An engineer is designing a railroad bridge to be built primarily from wrought iron, which is much stronger in tension than in compression. Which truss geometry is most appropriate, and why?
AA Howe truss, because its diagonals are in compression and iron handles compression well
BA Pratt truss, because its diagonals carry tension under typical downward loading, exploiting iron's tensile strength
CA Warren truss, because it has fewer members and uses less material overall
DAny truss works equally well, since geometry does not affect whether members are in tension or compression
The Pratt truss was historically dominant in iron bridge construction precisely because its geometry places the diagonal members (the longest, heaviest members) in tension under downward loading — matching iron's superior tensile strength. The Howe truss puts diagonals in compression, better suited for timber (which handles compression better). The Warren truss is geometrically efficient but the key insight here is that truss geometry is chosen to align the loading regime (tension vs. compression) with the material's strengths. Geometry is not arbitrary.
Question 2 Multiple Choice
A statically determinate truss (satisfying m = 2j − 3) has one member suddenly fail due to corrosion. What immediately happens structurally?
AThe remaining members redistribute the load and the truss continues to carry its design load safely
BThe truss becomes a mechanism — it can no longer maintain its shape and will collapse
CThe truss becomes redundant, providing an additional margin of safety
DOnly the members adjacent to the failed member are affected; the rest remain stable
A statically determinate truss has exactly the minimum number of members to maintain rigidity (m = 2j − 3). Remove one member and it has m − 1 = 2j − 4 members — one too few for a stable structure. It becomes a mechanism, meaning it can undergo finite motion (collapse) without stretching any members. This is the core tradeoff of determinacy: the analysis is simple (pure equilibrium), but there is no redundancy. Indeterminate (redundant) trusses have extra members that provide alternative load paths — if one fails, others absorb the load. Real bridge trusses are typically designed with redundancy for exactly this reason.
Question 3 True / False
Trusses achieve structural efficiency because their members carry loads through a combination of bending and axial forces.
TTrue
FFalse
Answer: False
This reverses the key principle. Trusses are efficient precisely because members carry *only* axial forces — pure tension or pure compression — with no bending. In a solid beam, the material near the neutral axis contributes almost nothing to bending resistance, so most of the material is underutilized. In a truss member loaded axially, every cross-sectional element is stressed at the same intensity, so the material is fully exploited. This is why a truss can span the same distance as a solid beam at a fraction of the weight — the elimination of bending is the source of the strength-to-weight advantage.
Question 4 True / False
A long, slender diagonal member in a truss that is in compression may fail at a load well below what its cross-sectional area alone would predict.
TTrue
FFalse
Answer: True
Slender compression members fail by buckling before yielding. Euler's buckling formula shows that the critical buckling load depends on the slenderness ratio (effective length divided by radius of gyration), not just on the cross-sectional area and material yield strength. A long, thin diagonal can buckle elastically at a stress far below the material's yield strength. This is why compression members must be checked against both yielding and buckling — and why truss design iterates between force analysis and member sizing, since the chosen cross-section affects the slenderness ratio which affects the buckling capacity.
Question 5 Short Answer
Why do truss members carry loads more efficiently than a solid beam of the same span and material, and what structural principle enables this?
Think about your answer, then reveal below.
Model answer: Truss members carry load in pure axial tension or compression — no bending. In a solid beam, the material near the neutral axis is nearly stress-free while material at the top and bottom flanges carries the full bending stress; most material is underutilized. In a truss member under axial load, the full cross-section is stressed uniformly, so every unit of material is doing useful work. This means a truss can be much lighter than a solid beam spanning the same distance, which is why trusses dominate long-span applications (bridges, roofs) where self-weight is a significant fraction of the total load.
The underlying principle is that triangular geometry constrains all deformation to axial deformation in members, eliminating the rotational freedom that produces bending. As long as loads are applied only at joints (nodes), no member experiences a transverse force along its length, so no bending moment develops. This assumption — pinned joints, loads at joints — is the idealization that makes simple truss analysis possible and that truss geometries are designed to approximate in practice.