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: Write the virtual work equation for the beam mechanism alone.
For the beam mechanism (hinges at both beam-column joints and at midspan), using rotation $\theta$ at the end hinges and $2\theta$ at midspan, deflection at midspan $= L\theta/2$:
$$ M_p\theta + M_p(2\theta) + M_p\theta = wL\left(\frac{L\theta}{4}\right) $$ $$ 4M_p\theta = \frac{wL^2\theta}{4} \quad \text{...(i)} $$
Step 2: Write the virtual work equation for the sway mechanism alone.
For the sway mechanism (hinges at the fixed base and at the left joint only, since the roller end needs none and the beam does not rotate), with rotation $\theta$ at each and sway $\Delta = L\theta$:
$$ M_p\theta + M_p\theta = P(L\theta) $$ $$ 2M_p\theta = PL\theta \quad \text{...(ii)} $$
Step 3: Add equations (i) and (ii) directly.
Adding the left sides and the right sides of (i) and (ii):
$$ (4M_p\theta) + (2M_p\theta) = \frac{wL^2\theta}{4} + PL\theta $$ $$ 6M_p\theta = PL\theta + \frac{wL^2\theta}{4} \quad \text{...(iii)} $$
Step 4: Correct for the shared hinge counted twice.
Equation (iii) treats the left beam-column joint as if it needs a SEPARATE hinge in both the beam mechanism and the sway mechanism, that is, it double counts the term $M_p\theta$ at that one joint.
But in the real combined mechanism, the beam-side rotation and the column-side rotation at that joint are in opposite senses and cancel each other, so that joint does not actually need to yield at all once the two mechanisms are combined into one.
So we must subtract the double-counted contribution, $2M_p\theta$ total (one $M_p\theta$ from each of (i) and (ii)), from the internal work side:
$$ 6M_p\theta - 2M_p\theta = PL\theta + \frac{wL^2\theta}{4} $$ $$ 4M_p\theta = PL\theta + \frac{wL^2\theta}{4} $$
Step 5: Solve for $M_p$.
Dividing by $\theta$:
$$ 4M_p = PL + \frac{wL^2}{4} $$ $$ M_p = \frac{PL}{4} + \frac{wL^2}{16} $$ Matching against $M_p = C_1PL + C_2wL^2$: $C_1 = 1/4$ and $C_2 = 1/16$.

Step 6: Compute the ratio.
$$ \frac{C_1}{C_2} = \frac{1/4}{1/16} = 4 $$ \[ \boxed{C_1/C_2 = 4} \]