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Urgent help required regarding combined stresses above yield
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- Subject: Urgent help required regarding combined stresses above yield
- From: "Dave Meney" <yenem(--nospam--at)iinet.net.au>
- Date: Wed, 23 Feb 2000 07:46:02 +0800
Hi, I am assisting a colleague who is involved as an expert witness in a case involving a structural failure. Essentially a rectangular hollow section (RHS) formed the end frame of a 3-D truss used as a stacker conveyor, and its 4" stub section below each bottom chord failed. The function of the stub is to transfer vertical (uplift) and horizontal forces from the truss to the base plate at the bottom of the stub. The truss structure has a large cantilever, so at the end frame in question the anchorage is resisting uplift. Thus the stub section contains axial tension, shear from the bottom chord axial force, bending at the top of the stub (due to the base shear and the height of the stub) and torsion (due to the shear from the bottom chord being applied to the side face of the RHS). _____________ |\ | | \ | | \ | |______\|_______ / stub To view, rotate truss 20° so that the little stub is vertical. I hope this suitably describes the problem. It's not a great detail - but we believe the omission of a stiffener included in the design caused the failure. The stiffener would have provided greater section capacity to the stub. The calculated stresses on the section *without* the stiffener were as follows - At top of stub- Bending X-X = 355.0 MPa (51.5 ksi) using plastic modulus Bending Y-Y = 44.4 MPa ( 6.4 ksi) using plastic modulus Tension = 32.0 MPa ( 4.6 ksi) Shear = 144.6 MPa (21.0 ksi) Torsion = 73.9 MPa (10.7 ksi) Combined stress using Von Mises formula = 574 MPa (83.3 ksi). At bottom of stub- Bending X-X = 0.0 MPa ( 0.0 ksi) assumed pinned Bending Y-Y = 44.4 MPa ( 6.4 ksi) using plastic modulus Tension = 32.0 MPa ( 4.6 ksi) Shear = 144.6 MPa (21.0 ksi) Torsion = 73.9 MPa (10.7 ksi) Combined stress using Von Mises formula = 386 MPa (56.0 ksi). The material is Grade 350 which has a yield strength of 350 MPa (50.8 ksi) and an ultimate strength of 430 MPa (62.4 ksi). Clearly the combined stress at the top of the stub section is above the ultimate strength and would result in failure. In fact, this is where the failure occurred. Thankyou for your patience so far! The issue is whether the stiffener would have prevented failure. Similar calculations indicate Von Mises stresses in the stiffened section reduce to 286 MPa (41.5 ksi) at the top of the stub. At the base of the section, the stiffener is ineffective and the combined stress remains at 386 MPa (56.0 ksi). This becomes the critical maximum stress in the section if it were stiffened. My question - even though the combined stress in the latter (stiffened) case exceeded yield by 36 MPa (5.2 ksi) and wouldn't satisfy the provisions of the code, could the section have *failed*? The Von Mises stress is still 44 MPa (6.4 ksi) below the ultimate strength of the section. Any contributions to this query or requests for further clarification would be appreciated today. Regards and thanks, Dave Meney <<<-------------------->>> Yenem Engineering Services 54 John Street Gooseberry Hill Western Australia 6076 Phone: +41 8 9257-2695 Fax: +41 8 9257-2264 Mobile: +41 0417-949-374 e-mail: yenem(--nospam--at)iinet.net.au <<<-------------------->>>
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