## Case Studies on Tensile Specimens without Bolts  
### Tensile Specim  
As a first example, the tensile specimen from Figure 3a was considered. The specimen has the dimensions \(\mathrm{h}=250\) mm, \(\mathrm{b}=100\) mm, \(\mathrm{t}=8\) mm (plate thickness), and a steel grade S355 (\(\mathrm{f}_{y}=355\) MPa). First, the analytical solutions were obtained. A _lower bound_ solution can be found by assuming a full plastic cross-section with:  
\[\mathrm{F}_{\mathrm{Tresca}}=\mathrm{F}_{\mathrm{Mises}}=\mathrm{f}_{y}\ \mathrm{b}\ \mathrm{t}=355\cdot 100\cdot 8=284.0\ \mathrm{kN} \tag{5}\]  
An _upper bound_ solution can be found by assuming the kinematic mechanism from Figure 3b, which consists of four identical slip lines. There is a small overlapping of the triangles in the middle  
Figure 3: Tensile specimen: (**a**) Loading and boundary conditions; (**b**) symmetrical slip line mechanism; (**c**) asymmetrical slip line mechanism; (**d**) yield zone mechanism. of the plate, which means there is a small yield zone. For simplicity, the work of the yield zone is neglected, and the solution can be obtained by formulating the external and internal work equations:  
\[\mathrm{W_{ex}=F\,u\,and\,W_{in}=4\,\tau_{y}\,L\,t\,v} \tag{6}\]  
The length of a single slip line can be expressed as \(L=b/(2\cos\alpha)\). With \(v=u/(2\sin\alpha)\), the work equation and the load function \(F=F(\alpha)\) results in:  
\[F\,u=4\,\tau_{y}\,\frac{b}{2\cos\alpha}\,t\,\frac{u}{2\sin\alpha}\Rightarrow F (\alpha)=\frac{2\,\tau_{y}\,b\,t}{\sin 2\alpha} \tag{7}\]  
The minimum value of the load function can be found with \(\alpha=\pi/4\). Depending on the used failure criterion (Tresca or Mises), the corresponding _upper bound_ solutions result in:  
\[F_{Tresca}=f_{y}\,b\,t=284.0\,\,\mathrm{kN}\,\,\mathrm{and}\,\,F_{Mises}=2\, \frac{f_{y}}{\sqrt{3}}\,b\,t=327.9\,\,\mathrm{kN} \tag{8}\]  
The same load function can be obtained for the asymmetrical slip-line mechanism from Figure 3c. The only difference is the horizontal degree-of-freedom at the load introduction:  
\[\left. \begin{array}{c}F\,u=\tau_{y}\,L\,t\,v\\ L=\frac{b}{\cos\alpha}\,\,,\,v=\frac{u}{\sin\alpha}\end{array} \right\} \Rightarrow F(\alpha)=\frac{2\,\tau_{y}\,b\,t}{\sin 2\alpha} \tag{9}\]  
The kinematic mechanism (yield zone mechanism) shown in Figure 3d can be described similarly, and the _upper bound_ solutions for the Tresca and the Mises criteria result both in the same value:  
\[\left. \begin{array}{c}F\,u=f_{y}\,L\,t\,v\\ L=b\,,\,v=u\end{array} \right\}\Rightarrow F_{Tresca}=F_{Mises}=f_{y}\,b\,t=284.0\,\,\mathrm{kN} \tag{10}\]  
The same tensile specimen was analyzed with the DLO procedure. Figure 4a shows the automatic generated node distribution (software MATLAB [23]). The kinematic mechanisms (_upper bound_ solutions), shown in Figure 4b-d, were obtained. The blue and red lines indicate a slip and a normal displacement behaviour, respectively. Figure 4: Collapse mechanisms (blue: Slip displacements, red: Normal displacements) from DLO for the tensile specimen: (**a**) Nodes; (**b**) slip line mechanism; (**c**) yield zone mechanism; (**d**) combined slip line and yield zone mechanism. The mechanism in Figure 4b was obtained for allowing slip line mechanisms only. The angle of the lines is precisely \(\alpha=45\). Three identical mechanisms were obtained in comparison to Figure 3b. This occurred because each of the shown slip lines would result in the same minimum internal work. The linear programming algorithm divides the internal displacement equally to the indicated slip lines. Therefore, the higher the number of possible slip lines, the lower the internal work dissipated at each slip line. The total internal work is always the same, independent of the number of slip lines. Depending on the used failure-criterion, the corresponding loads result in \(\text{F}_{\text{Tresca}}=284.0\) kN and \(\text{F}_{\text{Mises}}=329.7\) kN. For comparison, if only yield zone mechanisms were allowed, the mechanism shown in Figure 4c was obtained.