Winner of the Quiz

Marcelo DaSilva

Burns & McDonnell

Portal Frame Bridge

27 - 02 Apr 2025

Question 1: What is the primary structural advantage of using an integral portal frame with inclined legs for a pedestrian bridge compared to a conventional simply supported girder system?

  1. It eliminates the need for bearings and expansion joints at the piers, reducing maintenance requirements
  2. It improves moment distribution, reducing the required girder depth and enhancing structural efficiency
  3. It transfers horizontal forces from thermal expansion directly to the foundations, preventing deck movement
  4. It allows for independent movement of spans, reducing temperature-induced stresses
Explanation

Explanation: The primary structural advantage of an integral portal frame with inclined legs is its ability to redistribute bending moments, reducing peak stresses in the girder and allowing for a shallower, more efficient design. The rigid connection between the legs and girder enables frame action, which improves stiffness and load distribution. While eliminating bearings and transferring horizontal forces to the pier foundations are benefits, they are secondary effects rather than the main structural advantage. Unlike simply supported systems, where spans move independently, an integral frame introduces restraint, increasing internal stresses due to thermal effects rather than reducing them. Therefore, the improved moment distribution and structural efficiency make option B the correct answer.

Question 2: A pedestrian bridge consists of an integral portal frame with inclined legs that are 4m tall vertically. The distance between the tops of the legs is 8m, and the distance between the bases of the legs is 16m. The bridge has three spans of 4m - 8m - 4m (total length 16m), with the frame legs located at the two intermediate supports. The bridge carries a uniform dead load of 4 kN/m along its entire length, with the load distributed to the supports based on tributary lengths. Additionally, a concentrated live load of 120 kN is positioned directly above one of the frame legs. Using a load factor of 1.25 for dead load and 1.5 for live load, calculate the maximum compressive force in the leg under the point load. Ignore frame action effects and consider only the direct axial force transfer in the inclined leg.

  1. 297 kN
  2. 400 kN
  3. 424 kN
  4. 466 kN
Explanation

Solution:

  1. Calculate the factored uniform dead load:
    • Factored DL = 1.25 × 4 kN/m = 5 kN/m
  2. Calculate the dead load reactions at the interior support:
    • Tributary length for interior support = ½(4m) + ½(8m) = 2m + 4m = 6m
    • Dead load reaction = 5 kN/m × 6m = 30 kN
  3. Calculate the factored concentrated live load:
    • Factored LL = 1.5 × 120 kN = 180 kN
  4. Calculate the total vertical reaction at the loaded leg:
    • Total vertical reaction = 30 kN (DL) + 180 kN (LL) = 210 kN
  5. Calculate the axial compressive force in the loaded leg:
    • Angle with vertical = 45°
    • Axial force = Vertical reaction ÷ cos(angle with vertical)
    • Axial force = 210 kN ÷ cos(45°)
    • Axial force = 210 kN ÷ 0.7071 = 297 kN