# Urgent help required regarding combined stresses above yield

• To: <seaint(--nospam--at)seaint.org>
• 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

<<<-------------------->>>
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