# Plastic Analysis of Framed Structures The fully plastic moment of a section (Mp) is the maximum moment of resistance of a fully yielded cross-section. The yielded zone due to bending where infinite rotation can take place at constant plastic moment (Mp) is called a plastic hinge. In order to find the fully plastic moment of yielded section, we normally employ the force equilibrium equation by saying that the total force in tension and compression at that section are equal.

When a system of loads is applied to an elastic body, it will deform and show some resistance to deformation and such body is called a structure. On the other hand, if no resistance is set up against deformation, such as a body is said to have formed a mechanism. One of the fundamental conditions for plastic analysis is that the collapse load is reached when a mechanism is formed, in which the number of plastic hinges developed in the structure is sufficient to form a mechanism. The load factor (λ) at rigid plastic collapse is the lowest multiple of the design load which will cause the whole structure or part of it to form a mechanism. It is important to realise that the number of independent mechanisms in a structure is related to the number of plastic hinge locations and the degree of redundancy of the system.

In beams, identification of critical spans in terms of Mp or load factor can be obtained using the static or kinematic method by considering simple beam mechanisms. But in framed structures, other types of mechanisms such as joint, sway, and gable mechanisms are also considered. Each of these mechanisms can occur independently, but it also possible for a critical collapse mechanism to develop by combination of the independent mechanisms.

Solved Example
For the frame loaded as shown below, find the critical Mp value, and draw the collapse moment diagram. The loads are factored.

Solution

Degree of static indeterminacy
RD = (3m + r) – 3n
m (number of members) = 4
r (number of support reactions) = 5
n (number of nodes) = 5
R= (3 × 4 + 5) – 3(5) = 2
Therefore the frame is indeterminate to the second order.

A little consideration will show that the possible places where we could have plastic hinges in the structure are at node B, section C, section D (just to the left), section D (just to the right), section D (below) and section F. Nodes A, E, and F are all natural hinges. Therefore, total number of possible hinges = 6

Hence, number of independent collapse mechanisms = 6 – 2 = 4

These independent mechanisms are listed below;
– 2 Beam mechanisms (span B – D and span D – G)
– 1 sway mechanism of the entire frame due to the horizontal load
– 1 joint mechanism

Analysis of the independent mechanisms

Mechanism 1

δC = 3 × θ = 3θ
Internal work done due to rotations at internal hinges at B, D, and C
Total internal work done (Wi) = Mpθ + Mpθ + 2Mpθ = 4Mpθ

External work done (We) = 115 × 3θ = 345θ

Let external work done = Internal work done
4Mpθ = 345θ
On solving, Mp = 345/4 = 86.25 kNm

Mechanism 2

δF = 3 × θ = 3θ
Internal work done due to rotations at internal hinges at B, and F. No work is done at G because it is a natural hinge.
Total internal work done (Wi) = Mpθ + 2Mpθ = 3Mpθ

External work done (We) = 74 × 3θ = 222θ

Let external work done = Internal work done
3Mpθ = 222θ
On solving, Mp = 222/3 = 74 kNm

Mechanism 3 (Sway)

δ1 = 4 × θ = 4θ
Internal work done due to rotations at internal hinges at B, and D. No work is done at A, E, and G because they are all natural hinges.

Total internal work done (Wi) = Mpθ + Mpθ = 2Mpθ

External work done (We) = 25 × 4θ = 100θ

Let external work done = Internal work done
2Mpθ = 100θ
On solving, Mp = 100/2 = 50 kNm

Mechanism 4 (Joint Rotation)

Internal work done = Mpθ + Mpθ + Mpθ = 3Mpθ
External work done = 0 (no external work done)
Internal work done = Mpθ + Mpθ + Mpθ = 3Mpθ
External work done = 0 (no external work done)
Let us now consider the possible mechanism combination by preparing the table below

The critical Mp is therefore = 111.167 kNm (This was achieved after other combinations were considered but this was found to be the highest).

Note that by virtue of the combination, internal hinges at B, D2, and D3 were eliminated. So the critical collapse mechanism is as given below;

Analysis of Support Reactions
Note that the implication of our analysis above is that points C, F and D1 are plastic hinges, with full plastic moment of 111.167 kNm. While C and F are sagging, D1 is hogging. No other bending moment in the structure should exceed this value. Once a higher moment is discovered, it means that the critical collapse moment was not appropriately obtained.
∑MFR = 0 (anti clockwise positive)
3VG = 111.167
VG = 37.056 kN
∑MCL = 0 (clockwise positive)
3VA – 4H= 111.167 ———– (1)

∑MDL = 0 (clockwise positive)
6VA – 4H– (115 × 3) = -111.67
6VA – 4H= 233.833 ———– (2)
Solving (1) and (2) simultaneously;
VA = 40.721 kN
HA = 2.623 kN
∑MA = 0 (clockwise positive)
12VG + 6V– (74 × 9) – (115 × 3) – (25 × 4) = 0
6V= 666.328
VE = 111.055 kN
∑FX = 0
H= 25 + 2.623 = 27.623 kN
Bending Moment
You can verify that having obtained the support reactions, the bending moment diagram can easily be plotted without strenuous calculations.

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