Standard_Parts_Catalogue

Page 2132 | Information for Loads Computing the Strength of Indexing Plungers for Shear Loads / Flexure Loads of the Plunger Pin Provided that a miniscule gap remains between the guide of the indexing plunger and the indexing bore hole opposite, the load can be reduced to a clean shear action. As this is normally not the case, the “flexure” load case should preferably be considered on the following page. Approximately 80 % of the bolt’s tensile strength is assumed for the shear strength. This approach calculates against the tensile strength R m , i.e. against the indexing pin shearing off. Any pre-existing and re- maining deformation may, however, mean that the indexing plunger can be used no longer. To ensure the permanent and proper function of the indexing plunger, the yield limit R e must be considered in place of the tensile strength R m . Shear loads On principle, the design also needs an adequate safety coefficient to be taken into account. The usual safety coefficients under static load 1,2 to 1,5; pulsating 1,8 to 2,4 and alternating 3 to 4. Disclaimer: Our information and recommendations are given with non-binding effect and ruling out any liability, unless we have expressly committed ourselves in writing to provide information and recommendations. All products are standard elements for versatile uses and as such are subject to extensive standard tests. You should carry out your own test se- ries to verify whether a certain product is suitable for your specific applications. We cannot be held responsible for this. Safety information The tensile strength shown in the table opposite (R m ) and the yield or substitute yield limit (R e / R p 0,2 ) have been determine in tension tests involving tension specimen in accordance with DIN 50125- B6-30 These tests constitute the basis for the load bearing details given herein. Material characteristics d Bolt diameter Max. force F in N, acc. to material and strength value differs C45Pb / 1.0504 X 10 CrNiS 18 9 / 1.4305 at R e at R m at R e at R m  3  3160  3610 3270  4180  4  5620  6430 5830  7430  5  8790 10050 9110 11620  6 12660 14470 13120 16730  8 22510 25730 23320 29750 10 35180 40210 36440 46490 12 50660 57900 52470 66950 16 90070 102940 93290 119020 Computing examples, load values Example: Indexing plungers with a bolt diameter of 6 mm made of Stainless Steel with a yield limit of R e = 580 N/mm², com- putation against permanent deformation, the maximum per- missible shear stress is wanted. F per = (6 mm)² x π x 0,8 x 580 N/mm² = 13120 N 4 Formulas for computation Bolt cross-section Limit tension Shear force S = d² x π 4 τ a = 0,8 x R m F = S x τ a = d² x π x 0,8 x R m 4 Material R e R m Description Material no. in N/mm ² in N/mm ² C45Pb 1.0504 560 640 X 10 CrNiS 18 9 AISI 303 580 740