Periodica Polytechnica
Civil Engineering
x(y) pp. x-x, (year)
DOI: 10.3311/PPci.XXXX
Creative Commons Attribution b
RESEARCH ARTICLE
Two Theorems for Computation of
Virtual Displacements and It’s
Application in Structural Analysis
SU Zhenchao1, XUE Yanxia1
Received xx Xxxx 201x
Abstract
Two theorems are proposed to determine the projection of a
virtual displacement to the orientation under the case of
knowing the projections of a virtual displacement to the given
two or three orientations for object systems subject to
holonomic and scleronomic constraints. Applications to
structural static analysis are investigated using the two
theorems in this paper. Result reveals that these theorems are
easy to be used and the relating problems can be solved
quickly.
Keywords (Header1: Abs-Keyw_Ack)
projection of virtual displacement, principle of virtual
displacement, engineering mechanics, structural analysis.
1 Introduction
It is well known that the principle of virtual displacement
(or virtual work) is a main part in analytical statics, and the
important basis for analytical dynamics and structural analysis.
This principle provides an excellent tool for people to
investigate the equilibrium laws of object systems, and plays a
supporting role in classical mechanics.
Generally speaking, some current college textbooks[1-7] for
engineering mechanics or structural analysis course often
propose two main methods for building relations among virtual
displacements of different points for the system with
holonomic and scleronomic constraints, ie, analytical method
and geometrical method. Using analytical method, one can
determine the virtual displacements of interest points (usually
points of forces) by taking differentials of their position
coordinates, and geometrical method by introducing the
concept of instant center of rotation of virtual displacements.
1
Department of Civil Engineering, Tan Kah Kee College, Xiamen University,
363105, Zhangzhou city, Fujian Province, China (e-mail: [email protected],
[email protected])
These methods are very useful in application of principle of
virtual displacement. However, in actual experience we realize
that these methods are sometimes not sufficient for analyzing
complex structures.
In this study, we will gain an insight into the application of
the principle of virtual displacement and present two new
theorems on how to build up the relations among the
projections to different orientations of the virtual displacement
of a point. Several examples are given to illustrate the
application of two theorems and corresponding lemmas. The
article is organized as follows. In Section 2, a theorem is given
to find a projection under knowing the two projections to
different orientations of a point for planar systems, and two
lemmas for special cases are proposed. In Section 3, a theorem
for spatial systems is given to find a projection under knowing
Begin of Paper Title
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the three projections to different orientations of a point, and
two lemmas for special cases are proposed.
δr1 sin(    )  δr2 sin 

sin 
δr cos   δr1 cos(   )
sin(    )  2
sin 
δr sin(    )  δr2 sin 
 1
.
sin 
 cos(   ) 
2 The first theorem for plane cases
For a planar object system, if the projections of a virtual
displacement r of a certain dot to orientations n1 and n2 are
r1 and r2 separately, shown in fig.1, then the projection of the
virtual displacement to orientation n3 is
δr sin(    )  δr2 sin 
δr3  1
.
sin 
δr3  δr  e 3  δr  cos(   )i  sin(    ) j 
（1）
where, angles  and  are angles between n1and n2, n1 and n3
Lemma 1. When 
 0(180)，δr1  δr2  0 , then δr3  0 .
Lemma 2. When 
 90 , then
δr3  δr1 cos   δr2 sin  .
For demonstrating the effectiveness of this theorem in
separately.
solving the balancing problems of planar object systems, two
examples are given below.
Example 1. Determine the force in member CD of the truss,
shown in Fig.2. Assume all members are pin connected,
and AD  AE  ED  EC  CG  DG  DB  BG  a .
Fig. 1
Proof. Assume e, e1, e2 and e3 are the unit vectors of the
orientations of n, n1, n2 and n3, separately. Take the start point
A of the virtual displacement r as the origin of the coordinate
system xAy. Let  is the angle between n1 and x axis.
Fig. 2
Solution. To calculate the force in rod CD, it is isolated
According to the given conditions, one reads
from the system in Figure 2(b). Then, the system in Figure 2(b)
δr  e1  δr1 .
is a mechanism. Obviously, equilateral triangles AED and
δr  e 2  δr2 .
DBG can be regarded as rigid plates, and AED can be
assumed to rotate about A point with δ . Then,
Hence, one can get
δr  (cos i  sin j)  δr1.
（1）
δr  cos(   )i  sin(    ) j   δr2 .
δrE  δrD  aδ . Considering the orientation of δrD and
（2）
character of support B, point B is the virtual displacement
center of DBG, therefore δrG
where i, j are the standard unit vectors of x and y axis.
By taking the inner products
 δrD  aδ . Based on the
δr  i and δr  j as principle of virtual displacement, one reads
unknowns, and solving the equations of (1) and (2), one can
  δrD  FCD  δrC  0 .
F  δrG  FCD
（5）
get
Based on the theorem of projection of virtual
δr sin(    )  δr2 sin 
δr  i  1
.
sin 
（3）
δr cos   δr1 cos(   )
δr  j  2
.
sin 
（4）
displacement (i.e. the projections of the virtual displacements
of the points from a straight line belonging to a body, on that
line, are equal.), the projections of
EC and GC are zeros. Then, by employing the lemma 1 of
Then, the projection of the virtual displacement to
the above theorem,
orientation n3 is
2
Period. Polytech. Civil Eng.
δrC to the orientations
δrC must be zero. Therefore, based on the
First A. Author, Second B. Author
above analysis and (5), one can get
δrGH  δrH  cos  
  aδ  0.
 F  aδ  cos 30  FCD
Because of δ
 0 , then
For the bar CG, the projection of
  FCD   F  cos 30  
FCD
3
F.
2
( CD in compression).
Example 2. Determine the force in member DG of the truss,
4a cos 
δ .
sin 
δrG to the orientation C
G is
δrCG  δrC  cos(90   )  4a sin δ .
For the point G, the projections δrGH and δrCG are known.
By employing the formula (1) of the above theorem, the
shown in Fig.3. Assume all members are pin connected.
projection of
δrGD 
δrG to the orientation GD is
δrGH sin     290     δrCG sin   290   
sin 
  4a cos  sin 2 


 4a sin   δ
 sin  sin 

  8a cot  cos   4a sin  δ

2
4 
24
δ  
   8a  4 
 4a 
aδ .
2 5
2 5
5

By employing the principle of virtual displacement, one can
get
  δrG  0
F  δrD  FDG  δrD  FDG
（6）
Therefore, based on the above analysis and (6), one can get
 
F  4aδ  FDG  4aδ  cos(90   )  FDG
Because δ
Fig. 3
Solution. To calculate the force in rod DG, it is isolated
from the system in Figure 3(b). Obviously, triangles ACE and
FDG 
DBE can be regarded as rigid plates, and ACE can be
 , then
 0 , FDG  FDG
F
sin  
6

F
2
5
5

6

24
5
aδ  0.
5F
 0.279 F .
8
5
assumed to rotate about A point with δ . Then,
δrC  4aδ , δrE  4 2aδ . Considering that the
orientation of
δrB is horizontal, the virtual displacement
δrD must be horizontal, and
δrD  δrE  cos 45  4aδ ,
δrB  cos 45  δrE , δrB  8aδ .
For the bar BH, it is easy to get that
δrH  cos(90  2 )  δrB  cos ,
δrH 
Begin of Paper Title
If the projections of the virtual displacement δr of a dot
to three non-coplanar orientations n1 ,
δrG to the
n2 , n3 are δr1 , δr2 ,
δr3 , then the projection δr4 of the virtual displacement to
the orientation
n4 is
 e1x

δr4  (e4 x , e4 y , e4 z ) e2 x
e3 x

1
4a
δrB 
δ .
2 sin 
sin 
Then, for the bar GH, the projection of
orientation GH is
3 The second theorem for spatial cases
e1 y
e2 y
e3 y
e1z 

e2 z 
e3 z 
1
 δr1 
 
 δr2 
 δr 
 3 .
（7
）
where eix , eiy , eiz are projections of unit vector ei of spatial
orientation
ni to the Cartesian coordinate axes x, y, z,
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separately. ( i  1, 2, 3, 4)
displacement must be zero vector, and projection of the virtual
Proof. Assuming that i, j, k are the standard unit orthogonal
vectors of the spatial coordinate system, based on the given
conditions, one reads
displacement to n4 (every orientation) must be zero.
Lemma 5. If the unit vectors ei ( i  1, 2, 3) are standard
unit orthogonal vectors of the Cartesian coordinate axes x, y, z,
ei  eixi  eiy j  eiz k , i  1, 2, 3, 4
separately, the projection of the virtual displacement to
ei is δri , then the projection of the virtual displacement to the
and
 e1x δr  i  e1 y δr  j  e1z δr  k  δr1

e2 x δr  i  e2 y δr  j  e2 z δr  k  δr2
 e δr  i  e δr  j  e δr  k  δr
3y
3z
3
 3x
orientation
n4 is
δr4  e4 x δr1  e4 y δr2  e4 z δr3 .
（9）
Example 3. Determine the force in member AC of the
or
spatial truss, shown in Fig.4(a). Assume all members are pin
 e1x

e 2 x
 e3 x

e1 y
e2 y
e3 y
e1z  δr  i   δr1 
  

e2 z  δr  j    δr2 
e3 z  δr  k   δr3  .
connected. F1=5kN, F2=4kN, F3=2kN, F4=1.5kN, F5=1.5kN.
e1 , e2 , e3 are
Because the three orientations
non-coplanar， the coefficient matrix is reversible. Hence,
one can get
 δr  i   e1x

 
 δr  j   e2 x
 δr  k  e

  3x
e1 y
e2 y
e3 y
e1z 

e2 z 
e3 z 
-1
 δr1 
 
 δr2 
 δr 
 3 .
Thus, the projection of the virtual displacement
δr to
orientation n4 is
 δr  i 


δr4  (e4 x , e4 y , e4 z ) δr  j 
 δr  k 


 e1x

 (e 4 x , e 4 y , e 4 z ) e 2 x
 e3 x

e1z 

e2 z 
e3 z 
e1 y
e2 y
e3 y
1
 δr1 
 
 δr2 
 δr 
 3 .
Lemma 3. If the unit vectors ei ( i  1, 2, 3) are normal
orthogonal unit vectors, the projection
displacement to the orientation
e1x
δr4  (e4 x , e4 y , e4 z ) e1 y
e1z
δr4 of the virtual
e2 y
e2 z
C and B are fixed. Therefore, the projection of point A to the
e3 x  δr1 
 
e3 y  δr2 
e3 z  δr3  .
orientations BA and DA are zeros. Assuming that the
（8）
Lemma 4. If the unit vectors ei of orientations ni
( i  1, 2, 3) are non-planar vectors, and the projections of the
virtual displacement of a dot to ni are all zeros, then the virtual
4
Solution. To calculate the force in rod AC, it is isolated
from the system in Figure 2(b). Obviously, hinge points E, D,
n4 is
e2 x
Fig. 4
Period. Polytech. Civil Eng.
projection along the F4 direction is r, considering that the unit
vectors of the direction BA, A  C, DA and F4 are as
follows
4 3 
e BA   ,  , 0  , e AC   1, 0, 0 ,
5 5 
First A. Author, Second B. Author
eDA  0, 0, 1 , e F4  0,  1, 0 .
References
Then, utilizing the second theorem for spatial case, the
projection along the AC direction is
4
5
δr4  (1,0,0)  0

0

1
3 
0  0 
5
 
0 1  0 

 1 0  δr 


3
5
 4 0  4  0 
  3
 (1,0,0)  0 0  1  0   δr.


4
 0 1 0  δr 


By employing the principle of virtual displacement, one
reads
FAC  δrA  F4  δrA  0 .
（10）
Based on the above analysis and (10), one can get
[1] R. C. Hibbeler. “Engineering Mechanics: Statics”, 12th
Edition. Prentісе Hаll, 2010
[2] Andrew Pytel, Jaan Kiusalaas, Ishan Sharma.
“Engineering Mechanics: Statics”, third ed. SI,
CENGAGE Learning, 2010.
[3] Anthony Bedford, Wallace Fowler. “Engineering
Mechanics: Statics”, fifth ed., Pearson Education Inc. ,
2008.
[4] Dietmar Gross, Werner Hauger, Jorg Schroder ,
Wolfgang A. Wall, Nimal Rajapakse. “Engineering
Mechanics 1: Statics”, Springer-Verlag Berlin
Heidelberg, 2009.
[5] K. L. Kumar, “Engineering Mechanics”, 3ed.,
McGraw-Hill Publishing Company Limited, 2003.
[6] I. C. Jong, B. G. Rogers, “Engineering Mechanics:
Statics”, Saunders College Publishing, 1991.
[7] Gianluca Ranzi ， Raymond Ian Gilbert, “Structural
Analysis: Principles, Methods and Modelling”, CRC
Press, 2015.
3
FAC  δr  F4  δr  0 .
4
Because δ
 0 , then one can get
4
4
FAC   F4    1.5kN  2.0kN .
3
3
4 Conclusions
This paper mainly faces to the difficulty of computation of
virtual work, investigates on the projection of a dot’s virtual
displacement to a given orientation, propose two theorems and
corresponding lemmas, and discuss their application for
analyzing the forces of structural members. Computation
procession reveals that the formula for planar situation is easy
to use, and the formula for spatial situation is normalized and
easy to be remembered. Combining with current methods,
these results will be very helpful to analyze inner forces of
complicit structures. It’s necessary to say that the two
theorems can be used in dynamics when dynamical problems
convert into static problem in form by utilizing D’ Alembert’s
principle.
Acknowledgement
This paper is supported by planning scientific research
programs in education during the 13th five-year period in
Fujian (Grant No. FJJKCGZ16-152). The authors gratefully
acknowledge it’s supports.
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