1. Deflection: Virtual Work Method; Trusses
Theory of Structure - I

2. Contents External Work and Strain Energy Principle of Work and Energy
Principle of Virtual Work
Method of Virtual Work:
Trusses

3. External Work and Strain Energy
Most energy methods are based on the conservation of energy principle, which states that the work done by all the external forces acting on a structure, Ue, is transformed into internal work or strain energy, Ui.
Ue = Ui
External Work-Force. x F L D P Ue
Eigen work D
As the magnitude of F is gradually increased
from zero to some limiting value F = P, the final
elongation of the bar becomes D. F
Eigen work

4. x F D´
F ´ + P L P L F´
Eigen work D D´ D P
Displacement work
(Ue)Total = (Eigen Work)P + (Eigen Work)F´
+ (Displacement work) P

5. x (m) F
20 kN L L
1 cm
0.01 m
20 kN

6. x (m) F
0.01
20 kN L L
15 kN L
15 kN
0.75 cm
Eigen work
0.75 cm
0.25 cm
15 kN
0.0075
Displacement work
5 kN

7. External Work-Moment. q M q´
M ´ + M
Eigen work dq M q M
Displacement work
-----(8-12)
Eigen work
-----(8-13)
-----(8-14)

8. Strain Energy-Axial Force.D N

9. Strain Energy-Bendingx dx w P L M dx dq s e

10. Strain Energy-Torsiondx T c dq g t g J
For reference:
Strength of Material by Singer, Fourth Edition, Page 67-68

11. Strain Energy-Shear V g dx dy t g
For reference:
Strength of Material by Singer, Fourth Edition, Page

12. Principle of Work and Energyx
-PL
M diagram P x V M +
SMx = 0:

13. Principle of Virtual Work
Apply virtual load P´ first u L A
P´ = 1
Virtual loadings
1 • D = Su • dL
Then apply real load P1. A u L
Real displacements D
In a similar manner, dL
Virtual loadings
1 • q = Suq • dL P1
Real displacements

14. P1 P2 B n2 n1 n3 n4 n5 n6 n7 n8 n9 N2 N1 N3 N4 N5 N6 N7 N8 N9 B 1kN
Method of Virtual Work : Truss P1 P2 B
External Loading. n2 n1 n3 n4 n5 n6 n7 n8 n9 N2 N1 N3 N4 N5 N6 N7 N8 N9 B
1kN D
Where:
1 = external virtual unit load acting on the truss joint in the stated direction of D
n = internal virtual normal force in a truss member caused by the external virtual
unit load
D = external joint displacement caused by the real load on the truss
N = internal normal force in a truss member caused by the real loads
L = length of a member
A = cross-sectional area of a member
E = modulus of elasticity of a member

15. Temperature
Where:
D = external joint displacement caused by the temperature change
a = coefficient of thermal expansion of member
DT = change in temperature of member
Fabrication Errors and Camber
Where:
D = external joint displacement caused by the fabrication errors
DL = difference in length of the member from its intended size as
caused by a fabrication error

16. Example 8-15
The cross-sectional area of each member of the truss shown in the figure is
A = 400 mm2 and E = 200 GPa.
(a) Determine the vertical displacement of joint C if a 4-kN force is applied to the truss at C.
(b) If no loads act on the truss, what would be the vertical displacement of joint C if member AB were 5 mm too short?
(c) If 4 kN force and fabrication error are both accounted, what would be the vertical displacement of joint C. A B C
4 m
4 kN
3 m

17. SOLUTION
Part (a)
Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical 1 kN load is placed at joint C. The n force in each member is calculated using the method of joint.
Real Force N. The N force in each member is calculated using the method of joint.
1 kN A B C
4 kN
N(kN) A B C
n (kN)
0.667
-0.833 2
+2.5
-2.5
1.5 kN
4 kN
0.5 kN

18. = 1 kN C 4 kN -0.833 -0.833 C -2.5 +2.5 A A 0.667 B B 2 C n (kN)
L (m) 5 5 = A B C
nNL (kN2•m) 8
-10.41
10.41
10.67
DCv = mm,

19. Part (b): The member AB were 5 mm too shortC
n (kN)
1 kN
0.667
-0.833
5 mm
DCv = mm,
Part (c): The 4 kN force and fabrication error are both accounted.
DCv = = mm
DCv = mm,

20. Example 8-16
Determine the vertical displacement of joint C of the steel truss shown. The cross-section area of each member is A = 400 mm2 and E = 200 GPa.
4 m A B C D E F
4 kN

21. SOLUTION
Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical 1 kN load is placed at joint C. The n force in each member is calculated using the method of joint.
Real Force N. The N force in each member is calculated using the method of joint.
4 m A B C D E F
n (kN)
0.667
-0.471
-0.943
0.333 1
-0.333
4 m A B C D E F
4 kN
N(kN) 4
-5.66 -4
0.667 kN
0.333 kN
4 kN
1 kN

22. = A B C D E F n (kN) 1 kN A B C D E F 4 kN N(kN) A B C D E F L(m) A B
0.667
-0.471
-0.943
0.333 1
-0.333 A B C D E F
n (kN)
1 kN 4
-5.66 -4 A B C D E F
4 kN
N(kN) 4
5.66 A B C D E F
L(m) A B C D E F
nNL(kN2•m) =
10.67
15.07
30.18
5.33 16
DCv = 1.23 mm,

23. Example 8-17
Determine the vertical displacement of joint C of the steel truss shown. Due to radiant heating from the wall, members are subjected to a temperature change: member AD is increase +60oC, member DC is increase +40oC and member AC is decrease -20oC.Also member DC is fabricated 2 mm too short and member AC
3 mm too long. Take a = 12(10-6) , the cross-section area of each member is A = 400 mm2 and E = 200 GPa.
2 m A B C D
3 m
20 kN
10 kN
wall

24. SOLUTION Due to loading forces. 0.667 kN 1 kN 1 kN 13.33 kN 23.33 kN
2 m A B C D
3 m
n (kN)
2 m A B C D
3 m
20 kN
10 kN
N (kN)
0.667
-1.2 1
23.33
-24.04 20 2
3.61 3 A B C D
L (m)
31.13
104.12 60 A B C D
nNL(kN2•m)
DCv= 2.44 mm,

25. Fabrication error (mm) A B D C
1 kN
0.667
-1.2 1 A B C D
n (kN)
+40
-20
+60 A B D
DT (oC) C 2
3.61 3 A B C D
L (m)
Fabrication error (mm) -2
+ 3 A B D C
Due to temperature change.
= 3.84 mm,
Due to fabrication error.
= mm,
Total displacement .
= 1.35 mm,

26. Thank You