Bending of a cantilevered beam
The virtual work done by the external applied forces in moving through the virtual displacement
δ
w
(
x
)
{\displaystyle \delta w(x)\,}
is given by
δ
W
ext
=
∫
0
L
q
δ
w
d
x
+
P
δ
w
(
L
)
{\displaystyle \delta W_{\text{ext}}=\int _{0}^{L}q~\delta w~dx+P~\delta w(L)}
The work done by the internal forces are,
δ
W
int
=
∫
R
δ
U
d
V
=
∫
0
L
∫
S
δ
(
1
2
σ
i
j
ε
i
j
)
d
A
d
x
=
∫
0
L
∫
S
σ
i
j
δ
ε
i
j
d
A
d
x
{\displaystyle {\begin{aligned}\delta W_{\text{int}}&=\int _{\mathcal {R}}\delta U~dV\\&=\int _{0}^{L}\int _{\mathcal {S}}\delta \left({\frac {1}{2}}\sigma _{ij}\varepsilon _{ij}\right)~dA~dx\\&=\int _{0}^{L}\int _{\mathcal {S}}\sigma _{ij}\delta \varepsilon _{ij}~dA~dx\end{aligned}}}
From beam theory, the displacement field at a point in the beam is
given by
u
=
−
z
d
w
d
x
;
v
=
0
;
w
=
w
(
x
)
{\displaystyle u=-z{\frac {dw}{dx}}~~;~~v=0~~;~~w=w(x)}
The strains are, neglecting Poisson effects,
ε
x
x
=
∂
u
∂
x
=
−
z
d
2
w
d
x
2
;
ε
y
y
=
0
;
ε
z
z
=
0
{\displaystyle \varepsilon _{xx}={\frac {\partial u}{\partial x}}=-z{\frac {d^{2}w}{dx^{2}}}~~;~~\varepsilon _{yy}=0~~;~~\varepsilon _{zz}=0}
and the corresponding stresses are
σ
x
x
=
E
ε
x
x
=
−
E
z
d
2
w
d
x
2
;
σ
y
y
=
0
;
σ
z
z
=
0
{\displaystyle \sigma _{xx}=E\varepsilon _{xx}=-Ez{\frac {d^{2}w}{dx^{2}}}~~;~~\sigma _{yy}=0~~;~~\sigma _{zz}=0}
If we also neglect the shear strains and stresses, we get
δ
W
int
=
=
∫
0
L
∫
S
σ
x
x
δ
ε
x
x
d
A
d
x
=
∫
0
L
∫
S
E
ε
x
x
δ
ε
x
x
d
A
d
x
=
∫
0
L
E
d
2
w
d
x
2
d
2
(
δ
w
)
d
x
2
(
∫
S
z
2
d
A
)
d
x
=
∫
0
L
E
I
d
2
w
d
x
2
d
2
(
δ
w
)
d
x
2
d
x
{\displaystyle {\begin{aligned}\delta W_{\text{int}}=&=\int _{0}^{L}\int _{\mathcal {S}}\sigma _{xx}\delta \varepsilon _{xx}~dA~dx\\&=\int _{0}^{L}\int _{\mathcal {S}}E~\varepsilon _{xx}\delta \varepsilon _{xx}~dA~dx\\&=\int _{0}^{L}E{\frac {d^{2}w}{dx^{2}}}{\frac {d^{2}(\delta w)}{dx^{2}}}\left(\int _{\mathcal {S}}z^{2}~dA\right)~dx\\&=\int _{0}^{L}E~I{\frac {d^{2}w}{dx^{2}}}{\frac {d^{2}(\delta w)}{dx^{2}}}dx\end{aligned}}}
Therefore, from the principle of virtual work,
δ
W
=
∫
0
L
(
E
I
d
2
w
d
x
2
d
2
(
δ
w
)
d
x
2
+
q
δ
w
)
d
x
+
P
δ
w
(
L
)
=
0
{\displaystyle \delta W=\int _{0}^{L}\left(E~I{\frac {d^{2}w}{dx^{2}}}{\frac {d^{2}(\delta w)}{dx^{2}}}+q~\delta w\right)~dx+P~\delta w(L)=0}
Integrating by parts and after some manipulation, we get,
0
=
∫
0
L
[
d
2
d
x
2
(
E
I
d
2
w
d
x
2
)
+
q
+
P
δ
(
L
−
x
)
]
δ
w
d
x
+
[
(
E
I
d
2
w
d
x
2
)
δ
(
d
w
d
x
)
+
d
d
x
(
E
I
d
2
w
d
x
2
)
δ
w
]
|
0
L
{\displaystyle {\begin{aligned}0&=\int _{0}^{L}\left[{\frac {d^{2}}{dx^{2}}}\left(E~I{\frac {d^{2}w}{dx^{2}}}\right)+q+P\delta (L-x)\right]\delta w~dx+\left.\left[\left(E~I{\frac {d^{2}w}{dx^{2}}}\right)\delta \left({\frac {dw}{dx}}\right)+{\frac {d}{dx}}\left(E~I{\frac {d^{2}w}{dx^{2}}}\right)\delta w\right]\right|_{0}^{L}\end{aligned}}}
where
δ
(
L
−
x
)
{\displaystyle \delta (L-x)}
is the Dirac delta function,
∫
−
∞
∞
δ
(
L
−
x
)
f
(
x
)
d
x
=
f
(
L
)
{\displaystyle \int _{-\infty }^{\infty }\delta (L-x)~f(x)~dx=f(L)}
The Euler equation for the beam is, therefore,
d
2
d
x
2
(
E
I
d
2
w
d
x
2
)
+
q
+
P
δ
(
L
−
x
)
=
0
{\displaystyle {\frac {d^{2}}{dx^{2}}}\left(E~I{\frac {d^{2}w}{dx^{2}}}\right)+q+P\delta (L-x)=0}
and the boundary conditions are
E
I
d
2
w
d
x
2
(
L
)
=
0
d
d
x
(
E
I
d
2
w
d
x
2
)
|
x
=
L
=
0
{\displaystyle {\begin{aligned}E~I{\frac {d^{2}w}{dx^{2}}}(L)&=0\\\left.{\frac {d}{dx}}\left(E~I{\frac {d^{2}w}{dx^{2}}}\right)\right|_{x=L}&=0\end{aligned}}}
Application of the Hellinger-Prange-Reissner variational principle [ edit | edit source ]
The governing equations of the cantilever beam can be written as
κ
=
d
2
w
d
x
2
;
w
(
0
)
=
0
;
d
w
d
x
|
x
=
0
=
0
{\displaystyle \kappa ={\frac {d^{2}w}{dx^{2}}}~~;~~w(0)=0~~;~~\left.{\frac {dw}{dx}}\right|_{x=0}=0}
M
=
E
I
κ
{\displaystyle M=EI\kappa }
d
2
M
d
x
2
+
q
+
P
δ
(
L
−
x
)
=
0
;
M
(
L
)
=
0
;
d
M
d
x
|
x
=
L
=
0
{\displaystyle {\frac {d^{2}M}{dx^{2}}}+q+P\delta (L-x)=0~~;~~M(L)=0~~;~~\left.{\frac {dM}{dx}}\right|_{x=L}=0}
Recall that the Hellinger-Prange-Reissner functional is given by
H
[
s
]
=
∫
B
U
c
(
σ
)
−
∫
B
σ
:
ε
d
V
−
∫
B
f
∙
u
d
V
+
∫
∂
B
u
t
∙
(
u
−
u
^
)
d
A
+
∫
∂
B
t
t
^
∙
u
d
A
{\displaystyle {\mathcal {H}}[s]=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }})-\int _{\mathcal {B}}{\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}~dV-\int _{\mathcal {B}}\mathbf {f} \bullet \mathbf {u} ~dV+\int _{\partial {\mathcal {B}}^{u}}\mathbf {t} \bullet (\mathbf {u} -{\widehat {\mathbf {u} }})~dA+\int _{\partial {\mathcal {B}}^{t}}{\widehat {\mathbf {t} }}\bullet \mathbf {u} ~dA}
If we apply the strain-displacement constraints using the Lagrange multipliers
λ
{\displaystyle {\boldsymbol {\lambda }}\,}
and the displacement boundary conditions using the Lagrange multipliers
μ
{\displaystyle {\boldsymbol {\mu }}\,}
, we get a modified functional
H
¯
[
u
,
ε
,
λ
,
μ
]
=
∫
B
[
U
(
ε
)
+
λ
:
[
1
2
(
∇
u
−
∇
u
T
)
−
ε
]
−
f
∙
u
]
d
V
−
∫
∂
B
u
μ
∙
(
u
−
u
^
)
d
A
−
∫
∂
B
t
t
^
∙
u
d
A
{\displaystyle {\bar {\mathcal {H}}}[\mathbf {u} ,{\boldsymbol {\varepsilon }},{\boldsymbol {\lambda }},{\boldsymbol {\mu }}]=\int _{\mathcal {B}}\left[U({\boldsymbol {\varepsilon }})+{\boldsymbol {\lambda }}:[{\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})-{\boldsymbol {\varepsilon }}]-\mathbf {f} \bullet \mathbf {u} \right]~dV-\int _{\partial {\mathcal {B}}^{u}}{\boldsymbol {\mu }}\bullet (\mathbf {u} -{\widehat {\mathbf {u} }})~dA-\int _{\partial {\mathcal {B}}^{t}}{\widehat {\mathbf {t} }}\bullet \mathbf {u} ~dA}
For the cantilevered beam, the above functional becomes
H
¯
[
w
,
κ
,
λ
,
μ
1
,
μ
2
]
=
∫
0
L
[
E
I
2
κ
2
+
(
d
2
w
d
x
2
−
κ
)
λ
+
[
q
+
P
δ
(
L
−
x
)
]
w
]
d
x
−
M
(
L
)
d
w
d
x
(
L
)
−
d
M
d
x
(
L
)
w
(
L
)
+
μ
1
[
w
(
0
)
−
0
]
+
μ
2
[
d
w
d
x
(
0
)
−
0
]
{\displaystyle {\begin{aligned}{\bar {\mathcal {H}}}[w,\kappa ,\lambda ,\mu _{1},\mu _{2}]=&\int _{0}^{L}\left[{\frac {EI}{2}}\kappa ^{2}+\left({\frac {d^{2}w}{dx^{2}}}-\kappa \right)\lambda +[q+P\delta (L-x)]~w\right]~dx\\&-M(L){\frac {dw}{dx}}(L)-{\frac {dM}{dx}}(L)~w(L)+\mu _{1}[w(0)-0]+\mu _{2}[{\frac {dw}{dx}}(0)-0]\end{aligned}}}
Taking the first variation of the functional, we can easily derive the Euler equations and the associated BCs.
δ
κ
:
E
I
κ
−
λ
=
0
δ
w
:
d
2
λ
d
x
2
+
q
+
P
δ
(
L
−
x
)
=
0
δ
λ
:
d
2
w
d
x
2
−
κ
=
0
{\displaystyle {\begin{aligned}\delta \kappa :~~~&EI\kappa -\lambda =0\\\delta w:~~~&{\frac {d^{2}\lambda }{dx^{2}}}+q+P\delta (L-x)=0\\\delta \lambda :~~~&{\frac {d^{2}w}{dx^{2}}}-\kappa =0\end{aligned}}}
and
d
d
x
(
δ
w
)
:
λ
(
0
)
=
μ
2
,
λ
(
L
)
=
M
(
L
)
δ
w
:
d
λ
d
x
(
0
)
=
−
μ
1
,
d
λ
d
x
(
L
)
=
d
M
d
x
(
L
)
δ
μ
1
:
w
(
0
)
=
0
δ
μ
1
:
d
w
d
x
(
0
)
=
0
{\displaystyle {\begin{aligned}{\frac {d}{dx}}(\delta w):~~~&\lambda (0)=\mu _{2}~~,~~\lambda (L)=M(L)\\\delta w:~~~&{\frac {d\lambda }{dx}}(0)=-\mu _{1}~~,~~{\frac {d\lambda }{dx}}(L)={\frac {dM}{dx}}(L)\\\delta \mu _{1}:~~~&w(0)=0\\\delta \mu _{1}:~~~&{\frac {dw}{dx}}(0)=0\end{aligned}}}
The same process can be used to derive Euler equations using the
Hu-Washizu variational principle.