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Shear flow ? 1

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slickdeals

Structural
Joined
Apr 8, 2006
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Folks,

This may be a really simple, stupid question....

I have a situation where I need to connect 2 single angles (toe-toe) to create a U shape.

I think the weld should just be to keep the two members together, preventing them from separating. A skip weld maybe.

A single angle in bending unrestrained)....a case of two drunks leaning on one another? How do you design this?

 
This wouldn't be subject to lateral loading, as it only supports the beam above until cast. The section is symmetrical for gravity loading, so I don't see how shear flow is a concern.
 
The following summarizes the problem:

1. There is no lateral loading, since this member is only acting as formwork.

2. The member is symmetrical once made composite by joining the legs.

My concerns are:
What are the forces acting along this weld that will make this member composite. Once composite, this is akin to channel in weak axis bending.

Taking 2 angles, which want to rotate and trying to connect them together is the issue. What are the forces acting at the interface?

 
To simplify:
Assume two angles L4x4x3/8 created to form a 8" wide x 4" deep U shaped member.
Assume that the 2 angles are connected by a 3/16" thick x 0.5" wide member to mimic a weld.

Assume a loading of 0.5 klf and span of 10 ft.

 
I wouldn't call it a composite member. It is just two angles side by side, joined by enough weld to prevent rotation. But really, if you are just using it to cast a 200 x 200 beam across a lift shaft, how much rotation is that? A few short welds will suffice. If rotation remains a concern, weld a few spacers across the top.
 
Hi slickdeals

They will act like a normal beam when welded together because your loading through an axis of symmetry so there will be no twisting, therefore the weld will be subject to a moment and shear stresses similiar to that of a normal beam.

 
@slick, This "channel" would look like a structural member to many of the code officials that I banged heads with.



Michael.
Timing has a lot to do with the outcome of a rain dance.
 
For a section having one axis of symmetry and loaded in the plane of symmetry, the shear center would coincide with the center of gravity of the section. We could then use formula q = (V.Q) / I
where V = Shear at the support, I = M.I. of the two angles about minor axis of the (overall) section, and Q = moment of area of one angle about the c.g. of the (overall)section, i.e. about the weld axis.
 
Assume two angles L4x4x3/8 created to form a 8" wide x 4" deep U shaped member.
Assume that the 2 angles are connected by a 3/16" thick x 0.5" wide member to mimic a weld.

Assume a loading of 0.5 klf and span of 10 ft.

Assume:
.1 Vertical legs carry all of the load.
.2 Load is uniformly distributed over the horizontal legs.
.3 Span is 8"

M per foot = 500 * 8/8 = 500"#/'

S = 12(3/16)^2/6 = 0.0703 in^3

f = M/S = 7,111 psi in 3/16 plate.


BA
 
Correction to my previous post ....... The shear center would be that for a channel and not at the c.g. of the (overall)section.
Q = moment of area of one angle about the shear center / weld axis.
 
As BA has repeatedly hinted, the angles will burst apart sideways, and twist, in the absence of 'sufficient' weld. The vertical load is only being applied through the shear centre if the section is integral. It is not, in practice you will have two vertical loads applied, one to each angle, and it is not at the angle's shear centre.

Quantifying exactly how much weld is required depends on your definition of acceptable spread in the joint.


Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
@Greg,BA
I understand what you are saying, which is what I am concerned about too. In my OP, you will see the reference to two drunks leaning on each other.

 
And Happy New Year to all your wonderful folks and your families. Hope 2012 turns out to be better than 2011.

 
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