In radio astronomy, the word "beam" is used to refer to the response pattern of an antenna to a signal. In this context, astronomers speak of beam convolution,beam solid angle,beam width.
The more common use of the word "beam" is to refer to a bar (or rod, shaft, cantilever, etc.) under bending. A beam with one end clamped and the other end free is called a cantilever.
Let a beam have length L, width a, and thickness h. Also let the vertical displacement be
, the horizontal displacement be 
,and the angular displacement of the beam from the original horizontal (called the neutral surface or axis), and set up a Cartesian coordinate system along the neutral surface. The angular displacement is then given by
The longitudinal strain is
and the general strain component is
If q(x) is the downward force per unit area and V(x) is the shearing force per unit length, then force balance gives
or
Similarly, let M[i/] be the torque (a.k.a. bending moment) per unit length and [i]P the horizontal force (i.e.,tension) per unit area. Then torque balance gives
Taking the derivative of (7) and plugging in (5) then gives
But M is given by
where
is the z-component of the longitudinal stress,and
where E is Young's modulus and
is the Poisson ratio.Plugging in,
where I is the geometrical moment of inertia, given by
The geometrical moment of inertia per unit width is then
so
where the flexural rigidity is defined by
Plugging (14) into (8) gives
Similarly, plugging (14) into (7) gives
so
If the tension (i.e., horizontal force) vanishes, then
so
If
is furthermore a constant (equal to the weight of the bar per unit length), then solving the fourth-order ordinary differential equation (20) for 
gives
For a massless beam,
and the solution to (20) becomes
The following table summarizes the various constraints on placed by various boundary conditions on the bar.
[Ctrl+A 全部选择 提示：你可先修改部分代码，再按运行]
The more common use of the word "beam" is to refer to a bar (or rod, shaft, cantilever, etc.) under bending. A beam with one end clamped and the other end free is called a cantilever.
Let a beam have length L, width a, and thickness h. Also let the vertical displacement be
, the horizontal displacement be ,and the angular displacement of the beam from the original horizontal (called the neutral surface or axis), and set up a Cartesian coordinate system along the neutral surface. The angular displacement is then given by (1)
The longitudinal strain is
(2)
and the general strain component is
(3)
If q(x) is the downward force per unit area and V(x) is the shearing force per unit length, then force balance gives
(4)
or
(5)
Similarly, let M[i/] be the torque (a.k.a. bending moment) per unit length and [i]P the horizontal force (i.e.,tension) per unit area. Then torque balance gives
(6)
(7)
Taking the derivative of (7) and plugging in (5) then gives
(8)
But M is given by
(9)
where
is the z-component of the longitudinal stress,and(10)
where E is Young's modulus and
is the Poisson ratio.Plugging in,(11)
where I is the geometrical moment of inertia, given by
(12)
The geometrical moment of inertia per unit width is then
(13)
so
(14)
where the flexural rigidity is defined by
(15)
Plugging (14) into (8) gives
(16)
(17)
Similarly, plugging (14) into (7) gives
(18)
so
(19)
If the tension (i.e., horizontal force) vanishes, then
so(20)
(21)
If
is furthermore a constant (equal to the weight of the bar per unit length), then solving the fourth-order ordinary differential equation (20) for gives(22)
For a massless beam,
and the solution to (20) becomes(23)
The following table summarizes the various constraints on
placed by various boundary conditions on the bar.[Ctrl+A 全部选择 提示：你可先修改部分代码，再按运行]
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