Mechanical & Structural Engineering
Timoshenko Deep-Beam Deflection Formula: Bending Plus Transverse Shear
Add transverse shear deformation F*L / (4*kappa*G*A) to Euler-Bernoulli bending F*L^3 / (48*E*I) for short deep beams and sandwich cores in Excel and Arte Sheets.
Predict the exact 20% shear deflection missed by Euler-Bernoulli on a deep beam
Standard Euler-Bernoulli beam formulas w_EB = F * L^3 / (48 * E * I) assume infinite transverse shear rigidity, which works for slender beams (L / h > 15) but underpredicts deflection on short transfer girders (L / h <= 5) and low-G composite cores. Timoshenko (1921) adds transverse shear sag w_sh = F * L / (4 * kappa * G * A) with rectangular shear correction factor kappa = 5/6 (0.8333).
Verify by hand on a rectangular steel transfer block: width b = 0.08 m (B2), depth h = 0.20 m (B3), span L = 0.80 m (B4, aspect L/h = 4), central load F = 120,000 N (B5), E = 200 GPa (B6), G = 75 GPa (B7), and kappa = 5/6 (B8 = 0.8333333333). Area A = 0.016 m^2 and inertia I = 0.08 * 0.008 / 12 = 0.00005333 m^4 give Euler-Bernoulli w_EB = 0.1200 mm, transverse shear w_sh = 0.0240 mm (+20% extra!), and Timoshenko total w_tot = 0.1440 mm.
| Deflection Component (L/h = 4) | Formula (in mm) | Exact Output |
|---|---|---|
| Euler-Bernoulli Bending w_EB | 1000 * F * L^3 / (48 * E_GPa*10^9 * (b*h^3/12)) | 0.1200 mm |
| Timoshenko Shear w_shear | 1000 * F * L / (4 * kappa * G_GPa*10^9 * (b*h)) | 0.0240 mm |
| Total Timoshenko Deflection | w_EB + w_shear | 0.1440 mm |
| Shear-to-Bending Ratio | w_shear / w_EB = (E / (kappa*G)) * (h / L)^2 | 0.2000 (20.0% shear boost) |
Total Timoshenko Midspan Deflection (mm) Formula
Compute cross-sectional area A = b * h and second moment I = b * h^3 / 12 inside LET, converting GPa moduli to Pascals (* 10^9) and meter deflections to millimeters (* 1000).
=LET(b_m, B2, h_m, B3, l_m, B4, f_n, B5, e_pa, B6 * 10^9, g_pa, B7 * 10^9, kap, B8, IF(OR(b_m <= 0, h_m <= 0, l_m <= 0, e_pa <= 0, g_pa <= 0, kap <= 0), "Invalid Beam Input", LET(area_s, b_m * h_m, iner_i, b_m * (h_m^3) / 12, w_eb, (f_n * (l_m^3)) / (48 * e_pa * iner_i), w_sh, (f_n * l_m) / (4 * kap * g_pa * area_s), ROUND(1000 * (w_eb + w_sh), 4))))Dimensionless Shear-to-Bending Ratio & Slenderness Check
Notice that w_shear / w_EB simplifies algebraically to (E / (kappa * G)) * (h / L)^2 = (200 / 62.5) * (0.20 / 0.80)^2 = 3.2 * 0.0625 = 0.2000. Double the span L in B4 from 0.80 m to 1.60 m (L / h = 8): bending grows 8x to 0.9600 mm while shear grows only 2x to 0.0480 mm (ratio drops 4x to 0.0500).
=ROUND((B6 / (B8 * B7)) * ((B3 / B4)^2), 4)Try the example
Use the original values or change them to test your understanding.