Construction Technology::All about civil construction.: Beam deflection for various loads and supports

Beams can vary greatly in their geometry and composition. For instance, a beam may be straight or curved. It may be of constant cross section, or it may taper. It may be made entirely of the same material (homogeneous), or it may be composed of different materials (composite). Some of these things make analysis difficult, but many engineering applications involve cases that are not so complicated. Analysis is simplified if:

-The beam is originally straight, and any taper is slight

-The beam experiences only linear elastic deformation

-The beam is slender (its length to height ratio is greater than 10)

-Only small deflections are considered (max deflection less than 1/10 the span.

In this case, the equation governing the beam's deflection (

w

) can be approximated as:

\cfrac{\mathrm{d}^2 w(x)}{\mathrm{d} x^2}=\frac{M(x)}{E(x) I(x)}

where the second derivative of its deflected shape with respect to

x

is interpreted as its curvature,

E

is the Young's modulus,

I

is the area moment of inertia of the cross-section, and

M

is the internal bending moment in the beam.
If, in addition, the beam is not tapered and is homogeneous, and is acted upon by a distributed load

q

, the above expression can be written as:

EI~\cfrac{\mathrm{d}^4 w(x)}{\mathrm{d} x^4} = q(x)

This equation can be solved for a variety of loading and boundary conditions. A number of simple examples are shown below. The formulas expressed are approximations developed for long, slender, homogeneous, prismatic beams with small deflections, and linear elastic properties. Under these restrictions, the approximations should give results within 5% of the actual deflection.

Cantilever beams

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.

End-loaded cantilever beams

Cantilever beam with a force on the free end

The elastic deflection

\delta

and angle of deflection

\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]

\delta_B = \frac {F L^3} {3 E I}

\phi_B = \frac {F L^2} {2 E I}

where

F

= Force acting on the tip of the beam

L

= Length of the beam (span)

E

= Modulus of elasticity

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,

x

, along the span of an end loaded cantilevered beam can be calculated using:^[1]

\delta_x = \frac {F x^2} {6 E I} (3L - x)

\phi_x = \frac {F x} {2 E I} (2L - x)

Note that at

x = L

(the end of the beam), the

\delta_x

and

\phi_x

equations are identical to the

\delta_B

and

\phi_B

equations above.

Uniformly-loaded cantilever beams

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]

\delta_B = \frac {q L^4} {8 E I}

\phi_B = \frac {q L^3} {6 E I}

where

q

= Uniform load on the beam (force per unit length)

L

= Length of the beam

E

= Modulus of elasticity

I

= Area moment of inertia of cross section

The deflection at any point,

x

, along the span of a uniformly loaded cantilevered beam can be calculated using:^[1]

\delta_x = \frac {q x^2} {24 E I}(6L^2 - 4L x + x^2)

\phi_x = \frac {q x} {6 E I}(3L^2 - 3L x + x^2)

Simply-supported beams

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

Schematic of the deflection of a simply-supported beam.

Center-loaded simple beams

Simply-supported beam with a force in the center

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

\delta_C = \frac {F L^3} {48 E I}

where

F

= Force acting on the center of the beam

L

= Length of the beam between the supports

E

= Modulus of elasticity

I

= Area moment of inertia of cross section

The deflection at any point,

x

, along the span of a center loaded simply supported beam can be calculated using:^[1]

\delta_x = \frac {F x} {48 E I}(3L^2 - 4x^2)

for

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

Off-center-loaded simple beams

Simply-supported beam with a force off center

The maximum elastic deflection on a beam supported by two simple supports, loaded at a distance

a

from the closest support, is given by:^[1]

\delta_{max} = \frac {F a (L^2 - a^2)^{3/2}} {9\sqrt{3} L E I}

where

F

= Force acting on the beam

L

= Length of the beam between the supports

E

= Modulus of elasticity

I

= Area moment of inertia of cross section

a

= Distance from the load to the closest support (i.e.

a \leq L/2

)

This maximum deflection occurs at a distance

x_1

from the closest support and is given by:^[1]

x_1 = \sqrt{\frac{L^2 - a^2}{3}}

Uniformly-loaded simple beams

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]

\delta_C = \frac{5 q L^4} {384 E I}

where

q

= Uniform load on the beam (force per unit length)

L

= Length of the beam

E

= Modulus of elasticity

I

= Area moment of inertia of cross section

The deflection at any point,

x

, along the span of a uniformly loaded simply supported beam can be calculated using:^[1]

\delta_x = \frac{q x} {24 E I} (L^3 - 2L x^2 + x^3)

—–

Units

The formulas supplied above require the use of a consistent set of units. Most calculations will be made in SI or US customary units, although there are many other systems of units.