## Case Studies on Tensile Specimens without Bolts
### Tensile Specim with a Hole  
Figure 6a shows the same tensile specimen as used in Section 3.1, but now with a hole in the center of the specimen (diameter \(\text{d}=40\) mm). Figure 5: Failure mechanisms (equivalent plastic strain) from a finite element analysis (FEA) for the tensile specimen: (**a**) Mesh\({}_{2}\) (**b**) yield surface according to the Tresca criterion was used; (**c**) yield surface according to the Mises criterion was used. A possible _lower bound_ solution can easily be found by assuming a plastic net cross-section with  
\[\mathrm{F_{Tresca}}=\mathrm{F_{Mises}}=\mathrm{f_{y}}\;\mathrm{A_{net}}=\mathrm{ f_{y}}(\mathrm{b-d})\mathrm{t}=355\cdot(100-40)\cdot 8=170.4\;\mathrm{kN} \tag{11}\]  
As mentioned in the introduction, the used constant plastic stress distribution is quite different from the elastic stress distribution, due to the stress concentration at the hole. The stress magnification depends mainly on the form and the size of the hole (such as for mild or sharp notches). In the presented research, only typical plates with holes from bolts (mild notches) and ductile material behaviour were considered, so the use of the _lower bound_ theorem and the chosen constant plastic stress distribution are valid. The proof of the theorem and this example are also given in [3, 25]. A possible _upper bound_ slip line mechanism is shown in Figure 6b. The mechanism can be described similar to Equations (6) and (7):  
\[ \begin{array}{c}\mathrm{F\;u}=4\;\mathrm{\tau_{y}}\;\mathrm{L\;t\;v}\\ \mathrm{L}=\frac{\mathrm{b-d}}{2\cos\alpha}\;,\;\mathrm{v}=\frac{\mathrm{u}}{ 2\sin\alpha}\end{array} \right\}\Rightarrow\mathrm{F}(\alpha)=\frac{2\;\mathrm{ \tau_{y}(\mathrm{b-d})\mathrm{t}}}{\sin 2\alpha} \tag{12}\]  
Again, the minimum load can be obtained from the load function with \(\alpha=\pi/4\). Using the same dimensions and failure criteria as in the previous section, the ultimate loads result in:  
\[\mathrm{F_{Tresca}}=\mathrm{f_{y}}(\mathrm{b-d})\mathrm{t}=170.4\;\mathrm{kN }\;\mathrm{and}\;\mathrm{F_{Mises}}=2\frac{\mathrm{f_{y}}}{\sqrt{3}}(\mathrm{ b-d})\mathrm{t}=196.8\;\mathrm{kN} \tag{13}\]  
Another possible kinematic mechanism is the yield zone mechanism shown in Figure 6c. The mechanism can be described similar to Equation (10), with:  
\[ \begin{array}{c}\mathrm{F\;u}=\mathrm{f_{y}\;L\;t\;v}\\ \mathrm{L}=\mathrm{b-d}\;,\;\mathrm{v}=\mathrm{u}\end{array} \right\}\Rightarrow \mathrm{F_{Tresca}}=\mathrm{F_{Mises}}=\mathrm{f_{y}}(\mathrm{b-d})\mathrm{t} =170.4\;\mathrm{kN} \tag{14}\]  
Only the left side of the symmetric system was used in the DLO procedure, to reduce the number of discontinuities. Figure 7a shows the automatic generated distribution of the nodes. If only slip line mechanisms were considered, the mechanism in Figure 7b was obtained with corresponding loads of \(\mathrm{F_{Tresca}}=171.9\;\mathrm{kN}\) and \(\mathrm{F_{Mises}}=198.5\;\mathrm{kN}\) (for the whole system each). For the Tresca criterion, the load is 0.9% higher than the lower bound solution. The reason for this deviation is the inclination of the slip lines. The correct angle of \(\alpha=45\) cannot be reproduced exactly, with the used node distribution. Figure 6: Tensile specimen with a hole: (**a**) Loading and boundary conditions; (**b**) slip line mechanism; (**c**) yield zone mechanism. With a more refined node distribution, the angle of the slip line tends to be \(\alpha=45\) (see Figure 7c,d) with an ultimate load of \(\mathrm{F_{Tresca}}=171.2\) kN (still 0.4% higher). With a node distribution that includes slip lines with an angle of \(\alpha=45\), the minimum load of \(\mathrm{F_{Tresca}}=170.4\) kN can be found. An important fact is that this can also happen with a coarser node distribution. In general, the denser the node distribution, the higher the possibility to find a mechanism with lower resistance. The linear programming algorithm found the mechanism shown in Figure 7e for yield zone mechanisms only. The corresponding loads are \(\mathrm{F_{Tresca}}=\mathrm{F_{Mises}}=170.4\) kN.