1. The tied arch bridge eliminates the need for vertical hangers
2. The tied arch bridge can be constructed with a smaller rise-to-span ratio
3. The tied arch bridge transfers horizontal thrust forces to the foundations
4. The tied arch bridge contains the horizontal thrust forces within its structure, reducing foundation requirements

Explanation

Explanation: In a tied arch bridge, the deck acts as a tension tie that connects the ends of the arch, effectively containing the horizontal thrust forces that would otherwise be transferred to the foundations. This tension is carried by a dedicated tie girder typically located below the deck. The tie girder "ties" the ends of the arch together, creating a self-contained structural system where the horizontal forces are resolved within the superstructure itself. This key feature allows tied arch bridges to be built on weaker foundation conditions that would be unsuitable for traditional arch bridges, as the foundations primarily need to resist only vertical reactions rather than significant horizontal thrust forces. The structural efficiency of this system also permits more slender arch profiles and makes tied arch bridges particularly suitable for river crossings where foundation work is challenging or costly.

1. 5,400 kN
2. 7,200 kN
3. 8,550 kN
4. 9,120 kN

Explanation

Solution: 

1. Factored uniformly distributed load = (1.25 × 40) + (1.5 × 30) = 50 + 45 = 95 kN/m
2. For a parabolic arch with uniform load, the horizontal thrust H = wL²/(8f) Where: w = factored load, L = span, f = rise
3. H = 95 kN/m × (120m)²/(8 × 20m) = 95 × 14,400/(160) = 8,550 kN
4. The axial force in the tie girder (tension) equals this horizontal thrust
5. Therefore, maximum axial force in the tie girder = 8,550 kN

Explanation:  In a tied arch bridge, the tie member (a structural girder typically located below the deck) resists the horizontal thrust generated by the arch. The horizontal thrust depends on the span, rise, and loading of the arch. For a uniformly loaded parabolic arch, the formula H = wL²/(8f) provides the horizontal thrust, which equals the axial tensile force in the tie. This calculation demonstrates how the geometry of the arch (particularly the rise-to-span ratio) significantly affects the forces within the structure. A smaller rise results in larger horizontal forces, highlighting the importance of optimizing the arch geometry during design. In practice, additional considerations such as temperature effects, non-uniform loading patterns, and dynamic loading would also influence the final design forces.