In this case, we consider an infinitesimal slice of the bar and
perform a balance of forces on that slice (see Figure 1).
Figure 1. Continuum approach for the axial loading of a bar.
A balance of forces gives
A
σ
=
q
(
x
)
Δ
x
+
A
(
σ
+
Δ
σ
)
⟹
A
(
(
σ
+
Δ
σ
)
−
σ
Δ
x
)
+
q
(
x
)
=
0
.
{\displaystyle A{\boldsymbol {\sigma }}=\mathbf {q} (\mathbf {x} )~\Delta x+A({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})\qquad \implies \qquad A\left({\cfrac {({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})-{\boldsymbol {\sigma }}}{\Delta x}}\right)+\mathbf {q} (\mathbf {x} )=0~.}
Taking the limit as
Δ
x
→
0
{\displaystyle \Delta x\rightarrow 0}
A
lim
Δ
x
→
0
(
(
σ
+
Δ
σ
)
−
σ
Δ
x
)
+
q
(
x
)
=
0
⟹
A
d
σ
d
x
+
q
(
x
)
=
0
.
{\displaystyle A\lim _{\Delta x\rightarrow 0}\left({\cfrac {({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})-{\boldsymbol {\sigma }}}{\Delta x}}\right)+\mathbf {q} (\mathbf {x} )=0\qquad \implies \qquad {A{\cfrac {d{\boldsymbol {\sigma }}}{dx}}+\mathbf {q} (\mathbf {x} )=0~.}}
This is a differential equation written in terms of stress. We want to
express it in terms of displacements. How do we do this?
First, we convert the stresses into strains using the constitutive equation
(
σ
=
E
ε
{\displaystyle {\boldsymbol {\sigma }}=E{\boldsymbol {\varepsilon }}}
A
E
d
ε
d
x
+
q
(
x
)
=
0
.
{\displaystyle {AE{\cfrac {d{\boldsymbol {\varepsilon }}}{dx}}+\mathbf {q} (\mathbf {x} )=0~.}}
Then, we convert the strains into displacements using the
strain-displacement relations :
ε
:=
lim
l
→
0
Δ
l
l
⟹
ε
=
lim
Δ
x
→
0
(
u
+
Δ
u
)
−
u
Δ
x
≡
ε
=
d
u
d
x
.
{\displaystyle {\boldsymbol {\varepsilon }}:=\lim _{l\rightarrow 0}{\cfrac {\Delta l}{l}}\qquad \implies \qquad {\boldsymbol {\varepsilon }}=\lim _{\Delta x\rightarrow 0}{\cfrac {(\mathbf {u} +\Delta \mathbf {u} )-\mathbf {u} }{\Delta x}}\equiv {{\boldsymbol {\varepsilon }}={\cfrac {d\mathbf {u} }{dx}}}~.}
The differential equation in terms of the displacement
u
{\displaystyle \mathbf {u} }
A
E
d
2
u
d
x
2
+
q
(
x
)
=
0
.
{\displaystyle {AE{\cfrac {d^{2}\mathbf {u} }{dx^{2}}}+\mathbf {q} (\mathbf {x} )=0~.}}
Since
q
(
x
)
=
a
x
{\displaystyle \mathbf {q} (\mathbf {x} )=a\mathbf {x} }
inhomogeneous ordinary differential equation for the displacements in the bar.
A
E
d
2
u
d
x
2
+
a
x
=
0
.
{\displaystyle {AE{\cfrac {d^{2}\mathbf {u} }{dx^{2}}}+a\mathbf {x} =0~.}}
If we can find the displacements in the bar, then we can solve for the strains and therefore the stresses. But to do that, we have to solve the governing differential equation .
To get a unique solution, we need to provide boundary conditions .
In this case, these are
At
x
=
0
{\displaystyle \mathbf {x} =0}
u
=
0
{\displaystyle \mathbf {u} =0}
essential BC.
At
x
=
L
{\displaystyle \mathbf {x} =L}
f
=
R
{\displaystyle \mathbf {f} =\mathbf {R} }
natural BC. Since the ODE is in terms of
u
{\displaystyle \mathbf {u} }
u
{\displaystyle \mathbf {u} }
f
=
A
σ
=
A
E
ε
=
A
E
d
u
d
x
.
{\displaystyle \mathbf {f} =A{\boldsymbol {\sigma }}=AE{\boldsymbol {\varepsilon }}=AE{\cfrac {d\mathbf {u} }{dx}}~.}
Therefore, the BCs become
At
x
=
0
{\displaystyle \mathbf {x} =0}
u
=
0
{\displaystyle \mathbf {u} =0}
At
x
=
L
{\displaystyle \mathbf {x} =L}
A
E
d
u
d
x
=
R
{\displaystyle AE{\cfrac {d\mathbf {u} }{dx}}=\mathbf {R} }
The problem can then be stated as
Find
u
(
x
)
such that it satisfies
A
E
d
2
u
d
x
2
=
−
a
x
u
(
0
)
=
0
d
u
d
x
|
x
=
L
=
R
A
E
{\displaystyle {\begin{aligned}{\text{Find}}~&\mathbf {u} (\mathbf {x} )~{\text{such that it satisfies}}\\&AE{\cfrac {d^{2}\mathbf {u} }{dx^{2}}}=-a\mathbf {x} \\&\mathbf {u} (0)=0\\&\left.{\cfrac {d\mathbf {u} }{dx}}\right|_{\mathbf {x} =L}={\cfrac {\mathbf {R} }{AE}}\end{aligned}}}
The differential formulation is also called the strong form of the
problem.
The analytical solution strategy is as follows:
Set right hand side to zero and solve ( Homogeneous solution ).
Find one solution with RHS =
−
a
x
{\displaystyle -a\mathbf {x} }
Particular solution ).
General solution = homogeneous + particular.
Apply BCs. The homogeneous ODE is
A
E
d
2
u
d
x
2
=
0
.
{\displaystyle AE{\cfrac {d^{2}\mathbf {u} }{dx^{2}}}=0~.}
Integrate twice to get the homogeneous solution
A
E
u
=
C
1
x
+
C
2
⟹
u
(
x
)
=
C
1
x
+
C
2
A
E
.
{\displaystyle AE\mathbf {u} =C_{1}\mathbf {x} +C_{2}\quad \implies \quad \mathbf {u} (\mathbf {x} )={\cfrac {C_{1}\mathbf {x} +C_{2}}{AE}}~.}
The ODE is
A
E
d
2
u
d
x
2
=
−
a
x
.
{\displaystyle AE{\cfrac {d^{2}\mathbf {u} }{dx^{2}}}=-a\mathbf {x} ~.}
Integrate twice to get the particular solution (and assume that the constants of integration are zero)
A
E
u
=
−
a
x
3
6
⟹
u
(
x
)
=
−
a
x
3
6
A
E
.
{\displaystyle AE\mathbf {u} =-{\cfrac {a\mathbf {x} ^{3}}{6}}\quad \implies \quad \mathbf {u} (\mathbf {x} )=-{\cfrac {a\mathbf {x} ^{3}}{6AE}}~.}
Add the homogeneous and particular parts to get
u
(
x
)
=
−
a
x
3
+
D
1
x
+
D
2
6
A
E
.
{\displaystyle \mathbf {u} (\mathbf {x} )={\cfrac {-a\mathbf {x} ^{3}+D_{1}\mathbf {x} +D_{2}}{6AE}}~.}
At
x
=
0
{\displaystyle \mathbf {x} =0}
u
=
0
{\displaystyle \mathbf {u} =0}
D
2
=
0
{\displaystyle D_{2}=0}
At
x
=
L
{\displaystyle \mathbf {x} =L}
d
u
d
x
=
−
3
a
L
2
+
D
1
6
A
E
=
R
A
E
⟹
D
1
=
6
R
+
3
a
L
2
.
{\displaystyle {\cfrac {d\mathbf {u} }{dx}}={\cfrac {-3aL^{2}+D_{1}}{6AE}}={\cfrac {R}{AE}}\quad \implies \quad D_{1}=6R+3aL^{2}~.}
The displacement field in the bar is
u
(
x
)
=
−
a
x
3
+
(
6
R
+
3
a
L
2
)
x
6
A
E
.
{\displaystyle {\mathbf {u} (\mathbf {x} )={\cfrac {-a\mathbf {x} ^{3}+(6\mathbf {R} +3aL^{2})\mathbf {x} }{6AE}}~.}}
The strain in the bar is
ε
(
x
)
=
−
a
x
2
+
(
2
R
+
a
L
2
)
2
A
E
.
{\displaystyle {{\boldsymbol {\varepsilon }}(\mathbf {x} )={\cfrac {-a\mathbf {x} ^{2}+(2\mathbf {R} +aL^{2})}{2AE}}~.}}
The stress in the bar is
σ
(
x
)
=
−
a
x
2
+
(
2
R
+
a
L
2
)
2
A
.
{\displaystyle {\boldsymbol {\sigma }}(\mathbf {x} )={\cfrac {-a\mathbf {x} ^{2}+(2\mathbf {R} +aL^{2})}{2A}}~.}
A plot of this solution is shown in Figure 2. All known quantities have been taken to be 1 for the plot.
Figure 2. Exact solution for the axial loading of a bar.
