Problem 9.5. Failure load of a frame

Failure loads, F_R,pl [kN]=

Step 1.  Apply the kinematic theorem. Introduce plastic hinges into the frame to obtain a kinematically admissible mechanism.

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Step 2.  Moment in a plastic hinge is equal to the moment resistance. Draw plastic moment diagram. 

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Step 3.  Determine upper bound of the plastic failure load which belongs to the plastic moment diagram.

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Kinematic theorem is applied.

The degree of indeterminancy is one, two plastic hinges are introduced to obtain a kinematically admissible mechanism:

Moment in a plastic hinge is equal to the moment resistance. The plastic moment diagram is

B_y is obtained from moment equilibrium about the left support. The moment at the hinges can be calculated from the reaction forces:

$$\begin{gathered} M_{\mathrm{R}}^{–}=B_{x}h \\ {M}_{\mathrm{R}}^{+}=–B_{x}h+B_y\frac{l}{2} \\ \mathrm{These} \mathrm{equations} \mathrm{results} \mathrm{in} \\ {M}_{\mathrm{R}}^{+}=–M_{\mathrm{R}}^{–}+F\left(1+\frac{h}{l}\right)\frac{l}{2} \end{gathered}$$

Thus the failure load is

$$\begin{gathered} M_{\mathrm{R}}^{+}=–M_{\mathrm{R}}^{–}+B_y\frac{l}{2}=–M_{\mathrm{R}}^{–}+F_{\mathrm{R},\mathrm{pl}}\left(1+\frac{h}{l}\right)\frac{l}{2} \\ \to {\mathit{F}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}\mathbf{\le }2\frac{M_{\mathrm{R}}^{+}+M_{\mathrm{R}}^{–}}{l+h}=2\frac{24.0+36.0}{6.0+4.0}=\mathbf{12}\mathbf{ }\mathbf{kN} \end{gathered}$$