement. From $\Delta_y\times(4/5) = 1$, we get the vertical displacement required as $\Delta_y=5/4$.
The displacement of e perpendicular to eg is $(3/5)\times\Delta_y = 3/4$. The rotation of member eg is then $(3/4)/5 = 3/20\m^{-1}$.
The vertical displacement of d is the vertical displacement of e plus a portion due to the rotation of ed: $(5/4) + 2\m\times 3/20\m^{-1} = 31/20$.
Virtual work:
\[\begin{split} &N_{f}\times 1 - 12\kNpm\times2\m\times\frac{31}{60} + 12\kNpm\times3\m\times\frac{31}{40} + 60\kN\times\frac{5}{4} = 0\\ &N_{f} = -90.5\kN\\ \end{split}\]