Bending stresses in beams

Lesson 4 gave you the bending moment at every section of a beam. This lesson turns that moment into stress. The result, the flexure formula, is arguably the most important equation in structural design: it governs the size of almost every beam, joist, girder and slab. It also explains why beams are shaped the way they are, with deep webs and wide flanges that put material far from the middle of the section.

Assumptions of the elementary theory

The theory of bending developed here, often called Euler–Bernoulli beam theory, rests on a few assumptions that hold well for most practical beams.

• The beam is straight, and it has a plane of symmetry that contains the loads, so it bends in that plane without twisting.
• Plane sections remain plane: a cross-section that was flat and perpendicular to the axis before bending stays flat and perpendicular to the bent axis afterward.
• The material is linear elastic, homogeneous and has the same modulus in tension and compression.
• Deflections are small compared with the span, and the beam is long compared with its depth, so shear deformation can be neglected.

The second assumption is the key one. It can be shown exactly for pure bending (constant moment, no shear), and it is a very good approximation when shear is present, provided the beam is not short and deep.

Strain in a bent beam

When a beam sags, its top fibres shorten and its bottom fibres lengthen. Somewhere between them is a surface whose length does not change, the neutral surface. Its intersection with a cross-section is the neutral axis. If the beam bends into an arc of radius \(\rho\) (the radius of curvature), a fibre at a distance \(y\) above the neutral surface lies on an arc of radius \(\rho - y\). Comparing arc lengths gives the normal strain:

where \(\kappa = 1/\rho\) is the curvature. The strain varies linearly with distance from the neutral axis, compressive above it and tensile below it for a sagging beam. This follows from geometry alone; it does not depend on the material.

The flexure formula

For a linear elastic material, \(\sigma_x = E\varepsilon_x = -E\,y/\rho\), so the stress is also linear in \(y\). Two equilibrium conditions fix the rest. First, a beam in pure bending carries no net axial force, so the stresses over the section must add to zero:

The integral \(\int y\,dA\) is zero only if \(y\) is measured from the centroid. So the neutral axis passes through the centroid of the cross-section. Second, the moment of the stresses about the neutral axis must equal the internal bending moment:

where \(I\) is the second moment of area (moment of inertia) of the section about the neutral axis. This gives the moment–curvature relation \(1/\rho = M/(EI)\), the basis of deflection calculations in lesson 10. Eliminating \(\rho\) gives the flexure formula:

Here \(c\) is the distance from the neutral axis to the extreme fibre and \(S = I/c\) is the elastic section modulus. The negative sign simply says that positive moment causes compression above the neutral axis. In practice engineers find the magnitude from \(Mc/I\) and decide tension or compression by looking at which face is stretched.

Second moment of area and the parallel axis theorem

For a rectangle of width \(b\) and depth \(h\) about its centroidal axis, \(I = bh^3/12\), and the section modulus is \(S = bh^2/6\). The cube on the depth is why joists are placed on edge: a 38 × 235 mm joist is about 38 times stiffer and 6.2 times stronger in bending on edge than lying flat, for the same amount of wood.

Most sections are built from simple shapes. The second moment of area of a composite shape about its centroidal axis is found with the parallel axis theorem:

where \(I_i\) is each part's own centroidal moment of inertia, \(A_i\) its area and \(d_i\) the distance from its centroid to the centroid of the whole section. The \(A d^2\) terms usually dominate, which is the mathematical reason flanges are so effective: they place a lot of area far from the neutral axis. Rolled steel sections have their properties tabulated, so the theorem is mostly used for built-up girders, precast concrete sections and composite members.

The T-section example shows that unsymmetrical sections have two section moduli. For a material with different strengths in tension and compression, such as cast iron or concrete, the orientation matters: put the wide flange on the side whose material is weaker, so it is closer to the neutral axis. In a reinforced concrete T-beam, the slab acts as a compression flange at midspan, which is very efficient, while the steel at the bottom takes the tension.

Beyond first yield: the plastic moment

The flexure formula is valid only while the extreme fibre remains elastic. The moment at which the extreme fibre first reaches yield is the yield moment \(M_y = S\,\sigma_y\). A ductile steel beam does not fail there. As the moment increases, yielding spreads inward from the top and bottom until the whole section has yielded, with uniform compression above the neutral axis and uniform tension below. The moment at that state is the plastic moment:

where \(Z\) is the plastic section modulus. For a rectangle, \(Z = bh^2/4\), 1.5 times the elastic modulus; for a typical wide-flange shape the ratio \(Z/S\), called the shape factor, is about 1.1 to 1.2. Steel design codes for compact sections base bending resistance on \(Z\), which is one reason the mechanics of this lesson is the starting point, not the end point, of steel design.

Composite beams and the transformed section

Many beams combine two materials: timber with steel side plates, steel girders with a concrete deck, and reinforced concrete itself. Plane sections still remain plane, so the strain is still linear, but the stress jumps at the interface because the moduli differ. The transformed section method handles this neatly. Choose one material as the reference and define the modular ratio \(n = E_2/E_1\). Replace material 2 with an equivalent amount of material 1 whose width is multiplied by \(n\) (keeping the depth the same). The transformed section is homogeneous, so the flexure formula applies. Stresses in the transformed material are then multiplied by \(n\) to get the real stresses.

Section properties by code

Built-up sections are tedious to handle by hand. The program below takes any section made of rectangles, finds its centroid and second moment of area with the parallel axis theorem, and reports the two section moduli. It checks the T-section from the example and then evaluates a welded plate girder.

# Centroid, I and section moduli of a section made of rectangles
# Each rectangle: (width b, height h, y of its bottom edge), all in mm
def properties(name, rects):
    A = sum(b * h for b, h, y0 in rects)
    ybar = sum(b * h * (y0 + h / 2) for b, h, y0 in rects) / A
    I = sum(b * h**3 / 12 + b * h * (y0 + h / 2 - ybar) ** 2 for b, h, y0 in rects)
    top = max(y0 + h for b, h, y0 in rects)
    S_top, S_bot = I / (top - ybar), I / ybar
    print(f"{name}: A = {A:,.0f} mm2, ybar = {ybar:.1f} mm, I = {I/1e6:.2f} x10^6 mm4, "
          f"S_top = {S_top/1e3:.1f} x10^3 mm3, S_bot = {S_bot/1e3:.1f} x10^3 mm3")
    return I, ybar

properties("T-section", [(20, 180, 0), (150, 20, 180)])
# Welded girder: 300 x 25 flanges, 1000 x 12 web
I, yb = properties("Plate girder", [(300, 25, 0), (12, 1000, 25), (300, 25, 1025)])
M = 1500e6                                   # 1500 kN*m in N*mm
print(f"Girder stress under 1500 kN*m: {M * (1050 - yb) / I:.0f} MPa at the extreme fibre")

T-section: A = 6,600 mm2, ybar = 135.5 mm, I = 26.18 x10^6 mm4, S_top = 405.7 x10^3 mm3, S_bot = 193.3 x10^3 mm3
Plate girder: A = 27,000 mm2, ybar = 525.0 mm, I = 4940.62 x10^6 mm4, S_top = 9410.7 x10^3 mm3, S_bot = 9410.7 x10^3 mm3
Girder stress under 1500 kN*m: 159 MPa at the extreme fibre

The plate girder's flanges hold only about 55 % of the steel, yet they supply about 80 % of its second moment of area. That is the case for the I-shape in one number.