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Barlow’s Formula is a calculation used to show the relationship between internal pressure, allowable stress (also known as hoop stress), nominal thickness, and diameter. It is helpful in determining the maximum pressure capacity a pipe can safely withstand. The formula is expressed as P=2St/D, where: P pressure, psig t

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I want to create a custom field output in ABAQUS. For the purpose of proof on concept, I want to display the maximum shear stress as calculated from the Mohrs circle, as discussed here for a 2D shell. I have my code below for reference:

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This maximum normal stress is called principal stress and the plane related to it is called principal plane. But why should we find out the principal stress and maximum shear stress? Because by considering these stress values only we will go ahead with our further design calculations since these...

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Q13.11 Is it possible to find the location of a particular node? Q13.12 Is it possible to find the nodes associated with a particular element? Q13.13 When using the dynamic plotting capabilities of ABAQUS/Post using the mouse the mesh has disappeared from view. How can I restore the mesh view?

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To solve for the stress under fixed strain for a Maxwell material, we need to solve the ordinary differential equation in time. To do this, we can again use the symbolic toolbox in MATLAB. First, we divide through by p1 to obtain: We can then use the dsolve option in the MATLAB symbolic toolbox to solve the ordinary differential equation in time.

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Plot the y component of stress. Select Stress and Y-direction SY under Item to be con-toured. The maximum tangential stress is 50539 psi. If you desire to see the contours of the full model: > Utility Menu > PlotCtrls > Style > Symmetry Expansion > Periodic/Cyclic Symmetry > 1/4 Dihedral Sym Full model stress plot in X direction is shown below.

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Fig. 6. Maximum shear stress as a function of the wall thickness (The dots are the FEM results and the line is the design formula) 0.00365 N/mm2 0 N/mm2 Fig. 5. Finite element model of the tube with 20 mm wall thickness MAXIMUM SHEAR STRESS In Figure 6 the maximum shear stress is plotted as a function of the wall thickness.

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Example: Find the tensile stress of a 2 in. x 3 in. tension member subjected to a 45 kip axial load. Strain ... • The ratio of the maximum safe load to the

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To find out the maximum response ( nodal displacement, support reaction, section forces, beam end forces, beam stresses, plate stresses etc. ) out of a group of entities, you should first go to the Post processing mode. Select the entities ( could be nodes, beams or plates depending on what you would like to do).

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52100 yield stress capabilityσb＝120[kgf/mm2] Maximum allowable shear strengthτ ＝σb×0.8/（Safety factorα） ＝120×0.8/5 ＝19.2[kgf/mm2] Shear stress＝ Standard strength Safety factorα Standard strength: For ductile materials＝Yield stress For brittle materials＝Fracture stress Unwin safety factorαbased on tensile strength M P P

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Fig. 6. Maximum shear stress as a function of the wall thickness (The dots are the FEM results and the line is the design formula) 0.00365 N/mm2 0 N/mm2 Fig. 5. Finite element model of the tube with 20 mm wall thickness MAXIMUM SHEAR STRESS In Figure 6 the maximum shear stress is plotted as a function of the wall thickness.

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