Last Updated on September 15, 2026 by Maged kamel
Collapse Load for an indeterminate Beam.
What is the difference between the equilibrium or static method and the kinematics or mechanism method? From Prof. Capprani’s Structural Analysis III.
The Collapse Load for an indeterminate propped cantilever by the static method.
We want to estimate the collapse Load for an indeterminate Beam. which has a fixed support at point A and a roller support at point B, and is statically indeterminate of degree 1. A Load P is applied at point C, located L/2 from the left support.
Estimate the collapse Load P_collapse and the plastic Moment Mp, and determine their relationship using the static method.
For the static method, the Value of M-ve at the left support should be known; draw the bending Moment diagram, and subtract the sagging Moment of PL/4 to get M+ve. Value.
Estimate the collapse Load P-collapse and the plastic Moment Mp, and determine their relationship using the static method.
For the static method, the Value of M -ve at the left support should be known; the bending Moment is to be drawn, and the sagging Moment of PL/4 is to be subtracted from the Moment Value, as shown, to get the positive M+ve Value.
The Fixed-beam at A and with roller support at b is considered as a combination of two beams or Superposition of two cases, the first case where a simple Beam acted upon by Load P and the second case, due to fixation at A, Moment M will act at the left support A.
Then, we add these two cases to get the final Moment diagram. At support A, we have Ma = (3/16) P*L.
And zero Value at B. For the Load case, we have Moment Value =P*L/4, due to the Reaction at A=P/2, causing moment=(P/2*L/4)=P*L/4. The final Moment at c=P*L/4-(0.50)*(3/16)*P*L=(5/32)*P*L.
The proposed positions of plastic hinges.
The Load P is applied gradually until the collapse Load P_c is reached, which causes M-plasticity at the fixed end.
Let us check the collapse mechanism. We have A as a fixed support with 3 reactions, and a roller support with 1 reaction; we will have (3+1)=4.
We have one degree of indeterminacy. Add 1 to this one-degree indeterminacy. Then the number of hinges required to collapse it is 2.
The proposed location for the first hinge is at the Load location that creates a maximum Moment. The first hinge will be placed at point C, and the second at the fixed support, for a total of two hinges.
The direction of Mp at the fixed end is anti-clockwise, while the second Mp is at the middle of the span, at point C. Here is the sketch of the two moments.
The next sketch combines the two diagrams. At joint A, we have Mp, and Mp = 0 at the other joint; for the +ve Moment, we have 0 at point A, PL/4 at point C, and 0 at point B.
Check the Moment at C: (1/2 Mp) + Mp is positive at B; we add (1/2 Mp) + Mp = (P*L/4). While applying P, Mp is negative at support A; as the Load increases further, the positive Moment reaches Mp at C. The total positive Moment = P collapse*L/4.
Then (0.50Mp+Mp)+ Pp*L/4, 2 will go with 2, Mp=(1/6)*Pp*L. The Plastic Load, or collapse Load, Pp = 6 Mp/L.
The Collapse Load for an indeterminate Beam by the kinematic method.
To create a collapse, place two hinges: the first at point A and the second at point B. Due to the acting P-collapse Load, Deflection Δ will occur, and failure will occur.
The mechanism of failure is due to the two hinges, at the collapse, there will be angle θ, at the collapse, there will be angle θ at both points A, B. θ= tan(θ), since the angle is small. Accordingly, θ= tan θ=(Δ)/(L/2); there is symmetry at point C.
The external work PpΔ= internal work=Mpθ+Mp(2θ). PpΔ=3Mpθ.With θ = 2Δ/L, Pp = 6Mp/L, which matches the result from the static method.
To determine the reactions due to the acting Load P, two reactions of P/2 will exist at the two supports, but due to Mp, another reaction will be created: an upward reaction at A, Ra = M*P/L, to resist the end Moment at A.
Rb = M*P/L acting downwards to resist the anti-clockwise Moment Mp. Taking the Moment at joint B, Mp acts anti-clockwise; Ra = P/2 + M*P/L.
Using the incremental method, incremental loads.
To explain the incremental method, we assume the indeterminate Beam is 1 m long. We assume that the Yielding Moment, My, equals 7.50 kN · m, while the plastic Moment equals 9 kN · m. The source is quoted from Dr. C. Caprani, “Structural analysis III.”
We will start to use a workload that creates Moments, but with fewer values than My and Mp.
Due to the fixation at A, the slope of the field end is zero, so we will find the slope due to the negative Moment Ma and the positive Moment due to the Load using any approved method of structural analysis; we see the slope as the Area of the Moment values. Finally, we get Ma = 3/16)*PL while the positive Moment at point C equals 5*PL/32. Please refer to the following slide image for more details.
Consider the working Load to be 32 kN and estimate the values of the fixed-end Moment at A and the positive Moment at C. Compare these values with the Value of the Yield Moment. We find that the fixed Moment is larger than the positive Moment at C.
To find the working Load that causes the Yielding Moment at A, equate that Moment to 3Py*L/16; Py is found to be 40 kN. The positive Moment is 6.25 KN · m, which is less than My.
Add More Load until you have P = 48 kN; this Load creates the first plastic Hinge at A, since Ma = 9 kN·m = Mp. Still, Mc is less than Mp and equals 7.50 kN·m.
Increase the Load to 54 kN·m. Since the first hinge forms at a lower Load (48 kN·m), there will be no additional Moment at the first plastic hinge. The final moments are shown in the image below. We have two plastic hinges at Pn = 54 kN. We can use the equation we derived, Pp = 6 Mp/L, to verify our answer.
The following slide image shows how we can use the static method to find the Value and place of the second hinge and determine the plastic Load.
Expressions for the plastic theory.
The first Expression is the Load factor for a possible Load mechanism. λi lambda i) is the collapse Load divided by the working Load to reach Mp. The lambda is 54/32 = 1.6875.
To calculate the FOS, use the formula: First yield Load / Working Load. The Value equals 40/32=1.25.
The Factor of safety = P-first yield / P-working = 40/32 = 5/4 = 1.25.
You can view and download the PDF file for this Post from the following document.
This is valuable data regarding the structural analysis – III. colincaprani.com
The following Post discusses upper-bound definitions. The Post introduces the theory of upper bounds.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14th ed.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15th ed.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16th ed.