Question

A rigid-jointed portal frame, shown in the figure, consists of beam and columns of equal length \(L\). The frame has a fixed support at one end and a roller support at the other end. The frame is subjected to a uniformly distributed load \(w\) and a lateral load \(P\) as shown in the figure. The plastic moment capacity of the beam and column sections is \(M_p\). Consider a combined beam-column mechanism for plastic collapse. By applying the virtual work equation corresponding to the combined plastic collapse mechanism, \(M_p\) is expressed as
\[ M_p = C_1 P L + C_2 w L^2 \]
where \(C_1\) and \(C_2\) are constants.

The value of \((C_1/C_2)\) is ______ (in integer).

Hint:

In the combined mechanism, the hinge at the shared beam-column joint cancels between the sway and beam mechanisms, leaving hinges at the fixed base, the far joint, and midspan.

Answer 1

Correct Answer: 4

Step 1: List the possible plastic hinge locations.
The frame has three members of equal length \(L\): the left column, the beam, and the right column, all rigidly jointed and all with the same plastic moment capacity \(M_p\).
The left column base is a FIXED support, so it can develop a moment up to \(M_p\) before it hinges.
The right column base is a ROLLER support, which by definition carries no moment at all (it behaves like a free pin that can also slide), so it can never be a hinge location, it is already "moment-free".
The two rigid joints where the columns meet the beam can each develop a moment up to \(M_p\) before hinging, and the beam midspan, where the UDL \(w\) causes its own peak moment, is also a potential hinge location.
So the candidate hinge positions are: the fixed base, the left beam-column joint, the midspan of the beam, and the right beam-column joint.

Step 2: Set up the independent beam mechanism.
In a pure beam mechanism, only the beam deforms; the columns stay straight (no sway).
Since both ends of the beam are effectively "built-in" through the rigid joints, the beam behaves like a fixed-fixed beam under UDL \(w\), and it collapses with 3 hinges: one at each end joint, and one at midspan (where the sagging moment peaks).
Give the beam a virtual rotation \(\theta\) at each end hinge; by symmetry the midspan hinge then rotates by \(2\theta\), and the midspan deflects downward by \(\delta=\theta\,(L/2)\).
Internal work done by the 3 hinges:
\[ W_{int}=M_p\theta+M_p(2\theta)+M_p\theta=4M_p\theta \] External work done by the UDL (acting on a triangular deflection shape, base \(L\), peak \(\delta\), so the average deflection is \(\delta/2\)):
\[ W_{ext}=w\,L\,\left(\frac{\delta}{2}\right)=wL\cdot\frac{\theta L}{4}=\frac{wL^2\theta}{4} \] Equating: \(4M_p\theta=\dfrac{wL^2\theta}{4}\), which alone would give \(M_p=\dfrac{wL^2}{16}\); this is the separate beam mechanism result.

Step 3: Set up the independent sway mechanism.
In a pure sway mechanism, the whole frame leans sideways under \(P\) while the beam simply translates horizontally without rotating (to first order, both beam ends move sideways by the same amount, so the beam stays parallel to itself).
This requires the left column to rotate as a rigid bar about its fixed base, needing a hinge at that fixed base (rotation \(\theta\)) and a matching hinge at the left beam-column joint (also rotation \(\theta\), since the column rotates but the beam does not).
On the right side, the base is a roller, which is already free to rotate and slide, so it absorbs the sway without needing any plastic hinge there, and since the beam does not rotate either, there is no relative rotation, and hence no hinge, at the right beam-column joint.
So the sway mechanism uses only 2 hinges: the fixed base and the left joint, each rotating by \(\theta\), giving a sideways deflection at beam level of \(\Delta=\theta L\).
Internal work:
\[ W_{int}=M_p\theta+M_p\theta=2M_p\theta \] External work done by the lateral load \(P\):
\[ W_{ext}=P\Delta=PL\theta \] Equating: \(2M_p\theta=PL\theta\), which alone would give \(M_p=\dfrac{PL}{2}\); this is the separate sway mechanism result.

Step 4: Combine the two mechanisms.
The "combined beam-column mechanism" superposes the sway mechanism and the beam mechanism using the SAME virtual rotation \(\theta\) in both.
At the left beam-column joint, the sway mechanism needs a hinge rotating by \(\theta\) in one direction, and the beam mechanism needs a hinge rotating by \(\theta\) in the opposite direction at that same joint (the column tries to rotate the joint one way while the sagging beam tries to rotate it the other way); these two exactly cancel, so the combined mechanism needs NO hinge at the left joint at all.
What remains, after this cancellation, are the hinges at the fixed base (rotation \(\theta\), from sway), at the beam midspan (rotation \(2\theta\), from the beam action), and at the right beam-column joint (rotation \(\theta\), from the beam action, since this joint was not used by the sway mechanism).
Internal work for the combined mechanism:
\[ W_{int}=M_p\theta+M_p(2\theta)+M_p\theta=4M_p\theta \] External work is simply the sum of the external work from each contributing load, since both loads move through their respective displacements in this combined mechanism:
\[ W_{ext}=P(L\theta)+wL\left(\frac{\theta L}{4}\right)=PL\theta+\frac{wL^2\theta}{4} \]
Step 5: Equate internal and external work.
\[ 4M_p\theta=PL\theta+\frac{wL^2\theta}{4} \] Dividing through by \(\theta\) and by 4:
\[ M_p=\frac{PL}{4}+\frac{wL^2}{16} \] Comparing with \(M_p=C_1PL+C_2wL^2\):
\[ C_1=\frac{1}{4},\qquad C_2=\frac{1}{16} \]
Step 6: Find the ratio.
\[ \frac{C_1}{C_2}=\frac{1/4}{1/16}=\frac{16}{4}=4 \]
Final Answer:
The value of \(C_1/C_2\) is 4. \[ \boxed{C_1/C_2=4} \]