Problem 9.3. Plastic failure load – concentrated load

Problem a)

Plastic failure load, F_R,pl [kN]=

Problem b)

Plastic failure load, F_R,pl [kN]=

Problem a)

Static theorem is described in Section 9.2., Figure 9.11.

Step 1.  Draw an admissible moment distribution.

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Step 2.  Determine plastic lower bound of the failure load applying the static theorem.

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Problem b)

Kinematic theorem is described in Section 9.2, Figure 9.24.

Step 1. Introduce plastic hinges into the structure to get a kinematically admissible mechanism.

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Step 2.  Determine plastic upper bound of the failure load applying the kinematic theorem.

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Problem a)

Static theorem is described in Section 9.2, Figure 9.11.

First an admissible moment distribution is drawn. Degree of indeterminancy is one, one free parameter can be chosen arbitrarily. Let us choose the moment above the middle support to reach the resistance of the cross section, while the positive moment also reaches its maximum (same as in the case of elastic failure).

According to the static theorem the plastic failure load is equal or bigger than the load, which causes the above admissible moment distribution:

$M_R^{+}=–\frac{M_R^{–}}{2}+\frac{F_{R,\mathrm{pl}}L}{4} \to {\mathit{F}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}\mathbf{\ge }\frac{4}{L}\left(M_R^{+}+\frac{M_R^{–}}{2}\right)=\frac{4}{4.0}\left(36+\frac{36}{2}\right)=\mathbf{54}\mathbf{ }\mathbf{kN}$

Problem b)

Kinematic theorem is described in Section 9.2, Figure 9.24.

Plastic hinges are introduced into the structure to get a kinematically admissible mechanism. Degree of indeterminancy is one, thus two plastic hinges must be introduced, the chosen mechanism is shown in the Figure: 

At the plastic hinges the moment must be equal to the resistance of the cross section. According to the kinematic theorem the plastic failure load is not bigger than the load, which belongs to the above kinematically admissible mechanism:

$M_R^{+}=–\frac{M_R^{–}}{2}+\frac{F_{R,\mathrm{pl}}L}{4} \to {\mathit{F}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}\mathbf{\le }\frac{4}{L}\left(M_R^{+}+\frac{M_R^{–}}{2}\right)=\frac{4}{4.0}\left(36+\frac{36}{2}\right)=\mathbf{54}\mathbf{ }\mathbf{kN}$

Upper and lower bound resulted in the same value, thus the above force gives the plastic failure load of the structure.