Deflection (engineering)

Cantilever beams have one end fixed, so that the slope and deflection at that end must be zero.

Schematic of the deflection of a cantilever beam.

Cantilever beam with a force on the free end

The elastic deflection ${\displaystyle \delta }$ and angle of deflection ${\displaystyle \phi }$ (in radians) at the free end in the example image: A (weightless) cantilever beam, with an end load, can be calculated (at the free end B) using:[1]

${\displaystyle \delta _{B}={\frac {FL^{3}}{3EI}}}$

${\displaystyle \phi _{B}={\frac {FL^{2}}{2EI}}}$

where

${\displaystyle F}$ = Force acting on the tip of the beam

${\displaystyle L}$ = Length of the beam (span)

${\displaystyle E}$ = Modulus of elasticity

${\displaystyle I}$ = Area moment of inertia of the beam's cross section

Note that if the span doubles, the deflection increases eightfold. The deflection at any point, ${\displaystyle x}$, along the span of an end loaded cantilevered beam can be calculated using:[1]

${\displaystyle \delta _{x}={\frac {Fx^{2}}{6EI}}(3L-x)}$

${\displaystyle \phi _{x}={\frac {Fx}{2EI}}(2L-x)}$

Note: At ${\displaystyle x=L}$ (the end of the beam), the ${\displaystyle \delta _{x}}$ and ${\displaystyle \phi _{x}}$ equations are identical to the ${\displaystyle \delta _{B}}$ and ${\displaystyle \phi _{B}}$ equations above.

Cantilever beam with a uniform distributed load

The deflection, at the free end B, of a cantilevered beam under a uniform load is given by:[1]

${\displaystyle \delta _{B}={\frac {qL^{4}}{8EI}}}$

${\displaystyle \phi _{B}={\frac {qL^{3}}{6EI}}}$

where

${\displaystyle q}$ = Uniform load on the beam (force per unit length)

${\displaystyle L}$ = Length of the beam

${\displaystyle E}$ = Modulus of elasticity

${\displaystyle I}$ = Area moment of inertia of cross section

The deflection at any point, ${\displaystyle x}$, along the span of a uniformly loaded cantilevered beam can be calculated using:[1]

${\displaystyle \delta _{x}={\frac {qx^{2}}{24EI}}(6L^{2}-4Lx+x^{2})}$

${\displaystyle \phi _{x}={\frac {qx}{6EI}}(3L^{2}-3Lx+x^{2})}$

Simply-supported beams have supports under their ends which allow rotation, but not deflection.

Schematic of the deflection of a simply-supported beam.

Simply-supported beam with a force in the center

The deflection at any point, ${\displaystyle x}$, along the span of a center loaded simply supported beam can be calculated using:[1]

${\displaystyle \delta _{x}={\frac {Fx}{48EI}}(3L^{2}-4x^{2})}$

for

${\displaystyle 0\leq x\leq {\frac {L}{2}}}$

The special case of elastic deflection at the midpoint C of a beam, loaded at its center, supported by two simple supports is then given by:[1]

${\displaystyle \delta _{C}={\frac {FL^{3}}{48EI}}}$

where

${\displaystyle F}$ = Force acting on the center of the beam

${\displaystyle L}$ = Length of the beam between the supports

${\displaystyle E}$ = Modulus of Elasticity

${\displaystyle I}$ = Area moment of inertia of cross section

Simply-supported beam with a force off center

The maximum elastic deflection on a beam supported by two simple supports, loaded at a distance ${\displaystyle a}$ from the closest support, is given by:[1]

${\displaystyle \delta _{max}={\frac {Fa(L^{2}-a^{2})^{3/2}}{9{\sqrt {3}}LEI}}}$

where

${\displaystyle F}$ = Force acting on the beam

${\displaystyle L}$ = Length of the beam between the supports

${\displaystyle E}$ = Modulus of Elasticity

${\displaystyle I}$ = Area moment of Inertia of cross-section

${\displaystyle a}$ = Distance from the load to the closest support (i.e. ${\displaystyle a\leq L/2}$)

This maximum deflection occurs at a distance ${\displaystyle x_{1}}$ from the closest support and is given by:[1]

${\displaystyle x_{1}={\sqrt {\frac {L^{2}-a^{2}}{3}}}}$

Simply-supported beam with a uniform distributed load

The elastic deflection (at the midpoint C) on a beam supported by two simple supports, under a uniform load (as pictured) is given by:[1]

${\displaystyle \delta _{C}={\frac {5qL^{4}}{384EI}}}$

Where

${\displaystyle q}$ = Uniform load on the beam (force per unit length)

${\displaystyle L}$ = Length of the beam

${\displaystyle E}$ = Modulus of elasticity

${\displaystyle I}$ = Area moment of inertia of cross section

The deflection at any point, ${\displaystyle x}$, along the span of a uniformly loaded simply supported beam can be calculated using:[1]

${\displaystyle \delta _{x}={\frac {qx}{24EI}}(L^{3}-2Lx^{2}+x^{3})}$

The change in length ${\displaystyle \Delta L}$ of the beam is generally negligible in structures, but can be calculated by integrating the slope ${\displaystyle \theta _{x}}$ function, if the deflection function ${\displaystyle \delta _{x}}$ is known for all ${\displaystyle x}$.

Where:

${\displaystyle \Delta L}$ = change in length (always negative)

${\displaystyle \theta _{x}}$ = slope function (first derivative of ${\displaystyle \delta _{x}}$)

${\displaystyle \Delta L=-{\frac {1}{2}}\int _{0}^{L}(\theta (x))^{2}dx}$ [2]

If the beam is uniform and the deflection at any point is known, this can be calculated without knowing other properties of the beam.

Force: Newtons (${\displaystyle N}$)

Length: meters (${\displaystyle m}$)

Modulus of Elasticity: ${\displaystyle {\frac {N}{m^{2}}}(Pa)}$

Moment of Inertia: ${\displaystyle m^{4}}$

Force: Pounds force (${\displaystyle lb_{f}}$)

Length: Inches (${\displaystyle in}$)

Modulus of Elasticity: ${\displaystyle {\frac {lb_{f}}{in^{2}}}}$

Moment of Inertia: ${\displaystyle in^{4}}$

Other units may be used as well, as long as they are self-consistent. For example, sometimes the kilogram force (${\displaystyle kg_{f}}$) unit is used to measure loads. In such a case, the modulus of elasticity must be converted to ${\displaystyle {\frac {kg_{f}}{m^{2}}}}$.