2. Benchmark solutions#
2.1. Calculation method used for reference solutions#
2.1.1. Circular section#
2.1.1.1. Beam subjected to a punctual tensile force \(\mathit{Fx}\)#
The equilibrium equation is:
\(\frac{\partial }{\partial x}(\mathrm{EA}(x)\frac{\partial u}{\mathrm{dx}})=0\) with \(A(x)={A}_{1}(1+c\frac{x}{L})\) and \(c=\sqrt{\frac{{A}_{2}}{{A}_{1}}}-1\), \(N(L)={F}_{x}\)
By integrating [R3.08.01] twice, we get the displacements as a function of the force applied, that is:
\(u(x)=\frac{{\mathrm{LF}}_{x}}{{\mathrm{EA}}_{1}}(\frac{x}{L+\mathrm{cx}})\)
and therefore at the end L of the beam:
\(u(L)=\frac{L}{E\sqrt{{A}_{1}{A}_{2}}}{F}_{x}\)
Internal efforts are given by:
\(N(x)=\mathrm{EA}(x)\frac{\partial u}{\partial x}(x)={F}_{x}\)
and the constraints by:
\({\sigma }_{\mathrm{xx}}=\frac{N(x)}{A(x)}\)
2.1.1.2. Beam subjected to a point bending force \(\mathit{Fy}\)#
The equilibrium equation, under Euler’s hypothesis, is given by the equation:
\(\frac{{\partial }^{2}}{\partial {x}^{2}}\left[{\mathrm{EI}}_{z}(x)\frac{{\partial }^{2}v}{\partial {x}^{2}}\right]=0\) with \({I}_{z}(x)={I}_{{z}_{1}}{(1+c\frac{x}{L})}^{4}\) and \(c={(\frac{{I}_{{z}_{2}}}{{I}_{{z}_{1}}})}^{\frac{1}{4}}-1\), \({V}_{y}(L)={F}_{y}\)
We solve the equation by integration taking into account the modified law of behavior \({\mathrm{MF}}_{z}={\mathrm{EI}}_{z}\frac{{\partial }^{2}v}{\partial {x}^{2}}\) and the equilibrium equation \(\frac{\partial {\mathrm{MF}}_{z}}{\partial x}+{V}_{y}=0\)
The four successive integrations, taking into account for the calculation of the integration constants that:
\(\begin{array}{}\frac{\partial }{\partial x}\left[{\mathrm{EI}}_{z}(x)\frac{{\partial }^{2}v}{\partial {x}^{2}}\right](L)=-{V}_{y}(L)=-{F}_{y}\\ ({\mathrm{EI}}_{z}(L)\frac{{\partial }^{2}v}{\partial {x}^{2}})(L)=0\\ \frac{\partial v}{\partial x}(0)=0\\ v(0)=0\end{array}\)
lead to the expression of:
\(v(x)=+\frac{{F}_{y}{L}^{2}}{6E{I}_{{z}_{1}}}\frac{{x}^{2}(3L-x+2\mathrm{cx})}{{(L+\mathrm{cx})}^{2}}\)
and to the expression of \({\theta }_{z}(x)\)
\({\theta }_{z}(x)=+\frac{{F}_{y}\mathrm{L²}}{6E{I}_{{z}_{1}}}\frac{x(6\mathrm{L²}-3\mathrm{Lx}+6L\mathrm{cx}-\mathrm{cx²}+2\mathrm{c²}\mathrm{x²})}{{(L+\mathrm{cx})}^{3}}\)
Internal efforts are given by:
\(\begin{array}{}{V}_{y}(x)={F}_{y}\\ {\mathrm{MF}}_{z}(x)={F}_{y}(L-x)\end{array}\)
and the constraints by:
\({\sigma }_{\mathrm{xx}}(x)=∣{\mathrm{MF}}_{z}(x)∣\frac{R(x)}{{I}_{z}(x)}\)
\({\sigma }_{\mathrm{xy}}=\frac{{V}_{y}(x)}{A(x)}\) no shear correction coefficient under Euler’s hypothesis.
2.1.1.3. Beam subjected to a one-time torsional moment \(\mathrm{Mx}\)#
The movement is given by the equation:
\(\frac{\partial }{\partial x}\left[G{I}_{p}(x)\frac{\partial {\theta }_{x}}{\partial x}\right]=0\) with \({I}_{p}(x)={I}_{{p}_{1}}{(1+c\frac{x}{L})}^{4}\) and \(c={(\frac{{I}_{{p}_{2}}}{{I}_{{p}_{1}}})}^{\frac{1}{4}}-1\), \({M}_{x}(L)={M}_{x}\)
After integration, and taking into account the fact that:
\(G{I}_{p}(L)=\frac{\partial {\theta }_{x}}{\partial x}(L)={M}_{x}\), and \({\theta }_{x}(0)=0\)
we get the expression for \({\theta }_{x}(x)\):
\({\theta }_{x}(x)=\frac{L{M}_{x}}{3G{I}_{{p}_{1}}}\frac{x(3{L}^{2}+3L\mathrm{cx}+\mathrm{c²}\mathrm{x²})}{{(L+\mathrm{cx})}^{3}}\)
For internal efforts and constraints, we must also have:
\(\begin{array}{}{M}_{x}(x)={M}_{x}\\ {\sigma }_{\mathrm{xy}}(x)=\frac{{M}_{x}(x)}{{I}_{p}(x)}{R}_{T}(x)\\ {\sigma }_{\mathrm{xz}}(x)=\frac{{M}_{x}(x)}{{I}_{p}(x)}{R}_{T}(x)\end{array}\)
2.1.1.4. Beam subjected to a point bending moment \(\mathit{My}\)#
The reasoning for finding the analytical solution is the same as before. We use the law of behavior \({M}_{y}(x)=-{\mathrm{EI}}_{y}(x)\frac{{\partial }^{2}w}{\partial {x}^{2}}\) and the equilibrium equation \(\frac{\partial {\mathrm{MF}}_{y}}{\partial x}-{V}_{z}=0\) The calculation of integration constants differs: we have \({V}_{z}(L)=0\) and \({M}_{{F}_{y}}(L)={M}_{y}\).
We get the expression for \(w(x)\):
\(w(x)=\frac{L{M}_{y}}{6E{I}_{{y}_{1}}}\frac{\mathrm{x²}(3L+2\mathrm{cx})}{{(L+\mathrm{cx})}^{2}}\),
and the expression for \({\theta }_{y}(x)\):
\({\sigma }_{y}(x)=\frac{L{M}_{y}}{3E{I}_{\mathrm{y1}}}\frac{x(3{L}^{2}+3L\mathrm{cx}+{c}^{2}{x}^{2})}{{(L+\mathrm{cx})}^{3}}\)
For internal forces and stresses, we must also have:
\(\begin{array}{}{V}_{z}(x)=0\\ {\mathrm{MF}}_{y}(x)=\mathrm{My}\\ {\sigma }_{\mathrm{xx}}(x)=∣{\mathrm{MF}}_{y}(x)∣\frac{R(x)}{{I}_{y}(x)}\end{array}\)
2.1.1.5. Beam subjected to a consistently distributed tensile force \(\mathrm{fx}\)#
Equilibrium is described by the equation
\(\frac{\partial }{\partial x}(\mathrm{EA}(x)\frac{\partial u}{\partial x})=-{f}_{x}\) with \(A(x)={A}_{1}{(1+c\frac{x}{L})}^{2}\) and \(c={(\frac{{A}_{2}}{{A}_{1}})}^{\frac{1}{2}}-1\)
Integrating this equation for the first time, we get:
\(EA(x)\frac{\partial u}{\partial x}=-{f}_{x}x+{c}_{1}\)
Boundary condition \(N(L)=0\) implies \({c}_{1}={f}_{x}L\). So we have:
\(\frac{\partial u}{\partial x}=-{f}_{x}\frac{(L-x)}{EA(x)}\)
either:
\(u(x)={f}_{x}\int \frac{(L-x)}{EA(x)}\mathrm{dx}+{c}_{2}\)
\({c}_{2}\) is determined for \(u(0)=0\)
All calculations done, we have:
\(u(x)=\frac{{L}^{2}{f}_{x}}{E{A}_{1}{c}^{2}}\frac{cx+{c}^{2}x+(L+cx)\mathrm{log}\frac{L}{L+cx}}{L+cx}\).
Internal efforts are deduced from the \(N(x)=EA(x)\frac{\partial u}{\partial x}\) law of behavior:
\(N(x)={f}_{x}(L-x)\)
and the constraints are given by:
\({\sigma }_{\mathrm{xx}}(x)=\frac{N(x)}{A(x)}=\frac{{f}_{x}(L-x)}{{\left[\sqrt{{A}_{1}}+(\sqrt{{A}_{2}-\sqrt{{A}_{1}}})\frac{x}{L}\right]}^{2}}\)
2.1.1.6. Beam subjected to a consistently distributed bending force \(\mathit{fy}\)#
Starting from the equilibrium equation:
\(\frac{{\partial }^{2}}{\partial {x}^{2}}\left[E{I}_{z}(x)\frac{{\partial }^{2}v}{\partial {x}^{2}}\right]=-{f}_{y}\) with \({I}_{z}(x)={I}_{{z}_{1}}{(1+c\frac{x}{L})}^{4}\) and \(c={(\frac{{I}_{{z}_{2}}}{{I}_{{z}_{1}}})}^{\frac{1}{4}}-1\)
we perform four successive integrations. The integration constants are determined on the basis of the following limit conditions:
\(\begin{array}{}{V}_{y}(L)=0\\ {M}_{z}(L)=0\\ \frac{\partial v}{\partial x}(0)=0\\ v(0)=0\end{array}\)
The analytic expression for \(v(x)\) and \(\theta (z)\) in the presence of distributed loading is, any calculation done:
\(\begin{array}{}v(x)=\frac{-{f}_{y}{L}^{3}}{12{\mathrm{EI}}_{{z}_{1}}{c}^{4}{(L+\mathrm{cx})}^{2}}\left[\begin{array}{c}-6{L}^{2}\mathrm{cx}+{x}^{2}(-9{\mathrm{Lc}}^{2}-3{\mathrm{Lc}}^{4})+{x}^{3}(-2{c}^{3}+2{c}^{4}-2{c}^{5})\\ +\mathrm{log}(1+c\frac{x}{L})(6{L}^{3}+12{L}^{2}\mathrm{cx}+6L{c}^{2}{x}^{2})\end{array}\right]\\ {\theta }_{z}(x)=\frac{+{L}^{3}{f}_{y}x}{6E{I}_{{z}_{1}}{(L+\mathrm{cx})}^{3}}\left[3{L}^{2}-3\mathrm{Lx}+3\mathrm{Lcx}+{x}^{2}(1-c+{c}^{2})\right]\end{array}\)
Internal efforts are given by:
\({V}_{y}(x)={f}_{y}(L-x)\) and \({\mathrm{Mf}}_{z}(x)=\frac{1}{2}{f}_{y}{(L-x)}^{2}\)
constraints by:
\(\begin{array}{}{\sigma }_{\mathrm{xy}}(x)=\frac{{V}_{y}(x)}{A(x)}\\ {\sigma }_{\mathrm{xx}}(x)=∣{\mathrm{Mf}}_{z}(x)∣\frac{R(x)}{{I}_{z}(x)}\end{array}\)
2.1.2. Rectangular section#
2.1.2.1. Beam subjected to a punctual tensile force \(\mathit{Fx}\)#
The equilibrium equation is:
\(\frac{\partial }{\partial x}(\mathrm{EA}(x)\frac{\partial u}{\partial x})=0\) with \(A(x)={A}_{1}+({A}_{2}-{A}_{1})\frac{x}{L}\), \(N(L)={F}_{x}\)
By integrating twice, and taking into account the fact that:
\(EA(L)\frac{\partial u}{\partial x}(L)={F}_{x}\), \(u(0)=0\)
for the determination of the integration constants, we obtain the analytic expression for \(u(x)\), that is:
\(u(x)=\frac{{F}_{x}L}{{A}_{1}\mathrm{Ec}}\mathrm{log}(1+c\frac{x}{L})\)
For internal efforts and constraints, we have:
\(\begin{array}{}N(x)={F}_{x}\\ {\sigma }_{\mathrm{xx}}=\frac{N(x)}{A(x)}\end{array}\)
2.1.2.2. Beam subjected to a point bending force \(\mathit{Fy}\)#
The movement is given by the equation:
\(\frac{{\partial }^{2}}{\partial {x}^{2}}\left[E{I}_{z}(x)\frac{{\partial }^{2}v}{\partial {x}^{2}}\right]=0\) with \({I}_{z}(x)={I}_{{z}_{1}}{(1+c\frac{x}{L})}^{3}\) and \(c={(\frac{{I}_{{z}_{2}}}{{I}_{{z}_{1}}})}^{\frac{1}{3}}-1\), \({V}_{y}(L)={F}_{y}\)
The same reasoning as for the circular section leads to the following result:
\(v(x)=-\frac{{F}_{y}{L}^{2}}{2E{I}_{{z}_{1}}{c}^{3}}\frac{\left[2\mathrm{Lcx}+{c}^{2}{x}^{2}-{c}^{3}{x}^{2}+2L(L+\mathrm{cx})\mathrm{log}(\frac{L}{L+\mathrm{cx}})\right]}{(L+\mathrm{cx})}\)
\({\theta }_{z}(x)=\frac{{F}_{y}{L}^{2}}{2{\mathrm{EI}}_{{z}_{1}}}\frac{x(2L-x+\mathrm{cx})}{{(L+\mathrm{cx})}^{2}}\)
For internal efforts and constraints, we must have:
\(\begin{array}{}{V}_{y}(x)={F}_{y}\\ \\ {M}_{{F}_{z}}(x)={F}_{y}(L-x)\\ \\ {\sigma }_{\mathrm{xx}}(x)=\frac{{H}_{y}(x){M}_{{F}_{x}}(x)}{{\mathrm{2I}}_{z}(x)}\\ \\ {\sigma }_{\mathrm{xy}}(x)=\frac{{V}_{y}(x)}{A(x)}\end{array}\)
2.1.2.3. Beam subjected to a one-time torsional moment \(\mathit{Mx}\)#
The movement is given by the equation:
\(\frac{\partial }{\partial x}\left[G{I}_{p}(x)\frac{\partial {\theta }_{x}}{\partial x}\right]=0\) with \({I}_{p}(x)={I}_{{p}_{1}}{(1+c\frac{x}{L})}^{3}\) and \(c={(\frac{{I}_{{p}_{2}}}{{I}_{{p}_{1}}})}^{\frac{1}{3}}-1\), \({M}_{x}(L)={M}_{x}\)
Using the same reasoning as the circular beam, we get the analytic expression for \({\theta }_{x}(x)\):
\({\theta }_{x}(x)=\frac{L{M}_{x}x(2L+\mathrm{cx})}{2{I}_{{p}_{1}}G{(L+\mathrm{cx})}^{2}}\)
\({I}_{{p}_{1}}\) and \({I}_{{p}_{2}}\) are calculated according to the formulas given in the reference documentation [R3.08.01].
Internal forces and stresses are given by:
\(\begin{array}{}{M}_{x}(x)={M}_{x}\\ {\sigma }_{\mathrm{xy}}(x)=\frac{{M}_{x}(x)}{{I}_{p}(x)}{R}_{T}(x)={\sigma }_{\mathrm{xz}}\end{array}\)
2.1.2.4. Beam subjected to a point bending moment \(\mathit{My}\)#
We use the same reasoning as before, we obtain the following analytic expressions for \(w(x)\) and \({\theta }_{y}(x)\):
\(\begin{array}{}w(x)=-\frac{L{M}_{y}{x}^{2}}{2{\mathrm{EI}}_{{y}_{1}}(L+\mathrm{cx})}\\ \\ {\theta }_{y}(x)=\frac{L{M}_{y}x(2L+\mathrm{cx})}{2E{I}_{{y}_{1}}{(L+\mathrm{cx})}^{2}}\end{array}\)
for the efforts:
\(\begin{array}{}{V}_{z}(x)=0\\ {\mathrm{MF}}_{y}(x)={M}_{y}\end{array}\)
and for the constraints:
\({\sigma }_{\mathrm{xx}}(x)=\frac{{H}_{z}(x){\mathrm{MF}}_{y}(x)}{2{I}_{y}(x)}\)
2.1.2.5. Beam subjected to a consistently distributed tensile force \(\mathrm{Fx}\)#
The equilibrium equation is:
\(\frac{\partial }{\partial x}\left[\mathrm{EA}(x)\frac{\partial u}{\partial x}\right]=-{f}_{x}\) with \(A(x)={A}_{1}(1+c\frac{x}{L})\) and \(c=(\frac{{A}_{2}}{{A}_{1}})-1\).
After two integrations and taking into account the fact that \(N(L)=0\) to determine the first constant of integration, and \(u(0)=0\) to determine the second, we get the analytic expression for \(u(x)\):
\(u(x)=\frac{-L{f}_{x}}{E{A}_{1}{c}^{2}}\left[cx+(L+{L}_{c})\mathrm{log}(\frac{L}{L+cx})\right]\)
Internal forces are known by the following expression:
\(N(x)={f}_{x}(L-x)\)
and the constraints by:
\({\sigma }_{\mathrm{xx}}(x)=\frac{{f}_{x}(L-x)}{A(x)}\)
2.1.2.6. Beam subjected to a consistently distributed bending force \(\mathrm{Fy}\)#
The equilibrium equation is:
\(\frac{{\partial }^{2}}{\partial {x}^{2}}\left[{\mathrm{EI}}_{z}(x)\frac{{\partial }^{2}v}{\partial {x}^{2}}\right]=-{f}_{y}\) with \({I}_{z}(x)={I}_{{z}_{1}}{(1+c\frac{x}{L})}^{3}\) and \(c={(\frac{{I}_{{z}_{2}}}{{I}_{{z}_{1}}})}^{\frac{1}{3}}-1\)
We integrate this equation four times in succession. The integration constants are calculated taking into account the fact that:
\(\begin{array}{}{V}_{y}(L)=0\\ {\mathrm{MF}}_{z}(L)=0\\ \frac{\partial v}{\partial x}(0)=0\\ v(0)=0\end{array}\)
The analytical result for the arrow and the rotation in \(L\) is as follows:
\(\begin{array}{}v(x)=\frac{{L}^{3}{f}_{y}}{4E{I}_{{z}_{1}}{c}^{4}(L+cx)}\left[x(6\mathrm{Lc}+4{\mathrm{Lc}}^{2})+{x}^{2}({\mathrm{5c}}^{2}+{\mathrm{2c}}^{3}-{c}^{4})\right]\\ +({\mathrm{6L}}^{2}+{\mathrm{4L}}^{2}c+\mathrm{8Lcx}+4{\mathrm{Lc}}^{2}x+{\mathrm{2c}}^{2}{x}^{2})\mathrm{log}(\frac{L}{L+\mathrm{cx}})\end{array}\)
\(\begin{array}{}{\theta }_{z}(x)=\frac{{L}^{3}{f}_{y}}{4{\mathrm{EI}}_{{z}_{1}}{c}^{3}{(L+\mathrm{cx})}^{2}}\left[x(2\mathrm{Lc}+2{\mathrm{Lc}}^{3})+{x}^{2}(3{c}^{2}+2{c}^{3}-{c}^{4})\right]\\ +(2{L}^{2}+\mathrm{Lcx}+2{c}^{2}{x}^{2})\mathrm{log}(\frac{L}{L+\mathrm{cx}})\end{array}\)
Internal forces are given by the following expressions:
\({V}_{y}(x)={f}_{y}(L-x)\), \({\mathrm{Mf}}_{z}(x)=\frac{1}{2}{f}_{y}{(L-x)}^{2}\)
constraints by:
\(\begin{array}{}{\sigma }_{\mathrm{xy}}(x)=\frac{{V}_{y}(x)}{A(x)}\\ {\sigma }_{\mathrm{xx}}(x)=∣\frac{{\mathrm{Mf}}_{z}(x){h}_{y}}{{I}_{z}(x)2}∣\end{array}\)
2.1.3. General section#
2.1.3.1. Beam subjected to the forces of gravity#
Gravity forces are applied along the \(z\) axis. The movement of the beam induced by these forces is therefore a flexural movement in plane \((xoz)\).
The equilibrium equation is given by the expression:
\(\frac{{\partial }^{2}}{\partial {x}^{2}}(E{I}_{y}(x)\frac{{\partial }^{2}w}{\partial {x}^{2}})=\underset{\mathrm{poids}\mathrm{linéique}}{\rho A(x)g}\)
with \(A(x)={A}_{1}{(1+c\frac{x}{L})}^{2}\), \(c={(\frac{{A}_{2}}{{A}_{1}})}^{\frac{1}{2}}-1\), and \({I}_{y}(x)={I}_{\mathrm{y1}}{(1+d\frac{x}{L})}^{4}\) \(d={(\frac{{I}_{\mathrm{y2}}}{{I}_{\mathrm{y1}}})}^{\frac{1}{4}}-1\)
By integrating for the first time, we obtain the internal shear force:
\({V}_{z}(x)=-\int \rho A(x)g\mathrm{dx}+{C}_{1}\)
\({C}_{1}\) is set so that \({V}_{z}(L)=0\).
We get:
\({V}_{z}(x)=\frac{L{A}_{1}\rho g}{3c}\left[-{(1+c\frac{x}{L})}^{3}+{(1+c)}^{3}\right]\).
By integrating a second time, we obtain the internal bending moment:
\({M}_{y}(x)=\int {V}_{z}(x)\mathrm{dx}+{c}_{2}\)
\({C}_{2}\) is calculated so that \({M}_{y}(L)=0\)
We get:
\({M}_{y}(x)=\frac{{A}_{1}\rho g}{12{L}^{2}}{(L-x)}^{2}(6{L}^{2}+8{L}^{2}c+3{L}^{2}{c}^{2}+4\mathrm{Lcx}+2\mathrm{Lcx}+\frac{2{\mathrm{Lc}}^{2}x}{+{c}^{2}{x}^{2}})\)
We then calculate the rotation based on the law of behavior \({M}_{y}(x)=E{I}_{y}(x)\frac{\partial {\theta }_{y}}{\partial x}\)
So we have \({\theta }_{y}(x)=\int \frac{{M}_{y}(x)}{E{I}_{y}(x)}\mathrm{dx}+{C}_{3}\) with \({\theta }_{y}(0)=0\)
Arrow \(w(x)\) is determined from the Euler relationship: \({\theta }_{y}=-\frac{\partial w}{\partial x}\)
We calculate \(w(x)\) by integrating \({\theta }_{y}(x)\): \(w(x)=-\int {\theta }_{y}(x)+{C}_{4}\)
with \({C}_{4}\) such as \(w(0)=0\).
The analytical expressions for \({\theta }_{y}(x)\) and \(w(x)\) are not transcribed here because they are much too cumbersome. They were calculated, like the previous ones, by the formal calculation software MATHEMATICA.
2.2. Benchmark results#
Movements and rotations at the free end
Internal forces at both ends
Constraints at both ends
2.3. Uncertainty about the solution#
Analytical solution.
2.4. Bibliographical references#
[1] Report no. 2314/A of the Aerotechnical Institute « Proposal and implementation of new test cases missing the validation of ASTER beams »