Given:
A long rectangular beam with cross section
Find:
A solution for the displacement and stress fields, using strong boundary conditions on the edges
and
.
[Hint : Assume that the displacement can be expressed as a second degree polynomial (using the Pascal's triangle to determine the terms)
]
Step 1: Boundary conditions
![{\displaystyle {\begin{aligned}{\text{at}}~x_{1}&=0~;~~\sigma _{13}=0\\{\text{at}}~x_{1}&=a~;~~u_{3}=0\\{\text{at}}~x_{2}&=0~;~~\sigma _{23}=0\\{\text{at}}~x_{2}&=b~;~~\sigma _{23}=S\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34a2663a8e7a78562683447c489f16aed4dc428d)
Step 2: Assume a solution
Let us assume antiplane strain
![{\displaystyle u_{3}(x_{1},x_{2})=Ax_{1}^{2}+Bx_{2}^{2}+Cx_{1}x_{2}+Dx_{1}+Ex_{2}+F~;~~u_{1}=u_{2}=0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fd2df0b968fd3b50620f3cc02ce4576bad719ee)
Step 3: Calculate the stresses
The stresses are given by
, and
.
Therefore,
![{\displaystyle {\begin{aligned}\sigma _{13}&=\mu u_{3,1}=\mu (2Ax_{1}+Cx_{2}+D)\\\sigma _{23}&=\mu u_{3,2}=\mu (2Bx_{2}+Cx_{1}+E)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d29964f8bcf3d139feb0e44da543732040f3f4fe)
Step 4: Satisfy stress BCs
Thus we have,
![{\displaystyle {\begin{aligned}0&=\mu (Cx_{2}+D)\\0&=\mu (Cx_{1}+E)\\S&=\mu (2bB+Cx_{1}+E)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/326f3673757082bd1de5170b42fd8907df4874b1)
Since
and
can be arbitrary,
.
Hence,
which gives us
![{\displaystyle {\begin{aligned}u_{3}&=Ax_{1}^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F\\\sigma _{13}&=\mu (2Ax_{1})\\\sigma _{23}&=\mu (2{\frac {S}{2\mu b}}x_{2})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae1c0a4b742e2527ef57d895bd397d716f8f21bc)
Assume that the body force is zero. Then the equilibrium condition is
. Therefore,
![{\displaystyle {\begin{aligned}&u_{3,11}+u_{3,22}=0\\{\text{or,}}\quad &2A+2{\frac {S}{2\mu b}}=0\\{\text{or,}}\quad &A=-{\frac {S}{2\mu b}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c824ee0a90b8fec1ef18197274c45de758dd82ed)
Therefore, the stresses are given by
![{\displaystyle {\sigma _{13}=-{\frac {S}{b}}x_{1}~;~~\sigma _{23}={\frac {S}{b}}x_{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eef70dd411e4a6fd389d183035e8a6a9867ccac5)
Step 5: Satisfy displacement BCs
The displacement is given by
![{\displaystyle u_{3}=-{\frac {S}{2\mu b}}x_{1}^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/831045085592a870c8c6de9dae0d4d8e60bab50c)
If we substitute
, we cannot determine the constant
uniquely.
Hence the displacement boundary conditions have to be applied in a weak sense,
![{\displaystyle {\begin{aligned}&\int _{0}^{b}u_{3}(a,x_{2})dx_{2}=0\\{\text{or,}}\quad &\int _{0}^{b}\left(-{\frac {S}{2\mu b}}a^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F\right)dx_{2}=0\\{\text{or,}}\quad &\left.\left[{\frac {S}{2\mu b}}\left(-a^{2}x_{2}+{\frac {x_{2}^{3}}{3}}\right)+Fx_{2}\right]\right|_{0}^{b}=0\\{\text{or,}}\quad &{\frac {S}{2\mu b}}\left(-a^{2}b+{\frac {b^{3}}{3}}\right)+Fb=0\\{\text{or,}}\quad &{\frac {S}{2\mu }}\left(-a^{2}+{\frac {b^{2}}{3}}\right)+Fb=0\\{\text{or,}}\quad &{\frac {S}{2\mu }}\left(-{\frac {a^{2}}{b}}+{\frac {b}{3}}\right)+F=0\\{\text{or,}}\quad &F={\frac {S}{2\mu b}}\left(a^{2}-{\frac {b^{2}}{3}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b6884c02fd5c23402063d88105bd660effd4aad)
Therefore,
![{\displaystyle {u_{3}={\frac {S}{2\mu b}}\left(x_{2}^{2}-x_{1}^{2}+a^{2}-{\frac {b^{2}}{3}}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17d59148e087eb7a1b015a87943d87c9feb11760)