150 mm Weld thickness 6 mm Vdw = wtfyw/(√3 × Ymw) W

International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter9: Moments And Products Of Inertia Of Areas
Section: Chapter Questions
Problem 9.46P
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Te - Effective throat thickness well in mm = 6 sin45°

Consider the following drawing in which plate 1 is welded onto plate 2 as shown for tensile and shear loading. The plates are fillet welded with a weld thickness of 6 mm as shown for a length of 150 mm.

 

The plates have an ultimate strength of 275 MPa and the welding was done with F70 welding electrodes, which possesses an ultimate strength of 483 MPa.

 

Consider a partial factor of safety of 1.7 for the welded joint and determine the design strength of the weld joint, both in tension (Tdw) and in shear (Vdw) due to the forces acting on it.

 

a)      Determine the design strength of the following weld in tension ( ) due to vertical forces.

b)      Determine the design strength of the weld in shear ( ) due to horizontal forces.

 

Attached pic is shear equation for calculation

Tension equation for calculation is: Tdw = FyLwt/Ymw

 

150 mm
Weld
thickness
6 mm
Transcribed Image Text:150 mm Weld thickness 6 mm
Vdw = wtfyw/(√3 × Ymw)
W
Transcribed Image Text:Vdw = wtfyw/(√3 × Ymw) W
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