Step 1: Understanding the frame.
The given frame is a portal frame with two fixed columns and a rigid beam at the top. A horizontal load of 80 kN is applied at the top of the left column. Both columns have the same height (3 m), and the beam length is also 3 m. All members have the same moment of inertia.
Step 2: Distribution of horizontal load.
Since the frame is rigid and symmetrical in stiffness, the horizontal load will be shared equally by the two columns. Hence, each column will resist half of the horizontal shear:
\[
\text{Shear per column} = \frac{80}{2} = 40 \, \text{kN}.
\]
Step 3: Moment in the left column.
The maximum moment at the fixed base of the left column is given by:
\[
M = V \times h
\]
where \(V = 40 \, \text{kN}\) (shear in left column) and \(h = 3 \, \text{m}\).
\[
M = 40 \times 3 = 120 \, \text{kN.m}.
\]
Step 4: Analysis of options.
- (A) 120 kN.m: Correct, matches our calculation.
- (B) 240 kN.m: Would be true if the entire 80 kN load acted only on the left column, but due to frame action, it distributes.
- (C) 160 kN.m: Incorrect, does not match equilibrium conditions.
- (D) Zero: Not possible as the load generates moment.
Step 5: Conclusion.
The maximum moment in the left column is \(120 \, \text{kN.m}\).
