CIVL6264 Chap.2 Beam Strength and Plastic Analysis
Beam Strength and Plastic Analysis
Rigid-plastic beam analysis reduces the section to design compression and tension blocks whose forces must balance. This chapter shows how the relative slab and steel capacities locate the plastic neutral axis, how the lever arm creates moment resistance, and how partial shear connection caps the transferable force.
Every worked result is checked against the assumed neutral-axis location rather than accepted as arithmetic alone. The calculation method is a disciplined sequence. First select an assumed neutral-axis case and draw the design stress blocks. Next integrate each block into a resultant and enforce zero net axial force for pure bending. Then locate every resultant centroid and take moments using consistent units.
The assumed case is accepted only when its neutral axis and compression-block depth lie inside the material region used to derive them. If the slab block would extend beyond the effective depth, steel compression must enter the equilibrium or another case must be tried.
Full connection is also an assumption that must be demonstrated: the studs need enough resistance over the relevant transfer length to develop the slab force in the chosen block model. When connector capacity is smaller, the member becomes connection-limited and the internal forces and lever arms must be rebuilt.
The chapter uses simple original numbers so every result can be recomputed, but it keeps those illustrations separate from current-standard design resistance. The calculation clinic treats an impossible block depth or insufficient interface force as diagnostic evidence about the model, not as an invitation to force the arithmetic to match an expected answer.
What this chapter covers
- 01
Rigid-plastic stress blocks
- 02
Compression and tension equilibrium
- 03
Neutral axis in the slab
- 04
Neutral axis in the steel
- 05
Force-resultant lever arms
- 06
Partial shear connection
Worked example · free
Balance a simple force couple
- +1Use the smaller result as the balanced force: 1200 kN.
- +1Confirm C equals T for pure bending equilibrium.
- +1Keep the lever arm in metres.
- +1Calculate 1200 times 0.360 equals 432 kN m.
- +1State that 250 kN steel capacity remains unused in the simplified case.
Key terms
- Plastic neutral axis
- The boundary between compression and tension design blocks.
- Compression block
- An idealised region carrying design compression in rigid-plastic analysis.
- Force resultant
- A single force representing the integral of stress over an area.
- Lever arm
- The perpendicular distance between balanced compression and tension resultants.
- Partial connection
- A condition where connector resistance limits the force developed by composite action.
Beam Strength and Plastic Analysis FAQ
How is the plastic neutral axis selected?
Compare the available slab compression with the steel forces and enforce axial equilibrium. If a slab-contained block can balance the steel tension and fits within the slab, that case is possible. Otherwise a steel compression region or connection-limited case must be developed.
Why does the smaller resultant govern?
Pure bending requires compression and tension to balance. A larger available force on one side cannot contribute unless an equal opposite force exists. Using the smaller force preserves equilibrium and exposes unused capacity in the stronger side of the idealised section.
How does partial connection change beam strength?
Available connector resistance caps the longitudinal force that can enter the slab. The stress blocks and lever arms must then be rebuilt for that reduced transfer. Full-connection moment resistance cannot be retained after the interface becomes the weakest element.
Exam move
For every beam problem, sketch the section and draw force resultants before writing moment equations. Test the neutral-axis-in-slab case first, verify the required depth, and move to a steel-axis case only when equilibrium demands it. Finish by checking that the stud connection can transfer the force used in the blocks.
Practise the method with three blank sketches: neutral axis in the slab, neutral axis in the steel and partial connection. On each sketch, label the sign of every force and its centroid before writing any equation. Use a consistent convention for kilonewtons, millimetres and metres, and independently recompute the moment about two different points to confirm equilibrium.
Change one input at a time and predict whether the neutral axis moves upward or downward before calculating. Add a connector cap and rebuild the stress blocks rather than merely multiplying the full-connection moment by a percentage.
For review, create a short rejection checklist: block depth outside the section, unequal internal forces, missing lever-arm conversion, connector force below demand or insufficient ductility for the assumed plastic distribution. A complete practice answer closes by naming the controlling case and the physical check that supports it. Add one independent check after every worked moment.
Confirm that the internal axial forces sum to zero, that the lever arm has the correct dimension, and that the moment direction matches the applied sagging or hogging region. Then vary slab strength, steel strength or effective width and predict which force block changes first. A spreadsheet can support repetition, but retain a hand sketch and a transparent calculation so a reviewer can see why the selected case applies.
Never copy the source tutorial geometry into a new practice problem; preserve only the question type and author fresh dimensions. Keep one worked solution in which the first assumed case fails. Showing the rejected compression depth and the corrected neutral-axis case trains you to use equilibrium as a diagnostic tool rather than treating case selection as a memorised sequence.