3.5 Mechanics of Materials II: Beams, Torsion, Bending & Columns

Shear & Moment Diagrams

Before sizing a beam you must know the internal shear V and bending moment M along its length. They are governed by:

• dV/dx = −w (slope of shear diagram = negative of distributed load w)
• dM/dx = V (slope of moment diagram = shear)

Practical rules for drawing diagrams: a point load causes a jump in the shear diagram; a uniform load makes shear slope linearly and moment curve parabolically; the maximum moment occurs where shear crosses zero. For a simply supported beam of span L with uniform load w, the standard results (in the FE Reference Handbook beam-formula tables) are:

• Max shear V = wL/2 at the supports
• Max moment M = wL²/8 at midspan

For a central point load P on the same span, M_max = PL/4. Sign convention: sagging (concave up) moment is positive. Use the handbook's beam tables rather than re-deriving under time pressure.

To build the diagrams from scratch: first solve the support reactions, then start at the left end and move right, adding the effect of each load. Shear is the running sum of vertical forces from the left; the moment at any point is the area under the shear diagram up to that point. The moment is a local maximum or minimum exactly where the shear diagram passes through zero, and at a free end both shear and moment are zero — useful checkpoints for confirming a diagram is correct.

Flexural Stress, Shear Stress & Torsion

Bending (flexural) stress

The stress from a bending moment varies linearly across the depth:

• σ = M·c / I

where M is the bending moment, c is the distance from the neutral axis to the extreme fiber, and I is the centroidal moment of inertia. Stress is maximum at the top/bottom fibers and zero at the neutral axis. The section modulus S = I/c simplifies this to σ = M/S.

Transverse shear stress

Within a beam, vertical shear produces:

• τ = V·Q / (I·b)

where Q is the first moment of the area above the point about the neutral axis, and b is the section width there. τ is maximum at the neutral axis; for a rectangle, τ_max = 1.5·(V/A), 50% higher than the average.

Torsion

A torque T on a circular shaft creates shear stress:

• τ = T·r / J

with J the polar moment of inertia: solid shaft J = πr⁴/2 = πd⁴/32; hollow J = π(r_o⁴ − r_i⁴)/2. Maximum τ is at the outer surface (r = r_o). The angle of twist is φ = T·L / (J·G), with G the shear modulus and the product JG the torsional rigidity. Torsion appears in shafts, but also in the design of circular members and helps explain why hollow tubes are efficient: most of the polar moment of inertia comes from material far from the center, so a hollow section resists torsion (and bending) nearly as well as a solid one of the same outer diameter at a fraction of the weight.

Beam Deflection & Euler Column Buckling

Beam deflection comes from the elastic curve EI·(d²y/dx²) = M(x). The handbook tabulates standard cases; two to recognize:

• Simply supported, uniform load: δ_max = 5wL⁴/(384EI) at midspan
• Cantilever, end point load P: δ_max = PL³/(3EI) at the tip

Deflection scales with L⁴ or L³, so span dominates — doubling a uniformly loaded span increases deflection 16×.

Euler buckling of columns

A slender column fails by buckling before reaching its crushing stress. The Euler critical load is:

• P_cr = π²·E·I / (K·L)²

where I is the least moment of inertia (buckling occurs about the weak axis), L is the unbraced length, and K is the effective-length factor for end conditions:

The slenderness ratio is KL/r, where r = √(I/A); higher slenderness means a lower buckling load. Critical stress is σ_cr = P_cr/A = π²E/(KL/r)².

Worked Example — Bending Stress

A simply supported beam (span 20 ft) carries w = 2 kip/ft. Its rectangular section is 6 in wide × 12 in deep. Find the maximum bending stress.

• M_max = wL²/8 = 2·(20²)/8 = 2·400/8 = 100 kip·ft = 1,200 kip·in
• I = bh³/12 = 6·12³/12 = 864 in⁴; c = 6 in
• σ = Mc/I = (1,200·6)/864 = 7,200/864 = 8.33 ksi

The most common error here is a unit slip: the moment must be converted from kip·ft to kip·in (×12) so it is consistent with I in in⁴ and c in inches. Equivalently, using the section modulus S = I/c = 864/6 = 144 in³ gives σ = M/S = 1,200/144 = 8.33 ksi — a faster route when S is tabulated. For buckling problems, remember the column uses the least moment of inertia, whereas bending uses the I about the axis the moment acts on; mixing these up is a frequent trap.