The fully plastic moment of a section (M

_{p}) 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 (M

_{p}) is called a

**. 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.**

*plastic hinge*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 (Î») 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 M

_{p}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 M

_{p}value, and draw the collapse moment diagram. The loads are factored.

**Solution**

Degree of static indeterminacy

R

_{D}= (3m + r) - 3n

m (number of members) = 4

r (number of support reactions) = 5

n (number of nodes) = 5

R

_{D }= (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**

_{C}= 3 × Î¸ = 3Î¸

Internal work done due to rotations at internal hinges at B, D, and C

Total internal work done (W

*i*) = MpÎ¸ + MpÎ¸ + 2MpÎ¸ = 4MpÎ¸

External work done (W

*e*) = 115 × 3Î¸ = 345Î¸

Let external work done = Internal work done

4MpÎ¸ = 345Î¸

On solving, M

_{p}= 345/4 = 86.25 kNm

**Mechanism 2**

Î´

_{F}= 3 × Î¸ = 3Î¸

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 (W

*i*) = MpÎ¸ + 2MpÎ¸ = 3MpÎ¸

External work done (W

*e*) = 74 × 3Î¸ = 222Î¸

Let external work done = Internal work done

3MpÎ¸ = 222Î¸

On solving, M

_{p}= 222/3 = 74 kNm

**Mechanism 3 (Sway)**

Î´

_{1}= 4 × Î¸ = 4Î¸

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 (W

*i*) = M

_{p}Î¸ + M

_{p}Î¸ = 2MpÎ¸

External work done (W

*e*) = 25 × 4Î¸ = 100Î¸

Let external work done = Internal work done

2M

_{p}Î¸ = 100Î¸

On solving, M

_{p}= 100/2 = 50 kNm

**Mechanism 4 (Joint Rotation)**

Internal work done = M

_{p}Î¸ + M_{p}Î¸ + M_{p}Î¸ = 3M_{p}Î¸
External work done = 0 (no external work done)

Internal work done = M

_{p}Î¸ + M_{p}Î¸ + M_{p}Î¸ = 3M_{p}Î¸
External work done = 0 (no external work done)

Let us now consider the possible mechanism combination by preparing the table below

_{p}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, D

_{2}, and D

_{3}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 D

_{1}are plastic hinges, with full plastic moment of 111.167 kNm. While C and F are sagging, D_{1}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.**∑M**

_{F}^{R}= 0 (anti clockwise positive)
3V

_{G}= 111.167
V

_{G}= 37.056 kN**∑M**

_{C}^{L}= 0 (clockwise positive)
3V

_{A}- 4H_{A }= 111.167 ----------- (1)

**∑M**

_{D}^{L}= 0 (clockwise positive)
6V

_{A}- 4H_{A }- (115 × 3) = -111.67
6V

_{A}- 4H_{A }= 233.833 ----------- (2)
Solving (1) and (2) simultaneously;

V

_{A}= 40.721 kN
H

_{A}= 2.623 kN**∑M**

_{A}= 0 (clockwise positive)
12V

_{G}+ 6V_{E }- (74 × 9) - (115 × 3) - (25 × 4) = 0
6V

_{E }= 666.328
V

_{E}= 111.055 kN**∑F**

_{X}= 0
H

_{E }= 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.

Thank you for visiting Structville today, and God bless you.

wonderful post. its been a while. i want to ask, wat is d rational behind making theta positive or negative? is there a rule governing dat decision? again thank u very much fr enlightening me with such post. God bless u profusely.

ReplyDeleteHi Ovie,

DeleteThe rationale behind that is a matter of sign convention. You can likewise vary yours. For instance, you can call outward rotation positive, and inward rotation negative. You will arrive at the same answer provided you are consistent with the convention you adopted.

wonderful post. its been a while. i want to ask, wat is d rational behind making theta positive or negative? is there a rule governing dat decision? again thank u very much fr enlightening me with such post. God bless u profusely.

ReplyDeleteany reference book i would deeply appreciate

ReplyDelete