Uniqueness theorem, Static method and Kinematic method of analysis of beams and frames.

1. Presented by
Dr. Subhash V. Patankar
Department of Civil Engineering,
Sanjivani College of Engineering, Kopargaon,
Dist:Ahmednagar, Maharashtra, India.
E-mail: patankarsubhashcivil@sanjivani.org.in,
Mobile No. : 8087482971

2. CONTENTS
1) Uniqueness Theorem
2) Static method of analysis of beams
3) Kinematical method of analysis of beams
4) Examples

3. Uniqueness Theorem
 If a bending moment distribution can be obtained which
satisfy the three conditions of mechanism, equilibrium
and yield then collapse load corresponding to such
bending moment distribution will be true collapse load.
 Mechanism condition means sufficient number of plastic
hinges developed to transform the structure into
mechanism. It is also called as kinematical condition.
 Equilibrium condition means bending moment
distribution is in equilibrium with the collapse load. It is
called as static condition.
 Yield condition means bending moment does not exceed
the plastic moment Mp.

4. Mp
Curvature
Moment Kinematic Theorem, W ≥ Wu
Static Theorem, W ≤ Wu
True, W = Wu
Concept of Uniqueness Theorem

5. Lower Bound Theorem
 If the bending moment distribution can be found
which satisfy the condition of equilibrium and yield
then the corresponding load is less than or equal to
true collapse load.
 Equilibrium and yield conditions are satisfied but
not mechanism condition.
 i.e. W ≤ Wu
 It is a static method.
 This theorem is used to analyse simple cases of beam.
 It is also called as safe theorem.

6. Upper Bound Theorem
 If the load obtained from any assumed mechanism for
a given structure greater than or equal to the true
collapse load.
 Mechanism and Equilibrium conditions are satisfied
but not satisfied yield condition.
 i.e. W ≥ Wu
 It is a kinematical method.
 This theorem is used to analyse any type of beam.
 It is also called as unsafe theorem.

7. Analysis of beam for collapse load by static
method or Lower bound theorem
 Find the static redundancy of beam,
Dsi = r- e
 Fine the no. of plastic hinges developed in beam,
n = Dsi +1.
 Find the support reactions and draw B. M. diagram.
 Compare Plastic moment Mp with elastic moment Me
 Find the Collapse load

8. Analysis of beam for collapse load by Kinematic
method or Upper bound theorem
 Find the static redundancy of beam, Dsi = r- e
 Fine the no. of plastic hinges required for collapse
mechanism, n = Dsi +1.
 Find No. of possible mechanisms found = No. of
Plastic hinges developed – Dsi, i.e. Npm = Nph -Dsi
 Draw collapse mechanism.
 Using principle of virtual work find collapse load for
each mechanism.
 Safe Collapse load will be lowest of load found in all
mechanism.

9. W=Wc
L/2 L/2
L
Mp
Wc.L/4
Plastic hinge is developed at point load, so plastic moment, Mp
is compared with elastic moment Me
Mp = Me
Mp = Wc.L/4
Wc = 4Mp/L

10. Tips for analysis of structures using Kinematic
method
 Prediction of plastic hinges
1. at fixed support
2. at Rigid Joint
3. at interior support
4. at point load/ concentrated load
 Plastic hinges are not developed at roller support,
hinge support, free support.
 If static redundancy, Dsi is equal to No. of hinges
developed, n then it makes structure determinate.

11. W=Wc
L/2 L/2
L
Ø Ø
2ØMp
W = Wc
δ
Apply the principle of virtual work,
E. Work done = I. Work done
Wc.δ = Mp.(2Ø)
Wc.Ø.L/2 = 2Mp.Ø,
Wc = 4Mp/L
tanØ
= Ø = δ/(L/2)

12. Example: Find the safe load on a S.S. beam subjected to
point load W as shown in figures. If Mp is constant
throughout the span. Use static method and Kinematic
method
Wc
Wc
2Wc
W.L =Wc
Wc
a b
LL/4 L/2 L/4
L
L
Wc

13. WL=Wc
L/2 L/2
L
Mp
Wc.L/8
Ø Ø
2Ø
Mp
WL=Wc
δ

14. Static method
Plastic hinge developed at
center of beam,
Equate, Mp = Me
Mp = (W.L).L/8 = Wc.L/8
Wc = 8.Mp/L
Kinematic method
Apply Principle of virtual
work
E.W.D. = I.W.D.
W.L. δ/2 = Mp.(2Ø)
As tan Ø = Ø = δ/(L/2)
δ = Ø.L/2 & W.L = Wc
Wc.1/2.(Ø.L/2) = 2Mp.Ø
Wc = 8Mp/L

15. ϴ
Wc
Wc
L
Mp
Wc.L
Static Method
Mp = Wc.L
Wc = Mp/L
Kinematic Method
E.W.D. = I.W.D.
Wc. δ = Mp.ϴ
Wc ϴ.L = Mpϴ
Wc = Mp/L
δ
tanϴ = ϴ = δ/L
δ = ϴ.L

16. W=Wc
a b
L
Mp
W.a.b/L
ϴ Ø
(ϴ+Ø)
Mp
W=Wc
δ

17. Static method
Plastic hinge developed
below the point load
Equate, Mp = Me
Mp = Wc.a.b/L
Wc = Mp.L/a.b
Kinematic method
Apply Principle of virtual
work
E.W.D. = I.W.D.
Wc.δ = Mp.(ϴ + Ø)
As tan ϴ =ϴ =δ/a
tan Ø = Ø = δ/b
Wc. δ = Mp (δ/a + δ/b)
Wc = Mp (a + b)/a.b
Wc = Mp.L/a.b

18. WL=Wc
δ
ϴ
L
Static method
Equate, Mp = Me
Mp = Wc.L/2
Wc = 2Mp/L
Wc.L/2
Mp
Apply Principle of virtual work, E.W.D. = I.W.D.
Wc.δ/2 = Mp.(ϴ)
As tan ϴ =ϴ =δ/L, δ = ϴ.L
Wc. ϴ.L/2 = Mp (ϴ )
Wc = 2Mp/L

19. WL=Wc
2Wc
2WcWc
4Wc
4Wc
ϴ ϴ
(ϴ+Ø)
ϴØ
2ϴ ϴ
ϴ ϴ
2ϴ
δ
δ
δ
4m 3m 1m 2m 2m

20. WL=Wc
2WcWc 4Wc
ϴ ϴ
2ϴ
δ
1. To find the Dsi
Dsi = r – e = 6 – 2 = 4
2 Fine the no. of plastic
hinges required for collapse
mechanism,
n = Dsi +1 = 4 + 1 = 5
3. Find no. of possible
mechanisms found = No. of
Plastic hinges developed –
Dsi
Npm = Nph -Dsi
= 7 – 4 = 03
Apply Principle of virtual work
E.W.D. = I.W.D.
Wc.δ/2 = Mp.ϴ + Mp.2ϴ
Mp.ϴ
As tan ϴ =ϴ =δ/(L/2)
Wc. δ/2 = 4Mp (δ/(L/2)
Wc = 16Mp/L =4Mp --(1)
4m
2m 2m

21. 2Wc
2WcWc 4Wc
(ϴ+Ø)
Øϴ
δ
Apply Principle of virtual work
E.W.D. = I.W.D.
2Wc.δ = Mp.ϴ + Mp. (ϴ + Ø) + Mp. Ø
= 2Mp. ϴ + 2Mp.Ø
As tan ϴ =ϴ =δ/3, Similarly tan Ø = δ/1, δ = Ø
2Wc. δ = 2Mp.(δ/(3) + 2Mp. (δ )
Wc = 1.33 Mp---------(1)
3m 1m

22. WL=Wc
2WcWc
4Wc
4Wc
ϴ ϴ
2ϴ
δ
4m 3m 1m 2m 2m
Apply Principle of virtual work
E.W.D. = I.W.D.
4Wc.δ = Mp.ϴ + Mp. (2ϴ) + Mp.ϴ
= 4Mp.ϴ
As tan ϴ =ϴ =δ/2
4Wc. δ = 4Mp.(δ/(2)
Wc = 0.50 Mp---------(1)
So, Safe load or true collapse load on
beam is Wc = 0.50 Mp

23. Collapse Mechanism in Frame
 As frame consist of vertical and horizontal members
(i.e. Column & beam), there are different possible
collapse mechanisms.
 For the analysis of frame by kinematical method,
mainly three mechanisms are consider.
1. Beam mechanism (three plastic hinges are
necessary, two at rigid joints and one at load point)
2. Sway mechanism ( Four hinges are developed, two
at base and two at top of the columns are
necessary)
3. Combined mechanism (Combined effect of beam
and column or sway mechanism)

24. Frame
Beam Mechanism
Sway Mechanism Combined Mechanism
2Wc
Wc
2Wc
Wc
Wc
2Wc
Wc
ϴ ϴ
ϴ
ϴ
ϴ ϴ
ϴ ϴ
Ø
Ø
δ
δ
δ δ δ δ
δ
(ϴ+Ø)
(ϴ+Ø)
ϴ ϴ

25. Frame
Beam Mechanism
Sway Mechanism Combined Mechanism
50 kN
25kN
50kN
25kN
50kN
25kN
ϴ
ϴ
ϴ ϴ ϴ
ϴ ϴ
Ø
δ
δ δ δ δ
δ
4m
3m1m
(ϴ+Ø)
ϴ ϴ

26. A rectangular frame whose legs are fixed at base and
carried two point load, one on beam and second at
column beam joint as shown in Figure. Fine the plastic
moment if C/s of frame members is same.
Steps: 1. To find the Dsi
Dsi = r – e = 6 – 3 = 3
2 Fine the No. of plastic hinges required for collapse mechanism,
n = Dsi +1 = 3 + 1 = 4
3. Find no. of possible mechanisms found
= No. of Plastic hinges developed – Dsi
Npm = Nph -Dsi = 5 -3 = 02
( i.e. at 4 joints & below loads)

27. A] Beam Mechanism
Using Virtual work principle
E. Work Done = I. work Done
50 x δ = Mp.ϴ + Mp x (ϴ + Ø) + Mp. Ø
50 x δ = Mp.(2 ϴ + 2.Ø )
50xϴ = 2Mp. (ϴ + ϴ/3)
-------- δ = 1x ϴ = 3 Ø
Mp = 50/2.67 = 18.92 kNm----------------------(1)
ϴ Ø
(ϴ+Ø)
δ
50kN

28.  B] Sway Mechanism
Using Virtual work principle
E. Work Done = I. work Done
25.δ = Mp.ϴ + Mp.ϴ + Mp.ϴ + Mp.ϴ
25.δ = Mp.(2 ϴ + 2.Ø)
25.(4ϴ) = 4Mp. (ϴ )
---------- δ = 4.ϴ
Mp = 25 kNm------------------------------(2)
ϴ
ϴ
δ δ25kN
ϴ
ϴ

29. C] Combined Mechanism
Using Virtual work principle
E.W.D. = I.W.D.
50.Δ+ 25.δ = Mp.ϴ + Mp.ϴ
+ 2Mp.ϴ + Mp.ϴ + Mp.ϴ
50. Δ + 25.δ = 6Mp.(ϴ)
50.ϴ + 25x4ϴ = 6Mp. ϴ
-------------- δ = 4x ϴ & Δ = 1.ϴ
Mp = 150/6 = 25 kNm --------------------------(3)
 Ans.: Mp = 25 kNm (Maximum of 1, 2, & 3
mechanism)
ϴ
ϴ
Δ
δ25kN
ϴ
ϴ
50kN

30. Best of luck for your future life
Thank You