Theory of Structures II Method of Virtual Work (Beam) Reference: Structural Analysis 10th Edition in SI Units by R.C.

1. CED 426
Structural Theory II
Lecture 1
Principle of Virtual Work
Method of Virtual Work (Trusses)
Mary Joanne C. Aniñon
Instructor

2. Principle of Virtual Work
• Developed by John Bernoulli in 1717
• Sometimes referred to as the unit-load method
• Provides a general means of obtaining the displacement and slope at
a specific point on a structure, be it a beam, frame, or truss.

3. Principle of Virtual Work
• The external and internal loads are related by the equation of
equilibrium
• If we take a deformable structure of any shape or size and apply a
series of external loads P to it, it will cause internal load u at points
throughout the structure.
• As a consequences of these loadings, external displacements Δ will
occur at the P loads and internal displacements δ will occur at each
point of internal load u.

4. Principle of Virtual Work
• In general, the principle of work and energy states:
𝑃∆= 𝑢𝛿
Work of External Loads = Work of Internal Loads
The Principle of
Virtual Work was
developed based on
this concept.

5. Principle of Virtual Work
• Supposed it is necessary to determine the
displacement Δ of point A on the body caused by
the “real loads” P1 P2, and P3.
• The displacement Δ can be determined by first
placing on the body a “virtual” load such that this
force P’ acts on the same direction as Δ.
• For convenience, we will choose P’ to have a unit
magnitude, i.e., P’ =1
• The term “virtual” is used to described the load,
since it is imaginary and does not actually exist as
part of the real loading.

6. Principle of Virtual Work
• The unit load P’ does create an internal load u,
hence, P’ and u are related by the equations of
equilibrium.
• After the virtual loadings are applied, then the
frame is subjected to the “real loads” P1 P2, and P3.
• Point A will be displaced an amount Δ and, causing
the element to deform an amount 𝑑𝐿.

7. Principle of Virtual Work
• Therefore, we can write the virtual-work equation as
1 ∙ ∆= 𝑢 ∙ 𝑑𝐿
where
P’ = 1 = external virtual unit load acting in the direction of ∆
𝒖 = internal virtual load acting on the element in the direction
of 𝒅𝑳
Δ = external displacement caused by the real loads
𝒅𝑳 = internal deformation of the elements caused by the real
loads.

8. Principle of Virtual Work
• Since P’ = 1, therefore,
∆= 𝑢 ∙ 𝑑𝐿
• In similar manner, if the rotational displacement or
slope of the tangent at a point on a structure is to be
determined, a virtual couple moment M’ having a
“unit” magnitude is applied at the point.
• As a consequences, this couple moment causes a
virtual load 𝒖𝜽 in one of the elements of the body.

9. Principle of Virtual Work
• Assuming that the real loads deform the element an
amount 𝒅𝑳, the rotation θ can be found from the virtual-
work equation:
1 ∙ 𝜃 = 𝑢𝜃 ∙ 𝑑𝐿
where
M’ = 1 = external virtual unit couple moment acting in the direction
of θ
𝒖𝜽= internal virtual load acting on the element in the direction of 𝒅𝑳
θ = external rotational displacement caused by the real loads
𝒅𝑳 = internal deformation of the elements caused by the real loads.

10. Principle of Virtual Work
• This method for applying the principle of virtual work is often referred
to as the method of virtual forces, since a virtual force is applied
resulting in the calculation of a real displacement.

11. Method of Virtual Work: Trusses
EXTERNAL LOADING
• To compute the vertical displacement Δ of joint B of the truss
caused by external loadings P1 and P2:
1 ∙ ∆= 𝑢 ∙ 𝑑𝐿
where 𝑑𝐿 =
𝑁𝐿
𝐴𝐸
1 ∙ ∆ =
𝑛𝑁𝐿
𝐴𝐸
where
1 = external virtual unit load acting on the truss joint in the stated direction
of Δ

12. Method of Virtual Work: Trusses
𝑛 = internal normal force in a truss member caused by the external
virtual unit load
Δ = external join displacement caused by the real loads on the truss
𝑁 = internal normal force in a truss member caused by the real loads
𝐿 = length of the member
𝐴 = cross-sectional area of a member
𝐸 = modulus of elasticity of a member

13. Method of Virtual Work: Trusses
TEMPERATURE
• In some cases, truss members may change their length due to
temperature
• If α is the coefficient of thermal expansion for a member and ∆𝑇 is
the change in its temperature, the change in length of a member is
∆𝐿 = 𝑛𝛼 ∆𝑇 𝐿
• To compute the displacement of a selected truss joint due to this
temperature change
1 ∙ ∆= 𝑛𝛼 ∆𝑇 𝐿
where
1 = external virtual unit load acting on the truss joint in the stated direction of Δ

14. Method of Virtual Work: Trusses
𝑛 = internal normal force in a truss member caused by the external
virtual unit load
Δ = external join displacement caused by the temperature change
𝛼 = coefficient of thermal expansion of a member
∆𝑇 = change in temperature of member
𝐿 = length of the member

15. Method of Virtual Work: Trusses
FABRICATION ERRORS AND CAMBERS
• Occasionally, errors in fabricating the lengths of the members
of a truss may occur
• Also, in some cases truss members must be made slightly
longer or shorter in order to give the truss a camber
• To compute the displacement of a truss joint from its expected
position
1 ∙ ∆= 𝑛 ∆𝐿
where
1 = external virtual unit load acting on the truss joint in the stated direction of
Δ

16. Method of Virtual Work: Trusses
𝑛 = internal normal force in a truss member caused by the external
virtual unit load
Δ = external join displacement caused by the fabrication errors
∆𝐿 = difference in length of the member from its intended size as caused by
a fabrication error

17. Procedure for Analysis
Virtual Forces, 𝒏
• Place the unit load on the truss at the joint where the desired
displacement is to be determined. The load should be in the same
direction as the specified displacement, e.g., horizontal or vertical.
• With the unit load so placed, and all the real loads removed from the
truss, used the method of joints or the method of sections and
calculate the internal 𝒏 force in each truss member.
• NOTE: Assume that tensile forces are positive and compressive forces
are negative.

18. Procedure for Analysis
Real Forces, 𝑵
• Use the method of joints or the method of sections to determine the
𝑵 force in each member.
• NOTE: Assume that tensile forces are positive and compressive forces
are negative.

19. Procedure for Analysis
Virtual-Work Equation
• Apply the equation of virtual work, to determine the desired
displacement. It is importance to retain the algebraic sign for each of
the corresponding 𝒏 and 𝑵 forces when substituting these terms into
the equation.
• If the resultant sum 𝑛𝑁𝐿/𝐴𝐸 is positive, the displacement Δ is in
the same direction as the unit load. If a negative value results, Δ is
opposite to the unit load.

20. Procedure for Analysis
Virtual-Work Equation
• When applying 1 ∙ ∆= 𝑛𝛼 ∆𝑇𝐿, realize that if any of the members
undergoes an increase in temperature, ∆𝑇 will be positive, whereas a
decrease in temperature results in a negative value for ∆𝑇.
• For 1 ∙ ∆= 𝑛 ∆𝐿, when a fabrication error increases the length of a
member, ∆𝐿 is positive, whereas decrease in length is negative.

21. Procedure for Analysis
Virtual-Work Equation
• When applying any formula, attention should be paid to the units of
each numerical quantity. In particular, the virtual unit load can be
assigned any arbitrary unit (lb, kip, N, etc.), since the n forces will
have these sam units, and as a result of the units for both the virtual
unit load and the n forces will cancel from both sides of the equation.

22. EXAMPLE 1
Problem:
Determine the vertical displacement if joint C of the steel truss shown.
The cross-sectional area of each member is A = 300 mm2 and E = 200
Gpa.

23. EXAMPLE 1
Solution:
• Virtual Forces n
• The virtual force in each member is calculated
using the method of joints.
𝑀𝐴 = 0 :
1(6)-RD(9)=0
RD= 0.667 kN
𝐹𝑌 = 0:
0.667-1+ RAY =0
RAY= 0.333 kN
A
B C D
F E
1 kN
A
B C D
F E
1 kN
0.333 kN 0.667 kN

24. EXAMPLE 1
Solution:
• Virtual Forces n
𝐹𝑌 = 0 @ pt. D:
0.667+RDEsin(45)=0
RDE= -0.943 kN
𝐹𝑋 = 0 @ pt. D:
-(-0.943)cos(45)- RDC =0
RDC= 0.667 kN
𝐹𝑌 = 0 @ pt. C:
RCE-1=0
RCE= 1 kN
𝐹𝑋 = 0 @ pt. C:
0.667-RCB=0
RCB= 0.667 kN
0.667 kN
0.667 kN
-0.943 kN
45°
D
C 0.667 kN
1 kN
0.667 kN
1 kN
A
B C D
F E
1 kN
0.333 kN 0.667 kN

25. EXAMPLE 1
Solution:
• Virtual Forces n
𝐹𝑌 = 0 @ pt. E:
-1-(-0.943)cos(45)- REB cos(45)=0
REB= -0.471 kN
𝐹𝑋 = 0 @ pt. E:
-(-0.471)sin(45)+(-0.943)sin(45)-REF =0
REF= -0.333 kN
𝐹𝑋 = 0 @ pt. F:
-0.334-RFAsin(45)=0
RFA= -0.471 kN
𝐹𝑌 = 0 @ pt. D:
-(-0.471)cos(45)-REF =0
REF= 0.333kN
A
B C D
F E
1 kN
0.333 kN 0.667 kN
45°
E
0.667 kN
0.667 kN
1 kN
1 kN
-0.943 kN
45° 45°
-0.471 kN
-0.333 kN
-0.334 kN
F
E
45°
-0.471 kN
0.333kN

26. EXAMPLE 1
Solution:
• Virtual Forces n
𝐹𝑋 = 0 @ pt. A:
(-0.471)cos(45)+RAB =0
RAB= 0.333 kN
A
B C D
F E
1 kN
0.333 kN 0.667 kN
45°
0.667 kN
0.667 kN
1 kN
0.333 kN
A
45°
0.333kN
-0.471 kN
0.333kN
-0.333 kN
0.333 kN

27. EXAMPLE 1
Solution:
• Real Forces N
• The real forces in the members are calculated
using the method of joints (the same
procedure..)
A
B C D
F E
1 kN
0.333 kN 0.667 kN
45°
0.667 kN
0.667 kN
1 kN
0.333kN
-0.333 kN
0.333 kN
A
B C D
F E
20 kN
20 kN 20 kN
45°
0.667 kN
0.667 kN
20 kN
20 kN
-20 kN
0.333 kN
20 kN

28. EXAMPLE 1
Solution:
• Virtual-Work Equation
• Arrange the data in tabular form:
A
B C D
F E
1 kN
0.333 kN 0.667 kN
45°
0.667 kN
0.667 kN
1 kN
0.333kN
-0.333 kN
0.333 kN
A
B C D
F E
20 kN
20 kN 20 kN
45°
0.667 kN
0.667 kN
20 kN
20 kN
-20 kN
0.333 kN
20 kN
MEMBER n (kN) N (kN) L (m) nNL (kN2.m)
AB 0.333 20 3 20
BC 0.667 20 3 40
CD 0.667 20 3 40
DE -0.943 -28.284 4.243 113.20
FE -0.333 -20 3 20
EB -0.471 0 4.243 0
BF 0.333 20 3 20
AF -0.471 -28.284 4.243 56.60
CE 1 0 3 60

29. EXAMPLE 1
MEMBER n (kN) N (kN) L (m) nNL (kN2.m)
AB 0.333 20 3 20
BC 0.667 20 3 40
CD 0.667 20 3 40
DE -0.943 -28.284 4.243 113.20
FE -0.333 -20 3 20
EB -0.471 0 4.243 0
BF 0.333 20 3 20
AF -0.471 -28.284 4.243 56.60
CE 1 0 3 60
Σ = 369.7

30. EXAMPLE 1
Thus,
1 ∙ ∆ =
𝑛𝑁𝐿
𝐴𝐸
1𝑘𝑁 ∙ Δ𝑐𝑣 =
369.7 𝑘𝑁2 ∙ 𝑚
𝐴𝐸
Δ𝑐𝑣 =
369.7𝑥103 𝑁 ∙ 𝑚
(300𝑥10−6 𝑚2)(200𝑥109 𝑁
𝑚2)
Δ𝑐𝑣 = 0.00616 m or 6.16 mm