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: Understanding the Question and Frame Determinacy:
The portal frame has a hinged support (2 reaction components) and a roller support (1 reaction component). Total reactions \(R=3\). The equations of static equilibrium are 3 (\(\Sigma F_x=0, \Sigma F_y=0, \Sigma M=0\)). Since \(R=3\), the frame is statically determinate. We can find the reactions using these equations. Let the hinge be A (left) and the roller be D (right). The top joints are B (left) and C (right).
Step 2: Calculate Support Reactions:
We apply the equations of static equilibrium to the entire frame.

1. \(\Sigma F_x = 0\): Let \(A_x\) be the horizontal reaction at A.
   \[ 50 - A_x = 0 \implies A_x = 50 \text{ kN (acting towards the left)} \]
2. \(\Sigma M_A = 0\): Taking moments about the hinge support A (clockwise positive).
   \[ (50 \text{ kN} \times 3 \text{ m}) + (90 \text{ kN} \times 2 \text{ m}) - (D_y \times 4 \text{ m}) = 0 \] \[ 150 + 180 - 4D_y = 0 \] \[ 4D_y = 330 \implies D_y = 82.5 \text{ kN (upwards)} \]
3. \(\Sigma F_y = 0\): Let \(A_y\) be the vertical reaction at A.
   \[ A_y + D_y - 90 = 0 \] \[ A_y + 82.5 - 90 = 0 \implies A_y = 7.5 \text{ kN (upwards)} \]

Step 3: Bending Moment Calculation:
Now, we calculate the bending moments at critical points (joints and under loads).

• Support A (Hinge): \(M_A = 0\)
• Support D (Roller): \(M_D = 0\)
• Joint B (top of column AB): The bending moment at B is caused by the horizontal reaction \(A_x\). \[ M_{B} = -A_x \times (\text{height}) = -50 \text{ kN} \times 3 \text{ m} = -150 \text{ kNm} \] (The negative sign indicates a hogging moment, causing tension on the left side of the column).
• Joint C (top of column CD): Since there is no horizontal reaction at D, the shear force in column CD is zero. Therefore, the bending moment along column CD is zero. Thus, \(M_C = 0\).
• Midspan of beam BC (point E): The shear force just to the right of B is \(A_y = 7.5\) kN. The moment at E can be found by considering the beam segment BE. \[ M_E = M_B + (\text{Area of SFD from B to E}) = -150 \text{ kNm} + (7.5 \text{ kN} \times 2 \text{ m}) \] \[ M_E = -150 + 15 = -135 \text{ kNm} \]

Step 4: Determine Maximum Absolute Bending Moment:
The bending moments at the critical points are:

• \(M_A = 0\) kNm
• \(M_B = -150\) kNm
• \(M_E = -135\) kNm
• \(M_C = 0\) kNm
• \(M_D = 0\) kNm

Comparing the absolute values: \(|0|, |0|, |-150|, |-135|, |0|\). The maximum absolute value is 150 kNm.