Shape Factor For Circular Section: Derivation, Proof and Solved Examples (2026)

1. Introduction to Circular Cross-Section Plasticity

Solid and tubular circular sections serve as essential structural elements in civil, mechanical, and offshore engineering. From bridge pier columns and structural steel pins to offshore jacket piles and transmission poles, cylindrical members must withstand significant flexural actions. Evaluating the true ultimate capacity under limit state design requires establishing the shape factor for circular section.

           ELASTIC STRESS                         FULLY PLASTIC STRESS
           +------------+                            +------------+
          /      fy      \                          /      fy               |     /          |                        |    |=======|   |
        /    /             \                      /     |=======|           |   /                |                    |      |=======|     |
   ----+--------------------+---- N.A.       ----+--------------------+---- P.N.A.
       |   \                |                    |      |=======|     |
        \    \             /                      \     |=======|    /
         |     \          |                        |    |=======|   |
          \      fy      /                          \      fy      /
           +------------+                            +------------+
             M = My                                      M = Mp

The shape factor $S$ reflects the geometric reserve strength of a cross-section between first outer-fiber yield $M_y$ and full plastic collapse $M_p$:

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

For a solid circular cross-section, the flexural shape factor evaluates to $S = \frac{16}{3\pi} \approx 1.698$ (roughly $1.70$). This theoretical value shows that a solid circular section possesses nearly a $70\%$ reserve capacity beyond its elastic limit.

Because material in a circular cross-section is concentrated heavily near the neutral axis—where the cross-section is widest—yielding of the thin outermost fibers leaves substantial unyielded core material capable of carrying significant additional load.

2. Section Geometry and Elastic Bending Formulation

2.1 Elastic Second Moment of Area ($I_x$)

Consider a solid circular cross-section of radius $R$ and diameter $D = 2R$. Using polar coordinates or strip integration, differential area strip $dA = 2 x \, dy = 2 \sqrt{R^2 – y^2} \, dy$ at height $y$ yields second moment of area $I_x$:

$$I_x = \int_{-R}^R 2 y^2 \sqrt{R^2 – y^2} \, dy = \frac{\pi R^4}{4} = \frac{\pi D^4}{64}$$

                       CIRCULAR SECTION INTEGRATION STRIP
                                 +---------------+
                              /                                                 /     x = sqrt(R^2-y^2)                              |   |<------ 2x ------>|    |  ^
                           |===+==================+====|  | dy  (at height y)
                       ----+---------------------------+--v- Neutral Axis (y = 0)
                           |                           |
                            \                         /
                              \                     /
                                 +---------------+
                                 |<----- D ----->|

In structural mechanics, this formulation confirms that circular members offer identical moment of inertia in every radial direction, preventing weak-axis bias.

2.2 Elastic Section Modulus ($Z_e$) and First Yield Moment

Maximum distance from neutral axis to extreme outermost fiber is $y_{\max} = R = D/2$. The elastic section modulus $Z_e$ is:

$$Z_e = \frac{I_x}{y_{\max}} = \frac{\frac{\pi D^4}{64}}{\frac{D}{2}} = \frac{\pi D^3}{32} \approx 0.098175 D^3$$

Bending moment corresponding to initial yielding at the outermost fiber is yield moment $M_y$:

$$M_y = f_y \cdot Z_e = f_y \cdot \frac{\pi D^3}{32}$$

2.3 Progressive Plastic Penetration in Cylindrical Members

When applied bending moment surpasses $M_y$, yielding penetrates inward from top and bottom edges toward the neutral axis. Unlike rectangular beams with constant width, circular sections widen as the plastic boundary approaches $y=0$. This expanding width increases the resisting moment rate, delaying full plastification and providing exceptional rotational ductility.

3. Mathematical Derivation of Plastic Modulus and Shape Factor

3.1 Equal Area Axis and Semicircular Centroid Derivation

Under pure flexural plasticity without axial force, longitudinal equilibrium demands equal tension and compression areas $A_c = A_t = A/2 = \pi R^2 / 2$. Due to symmetry, plastic neutral axis (PNA) sits along the horizontal diameter.

                     SEMICIRCULAR COMPRESSION BLOCK
                                 +---------------+
                              /                                                 /        Area = πR^2/2                               |                           |
                           |             * y_bar = 4R/(3π)
                       ----+---------------------------+---- Plastic Neutral Axis
                           |                           |

The centroid distance $\bar{y}$ of a semicircle from the diametral axis is derived via first moment of area integration:

$$\bar{y} = \frac{\int_0^R y \cdot \left(2 \sqrt{R^2 – y^2}\right) dy}{\frac{\pi R^2}{2}} = \frac{\frac{2}{3} R^3}{\frac{\pi R^2}{2}} = \frac{4 R}{3 \pi} = \frac{2 D}{3 \pi}$$

3.2 Plastic Section Modulus ($Z_p$) Proof

The plastic section modulus $Z_p$ is the sum of first moments of area for compression and tension halves about the PNA:

$$Z_p = 2 \left( A_c \cdot \bar{y}_c \right) = 2 \left[ \left( \frac{\pi R^2}{2} \right) \left( \frac{4 R}{3 \pi} \right) \right] = \frac{4 R^3}{3} = \frac{D^3}{6} \approx 0.166667 D^3$$

Fully plastic moment capacity $M_p$ of a solid circular section is $M_p = f_y Z_p = f_y D^3 / 6$.

3.3 Exact Shape Factor Calculation ($16/3\pi$)

The exact analytical value is established by taking the ratio of $Z_p$ to $Z_e$:

$$S = \frac{Z_p}{Z_e} = \frac{\frac{D^3}{6}}{\frac{\pi D^3}{32}} = \frac{32}{6 \pi} = \frac{16}{3 \pi} \approx 1.69765$$

This proof shows that a solid circular section provides a $69.8\%$ flexural strength increase over first yield.

4. Hollow Circular Tubes and Pipe Cross-Sections

4.1 Thick-Walled Circular Tube Derivations

For a hollow circular tube with outer diameter $D_o$ and inner diameter $D_i$:

                       HOLLOW CIRCULAR CROSS-SECTION
                                 +---------------+
                              /     t = Ro - Ri                               /   +---------------+                              |   /                 \   |
                       ----+--+-------------------+--+---- Neutral Axis
                           |   \                 /   |
                            \   +---------------+   /
                              \                     /
                                 +---------------+
                                 |<----- Do ---->|

The elastic and plastic section moduli are:

$$Z_e = \frac{\pi (D_o^4 – D_i^4)}{32 D_o}, \quad Z_p = \frac{D_o^3 – D_i^3}{6}$$

With diameter ratio $\alpha = D_i / D_o$, the section ratio becomes:

$$S_{\text{ tube}} = \frac{16}{3 \pi} \cdot \frac{1 – \alpha^3}{1 – \alpha^4}$$

4.2 Thin-Walled Circular Tube Approximation

For thin-walled pipes where $t \ll R_m$: $Z_e \approx \pi R_m^2 t$, $Z_p \approx 4 R_m^2 t$, giving $S_{\text{ thin}} = 4 / \pi \approx 1.2732$. As wall thickness decreases, reserve strength shifts smoothly from $1.698$ to $1.273$.

4.3 Parametric Influence of Wall Thickness Ratio

As the ratio $t/D_o$ decreases from $0.50$ (solid bar) to $0.01$ (thin tube), material shifts away from the neutral axis to the outer perimeter. Consequently, the elastic modulus increases faster than the plastic modulus, causing the shape factor to drop monotonically from $1.698$ toward $1.273$.

5. Plastic Hinge Length and Rotational Capacity

5.1 Hinge Length Under Concentrated Mid-Span Load

For a simply supported circular beam under central point load $P$:

$$L_p = L \left( 1 – \frac{1}{S} \right) = L \left( 1 – \frac{3\pi}{16} \right) \approx 0.411 L$$

The plastic hinge spreads across $41.1\%$ of the span, providing excellent energy dissipation.

5.2 Hinge Spread Under Uniformly Distributed Loading

Under uniformly distributed load $w$, setting moment equal to $M_y = M_p / S$ yields:

$$L_p = L \sqrt{1 – \frac{1}{S}} = L \sqrt{1 – \frac{3\pi}{16}} \approx 0.641 L$$

Plastic deformation covers $64.1\%$ of the total span before a mechanism forms.

6. Detailed Step-by-Step Worked Numerical Examples

6.1 Solid Circular Shaft Parameters and Material Properties

Consider a high-strength solid steel pin in a bridge bearing assembly:

• Diameter: $D = 200\text{ mm} = 0.200\text{ m}$

• Yield strength: $f_y = 400\text{ MPa}$

• Effective span: $L = 1.50\text{ m}$ (Simply supported under central load $P$)

6.2 Yield Moment, Plastic Moment, and Section Ratio

Step 1: Compute Area and Moment of Inertia
$$A = \frac{\pi (200)^2}{4} = 31{,}415.93\text{ mm}^2, \quad I_x = \frac{\pi (200)^4}{64} = 7.85398 \times 10^7\text{ mm}^4$$

Step 2: Section Moduli Calculations
$$Z_e = \frac{\pi (200)^3}{32} = 785{,}398.16\text{ mm}^3 = 7.85398 \times 10^{-4}\text{ m}^3$$
$$Z_p = \frac{200^3}{6} = 1{,}333{,}333.33\text{ mm}^3 = 1.33333 \times 10^{-3}\text{ m}^3$$

Step 3: Moment Capacities and Verification
$$M_y = 400\text{ MPa} \times 7.85398 \times 10^{-4}\text{ m}^3 = 314.16\text{ kN}\cdot\text{ m}$$
$$M_p = 400\text{ MPa} \times 1.33333 \times 10^{-3}\text{ m}^3 = 533.33\text{ kN}\cdot\text{ m}$$
$$S = \frac{533.33}{314.16} = 1.69765 \approx 1.70$$

6.3 Hollow Tube Comparison and Weight Optimization

Evaluating a hollow tube with $D_o = 200\text{ mm}$ and thickness $t = 20\text{ mm}$ ($D_i = 160\text{ mm}$):

• $A_{\text{ tube}} = 11{,}309.73\text{ mm}^2$ ($64\%$ weight saving)

• $Z_{e,\text{ tube}} = 4.6369 \times 10^5\text{ mm}^3$

• $Z_{p,\text{ tube}} = 6.5067 \times 10^5\text{ mm}^3$

• $S_{\text{ tube}} = 1.403$

6.4 Collapse Load Multipliers Under Point and Uniform Loading

Point load capacities evaluate to $P_y = 4 M_y / L = 837.76\text{ kN}$ and $P_u = 4 M_p / L = 1{,}422.21\text{ kN}$, giving $P_u / P_y = 1.698$. Under uniform distributed loading $w$, elastic load limit is $w_y = 8 M_y / L^2 = 1{,}117.01\text{ kN/m}$ and plastic collapse load is $w_u = 8 M_p / L^2 = 1{,}896.28\text{ kN/m}$.

7. Engineering Standard Codes and Design Implications

Practical design codes incorporate circular plastic reserve with specific limits:

1. AISC 360-22 Provisions: Per [AISC 360-22 Specification for Structural Steel Buildings at AISC], solid round bars are covered in Section F11 with nominal plastic strength $M_n = M_p = f_y Z_p \le 1.6 M_y$. Because the theoretical shape factor is $1.698$, AISC caps nominal strength at $1.60 M_y$ to limit excessive plastic deformation prior to mechanism mobilization. Round HSS tubular members are governed by Section F8.
2. Eurocode 3 (EN 1993-1-1): In [Eurocode 3 EN 1993-1-1 Structural Steel Design at Eurocodes Building the Future], circular hollow sections (CHS) are classified based on diameter-to-thickness ratio $d/t \le 50\epsilon^2$ for Class 1 (plastic).
3. Offshore and Subsea Standards (API RP 2A / DNV-ST-F101): Tubular piles and subsea pipelines rely heavily on the $1.27 – 1.40$ shape factor in strain-based design.

8. Summary of Circular Cross-Section Mechanics

The derivation of the flexural shape factor reveals an impressive $69.8\%$ flexural reserve capacity for solid circular members ($S = 16 / 3\pi \approx 1.70$). Moving from solid profiles to hollow tubes transitions the shape factor smoothly toward the thin-walled limit of $4/\pi \approx 1.27$. Understanding the flexural shape factor empowers engineers to design safe, cost-efficient, and resilient structural members across civil and offshore applications.

References & Standards Cited

1. AISC 360-22: Specification for Structural Steel Buildings, American Institute of Steel Construction, Chicago, IL, 2022.
2. EN 1993-1-1:2005+A1:2014: Eurocode 3: Design of steel structures – Part 1-1: General rules and rules for buildings, CEN, Brussels, 2014.
3. API RP 2A-WSD: Recommended Practice for Planning, Designing and Constructing Fixed Offshore Platforms, 22nd Edition, American Petroleum Institute, Washington, D.C., 2014.
4. Calladine, C. R. (1985). Plasticity for Engineers, Ellis Horwood Ltd., Chichester.
5. Chen, W. F., & Sohal, I. (1995). Plastic Design and Second-Order Analysis of Steel Frames, Springer-Verlag, New York.

Frequently Asked Questions (FAQ)

Why is the shape factor for circular section higher than a rectangular section?

A circular section has a shape factor of $1.70$, compared to $1.50$ for a rectangle. This happens because the circle has its maximum width at the neutral axis and tapers to narrow points at outer fibers. When outer fibers yield, most cross-sectional material remains unyielded near the center.

What is the exact formula for the plastic section modulus of a solid circle?

The exact formula is $Z_p = D^3 / 6 = 4 R^3 / 3$, where $D$ is diameter and $R$ is radius.

What is the shape factor of a thin-walled circular pipe?

For a thin-walled circular tube where thickness $t ll R$, the shape factor is $S = 4 / pi approx 1.273$.

Does the shape factor of a circle change under combined axial force and bending?

Yes. When axial load $N$ is present, the equal-area neutral axis shifts away from the diametral axis, reducing plastic moment capacity $M_{pc}$.

Why does AISC 360-22 cap nominal flexural strength of round bars at $1.60 M_y$?

Although theoretical shape factor is $1.698$, AISC caps design flexural strength at $1.60 M_y$ to prevent severe plastic deformation and serviceability problems.