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:

For statically determinate portal frames, first find all support reactions. Then calculate bending moments at each joint and under any point loads. The maximum moment will occur at one of these critical locations.

Answer 1

Step 1: Calculate Global Support Reactions.
Let the hinge support be $A$ and the roller be $D$. The top joints are $B$ (above $A$) and $C$ (above $D$). - Sum of horizontal forces ($\sum F_x = 0$): $H_A = 50$ kN (acting towards the left). - Take moment about A ($\sum M_A = 0$): $(50 \times 3) + (90 \times 2) - (R_D \times 4) = 0 \implies 150 + 180 = 4R_D \implies R_D = 82.5$ kN (upward). - Sum of vertical forces ($\sum F_y = 0$): $R_A + 82.5 - 90 = 0 \implies R_A = 7.5$ kN (upward).
Step 2: Calculate Bending Moments at Critical Points.
- At joint $B$ (top of column $AB$): $M_B = H_A \times 3 = 50 \times 3 = 150$ kNm. - At joint $C$ (top of column $CD$): Since $D$ is a roller and there are no horizontal loads on column $CD$, the shear in $CD$ is zero. Thus, $M_C = 0$. - Under the 90 kN load (point $E$, midspan of $BC$): Calculating from the left: $M_E = (R_A \times 2) + M_{joint\_contribution}$. Using the beam equilibrium: $M_E = (7.5 \times 2) - 150$ (considering $M_B$ as hogging) $= 15 - 150 = -135$ kNm.
Step 3: Conclusion.
The absolute values of moments are $|150|$, $|135|$, and $|0|$. The maximum is 150 kNm.