A suspended-span bridge on two shear connections
The Gerber drop-in span, scaled up to a whole bridge. Two shore girders cantilever past their piers, and an entire span hangs between their tips on a bolted fin plate at each end. Two plates means two hinges, and two hinges are exactly the extra equations four supports demand — the whole bridge falls to plain statics. We count our way to determinacy, lift the suspended span out, hand its 30 kN to each shore girder, and stitch the diagrams back together.
- Shore girders
- 610UB1250–9 m and 15–24 m, cantilevering past the piers
- Drop-in span
- 310UB40.49–15 m, hung at both ends
- Joints
- Fin plates at 9 m and 15 mshear only — each is a hinge
- Supports
- Pin at 0 · rollers at 6, 18, 24 mA, B, C, D
- 10 kN/mUDL over the full 24 m
- 200 GPa
Step 1 — Read the bridge before you touch the numbers
Count the unknowns, count the equations: two hinges are exactly what four supports need.
Four supports along one beam — a pin at A and rollers at B, C and D — should ring an alarm. Count the reaction components: the pin holds two, each roller one, so five unknowns against the three equations of plane statics. Two short of solvable. Now look at the joints at 9 m and 15 m: at each one the drop-in span hangs off the shore girder’s cantilever tip through a fin plate bolted to the webs. No flange connection means no path for bending moment — each joint is a hinge, and each hinge donates one extra equation: at the pin. Two hinges, two equations: 3 + 2 = 5. The bridge is statically determinate.
The steelwork has already written the structural model: two shore girders A–B–G₁ and G₂–C–D cantilevering past their piers, and a suspended span G₁–G₂ riding between them on two shear plates. Everything that follows is statics.
Step 2 — Lift out the drop-in span
Cut at both hinges and the suspended span is a 6 m simply supported beam.
Cut at G₁ and G₂. The only thing each fin plate passes is a vertical force, so the suspended span becomes a 6 m simply supported beam sitting on a plate at each end, carrying its own 10 kN/m. Symmetry does the rest:
Each hinge carries 30 kN — hold that number; it is about to load both shore girders and later it becomes the design force for both bolt groups. The span also keeps its own private moment diagram: the familiar parabola peaking at kN·m at its middle, which sits at 12 m of the bridge.
Step 3 — The shore girder carries 30 kN at its tip
An overhanging beam with a point load on the cantilever — and symmetry solves the far side for free.
By Newton’s third law, the girder A–B–G₁ carries what its plate holds up: a 30 kN point load at the tip, on top of its own 9 m of 10 kN/m. Moments about A:
The bridge is symmetric, so the far girder is the same problem in a mirror: kN and kN for free. Global check: ✓. Four reactions, all from statics — the two hinges did exactly what Step 1 promised.
Step 4 — Assemble the diagrams
Hogging −135 over both piers, zero at both hinges, sagging 45 in the middle — symmetric.
Stitch the three pieces back together. The moment at each pier comes from the left free body: kN·m of hogging — the price of each cantilever — and by symmetry the same −135 sits over C at 18 m. At the first hinge:
Zero — by arithmetic, not by assumption, and its mirror image says the same at 15 m. Between the two zeros the drop-in span’s parabola peaks at 45 kN·m at 12 m. The shear diagram tells the connection story: walking left to right it passes through +30 kN as it crosses the plate at 9 m and the mirror-image −30 kN at 15 m — the same 30 kN magnitude at both plates.
Step 5 — What the connections feel
V(9) = V(15) = 30 kN: one design shear covers both bolt groups.
The 30 kN crossing each plate is each bolt group’s design shear — exactly the number the single drop-in example turned into a bolt, plate and coped-web check, done here twice from one calculation because the bridge is symmetric. Two identical connections, one design, fabricated as a pair: this repetition is half of why the suspended-span layout was so economical to build.
Step 6 — The shape betrays both hinges
Two slope kinks, one at each plate, with the drop-in span riding down between them.
The deflected shape is the drawing that makes the whole system visible. Over the stiff 610UB125 shore girders the curve barely moves; at G₁ it kinks — a real hinge transmits no moment, so nothing forces the two sides to leave the joint at the same slope — and at G₂ it kinks again. Between the two kinks the shallow drop-in span rides down as a piece, sagging on its two moving supports. One kink said “hinge” in the previous example; two kinks say “suspended span”.
The proof
Six hand numbers vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Abutments R_A = R_D | 7.5 kN (statics) | 7.5 kN | |
| Piers R_B = R_C | 112.5 kN (ΣM_A on the shore girder) | 112.5 kN | |
| Moment at the piers | −135 kN·m at 6 m and 18 m | −135 kN·m | |
| Moment at both hinges | 0 (by statics, Step 4) | 0 | |
| Shear at each connection | 30 kN (the fin-plate force) | 30 kN | |
| Drop-in midspan | 45 kN·m at 12 m (wL²/8) | 45 kN·m at 12 m |
Every value was worked by hand with the classical method, then checked against this site’s solver — the same engine the Try it button opens. This agreement is re-run automatically on every build.
Now make it yours
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