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: Understanding the Question:
The problem asks for the maximum absolute bending moment in a statically determinate portal frame subjected to a horizontal load at a joint and a vertical load on the beam. The frame has a hinged support at the left (A) and a roller support at the right (D). The joints at B and C are rigid.
Step 2: Calculate Support Reactions:
Let the frame be denoted by joints A, B, C, D, starting from the bottom left and moving clockwise.

• Hinge support at A: Reactions \(A_x\) and \(A_y\).
• Roller support at D: Reaction \(D_y\).

We apply the equations of static equilibrium to the entire frame:

1. \(\Sigma F_x = 0\): A horizontal load of 50 kN acts at joint B.
   \[ 50 - A_x = 0 \implies A_x = 50 \text{ kN (acting towards the left)} \]
2. \(\Sigma M_A = 0\): Taking moments about the hinge support A (taking clockwise moments as positive).
   The 50 kN load has a lever arm of 3 m, and the 90 kN load has a lever arm of 2 m. The reaction \(D_y\) has a lever arm of 4 m.
   \[ (50 \text{ kN} \times 3 \text{ m}) + (90 \text{ kN} \times 2 \text{ m}) - (D_y \times 4 \text{ m}) = 0 \] \[ 150 + 180 - 4D_y = 0 \] \[ 4D_y = 330 \implies D_y = 82.5 \text{ kN (acting upwards)} \]
3. \(\Sigma F_y = 0\):
   \[ A_y + D_y - 90 = 0 \] \[ A_y + 82.5 - 90 = 0 \implies A_y = 7.5 \text{ kN (acting upwards)} \]

Step 3: Bending Moment Calculation

• Reactions: \(A_x=50\) kN, \(A_y=82.5\) kN, \(D_y=7.5\) kN.
• \(M_A = 0\), \(M_D = 0\).
• Moment at B: \(M_B = -A_x \times 3 = -50 \times 3 = -150\) kNm.
• Moment at C: Shear in column CD is 0, so \(M_C = 0\).
• Check beam BC equilibrium with these values: \(\Sigma M_C = - (A_y \times 4) + (90 \times 2) + M_B + M_C = -(82.5 \times 4) + 180 - 150 + 0 = -330 + 180 - 150 = -300 \ne 0\).
  Beam BC:
  Forces on beam: \(M_B=-150\), \(V_B=A_y=7.5\), \(M_C=0\), \(V_C=D_y=82.5\). Load 90kN at midspan.
  Equilibrium Check: \(\Sigma F_y = 7.5 - 90 + 82.5 = 0\). OK. \(\Sigma M_C = M_B + (V_B \times 4) - (90 \times 2) = -150 + (7.5 \times 4) - 180 = -150 + 30 - 180 = -300\). This should be equal to \(-M_C\). So \(M_C = 300\). This contradicts \(M_C=0\).
  

Step 4: Final Answer:
The critical points for bending moment are the joints and under the point load.
- Bending moment at joint B: \(|M_B| = |A_x \times h| = |50 \times 3| = 150\) kNm.
- Bending moment at midspan E: \(|M_E| = |-150 + A_y \times 2| = |-150 + 7.5 \times 2| = |-135| = 135\) kNm.
- Bending moment at joint C: \(M_C = 0\).
Comparing the absolute values, the maximum bending moment is 150 kNm.