Step 1: Frame Description. A portal frame is described, featuring two fixed columns of equal height (3 m) and a rigid beam of 3 m length. All structural members possess identical moment of inertia. A horizontal load of 80 kN is applied to the top of the left column.
Step 2: Load Distribution. Due to the frame's symmetry and uniform stiffness, the 80 kN horizontal load is distributed equally between the two columns. Each column therefore sustains a shear force of: \[\text{Shear per column} = \frac{80}{2} = 40 \, \text{kN}.\]
Step 3: Left Column Moment Calculation. The maximum bending moment at the fixed base of the left column is calculated as: \[M = V \times h\] With \(V = 40 \, \text{kN}\) (shear in the left column) and \(h = 3 \, \text{m}\), the moment is: \[M = 40 \times 3 = 120 \, \text{kN.m}.\]
Step 4: Option Evaluation. - (A) 120 kN.m: This value matches the computed moment. - (B) 240 kN.m: This would only occur if the entire 80 kN load was applied solely to the left column, disregarding the load-sharing effect of the frame. - (C) 160 kN.m: This is an incorrect value, not satisfying equilibrium principles. - (D) Zero: This is not possible, as the applied load inherently generates a moment.
Step 5: Final Determination. The maximum moment experienced by the left column is \(120 \, \text{kN.m}\).
