GATE 2026 CE (CE2) – Question 49
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
$$M_p = C_1 PL + C_2 wL^2$$
where $C_1$ and $C_2$ are constants.
[Figure: A portal frame with two columns and a beam, each of length L. The left column is fixed at its base and the right column stands on a roller support. The load w per unit length acts downward on the whole beam and a horizontal load P acts to the right at the top of the left column.]
The value of $(C_1/C_2)$ is ______ (*in integer*).
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Correct answer: 4
Explanation
In the combined mechanism the left column turns through $\theta$ about its fixed base, which moves the top of the frame sideways by $L\theta$. The plastic hinges form at the fixed base (rotation $\theta$), at the middle of the beam (rotation $2\theta$) and at the top of the right column (rotation $\theta$), and the hinge at the left joint cancels out. The internal work is $M_p(\theta + 2\theta + \theta) = 4M_p\theta$. The external work is $P(L\theta)$ for the lateral load plus $wL \times \frac{L\theta}{4} = \frac{wL^2\theta}{4}$ for the distributed load, since the beam's middle drops by $\frac{L\theta}{2}$ and the average drop is half of that. Equating gives $4M_p = PL + \frac{wL^2}{4}$, so $M_p = \frac{1}{4}PL + \frac{1}{16}wL^2$. Then $C_1 = \frac{1}{4}$, $C_2 = \frac{1}{16}$ and $\frac{C_1}{C_2} = 4$.