Beam Deflection

Last updated May 17, 2023
By Ian Story

The governing equation for beams under Euler-Bernoulli beam theory is:

    \[EIv''''(x) = q(x)\]

Integrating four times (and explicitly pulling out the constants of integration) gives:

    \[EIv'''(x) = \int_0^x q(x) + C_1\]

    \[EIv''(x) = \iint_0^x q(x) + C_1x + C_2\]

    \[EIv'(x) = \iiint_0^x q(x) + \frac{C_1x^2}{2} + C_2x + C_3\]

    \[EIv(x) = \iiiint_0^x q(x) + \frac{C_1x^3}{6} + \frac{C_2x^2}{2} + C_3x + C_4\]

Recognizing that:

    \[EIv'''(x) = V(x)\]

    \[EIv''(x) = M(x)\]

    \[v'(x) = \theta(x)\]

We can rewrite the above equations as follows:

    \[V(x) = \int_0^x q(x) + C_1\]

    \[M(x) = \iint_0^x q(x) + C_1x + C_2\]

    \[\theta(x) = \frac{\iiint_0^x q(x)}{EI} + \frac{C_1x^2}{2EI} + \frac{C_2x}{EI} + \frac{C_3}{EI}\]

    \[v(x) = \frac{\iiiint_0^x q(x)}{EI} + \frac{C_1x^3}{6EI} + \frac{C_2x^2}{2EI} + \frac{C_3x}{EI} + \frac{C_4}{EI}\]

By setting x = 0, we find that the constants of integration are related to the initial values for shear, moment, slope, and deflection as follows:

    \[V(0) = \int_0^0 q(x) + C_1 \Rightarrow C_1 = V_0\]

    \[M(0) = \iint_0^0 q(x) + C_1(0) + C_2 \Rightarrow C_2 = M_0\]

    \[\theta(0) = \frac{\iiint_0^0 q(x)}{EI} + \frac{C_1(0)^2}{2EI} + \frac{C_2(0)}{EI} + \frac{C_3}{EI} \Rightarrow \frac{C_3}{EI} = \theta_0\]

    \[v(0) = \frac{\iiiint_0^0 q(x)}{EI} + \frac{C_1(0)^3}{6EI} + \frac{C_2(0)^2}{2EI} + \frac{C_3(0)}{EI} + \frac{C_4}{EI} \Rightarrow \frac{C_4}{EI} = v_0\]

Assume that we know the moments and deflections at each end of the beam. This gives us two of the constants of integration, M_0 and v_0. To solve for remaining constants of integration, we can apply the additional known boundary conditions at the end of the beam. This gives us a system of 2 equations to solve for the 2 unknown constants:

    \[M(L) = \iint_0^L q(x) + V_0L + M_0\]

    \[v(L) = \frac{\iiiint_0^L q(x)}{EI} + \frac{V_0L^3}{6EI} + \frac{M_0L^2}{2EI} + \theta_0L + v_0\]

Solving this system of equations gives:

    \[V_0 = \frac{M(L) - \iint_0^L q(x) - M_0}{L}\]

    \[\theta_0L = v(L) - \frac{\iiiint_0^L q(x)}{EI} - \frac{V_0L^3}{6EI} - \frac{M_0L^2}{2EI} - v_0\]

    \[\theta_0 = \frac{v(L) - \frac{\iiiint_0^L q(x)}{EI} - \frac{V_0L^3}{6EI} - \frac{M_0L^2}{2EI} - v_0}{L}\]

[
\theta_0 = \frac{v(L) – \frac{\iiiint_0^L q(x)}{EI} – \frac{[\frac{M(L) – \iint_0^L q(x) – M_0}{L}]L^3}{6EI} – \frac{M_0L^2}{2EI} – v_0}{L}
] [
\theta_0 = \frac{v(L) – \frac{\iiiint_0^L q(x)}{EI} – \frac{[M(L) – \iint_0^L q(x) – M_0]L^2}{6EI} – \frac{M_0L^2}{2EI} – v_0}{L}
]

Putting this all together into a single equation for v(x):

    \[v(x) = \frac{\iiiint_0^x q(x)}{EI} + \frac{[\frac{M(L) - \iint_0^L q(x) - M_0}{L}]x^3}{6EI} + \frac{M_0x^2}{2EI} + C_3x + v_0\]

sdaf

    \[\begin{bmatrix}0 & 0 & 0 & \frac{1}{EI} \\0 & 1 & 0 & 0 \\\frac{L^3}{6EI} & \frac{L^2}{2EI} & \frac{L}{EI} & \frac{1}{EI} \\L & 1 & 0 & 0\end{bmatrix}\begin{Bmatrix}C_1 \\C_2 \\C_3 \\C_4\end{Bmatrix} = \begin{bmatrix}v(0) - \frac{\iiiint q(0)}{EI} \\M(0) - \iint q(0) \\v(L) - \frac{\iiiint q(L)}{EI} \\M(L) - \iint q(L)\end{bmatrix}\]