Last Updated on September 15, 2026 by Maged kamel
- Location of plastic hinges and effect on the lambda Value
Location of plastic hinges and effect on the lambda Value
The location of a plastic hinge significantly affects the Load factor graph, typically creating a U-shaped or concave curve where the minimum point represents the actual collapse Load factor (λ). The following slide illustrates the procedure for determining different values of λ based on the assumed plastic hinges.
Different selection of the hinge Locations
The plastic hinges between joints A &C.
We will check an example to examine a different selection of plastic-hinge locations using the upper bound.
In our previous example (see Post 33), we selected a plastic hinge at point C, where the Load acts. If somebody else selects another point, say point d between points A and C.
The distance d is 0.20 m from joint A. Let us check the mechanism due to that selection.
The number of indeterminacy = 3 + 1- 3 = 1; we need two joints to create a mechanism. We have two hinges: one at A and the other at D.
We will use either the kinematic or the virtual work method. We have a negative Mp at joint A and a positive Mp at joint D. The Deflection at point D is Δ. We can relate the Deflection to angle β.
For the angle α at joint A, the slope tan α= α=Δ/0.20. Let the slope at joint B be β; the distance d-b is 1-0.2=0.80 m. Angle β = tan β = Δ/0.80. We have Mp acting at d with an angle = α + β.
The external work at C = the internal work; note that 32*λ*(Deflection underneath the Load).=32*λ1*(0.50/0.80)* Δ =32*λ1*(5/8)* Δ = the internal work= Mp*α+Mp*(α+ β )=Mp*(Δ/0.20)+Mp*((Δ/0.20)+(Δ/0.80)). Please refer to the following slide.
32*λ*5/8*Δ=20*λ*Δ=Mp*( 0.20)+Mp*(5*Δ/0.80)=Mp*(4Δ+5Δ)/0.80)=9*Mp*Δ/0.80.20*λ=9*Mp/0.80, Mp is given as 9.0 KN.m 20*λ=81/0.80=101.26, λ1=5.0625.
Based on the estimated lambda, we will determine the Shear values at A and B and check whether any point has a Moment greater than Mp. For the equilibrium at joint D for Part DB, the sum of moments = 0; we will equate Mp to the Moment from the right side, which is 0.80*Vb- 9.60*λ1
The Load (32*λ1) = 32*5.0625 = 162.0 KN; VB = 72 KN. Estimate the Moment at C, Mc = 72*0.50 = 36.0 KN · m. This Moment is greater than Mp = 9 kN·m. The solution is not correct.
From the left side, we will find that MD equals 9 KN, which is exactly the Value of Mp. Load P, which is equal to λ1*Pw = 5.0625*32 = 162 kN. We use a (statistical method) and draw the Moment diagram on one side and check the Value of the Moment at C, which is equal to PL/4-0.50 Mp=160*1/4-0.5*9=(162*1/4)- 0.5*9 = 36 kN · m. This is the same Value we estimated earlier.
It is not permitted to have a higher Value at C = 36.0 kN · mm, so we must verify the assumption at the point. We must ensure that no Value at the moment exceeds Mp; that is why it is an upper-bound theorem. Conduct a thorough check for another point.
Derive a general Expression for the Value of lambda based on the location of the hinged support.
Case 1-Hinge at D with a distance of a from the right support.
We can derive an Expression for the case where the plastic hinge is at D by giving a virtual rotation at B equal to 1. For β = 1, we obtain the Deflection values at C and D. The Deflection at C is L/2, while the Deflection at D is a, where a is the distance from b to the Plastic hinge at D. The span length is L.
Equate the external energy and internal energy. We have λ1*pw*L/2 = Mp(2*α + β). The α=(a/L-a) and β =1.
Readjust the term for internal energy, Wi=Mp*(2a/(L-a)+1)
Equating internal energy to external energy yields λ1 = (L-a) for a > L/2 but < L. Please refer to the following slide.
For our case, where L=1, Mp=9 kN·m, Pw=32 kN, and a=0.80 m, we find λ1=5.0625; if we consider a=0.50 m, then λ1=1.6875.
The plastic hinges are between joints C &B.
Another selection for a plastic hinge: if someone selects the plastic hinge location at d between C and B, 0.20 m from Joint B. Again, the angle at Joint A is α, and the angle at Joint B is β; the Deflection at point d is Δ.
The Deflection under the Load will be a fraction of Δ, with A-C = 0.50 m. tan α = α = Δ/0.80; the Deflection at C = (0.50/0.80)*Δ = (5/8)*Δ; tan β = Δ/0.20.
The external work at D= internal work, then 32*λ2*(deflection underneath the load)=32*λ2*((5/8)*Δ)=Mp*α+Mp*(α+β).
20*λ*Δ=Mp*(Δ/0.80+(25*Δ/4)=Mp*Δ*(6/0.80), Mp=9.0 KN.M. λ2=3.375.
Based on the estimated lambda, we will determine the Shear values at A and B and check whether any point has a Moment greater than Mp. For equilibrium at joint E for Part EB, the sum of moments = 0; we equate Mp to the Moment from the right side, which is 0.20*Vb=Mp. Since Mp = 9 kN · m, Vb = 45 kN. The VA value equals 32*λ2-Vb=32*53.375-45=108-45=63 Kips.
The Load at C is 32λ = 32 (3.375) = 108 kN; the Moment at C is 63 (0.50) = 23.50 kN·m, while Mp, the maximum Value, is 9.0 kN·m.
Selecting the hinge to the right of point C is incorrect because it results in a Moment greater than Mp at point C.
We use a (static method) and draw the Moment diagram on one side and check the Value of the Moment at C, which is equal to PL/4-0.50 Mp=160*1/4-0.5*9=(108*1/4)- 0.5*9 = 22.5 kN. This is the same Value we estimated earlier.
Case 2-Hinge at E with a distance of a from the right support.
We can derive an Expression for the case where the plastic hinge is at E by giving a virtual rotation at A equal to 1. For α=1, we obtain the Deflection values at C and D. The Deflection at C is L/2, while the Deflection at E is L-a, where a is the distance from B to the Plastic hinge at E. The span length equals L. The Value of angle β = =L-a-a)/a/a.
Equate the external energy and internal energy. We have λ2*pw*L/2 = Mp(2*α + β). The α=(1) and β =L-a/a.
Readjust the term for internal energy: Wi = M_p*(2 + (L-a)/a).
Equating internal energy to external energy will yield an Expression of λ2= Mp*(L+a)/(Pw*L/2)*(a) for a <L/2 but >0. For our case, where L=1, Mp=9 kN·m, Pw=32 kN, and a=0.20 m, we find λ2=1.6875; if we consider a=0.50 m, then λ2=1.6875.Please refer to the following slide.
The gMp gragraph isssed on the locations of the plastic hinges.
The next page shows a graph where the horizontal axis is the distance in m from support B, denoted a, and the y-axis represents λ. The first choice is a hinge on the left side, with a = 0.80 m, the distance from the right support at B.
For a plastic hinge at the left of point C, λ=(9/16) *(a+1)/(1-a); this relation gives a curved shape, and each change in a corresponds to a corresponding λ Value. for 1-a=0.20 m, a=0.80 m.
The plastic hinge location was chosen to be between C and B. Another curve is given, a is also estimated as the distance from the right support B a distance=0.20m. The Value of λ2= (9/16)*(1+a)/a=(9/16)*(1.0+0.20)/0.20)=(9/16)*(1.0+0.20)/0.20 =54/16 =3.375, which is represented by this point on the graph.
What is the correct selection?
It is the minimum λ, or the minimum Value of P; collapse occurs at λ = 1.6875. Another way to express Mp is as a function of x, where x is the distance from the left support.
The remaining distance is L-x, where L is the span distance; in the proposed collapse, we either have λ = λc = 1.6875 (correct) or λ> λc (unsafe), which is why plastic analysis is an upper-bound method. Another problem for a different Load is included in the PDF document.
Thanks a lot. I will attach the PDF document for this Post. You can view or download it.
The Content of this Post is quoted from Structural Analysis III.
For the next Post, refer to a plastic Moment for continuous beams.
Here is the Link to Chapter 8, Bending Members, in A Beginner’s Guide to the Steel Construction Manual, 14th ed.
Here is the Link to Chapter 8, Bending Members, in A Beginner’s Guide to the Steel Construction Manual, 15th ed.
Here is the Link to Chapter 8, Bending Members, in A Beginner’s Guide to the Steel Construction Manual, 16th ed.