1. Deflections using energy methods

2. Outline External work & strain energy Principle of work & energy
Principle of virtual work
Method of virtual work: trusses
Method of virtual work: beams & frames
Virtual strain energy caused by axial load, shear, torsion & temperature
Castigliano’s theorem
Castigliano’s theorem for trusses
Castigliano’s theorem for beams & frames

3. 4-1 External Work & Strain Energy
For more complicated loadings or for structures such as trusses & frames, it is suggested that energy methods be used for the computations.
Most energy methods are based on the conservation of energy principal.
Work done by all external forces acting on a structure, Ue is transformed into internal. work or strain energy Ui
Ue = Ui equation 9.1

4. 4-1 External Work & Strain Energy
If the material’s elastic limit is not exceeded, the elastic strain energy will return the structure to its undeformed state when the loads are removed.
When a force F undergoes a disp dx in the same direction as the force, the work done is
d Ue = F dx
If the total disp is x, the work becomes:

5. 4-1 External Work & Strain Energy
Consider the effect caused by an axial force applied to the end of a bar as shown in Fig. 9.1(a).
F is gradually increased from 0 to some limiting value F = P.
The final elongation of the bar becomes .
If the material has a linear elastic response, then F = (P/ )x.

6. 4-1 External Work & Strain Energy
Fig 9.1

7. 4-1 External Work & Strain Energy
Substituting into equation 9.2 & integrating from ) to , we get:
Suppose P is already applied to the bar & that another force F’ is now applied, so that the bar deflects further by an amount ‘ as shown in Fig. 9.1(b).

8. 4-1 External Work & Strain Energy
The work done by P when the bar undergoes the further deflection is then
d Ue’ = P’ eqn 9.4
Here the work rep the shaded rectangular area in Fig. 9.1(b).
In this case, P does not change its magnitude since ’ is caused only by F’.
Work = force x disppacement

9. 4-1 External Work & Strain Energy
When a force P is applied to the bar, followed by an application of the force F’, the total work done by both forces is rep by the triangular area ACE in Fig. 9.1(b).
The triangular area ABG rep the work of P that is caused by displacement .
The triangular area BCD rep the work of F’ since this force causes a displacement ’.
Lastly the shaded rectangular area BDEG rep the additional work done by P.

10. 4-1 External Work & Strain Energy
The work of a moment = magnitude of the moment (M) x the angle (d) through which it rotates, Fig. 9.2.
d Ue = M d
If the total angle of
rotation is  radian,
the work becomes

11. 4-1 External Work & Strain Energy
If the moment is applied gradually to a structure having a linear elastic response from 0 to M, then the work done is
However, if the moment is already applied to the structure & other loadings further distort the structure an amount ’, then M rotates ’ & the work done is

12. 4-1 External Work & Strain Energy
When an axial force N is applied gradually to the bar in Fig. 9.3, it will strain the material such that the external work done by N will be converted into strain energy.
Provided the material is linearly
elastic, Hooke’s Law is valid.
 = E
If the bar has a constant x-
sectional area A and length L.

13. 4-1 External Work & Strain Energy
The normal stress is  = N/A
The final strain is  = /L
Consequently, N/A = E(/L)
Final deflection:
Substituting into equation 9.3 with P = N,

14. 4-1 External Work & Strain Energy
Consider the beam shown in Fig. 9.4(a)
P & w are gradually apply.
These loads create an internal moment M in the beam at a section located a distance x from the left support.
The resulting rotation of the DE dx, Fig. 9.4(b) can be found from equation 8.2.
Consequently, the strain energy or work stored in the element is determined from equation 9.6 since the internal moment is gradually developed.

15. 4-1 External Work & Strain Energy
Hence,
The strain energy for the beam is determined by integrating this result over the beam’s length.

16. 4-2 Principle of Work & Energy
Consider finding the displacement at a point where the force P is applied to the cantilever beam in Fig 9.5.
From equation 9.3, the external work:
To obtain the resulting strain energy, we must first determine the internal moment as a function of position x in the beam and apply equation 9.11.

17. 4-2 Principle of Work & Energy
In this case, M = - Px so that:
Equating the external work to internal strain energy and solving for the unknown displacement, we have:

18. 4-2 Principle of Work & Energy
Limitations
It will be noted that only one load may be applied to the structure.
Only the displacement under the force can be obtained.

19. 4-3 Principle of virtual work
If we take a deformable structure of any shape or size & apply a series of external loads P to it, it will cause internal loads u at points throughout the structure.
It is necessary that the external and internal loads be related by the equation of equilibrium.

20. 4-3 Principle of virtual work
As a consequence of these loadings, external displacement  will occur at the P loads and internal displacement  will occur at each point of internal loads u.
In general, these displacement do not have to be elastic, and they may not be related to the loads.

21. 4-3 Principle of virtual work
In general, the principle states that:
Consider the structure (or body) to be of arbitrary shape as shown in Fig. 9.6(b).
Suppose it is necessary to determine the displacement  of point A on the body caused by the “real loads” P1, P2 and P3.

22. 4-3 Principle of virtual work
It is to be understood
that these loads cause
no movement of the
supports.
They can strain the
material beyond the
elastic limit.

23. 4-3 Principle of virtual work
Since no external load acts on the
body at A and in the direction of ,
the displacement , the displacement
can be determined by first placing on
the body a “virtual” load such that this
force P’ acts in the same direction
as  shown in Fig. 9.6(a).

24. 4-3 Principle of virtual work
We will choose P’ to have a unit magnitude, P’ =1.
Once the virtual loadings are applied, then the body is subjected to the real loads P1, P2 and P3 as shown in Fig. 9.6(b).
Point A will be displaced an amount  causing the element to deform an amount dL.

25. 4-3 Principle of virtual work
As a result, the external virtual force P’ & internal load u “ride along” by  and dL & therefore, perform external virtual work of 1.  on the body and internal virtual work of u.dL on the element.
By choosing P’ = 1, it can be seen from the solution for  follows directly since  = udL.
A virtual couple moment M’ having a unit magnitude is applied at this point.

26. 4-3 Principle of virtual work
This couple moment causes a virtual load u in one of the elements of the body.
Assuming that the real loads deform the element an amount dL, the rotation  can be found from the virtual –work equation.

27. 4-4 Method of virtual work: Trusses
External loading
Consider the vertical displacement  of joint B in Fig. 9.7.
If the applied loadings P1 & P2 cause a linear elastic material response, the element will deform.

28. 4-4 Method of virtual work: Trusses
External loading (cont’d)
Applying equation 9.13, the virtual work equation of the truss is:

29. 4-4 Method of virtual work: Trusses
External loading (cont’d)
The external virtual load creates internal virtual forces n in each of the members.
The real loads caused the truss joints to be displaced in the same direction as the virtual unit load.
Each member is displacement NL/AE in the same direction as its respective n force.
Hence, ext virtual work = sum of int. (virtual) strain energy stored in truss members.

30. 4-4 Method of virtual work: Trusses
Temperature
In some cases, truss members may change their length due to temperature.
The displacement of a selected truss joint may be written as

31. 4-4 Method of virtual work: Trusses
Fabrication errors & camber
Errors in fabricating the lengths of the members of a truss may occur.
Truss members may also be made slightly longer or shorter in order to give the truss a camber.
Camber is often built into bridge truss so that the bottom cord will curve upward by the same amount equivalent to the downward deflection when subjected to the bridge’s full dead weight.

32. 4-4 Method of virtual work: Trusses
Fabrication errors & camber (cont’d)
The displacement of a truss joint from its expected position can be written as:
A combination of right sides of equations 9.15 to 9.17 may be necessary if both external loads, thermal change & fabrication errors are taking place.

33. Example 4.1 Determine the vertical displacement of joint C
of the steel truss shown
in Fig. 9.8(a).
The cross-sectional area
of each member = 300mm2
E = 200GPa

34. Example 4.1 - solution Virtual force Real forces
Only a vertical 1kN load is placed at joint C.
The force in each member is calculated using method of joints.
Results are shown in Fig. 9.8(b).
Real forces
The real forces are calculated using method of joints.
Results are shown in Fig. 9.8(c).

35. Example 4.1 - solution Virtual work equation
Arranging the data in tabular form, we have

36. Example 4.2
The cross-sectional area of each member shown in Fig. 9.9(a).
A = 400mm2
E = 200GPa
Determine the vertical displacement of joint C if no loads act on the truss, what would be the vertical displacement of joint C if member AB is 5mm too short.

37. Example 4.2
Fig 9.9

38. Example 4.2 - solution Virtual forces Applying equation 9.17
The support reactions at A & B are calculated.
The n force in each member is determined .using method of joints as shown in Fig. 9.9(b).
Applying equation 9.17

39. Example 4.3
Determine the vertical displacement of joint C of the steel truss shown in Fig. 9.10(a).
Due to radiant heating from the wall, member AD is subjected to increase in temp = +60oC.
Take  = 1.08(10-5)/oC and E = 200GPa
The cross-sectional area of each member is indicated in the figure.

40. Example 4.3
Fig 9.10

41. Example 4.3 - solution Virtual forces n Real forces N
The forces in members are computed, Fig. 9.10(b).
Real forces N
Since n forces in AB and BC are zero, N forces need not be computed.
Virtual work equation
Both loads and temp affect the deformation.
Equations 9.15 and 9.16 are combined.

42. Example solution
Virtual work equation (cont’d)

43. 4-5 Method of virtual work: Beams & Frames
The Principle of virtual work may be formulated for beam & frame deflections by considering the beam shown in Fig. 9.11(a).
To compute  a virtual unit load acting in the direction of  is placed on the beam at A.
The internal virtual moment m is determined by the method of sections at an arbitrary location x from the left support, Fig 9.11(b).
When point A is displaced , the element dx deforms or rotates d = (M/EI)dx.

44. 4-5 Method of virtual work: Beams & Frames
Fig 9.11

45. 4-5 Method of virtual work: Beams & Frames

46. 4-5 Method of virtual work: Beams & Frames
If the tangent rotation or slope angle  at a point on the beam’s elastic curve is to be determined, a unit couple moment is applied at the point.
The corresponding internal moment m have to be determined.

47. 4-5 Method of virtual work: Beams & Frames
If concentrated forces or couple moments act on the beam or the distributed load is discontinuous, separate x coordinates will have to chosen within regions that have no discontinuity of loading.

48. Example 4.4 Determine the displacement of point B of the steel
beam shown in
Fig. 9.13(a).
Take
E = 200GPa
I = 500(106) mm4

49. Example 49.4 - solution Virtual moment m Real moment M
The vertical displacement of point B is obtained by placing a virtual unit load of 1kN at B, Fig. 9.13(b).
Using method of sections, the internal moment m is formulated as shown in Fig. 9.13(b).
Real moment M
Using the same x coordinate, M is formulated as shown in Fig. 9.13(c).

50. Example solution
Virtual work equation

51. Example 4.5 Determine the tangential rotation at point A of the steel
beam shown
in Fig. 9.14(a).
Take
E = 200GPa
I = 60(106) mm4

52. Example 4.5 - solution Virtual moment m Real moment, M
The tangential rotation of point A is obtained by placing a virtual unit couple of 1kNm at A, Fig. 9.14(b).
Using method of sections, the internal moment m is formulated as shown in Fig. 9.14(b).
Real moment, M
The internal moment is formulated as shown in Fig. 9.14(c).

53. Example solution
Virtual work equation

54. Example 4.9 Determine the tangential rotation at point C of the frame
shown in Fig. 9.18(a).
Take
E = 200GPa
I =15(106) mm4

55. Example 4.9 - solution Virtual moments m Real moments M
A unit moment is applied at C and the internal moments m are calculated, Fig. 9.18(b).
Real moments M
In a similar manner, the real moments are are calculated as shown in Fig. 9.18(c).

56. Example 4.9 - solution Virtual work equation
Using the data in Fig 9.18(b) & 9.18(c), we have: