Problem 4 The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress is Tult = 370 MPa and a bearing stress ultimate is 400 MPa. Use a factor of safety for shear of F.S =2.5. Thickness of the plate is 13mm. 30 mm 80 kN 30 mm 40 kN 40 kN

Problem 4 The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress is Tult = 370 MPa and a bearing stress ultimate is 400 MPa. Use a factor of safety for shear of F.S =2.5. Thickness of the plate is 13mm. 30 mm 80 kN 30 mm 40 kN 40 kN

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.12.2P

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Concept explainers

Types and properties of steel

Steel is an alloy of iron. It is formed by heating iron and combining carbon (0.1-1.7%) with it and minor amounts of other elements such sulfur, chromium, manganese, nickel, etc. Thousands of steel grades have been published, registered, or regulated around the world, each with its chemical composition, and special numbering systems have been devised in various countries to classify the vast number of alloys.

Question

Problem 4
The joint is fastened together using two bolts. Determine the required diameter of the bolts
if the failure shear stress is Tult = 370 MPa and a bearing stress ultimate is 400 MPa. Use a
factor of safety for shear of F.S =2.5. Thickness of the plate is 13mm.
30 mm
80 kN
30 mm
40 kN
40 kN

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