Bending, torsion, and beam mechanics

Start with a clamped beam

Imagine a straight beam held firmly at its left end, with a downward force at its free right end. The clamp prevents that end from moving or rotating. The beam can be at rest while its fibres stretch, shorten and carry stress. We will first find the internal moment needed for balance, then find the deformation predicted by one simple elastic model.

Supply length L = 1.5 m, rectangular width b = 0.04 m, height h = 0.03 m, Young’s modulus E = 100 GPa and downward tip-force magnitude P = 24 N. The height runs vertically; the width runs perpendicular to the plane of bending. These are stipulated teaching values. Assume a straight, prismatic, homogeneous beam, small deflection and slope, a linear elastic response and negligible self-weight and distributed loading.

For preparation, use Units, standards, and physical constants and Algebra, functions, graphs, and logarithms. Vectors and vector spaces and Newton’s laws and free-body diagrams help with signed forces and the chosen body. The necessary moment balance and material laws are introduced locally. On a first reading, follow the load, stress signs, graph and comparisons. Derivatives, Taylor series, and local approximation and Integrals and multivariable calculus are optional return paths for the short curvature and integration derivations.

Balance a cut section

Measure the reference coordinate x from the clamp toward the tip, with 0 ≤ x ≤ L. Let transverse height y and displacement w be positive upward. The applied tip force is therefore −P in the upward direction. A couple is a turning action with no net force; take counterclockwise external couples as positive about the axis out of the page. Whole-beam equilibrium requires an upward clamp force P and a counterclockwise clamp couple PL.

Cut the beam at x and choose the right-hand portion as the body. Define the internal scalar M so that a positive M acts clockwise on this portion. Its signed couple about the cut is −M. The downward tip force acts a distance L−x away and gives −P(L−x). Moment balance is −M−P(L−x) = 0, hence M(x) = −P(L−x). Here M(0) = −36 N m and M(L) = 0.

The external clamp couple is +36 N m, whereas the internal scalar at x = 0 is −36 N m under our cut convention. Keep the chosen body and action attached to each sign. No tip couple is applied, so the free-end moment is zero; the end force still requires a nonzero transverse shear force. Balanced external forces do not imply zero internal stress.

Connect curvature to fibre stress

Longitudinal strain ε_x is the small fractional length change of a fibre along the beam. Tensile strain and tensile normal stress σ_x are positive. In this model σ_x = Eε_x. E is a material modulus, with unit Pa = N/m²; it describes the supplied linear stress–strain relation.

Euler–Bernoulli beam kinematics assumes that cross-sections remain plane and perpendicular to the deflected neutral axis. With no axial extension of that axis, the longitudinal displacement is u_x = −y w′(x), giving ε_x = −y w″(x). A prime means a derivative with respect to reference position x: w′ is slope, and w″ is curvature in the small-slope approximation. Their units are 1 and m⁻¹. This is a spatial description of a static beam.

For the centred symmetric section, the axial normal-force resultant vanishes. Define I = ∫_A y² dA, the second moment of area about the bending axis. The signed moment is M = −∫_A yσ_x dA = EIw″. Therefore σ_x(x,y) = −M(x)y/I. Where M is negative, the upper fibre y > 0 is in tension and the lower fibre is in compression. The neutral fibre y = 0 has zero bending normal stress.

This relation describes the bending normal stress. Neglecting the shear-deformation contribution to deflection does not make shear force or shear stress vanish. Local three-dimensional stresses close to an ideal clamp or a point load are outside this reduced field description.

Separate material and section geometry

For our rectangle, a horizontal strip has area b dy, so I = b∫_(−h/2)^(h/2) y² dy = bh³/12 = 9 × 10⁻⁸ m⁴. Area farther from the neutral axis contributes more because of the y² weighting. The cross-sectional area bh alone does not specify the bending response; the orientation of the rectangle matters.

Bending rigidity is EI = 9000 N m². E describes the material law, I describes section geometry, and EI combines them. None is the beam’s tip force stiffness: for this cantilever that is P/|w(L)| = 3EI/L³, with unit N/m. I is an area moment, not a mass moment of inertia, whose unit would be kg m².

Find the cantilever deflection

Use EIw″ = −P(L−x). The clamp imposes w(0) = 0 and w′(0) = 0: no transverse displacement and no rotation. Two integrations give w′(x) = −P(Lx−x²/2)/(EI) and w(x) = −Px²(3L−x)/(6EI). The integration constants are zero because of those two clamp conditions.

At the free tip, w(L) = −PL³/(3EI) = −0.003 m = −3 mm. The negative sign means downward. The tip slope is −PL²/(2EI) = −0.003. Deflection grows in magnitude toward the tip, while |M| and the bending surface stress are largest at the clamp boundary of the reduced model. The upper-fibre value there is +6 MPa and the lower-fibre value is −6 MPa.

Here |w(L)|/L = 0.002, the largest slope magnitude is 0.003, and the largest bending strain magnitude is 0.00006. These check consistency with the stipulated small regime. They do not establish an elastic limit or allowable load for a real material. The polynomial is exact within the supplied static linear beam model, whose end-detail and material-calibration limits still apply.

Twist a separate circular shaft

Now choose a separate solid circular shaft with length 1 m, radius R = 0.01 m and shear modulus G = 40 GPa. Fix its end at reference z = 0 and apply a torque T = 4π N m about the positive shaft axis at z = 1 m. The support applies the opposite torque. Let φ(z) be the twist angle, positive by the right-hand rule about increasing z. We assume a homogeneous linear elastic shaft, small strain and twist, uniform transmitted torque, and a circular section that stays plane without warping, away from end details.

A radial line at distance r moves tangentially as neighbouring sections rotate relative to one another. The engineering shear strain is γ(r,z) = rφ′(z), and the material law is τ = Gγ. Here φ′ means the spatial twist rate dφ/dz, not angular velocity. Engineering shear strain is twice the tensor shear component; keep the convention when using G.

The torque resultant is T = ∫_A rτ dA = Gφ′J, where J = ∫_A r² dA is the polar second moment of area. For a solid circle, J = ∫_0^(2π)∫_0^R r³ dr dθ = πR⁴/2. Thus φ′ = T/(GJ), τ(r) = Tr/J, and the fixed-end condition gives φ(z) = Tz/(GJ) for the uniform member.

Here J = (π/2) × 10⁻⁸ m⁴ and GJ = 200π N m². The tip twist is 0.020 rad. Surface shear stress is 8 MPa, and the surface engineering shear strain is 0.0002. Shear stress is zero on the axis and rises linearly to the surface. At half-radius it is 4 MPa. For this uniform interior field the surface stress is constant along z, even though φ increases with z.

G is a material modulus; GJ is torsional rigidity. The whole shaft’s torque–angle stiffness is GJ/L, with unit N m/rad when the angle kind is retained. These separate beam and shaft examples do not supply a universal relation E = G. The circular-shaft formula is not a prescription for arbitrary noncircular sections.

Compare bending and torsion

Read the blue circles and curve as beam displacement w(x), violet diamonds and line as bending moment M(x), and orange squares and line as shaft twist φ(z). The marks identify the same five selected fractions of each body’s reference length used in the nearby tables. The curves are the analytic static fields between those rows; the marks are neither measurements nor moving particles.

The vertical quantities and scales differ. The top plot uses millimetres, the middle N m, and the bottom radians. Their slopes on the screen cannot compare a displacement with a moment or a twist. The first two bottom horizontal lines frame lower bounds of −3.5 mm and −40 N m; the dashed guides mark their physical zero values. The shaft axes meet at zero. On narrow screens, scroll the graph locally while the explanation and tables wrap.

The tables round displacement, stress and twist displays to three fractional places. The calculations and graph retain the exact stipulated model values; these displayed digits do not imply measurement precision.

Predict a change before calculating

At fixed geometry and material law, doubling P doubles M, w and bending stress. Changing the section is different: at fixed width, I scales as h³, so tip displacement scales as h⁻³, while surface stress scales as h⁻² because the surface distance h/2 also changes. Doubling the height gives one-eighth the deflection magnitude and one-quarter the surface-stress magnitude.

Rotating the same rectangle exchanges b and h. The area remains 0.0012 m², but I becomes 1.6 × 10⁻⁷ m⁴ and EI becomes 16000 N m². The deflection magnitude is 9/16 of the original and the bending surface stress magnitude is 3/4 of the original. Section orientation changes response even with the same material and area.

For the circular shaft at fixed T, G and length, J scales as R⁴, tip twist as R⁻⁴ and surface shear as R⁻³. Doubling radius gives one-sixteenth the twist and one-eighth the surface stress. These predictions use the same stipulated linear models; none establishes a safe load or extends the formulas to a different section family.

Keep the model boundary visible

The beam coordinates x and y and shaft coordinate z refer to their undeformed reference geometry. These static position-dependent fields are not time histories. Radians retain the angle meaning even though radian measure is dimensionless in SI algebra; strains are fractional changes, and stresses have units Pa.

The examples exclude local clamp and end-load details, large deflection, yielding, buckling, fracture, fatigue, creep, viscoelasticity and dynamics. A real prediction also needs the material data, supports, geometry and relevant failure mechanisms. Here equilibrium, elastic response and section geometry explain the supplied bending and torsion fields within their stated limits.

Worked examples

Try it yourself

1. For the supplied beam, evaluate M and the bending normal stress at y = −h/2, 0 and +h/2 at x = 0 and x = L. State the clamp displacement/slope conditions and distinguish its external reaction couple from M(0).

   Show the answer and reasoning

   At x = 0, M = −36 N m and bottom/neutral/top stresses are −6, 0 and +6 MPa. At x = L, M = 0 and all three bending normal stresses are zero. The clamp has w(0) = w′(0) = 0; its external counterclockwise reaction couple is +36 N m.

   Insert M = −P(L−x) into σ_x = −My/I with tensile stress positive. A positive internal scalar M acts clockwise on the chosen right-hand beam portion at the cut, whereas the external support couple acts on the whole beam. The tip force still requires shear; zero bending normal stress does not set all stress components to zero.

2. Keep the original beam geometry and E but change P to 48 N. Compare the tip displacement and largest absolute bending surface stress under the same stipulated linear assumptions.

   Show the answer and reasoning

   The tip displacement is −6 mm, and the largest absolute bending surface stress is 12 MPa. Both are twice their original magnitudes.

   M, w and σ_x are proportional to P when L, E and the section stay fixed. The upper surface remains tensile and the lower compressive. The ratio is a model prediction, not evidence that the doubled load is safe or that a real material stays elastic.

3. Keep b = 0.04 m and the original P, L and E, but increase h from 0.03 m to 0.06 m. Find the ratios and new tip displacement and largest absolute bending surface stress.

   Show the answer and reasoning

   I increases by 8. Tip displacement magnitude becomes 1/8, giving −0.375 mm. Surface-stress magnitude becomes 1/4, giving 1.5 MPa.

   I = bh³/12 supplies the factor 2³ = 8. Deflection is proportional to 1/I. At the surface, |σ_x| = |M|h/(2I), so the doubled surface distance leaves a net factor 2/8 = 1/4. E is unchanged.

4. Rotate the original rectangle to b = 0.03 m and h = 0.04 m, retaining the same P, L and E. Compare I, EI, tip displacement and bending surface stress. Why is area alone insufficient?

   Show the answer and reasoning

   I = 1.6 × 10⁻⁷ m⁴ and EI = 16000 N m². The tip displacement is −1.6875 mm, with magnitude 9/16 of the original. The largest surface-stress magnitude is 4.5 MPa, or 3/4 of the original.

   Area remains 0.0012 m², but the y² area weighting changes. I_new/I_old = (0.04/0.03)² = 16/9. The new half-height is 0.020 m, so surface stress has the additional distance factor and ratio (4/3)/(16/9) = 3/4.

5. For the supplied circular shaft, find φ at z = 1 m and engineering shear strain and stress at r = R/2. Distinguish their units and meanings.

   Show the answer and reasoning

   Tip φ = 0.020 rad. At half-radius γ = 0.0001 and τ = 4 MPa. The tensor shear component is 0.00005.

   φ′ = T/(GJ) = 0.020 rad/m. Multiplying by length gives the angle; multiplying by radius gives dimensionless engineering shear strain. Gγ gives stress in Pa. Half-radius strain and stress are half their surface values. No angular velocity has been specified.

6. Double the supplied circular shaft’s radius to 0.02 m, keeping T, G and length fixed. Find the new tip angle and surface shear stress, including their ratios.

   Show the answer and reasoning

   J increases by 16. Tip angle is 0.00125 rad, or 1/16 of the original. Surface shear stress is 1 MPa, or 1/8 of the original.

   J = πR⁴/2 gives 2⁴ = 16. Twist is proportional to 1/J. Surface stress T R/J has one extra factor of R, leaving 2/16 = 1/8. The radius powers are conditional on the same circular-shaft model.

What to carry forward

• Choose the cut body and signs before calculating an internal moment or stress.
• E and G are material moduli; EI and GJ also include section geometry.
• The supplied cantilever has a downward deflection and a tensile upper fibre where its signed moment is negative.
• Circular-shaft shear increases with radius, while twist accumulates along the member.
• Height, orientation and radius comparisons depend on the stipulated section family and linear model.