Beams, Deflection, Columns, Mohr's Circle, and Combined Loading

Beam bending and shear

A transverse load on a beam creates internal shear V and bending moment M. They are linked to the distributed load w by the Handbook relations dV/dx = −w(x) and dM/dx = V(x), so moment is the area under the shear diagram and shear is the area under the load diagram. Maximum moment occurs where shear crosses zero — the fastest way to locate it on a shear diagram.

The flexure (bending) stress formula gives normal stress that varies linearly across the depth:

$\sigma = \frac{M c}{I}$

where c is the distance from the neutral (centroidal) axis to the extreme fiber and I is the area moment of inertia of the cross-section. The section modulus S = I/c packages this as σ_max = M/S. Transverse shear stress is

$\tau = \frac{VQ}{I b}$

with Q the first moment of the area above (or below) the point and b the width there. For a rectangular section, τ peaks at the neutral axis at 1.5 V/A — 50% higher than the average shear V/A, a frequent FE distinction.

Worked example: a simply supported beam, length 4 m, central point load 12 kN. Max moment at center M = PL/4 = 12(4)/4 = 12 kN·m. For a 50 mm × 150 mm rectangle (tall), I = bh³/12 = (0.05)(0.15)³/12 = 1.406×10⁻⁵ m⁴, c = 0.075 m. σ_max = Mc/I = 12 000(0.075)/1.406×10⁻⁵ = 64.0 MPa.

Beam deflection

Deflection comes from EI y'' = M(x) (the elastic curve), but FE expects you to use tabulated formulas from the Handbook rather than integrate. Memorize the few highest-yield cases:

Stiffer (larger EI) or shorter beams deflect less; deflection scales with L³ or L⁴, so doubling span increases deflection 8× or 16×.

Column buckling

Long slender columns fail by elastic buckling below the yield stress. Euler's formula gives the critical load:

$P_{cr} = \frac{\pi^2 E I}{(KL)^2}$

KL is the effective length, and K depends on end conditions: K = 1 pinned-pinned, K = 0.5 fixed-fixed, K = 0.7 fixed-pinned, and K = 2 fixed-free (cantilever column). Use the smallest I (weak axis) because the column buckles about the axis of least resistance. The critical stress is σ_cr = P_cr/A = π²E/(KL/r)², where r = √(I/A) is the radius of gyration and KL/r is the slenderness ratio. Euler's equation is valid only above a transition slenderness; below it, columns fail by yielding/crushing, so check σ_cr against S_y.

Worked example: a 3 m pinned-pinned steel column (E = 200 GPa, I = 4×10⁻⁶ m⁴). P_cr = π²(200×10⁹)(4×10⁻⁶)/(1·3)² = 7.90×10⁶/9 = 877 kN. Fix both ends (K = 0.5) and the same column carries 4× more: 3.51 MN.

Mohr's circle and combined loading

For a plane stress state (σ_x, σ_y, τ_xy), the principal stresses and maximum shear come from the transformation equations, neatly captured by Mohr's circle:

$\sigma_{avg} = \frac{\sigma_x + \sigma_y}{2}, \qquad R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$ $\sigma_{1,2} = \sigma_{avg} \pm R, \qquad \tau_{max} = R$

The circle's center sits on the σ-axis at σ_avg; its radius R is the in-plane maximum shear stress. Worked example: σ_x = 80 MPa, σ_y = 20 MPa, τ_xy = 30 MPa. σ_avg = 50; R = √(30² + 30²) = 42.4 MPa. σ₁ = 92.4 MPa, σ₂ = 7.6 MPa, τ_max = 42.4 MPa. Combined loading superposes axial (P/A), bending (Mc/I), and torsional (Tr/J) stresses at a point, then feeds the resulting σ_x, σ_y, τ_xy into Mohr's circle to get the true principal stress used in a failure check — the capstone integration of the whole mechanics-of-materials chain.

Shear and moment diagrams in practice

Building shear (V) and bending-moment (M) diagrams is a high-yield FE skill. Start at one end with the support reaction, step along the beam, and add/subtract loads: a point load causes a vertical jump in V; a uniform load makes V slope linearly (dV/dx = −w); the M-diagram slope equals V (dM/dx = V), so M is the running area under the V-diagram. A point load gives a triangular M-diagram peaking under the load; a uniform load gives a parabolic M-diagram peaking at midspan. For a simply supported beam with central load P, V steps from +P/2 to −P/2 at center and M peaks at PL/4. For a full-span uniform load w, M_max = wL²/8 at midspan.

Locating M_max where V = 0 is the fastest route to the bending-stress check.

Failure theories for combined stress

Once principal stresses are known, FE tests two ductile-failure criteria. The maximum-shear-stress (Tresca) theory predicts yield when τ_max = S_y/2, i.e. (σ₁ − σ₃) reaches S_y. The distortion-energy (von Mises) theory uses the von Mises stress σ' = √(σ₁² − σ₁σ₂ + σ₂²) for plane stress, predicting yield when σ' = S_y; it is less conservative and generally more accurate for ductile metals.

For the earlier state σ₁ = 92.4, σ₂ = 7.6 MPa, von Mises gives σ' = √(92.4² − 92.4·7.6 + 7.6²) = 88.9 MPa, so a material with S_y = 250 MPa carries FS = 250/88.9 = 2.8. Combining a combined-loading stress state with a failure theory and a factor of safety is the complete FE design-check sequence — and the reason this chapter's topics are so heavily interlinked.