If this is true, how do you design the bolts in a built up girder (plate for
web and angles for flanges that are bolted to the web plate)?
Chris Towne, E.I.T.
----- Original Message -----
From: "Rick Byrd" <Rbyrd(--nospam--at)rmbyrd.com>
Sent: Monday, February 14, 2000 4:03 PM
Subject: Re: Shear Flow
If you are fastening the channels to the side of the GLB, then shear flow is
not necessarily the calculation you require. Shear flow calculates the
horizontal shear through the width of the, as an example, the glue joint
between laminations of a GLB resists the shear calculated by shear flow.
Side plates will strengthen the beam. It will add capacity based on
stiffness of the origional member vs the stiffness of the side plates added.
The amount of load carried through the steel side plates needs to have that
reaction transferred back into the GLB at the end of the beam, unless the
plates bear on the GLB support. This load is a vertical load, not a shear
flow load. If the plates bear on the GLB support then the bolts only need be
designed for the transfer of the vertical load to the plates.
If you are bolting the plates to the top and the bottom of the beam, then
yes, shear flow is the required calculation.
>>> "Gobo, Gina" <ggobo(--nospam--at)DLRGROUP.com> 02/14 11:42 AM >>>
I have question about composite beams:
I have a wood glulam beam that I would like to reinforce by adding a steel
channel to both sides. I am trying to find the shear flow at the channel to
wd beam contact area so that I may figure our how many bolts I need to
faster the composite beam together. I am using the formula f = QV/I, where V
is the maximum shear load on the beam, I is the moment of inertia for the
composite section, and Q the first moment of inertia of the glulam beam that
is in contact with the channel = yA where, y is the distance from the
centroid of the block of wd in contact with the channels to the neutral axis
and A is the area of the same block of wd. If I am using a channel that is a
C12X20.7 and a 6.75X31.5 glulam, Q = ((12/2)-dist to neutral axis of entire
section)*6.75*12=248in3. I = 17964.83 in4 and V = 48kips. f = 8kip/ft which
seems very high.
The shear flow that I get seems very high. Do I have to use the transformed
section in order to get the shear flow? (i.e. transform the wd beam into an
equivalent steel member in order to get the correct shear flow). Or, is
using I of the composite section based on the geometry of the cross section
Gina T. Gobo, E.I.T.