Question

A portal frame has a span of 4 m (2 m + 2 m) and a height of 3 m. The left base is hinged and the right base is roller supported. A horizontal load of 50 kN acts at the top left joint. A vertical concentrated load of 90 kN acts at the midspan of the top beam. Determine the absolute value of the maximum bending moment in the frame.

Hint:

In statically determinate frames, the maximum bending moments often occur at the rigid joints due to the lever action of the columns or beams. Always check the joints and points under concentrated loads.

Answer 1

Step 1: Determine Reaction Forces.
$\sum F_x = 0 \implies H_{hinge} = 50$ kN. Summing moments at the hinge: $50(3) + 90(2) = R_{roller}(4) \implies 150 + 180 = 4R_{roller} \implies R_{roller} = 82.5$ kN. Vertical hinge reaction: $R_{hinge} = 90 - 82.5 = 7.5$ kN.
Step 2: Bending Moment Diagram (BMD) Analysis.
- At the top-left joint ($M_B$): $M_B = 50 \times 3 = 150$ kNm (Column side). - Along the beam at midspan: $M(x) = R_{hinge}x - M_B$. At $x=2$: $M = 7.5(2) - 150 = -135$ kNm. - At the top-right joint ($M_C$): Sum of forces on column $CD$ is zero horizontally, so $M_C = 0$.
Step 3: Max Moment.
The maximum absolute value is $\max(|150|, |135|, |0|) = 150$ kNm.