Interactive Lecture
Elastic–Plastic Behaviour and Residual Stresses in Structural Members
by Dr Arnaldo Delli Carri
When structural members are subjected to bending moments beyond their elastic limit, they undergo plastic deformation. This is the basis of modern limit state design — it lets engineers exploit the full strength of the material, not just its elastic capacity.
\[\sigma = \frac{My}{I}\]
\[M_e = \frac{\sigma_y I}{y_{max}},\quad M_p = \sigma_y Z,\quad f=\frac{M_p}{M_e}=\frac{Z}{S}\]
M < Me
M ≈ Me
Me < M < Mp
M ≈ Mp
Loading State: Unloaded
Maximum Moment: 0 kN·m
Residual Stress: 0 MPa
Plastic bending theory enables more economical and safer design than elastic analysis alone allows. Here's where you'll encounter it:
| Concept | Equation | Description |
|---|---|---|
| Elastic Section Modulus | \(S = \frac{I}{y_{max}}\) | Relates second moment of area to elastic bending resistance |
| Elastic Limit Load | \(M_e = \sigma_y S\) | Moment when extreme fibre first reaches yield stress |
| Plastic Section Modulus | \(Z = \sum A_i \bar{y}_i\) | First moment of area about plastic neutral axis |
| Plastic Limit Load | \(M_p = \sigma_y Z\) | Moment when entire section reaches yield stress |
| Shape Factor | \(f = \frac{M_p}{M_e} = \frac{Z}{S}\) | Ratio indicating reserve capacity (rectangle: 1.5, I-beam: ~1.15) |
A rectangular steel beam has width \(b = 150\) mm and depth \(h = 300\) mm. The yield stress is \(\sigma_y = 275\) MPa. Calculate:
Show Solution
(a) Elastic Section Modulus:
\[I = \frac{bh^3}{12} = \frac{150 \times 300^3}{12} = 337.5 \times 10^6 \text{ mm}^4\]
\[S = \frac{I}{y_{max}} = \frac{337.5 \times 10^6}{150} = 2.25 \times 10^6 \text{ mm}^3\]
(b) Plastic Section Modulus:
For a rectangle, the plastic neutral axis is at mid-depth. The plastic modulus is:
\[Z = 2 \times \left(\frac{bh}{2} \times \frac{h}{4}\right) = \frac{bh^2}{4} = \frac{150 \times 300^2}{4} = 3.375 \times 10^6 \text{ mm}^3\]
(c) Elastic Limit Load:
\[M_e = \sigma_y S = 275 \times 2.25 \times 10^6 = 618.75 \times 10^6 \text{ N·mm} = 618.75 \text{ kN·m}\]
(d) Plastic Limit Load:
\[M_p = \sigma_y Z = 275 \times 3.375 \times 10^6 = 928.1 \times 10^6 \text{ N·mm} = 928.1 \text{ kN·m}\]
(e) Shape Factor:
\[f = \frac{M_p}{M_e} = \frac{928.1}{618.75} = 1.50\]
This confirms the theoretical shape factor of 1.5 for rectangular sections.
Consider the same rectangular section from Problem 1 (\(b = 150\) mm, \(h = 300\) mm, \(\sigma_y = 275\) MPa). If a bending moment of \(M = 750\) kN·m is applied, determine:
Show Solution
(a) Classification:
From Problem 1: \(M_e = 618.75\) kN·m and \(M_p = 928.1\) kN·m
Since \(M_e < M = 750 < M_p\), the section is in the elastic-plastic state.
(b) Elastic Core Depth:
Let \(c\) be the half-depth of the elastic core (measured from the neutral axis). The stress distribution has three zones:
Integrating the moment contribution from each zone about the neutral axis gives:
\[M = \underbrace{2\sigma_y b \left(\frac{h^2}{8} - \frac{c^2}{2}\right)}_{\text{plastic zones}} + \underbrace{\frac{2\sigma_y b c^2}{3}}_{\text{elastic core}} = \sigma_y b \left(\frac{h^2}{4} - \frac{c^2}{3}\right)\]
Substituting values:
\[750 \times 10^6 = 275 \times 150 \times \left(\frac{300^2}{4} - \frac{c^2}{3}\right)\]
\[750 \times 10^6 = 41250 \times \left(22500 - \frac{c^2}{3}\right)\]
\[\frac{750 \times 10^6}{41250} = 18181.8 = 22500 - \frac{c^2}{3}\]
\[c^2 = 3 \times (22500 - 18181.8) = 12954.5\]
\[c = 113.8 \text{ mm}\]
Elastic core depth: \(2c = 227.6\) mm
(c) Stress Distribution:
The plastic zone thickness on each side is \(h/2 - c = 150 - 113.8 = 36.2\) mm:
A symmetrical I-section has the following dimensions: flange width \(b_f = 200\) mm, flange thickness \(t_f = 25\) mm, web thickness \(t_w = 15\) mm, total depth \(h = 300\) mm. The yield stress is \(\sigma_y = 355\) MPa.
Calculate the shape factor \(f\). Comment on why I-sections have lower shape factors than rectangular sections.
Show Solution
Step 1: Calculate Second Moment of Area \(I\):
For a symmetric I-section, we subtract the two rectangular voids from the enclosing rectangle:
\[I = \frac{b_f h^3}{12} - \frac{(b_f - t_w)(h - 2t_f)^3}{12}\]
\[I = \frac{200 \times 300^3}{12} - \frac{185 \times 250^3}{12}\]
\[I = 450.0 \times 10^6 - 240.9 \times 10^6 = 209.1 \times 10^6 \text{ mm}^4\]
Step 2: Calculate Elastic Section Modulus \(S\):
\[S = \frac{I}{y_{max}} = \frac{209.1 \times 10^6}{150} = 1.394 \times 10^6 \text{ mm}^3\]
Step 3: Calculate Plastic Section Modulus \(Z\):
For a symmetric I-section, the plastic neutral axis (PNA) is at mid-depth. We sum the first moment of area of each half about the PNA, considering the sub-areas and their centroidal distances:
\[Z = 2(A_f \bar{y}_f + A_{hw} \bar{y}_{hw}) = 2(5000 \times 137.5 + 1875 \times 62.5)\]
\[Z = 2(687500 + 117188) = 1.609 \times 10^6 \text{ mm}^3\]
Step 4: Calculate Shape Factor:
\[f = \frac{Z}{S} = \frac{1.609 \times 10^6}{1.394 \times 10^6} = 1.154\]
Comment: I-sections have lower shape factors (\(f \approx 1.12–1.15\)) compared to rectangular sections (\(f = 1.5\)) because most of their material is concentrated in the flanges, far from the neutral axis. This means the material is already efficiently placed for elastic bending, leaving less "reserve" capacity to exploit through plasticity. Rectangular sections waste more material near the neutral axis in elastic bending, so they gain more from plastic behaviour.