D
n
33 Ib n ,33 À D
n
11 Ab n À I _
b n ¼
n 0 μ
n
33
D
n
33
I _
ϕ
1
ð Þ þ D
n
11
μ
n
33
D
n
33
À μ
n
11
An 0 ϕ
1
ð Þ ,
b n ,3 ¼ 0, x 3 ¼ 0, L,
b n ¼ 0, t ¼ 0:
ð4:80Þ
Equation (4.80) is a standard mathematical problem. Its solution can be obtained by
separation of variables without any challenge, and is not presented here.
As an example, consider a circular ZnO beam with L ¼ 600 nm and a ¼ 25 nm.
The suddenly applied end shear force f y ¼ 0.2 nN. n 0 ¼ N
þ
D ¼ 10
23 m
À3 . Right after
the load is applied, the beam rises from the horizontal reference state, gradually
reaches its largest deflection, then falls back. The initial rising part of the motion is
shown in Fig. 4.9 for a few time instants. Five terms of the series are used in Fig. 4.9.
This is very accurate for predicting the deflection curve. However, for the potential
and carrier concentration to be calculated next, it is less accurate [4]. We note that
there is a fundamental difference between the extensional waves in Sect. 3.12 and the
bending waves here. When a load is suddenly applied at the right end of the beam, it
generates component waves or modes with different frequencies and wavelengths.
Extensional waves are nondispersive. All waves propagate at the same finite speed.
Therefore the left end feels the initial disturbance at the right end after a finite time
interval. On the contrary, bending waves are dispersive. Long waves propagate
slowly, but short waves propagate quickly. Therefore, for bending, the left end of
the beam quickly feels the initial disturbance at the right end, essentially taking no
time.
We are particularly interested in the development of the electron distribution
under ϕ
(1) . In Fig. 4.10, we show Δn ¼ x 2 n
(1) (x 3 , t) at the four time instants during
the initial rising part of motion in the region of interest, i.e., away from the left end
where there are field concentrations. Figure 4.10 is in fact from a COMSOL
numerical solution of Eq. (4.80) [4]. The analytical solution of Eq. (4.80) looks
Fig. 4.9 Deflection curve at
different time instants.
m ¼ 1–5
106
4 Bending of Beams
n
33 Ib n ,33 À D
n
11 Ab n À I _
b n ¼
n 0 μ
n
33
D
n
33
I _
ϕ
1
ð Þ þ D
n
11
μ
n
33
D
n
33
À μ
n
11
An 0 ϕ
1
ð Þ ,
b n ,3 ¼ 0, x 3 ¼ 0, L,
b n ¼ 0, t ¼ 0:
ð4:80Þ
Equation (4.80) is a standard mathematical problem. Its solution can be obtained by
separation of variables without any challenge, and is not presented here.
As an example, consider a circular ZnO beam with L ¼ 600 nm and a ¼ 25 nm.
The suddenly applied end shear force f y ¼ 0.2 nN. n 0 ¼ N
þ
D ¼ 10
23 m
À3 . Right after
the load is applied, the beam rises from the horizontal reference state, gradually
reaches its largest deflection, then falls back. The initial rising part of the motion is
shown in Fig. 4.9 for a few time instants. Five terms of the series are used in Fig. 4.9.
This is very accurate for predicting the deflection curve. However, for the potential
and carrier concentration to be calculated next, it is less accurate [4]. We note that
there is a fundamental difference between the extensional waves in Sect. 3.12 and the
bending waves here. When a load is suddenly applied at the right end of the beam, it
generates component waves or modes with different frequencies and wavelengths.
Extensional waves are nondispersive. All waves propagate at the same finite speed.
Therefore the left end feels the initial disturbance at the right end after a finite time
interval. On the contrary, bending waves are dispersive. Long waves propagate
slowly, but short waves propagate quickly. Therefore, for bending, the left end of
the beam quickly feels the initial disturbance at the right end, essentially taking no
time.
We are particularly interested in the development of the electron distribution
under ϕ
(1) . In Fig. 4.10, we show Δn ¼ x 2 n
(1) (x 3 , t) at the four time instants during
the initial rising part of motion in the region of interest, i.e., away from the left end
where there are field concentrations. Figure 4.10 is in fact from a COMSOL
numerical solution of Eq. (4.80) [4]. The analytical solution of Eq. (4.80) looks
Fig. 4.9 Deflection curve at
different time instants.
m ¼ 1–5
106
4 Bending of Beams