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Plastic Analysis (Collapse Mechanism)

When all elements of the beam's cross-section have yielded, any further increase in load will cause collapse. At this stage, the beam has reached the yield plateau of the stress→strain curve, and unrestricted plastic flow will occur. A plastic hinge forms, and together with the actual hinges at the beam ends, they constitute an unstable mechanism.

Structural analysis based on collapse mechanism is called plastic analysis. This method applies the principle of virtual work, where an assumed mechanism is subjected to virtual displacements consistent with the mechanism's motion, and the external work is equated to the internal work.

Before solving plastic analysis problems, recall some principles of the virtual work method:

External Work:

1. Work done by a force $P$ that varies linearly with displacement $y$ from zero to its final value:
   $$ W = \tfrac{1}{2} P y \quad (1) $$
2. Work done by a constant force $P$ while its point of application undergoes displacement $y$:
   $$ W = P y \quad (2) $$

Internal Work:

Virtual work equation at collapse:

Real hinges and plastic hinges in virtual work:

• Real hinge: an actual support or connection that can rotate freely and cannot resist moment. It helps define the collapse mechanism, but it does not dissipate plastic energy because its resisting moment is zero.
• Plastic hinge: a yielded section in the member where the bending moment has reached the plastic moment capacity M_p. It still resists moment while rotating, so it contributes internal work.
• External work: include applied loads and moments moving through their compatible virtual displacements or rotations, such as PΔ for a force and Mθ for an applied couple.
• Internal work: include only the plastic hinges that form in the member, using M_pθ for each plastic hinge rotation. Do not include real support hinges as internal work terms.

Why the lesser collapse load is chosen:

For the same structure, several possible collapse mechanisms may be drawn. Each mechanism gives a trial collapse load from the virtual-work equation. The governing load is the least of these values because the structure will collapse as soon as any valid mechanism can form. A larger computed load belongs to a mechanism that would require the structure to remain stable past an earlier collapse mechanism, so it is not the controlling collapse load.

Plastic moment reminders:

Major Cases in Plastic Analysis

Recall:
When all the elements of the cross-section of the beam has yielded, any further increase in the load will cause collapse, since all elements of the cross-section reached the yield plateau of the stress-strain curve and unrestricted plastic flow will occur.

Problem:

A propped beam having a span of 6m carries two concentrated loads of 0.6Pu and Pu at its middle thirds. The beam has a plastic section modulus of 1852x10³mm³. Fy=248MPa. Determine the value of Pu so that the beam is about to collapse.
A. 324kN
B. 353kN
C. 422kN
D. 521kN

$$ \frac{y}{4} = \frac{y_2}{2} $$ $$ y_2 = \frac{2}{4}y $$ $$ y_2 = 0.5y $$ $$ 2\theta = 4\alpha $$ $$ \alpha = \frac{2}{4}\theta $$ $$ \alpha = 0.5\theta $$ $$ \tan\theta = \frac{y}{2} $$ $$ 2\theta = y $$ $W_{ext} = W_{int}$
$ 0.6P_u(y) + P_u(0.5y) = M_p(\theta + \theta + \theta_a) $
$ 0.6P_u(2\theta) + P_u(0.5(2\theta)) = M_p(\theta + \theta + 0.5\theta) $
$ 1.2P_u\theta + P_u\theta = M_p(2.5\theta) $
$ M_p = F_y Z_x = 248(1852 \times 10^3) $
$ 2.2P_u = 248(1852 \times 10^3)(2.5) $
$ P_u = 521.9 \;\text{kN} $ $$\frac {y}{4}=\frac {y_2}{2}$$ $$y_2=0.5y$$ $$2\alpha=4\theta$$ $$\alpha=2\theta$$ Therefore, $$y=4\theta$$ $W_{ext} = W_{int}$
$0.6P_u(y_2) + P_u(y) = M_p(2\theta + \alpha)$
$0.6P_u(0.5y) + P_u(y) = 248(1852 \times 10^3)(2\theta + \alpha)$
$0.6P_u(0.5(4\theta)) + P_u(2\theta) = 248(1852 \times 10^3)(2\theta + 2\theta)$
$P_u=353.3kN$

Why choose the lesser load? The two trial mechanisms are possible collapse patterns for the same beam. The beam will collapse as soon as the first valid mechanism can form, so the smaller computed load controls. The larger value would require the beam to remain stable after the lower-load mechanism has already formed, so it is not governing.

Real hinge versus plastic hinge: A real hinge is an actual support or connection that is already free to rotate and has no moment resistance. It helps define the shape of the mechanism, but it is not counted as internal work. A plastic hinge is a yielded section of the beam where the bending moment has reached M_p; it rotates while still resisting moment, so each plastic hinge contributes internal work equal to M_pθ.

External work and internal work for this problem: The applied loads 0.6P_u and P_u are included in external work because they move through vertical displacements during the mechanism. The plastic hinges formed in the beam are included in internal work because they rotate while resisting M_p. The actual hinge support is not added to internal work because its resisting moment is zero.

$$ P_u = \min(353.3, \; 521.9) = 353.3 \;\text{kN} $$