Timoshenko Beam Formulas

Last updated July 27, 2023
By Ian Story

Uniformly Distributed Load

$\Delta_m_a_x$ (at center) $= \frac{5wL^4}{384EI}+\frac{wL^2}{8kAG}$

$\Delta_x = \frac{wx}{24EI}\left(L^3-2Lx^2+x^3\right)+\frac{wx(L-x)}{2kAG}$

Point Load at x

$\Delta_m_a_x$ (does not necessarily occur at point of load application) $=$ TBD

$\Delta_a$ (at point of load) $= \frac{Pa^2b^2}{3EIL}+\frac{Pab}{LkAG}$

$\Delta_x$ (when x < a) $= \frac{Pbx}{6EIL}(L^2-b^2-x^2)+\frac{Pbx}{LkAG}$

$\Delta_x$ (when x > a) $= \frac{Pa(L-x)}{6EIL}(2Lx-a^2-x^2)+\frac{Pa(L-x)}{LkAG}$

Derivations

Step 1

$EI\Phi(x)''' = q(x)$

$\Phi(x)' = \frac{-M(x)}{EI} = \frac{\int\int q(x)dx}{EI} + C_1x + C_2$

Use boundary conditions (typically 2 known moments) to solve for $C_1$ and $C_2$ .

$\int\Phi(x) = \frac{\int\int\int\int q(x)dx}{EI} + \frac{C_1x^3}{6} + \frac{C_2x^2}{2} + C_3x + C_4$

Step 2

$\Delta(x)'' = \Phi(x)' - \frac{q(x)}{k_sAG}$

$\Delta(x) = \int\Phi(x) - \frac{\int\int q(x)dx + C_5x + C_6}{k_sAG}$

$\Delta(x) = \frac{\int\int\int\int q(x)dx}{EI} + \frac{C_1x^3}{6} + \frac{C_2x^2}{2} + C_3x + C_4 - \frac{\int\int q(x)dx + C_5x + C_6}{k_sAG}$

Define: $C_7 = C_3 - \frac{C_5}{k_sAG}$

Define: $C_8 = C_4 - \frac{C_6}{k_sAG}$

$\Delta(x) = \frac{\int\int\int\int q(x)dx}{EI} - \frac{\int\int q(x)dx}{k_sAG} + \frac{C_1x^3}{6} + \frac{C_2x^2}{2} + C_7x + C_8$

Use boundary conditions (typically 2 known deflections) to solve for $C_7$ and $C_8$ .

All 4 constants of integration are now known. Simplify to find clean formulas.

Sources:

Derivation of base equations: Euler-Bernoulli and Timoshenko Beam Theories Analytical and Numerical Comprehensive Revision, by Abdarrhim M. Ahmed and Abdussalam M. Rifai