A drop-in span on a shear connection
A deep girder cantilevers past its support, and a shallower beam simply hangs from its tip on a bolted fin plate. That little plate is the whole story: it carries shear but no moment, which makes it a hinge — and the hinge is what lets plain statics solve a beam that looks indeterminate. We read the connection, split the beam at it, and solve each piece with tools from the earlier examples.
- Girder
- 610UB125A (pin, 0) → B (roller, 6 m) → tip J (8 m)
- Drop-in
- 310UB40.4J (8 m) → C (roller, 13 m)
- Joint at J
- Fin-plate shear connectiontransfers shear, releases moment
- 12 kN/mUDL over the full 13 m
- 200 GPa
- 988.3 / 86.6 × 10⁶ mm⁴AS4100 library values
Step 1 — Read the connection before you touch the numbers
The drawing tells you the structural model: a fin plate carries shear, not moment.
Look closely at the joint at J (8 m), where the shallow 310UB40.4 hangs off the tip of the deep 610UB125. The two beams are joined only through a small plate bolted to their webs. The flanges — the parts of an I-section that carry bending — are not connected at all; the smaller beam is even coped (notched) at the top so it can tuck in under the girder’s flange level.
No flange connection means no path for a bending moment to cross the joint. Whatever else happens, the moment at J must be zero — which is exactly what engineers call a hinge. One glance at the steelwork has already written our structural model: two spans, pinned together at J.
Step 2 — Split the beam at the hinge
A hinge transfers a force but no moment — so cutting there costs nothing.
Three supports on a continuous beam would normally be statically indeterminate — statics alone gives three equations, and this beam has four unknown reaction components. The hinge is the missing fourth equation:
Better still, it lets us take the structure apart. Cut at J: the only thing the fin plate passes between the two pieces is a vertical force . The suspended span J–C becomes a simply supported beam sitting on the plate at one end and the roller at the other — the problem we already know how to solve.
Step 3 — Solve the drop-in span
Symmetry again: each end of the 5 m span carries half the load.
The suspended span carries = 12 × 5 = 60 kN, symmetrically:
= 30 kN is our first final answer, and = 30 kN is the force the fin plate must carry — hold on to that number; it comes back twice, first as a load on the girder and later as the design force for the bolts.
Step 4 — Carry the 30 kN back to the girder
The main beam sees the drop-in span as a point load on its cantilever tip.
By Newton’s third law, the girder carries what the plate holds up: a 30 kN point load at its tip J, on top of its own 12 kN/m. The girder A–B–J is just an overhanging beam. Moments about A:
Global check: 22 + 104 + 30 = 156 kN = 12 × 13 ✓. Every reaction came from statics — the hinge turned an indeterminate-looking beam back into two textbook problems.
Step 5 — The shear diagram, and the number the bolts feel
Walk the whole 13 m; the value crossing the connection is the connection’s design force.
Now stitch the pieces back together and walk left to right: start at +22, fall under the UDL to −50 just before B, jump to +54 over the big reaction, fall again to +30 at J — exactly the 30 kN crossing the fin plate — then continue down through the drop-in span to −30 at C:
Step 6 — The moment diagram passes through zero at the hinge
Hogging −84 over the interior support, zero at J, sagging 37.5 in the drop-in span.
The moment at B comes from the left free body: kN·m of hogging — the price of the cantilever. From B the moment climbs, and at the connection:
Zero — by arithmetic, not by assumption. Statics has confirmed what the steelwork promised in Step 1. Inside the drop-in span the moment is the familiar parabola peaking at kN·m at 10.5 m.
Step 7 — The shape gives the hinge away
A slope kink at J: the two spans rotate independently on the pin.
The deflected shape is the drawing that makes the whole example click. Over the girder the beam barely moves — a 610UB125 is stiff. At J the curve kinks: a real hinge transmits no moment, so nothing forces the two sides to leave the joint at the same slope. The drop-in span then sags freely to about 6.2 mm — it is both the shallower section and the piece hanging off a moving support.
The proof
Six hand numbers vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Reaction R_A | 22 kN (statics) | 22 kN | |
| Reaction R_B | 104 kN (ΣM_A on the girder) | 104 kN | |
| Reaction R_C | 30 kN (wL/2 of the drop-in) | 30 kN | |
| Shear at the connection | 30 kN (the fin-plate force) | 30 kN at 8 m | |
| Moment at the hinge M_J | 0 (by statics, Step 6) | 0 at 8 m | |
| Hogging M_B | −84 kN·m at 6 m | −84 kN·m at 6 m | |
| Max sagging | 37.5 kN·m at 10.5 m (wL²/8) | 37.5 kN·m at 10.5 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
Open this exact model in the calculator — then change a load, drag a support, and watch every diagram update in real time. The best way to build intuition is to break it and see what happens.
Take it with you
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