Structures

Statically determinate beam

n_r = 3 reactions; V_2 and M_3 from equilibrium alone; do not depend on E or I.

Statically hyperstatic beam

n_r > 3 reactions; one or more redundants. Close the system with compatibility BCs.

Roller support

R_2 only (vertical force). Kinematic BC: u_2 = 0.

Pin support

R_1 and R_2 (both translations restrained). Kinematic BC: u_2 = 0.

Clamp (fixed) support

R_1, R_2, M_A. Kinematic BCs: u_2 = 0 and u_2' = 0.

Cable at angle α

tan α = R_2/R_1; one direction restrained; compatibility u_2(tip) = 0.

Transverse spring at support

R_2 = k u_2 (elastic). NOT u_2 = 0. Compatibility: u_2(spring) = -R/k.

Rotational spring at support

M = k_θ u_2'. Elastic rotation, NOT u_2' = 0.

Differential relation for shear

dV_2/dx_1 = -p_2, where p_2 is distributed load in +x_2 direction.

Differential relation for moment

dM_3/dx_1 = V_2.

Positive V_2 on positive face

Acts in +x_2 direction (up). On negative face, sign flips.

Positive M_3 convention (Bauchau)

Tension on -x_2 side. Sagging under downward load gives M_3 > 0.

Jump in V_2 across point force P

ΔV_2 = +P if force acts in +x_2 direction.

Jump in M_3 across applied couple M_0

ΔM_3 = ±M_0 (sign from FBD).

Internal hinge condition

M_3 = 0 at hinge. u_2' discontinuous. u_2 and V_2 continuous.

Where does max |M_3| occur?

Where V_2 = 0 (smooth interior) or at a discontinuity.

Cut-section method

Cut at x_1. Sum forces on one portion: ΣF_x2 = 0 → V_2. Sum moments at cut → M_3.

Centroid of composite section

x̄_2 = Σ A_i x_2,i / Σ A_i. Same form for x̄_3.

Modulus-weighted centroid

Replace A_i with E_i A_i. Required for multi-material E-B.

Definition of I_22

∫_A x_3² dA. Bending about ī_2 axis.

Definition of I_33

∫_A x_2² dA. Bending about ī_3 axis.

Definition of I_23

∫_A x_2 x_3 dA. Cross-product term; signs matter.

Parallel-axis theorem

I_22^global = I_22^(c) + A · x_3,i². For I_23 use x_2,i · x_3,i (with sign).

When is I_23 = 0?

When section has at least one axis of symmetry. T, I, C, doubly-symmetric sections.

Why I_23 ≠ 0 for Z-section

Point-symmetric but no line symmetry. Both flanges contribute same-sign because (-h/2)(-b/2) = +hb/4.

Centroidal bending stiffness H^c_ij

H^c_ij = ∫_A E x_i x_j dA at modulus-weighted centroid. Single-material: H^c_33 = E·I_33.

Principal axes rotation formula

tan(2θ_p) = 2 H^c_23 / (H^c_33 - H^c_22). Mohr's-circle form.

Why rotate to principal axes?

Decouples bending about ī_2 and ī_3. Constitutive law becomes diagonal.

EB1

Plane sections remain plane.

EB2

Plane sections normal to deformed neutral axis. No transverse shear: γ_12 = 0.

EB3

Small displacements / rotations. Undeformed equilibrium. u_2'' ≈ κ.

EB4

Lateral stresses σ_2, σ_3 negligible vs σ_1. Uniaxial stress at each fiber.

When does EB2 fail?

Short, stubby beams (L/h ≲ 5). Use Timoshenko.

When does EB3 fail?

Large rotations, lateral buckling. Use nonlinear beam-column theory.

Does E-B require linear elasticity?

No. E-B is kinematic only. Beam can be plastic, composite, or damaged.

E-B beam ODE

H^c_33 u_2'''' = -p_2(x_1), or equivalently H^c_33 u_2'' = M_3.

BC for clamp

u_2 = 0 and u_2' = 0.

BC for pin or roller

u_2 = 0 and M_3 = H^c_33 u_2'' = 0.

BC for free end (no tip load)

V_2 = H^c_33 u_2''' = 0 and M_3 = H^c_33 u_2'' = 0.

BC for tip transverse spring k

V_2 = -k u_2 at tip.

BC for tip rotational spring k_θ

M_3 = -k_θ u_2' at tip.

Cantilever, tip load P (down)

u_2(L) = -PL³/(3 EI). Slope u_2'(L) = -PL²/(2 EI).

Cantilever, UDL p_0

u_2(L) = -p_0 L⁴/(8 EI). Slope u_2'(L) = -p_0 L³/(6 EI).

SS beam, center load P

u_2(L/2) = -PL³/(48 EI). Slope at supports = ±PL²/(16 EI).

SS beam, UDL p_0

u_2(L/2) = -5 p_0 L⁴/(384 EI). Slope at supports = ±p_0 L³/(24 EI).

Propped cantilever, UDL p_0: reactions

R_B = 3 p_0 L/8 (tip roller). R_A = 5 p_0 L/8 (clamp). M_A = -p_0 L²/8.

Propped cantilever, UDL p_0: peak positive moment

M_3 = +9 p_0 L²/128 at x_1 = 5L/8.

6 generalized strains (3D beam)

ε_1^c = u_1^c', κ_1 = φ_1', κ_2 = -u_3'', κ_3 = -u_2''. Plus 2 transverse shears (zero in E-B).

6 stress resultants (3D beam)

N_1 = ∫σ_1 dA; M_2 = ∫x_3 σ_1 dA; M_3 = -∫x_2 σ_1 dA; V_2 = ∫τ_12 dA; V_3 = ∫τ_13 dA; M_1 = ∫(x_2 τ_13 - x_3 τ_12) dA.

Sectional constitutive law (principal frame)

N_1 = S^c ε_1^c; M_2 = H^c_22 κ_2; M_3 = H^c_33 κ_3. Diagonal when H^c_23 = 0.

Axial-bending decoupling condition

Origin at modulus-weighted centroid (∫E x_2 dA = ∫E x_3 dA = 0).

Two-bending decoupling condition

Axes are principal: H^c_23 = 0.

Bending-torsion decoupling condition

Loaded through the shear center. Coincides with centroid for doubly-symmetric sections.

Axial stiffness S^c

S^c = ∫_A E dA. Single-material: S^c = EA.

Coupling stiffness H^c_23

H^c_23 = ∫_A E x_2 x_3 dA. Nonzero for unsymmetric sections (e.g., L, Z).

Unsymmetric bending stress formula

σ_1 = N_1/S^c + [(H^c_33 M_2 + H^c_23 M_3) x_3 - (H^c_22 M_3 + H^c_23 M_2) x_2] / (H^c_22 H^c_33 - (H^c_23)²).

Neutral axis definition

Locus where σ_1 = 0.

Neutral axis angle (principal frame, pure bending)

tan β = (M_3/H^c_33) / (M_2/H^c_22). Angle from ī_2.

Why neutral axis ≠ moment vector axis?

Equal only when H^c_22 = H^c_33 (e.g., circle). Otherwise, axes tilt differently.

Where is max stress in unsymmetric bending?

At the point with greatest PERPENDICULAR distance from neutral axis (NOT from centroid).

What does Timoshenko relax?

EB2. Cross-section rotation φ becomes independent of u_2'.

Timoshenko kinematics

u_1(x_1, x_2) = u_1^c(x_1) - x_2 φ(x_1). Plane sections still plane.

Timoshenko shear strain γ

γ = u_2' - φ. In E-B, γ ≡ 0 forces φ = u_2'.

Timoshenko constitutive: M_3

M_3 = -EI φ'.

Timoshenko constitutive: V_2

V_2 = kGA γ = kGA(u_2' - φ).

Shear correction factor k: rectangle

k = 5/6.

Shear correction factor k: I-section

k ≈ 0.9.

Timoshenko combined ODE

EI u_2'''' = p_2 - (EI/kGA) p_2''. Reduces to E-B as kGA → ∞.

Timoshenko midspan deflection (SS, UDL)

δ_Timo = δ_EB + qL²/(8 kGA). The added term is the shear correction.

Ratio δ_shear/δ_EB (rectangle, SS UDL)

(4/5)(E/kG)(h/L)². Becomes significant for short beams.

L/h threshold for 5% Timoshenko correction

L/h ≈ 7 for steel rectangle (E/kG = 3). Below this, E-B underestimates by >5%.

Cantilever shear force under uniform distributed load p₀

V(x₁) = p₀(L − x₁)

Cantilever bending moment under uniform distributed load p₀

M(x₁) = (p₀/2)(L − x₁)²

Centroid condition for a thin-walled section

∫ x₃ t ds = 0

Semicircular arc contribution to first moment about centroidal x₂-axis

∫arc x₃ ds = −2R²

Two flange contribution to first moment

2∫₀ᵈ x₃ dx₃ = d²

Condition for centroid to lie on the vertical web

d² − 2R² = 0

Required flange length for centroid location shown

d = √2 R

Bending stiffness about x₂-axis

H₂₂ᶜ = E∫ x₃² t ds

Semicircular arc contribution to H₂₂ᶜ

(π/2)EtR³

Flange contribution to H₂₂ᶜ

(2/3)Etd³

Total bending stiffness before substitution

H₂₂ᶜ = Et[(πR³)/2 + (2d³)/3]

Total bending stiffness using d = √2R

H₂₂ᶜ = EtR³[(π/2) + (4√2)/3]

Axial bending stress

σ₁₁ = (M/H₂₂ᶜ)Ex₃

Axial flow definition

n = tσ₁₁

Axial flow expression

n = (EtM/H₂₂ᶜ)x₃