Problem 9.4. Plastic failure load – distributed load

Ratio of the plastic and elastic failure loads, p_R,pl/p_R,el=

Step 1.  Determine elastic failure load.

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Step 2.  Draw plastic moment distribution.

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Both static and kinematic theorem can lead to the following moment diagram:

See Problem 9.3.

When the static theorem is applied the above diagram is one of the admissible moment distributions. When the kinematic theorem is applied three plastic hinges are introduced (at the ends and at the middle of the beam) to obtain a kinematically admissible mechanism, where the moments in the plastic hinges are equal to the resistances.

Step 3.  Determine plastic failure load.

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Step 4.  Give the ratio of the plastic and elastic failure loads.

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Elastic analysis

In case of elastic failure the maximum moment (here the negative moment at the support) reaches the moment resistance. The elastic failure load is

$M_{\mathrm{R}}^{–}=\frac{p{L}^2}{12} \to p_{\mathrm{R},\mathrm{el}}=\frac{12M_R^{–}}{L^2}=\frac{12\times 8}{3^2}=10.67 \mathrm{kN}/\mathrm{m}$

Plastic analysis

Both static and kinematic theorem can lead to the following moment diagram:

See Problem 9.3.

For applying static theorem the above diagram is one of the admissible moment distributions. For applying the kinematic theorem three plastic hinges are introduced (at the ends and at the middle of the beam) to obtain a kinematically admissible mechanism, where the moments in the plastic hinges are equal to the resistances.

The equality of the upper and the lower bound gives following plastic failure load, which belongs to the above moment diagram:$M_R^{+}=–M_R^{–}+\frac{p_{R,\mathrm{pl}}{L}^2}{8} \to {\mathit{p}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}\mathbf{=}\frac{8}{L^2}\left(M_R^{+}+M_R^{–}\right)=\frac{8}{3^2}\left(8.0+8.0\right)=\mathbf{14}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{kN}\mathbf{/}\mathbf{m}$

The ratio of the elastic and plastic failure loads is

$\frac{{\mathit{p}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}}{{\mathit{p}}_{\mathbf{R}\mathbf{,}\mathbf{el}}}\mathbf{=}=\frac{L^2}{12M_R^{–}}\frac{8(M_R^{–}+M_R^{+})}{L^2}=\frac{16}{12}=\mathbf{1}\mathbf{.}\mathbf{33}$