Strength - Bottom Flange (continued)
The shear stress in the flange, fd, caused by the internal diaphragm vertical shear due to factored loads is
approximated as:
27
ëø öø(1.5)(31.05)
1406
ìø
VQ 2
íø øø
fd = = ksi Eq (C6.11.8.1.1-2)
= 5.24
I tfc 112375(1.5)
( )
fv tot = ksi
1.08 + 5.24 = 6.32
The effect of bending in the plane of the diaphragm for boxes supported on two bearings is insignificant
and was, therefore, ignored in the design of the example girder. The effect of these forces on a box
supported on a single bearing is likely to be more significant and should be considered. The effective
section specified in Article C6.11.8.1.1 may be used to compute the flange bending stress about the
tangential z-axis due to bending of the internal diaphragm over the sole plate. In this case, the resulting
minimum and maximum principal stresses in the flange should be input into the more general form of the
Huber-von Mises-Hencky yield criterion given as follows:
2 2
Eq (C6.11.8.1.1-1)
fbu - fbu fby + fby + 3 fd + fv d" Æf Rb Rh Fyc
( )2
The factored vertical bending stress in the bottom flange, fbu, was computed earlier to be -41.6 ksi.
fby = stress in the flange due to the factored loads caused by major-axis bending of the internal
diaphragm over the bearing sole plate
= taken as 0.0 ksi for a box supported on two bearings
Rh = 1.0 (Article 6.10.1.10.1)
Rb = 0.997 (previously calculated)
2 2 2
ksi
(-41.6) - (-41.6)(0) + (0) + 3 (5.24 + 1.08) = 43.02 1.0(0.997)(1.0) 50 = 49.85
The combined principal stresses in the diaphragm due to the factored loads is checked using the general
form of the Huber-von Mises-Hencky yield criterion.
2 2
Ã1 - Ã1Ã2 + Ã2 d" Fy
where , are the maximum and minimum principle stresses in the diaphragm
Ã1 Ã2
D-31
Girder Stress Check Section 5-5 G2 Node 36
Strength - Bottom Flange (continued)
2
Ãy + Ãz Ãy - Ãz 2
ëø öø ëø öø
, =
Ã1 Ã2
ìø ± ìø + fv
2 2
íø øø íø øø
= stress in the diaphragm due to vertical bending of the diaphragm over the bearing sole plate
Ãy
= stress in the diaphragm due to bending of the diaphragm about its longitudinal axis
Ãz
fv = shear stress in the diaphragm
Fy = specified minimum yield stress of the diaphragm
Since the example box is supported on two bearings, the stress in the diaphragm due to vertical bending
of the diaphragm over the bearing sole plate is typically relatively small and will be neglected for simplicity
in this example. is also typically neglected. If no bending is assumed, the two principal stresses are
Ãz
simply equal to the tensile and compressive stresses with a magnitude equal to the shear stress.
2 2
= 0
Ã1,2 ± = 6.32 ksi
(0) + 6.32 ±
Check the combined principal stresses.
2 2
ksi
6.32 - (6.32)(-6.32) + (-6.32) = 10.95
D-32
Girder Stress Check Section 5-5 G2 Node 36
Longitudinal Flange Stiffener
Try a WT 8x28.5 structural tee for the longitudinal stiffener with the stem welded to the bottom flange.
The projecting width, bl, of the stiffener must satisfy the following requirement:
E 29000
= in. Eq (6.11.11.2-1)
bl d" 0.48ts 0.48(0.715) = 8.27
Fyc 50
where ts is taken as the flange thickness of the structural tee since each half-flange buckles
similarly to a single plate connected to a web.
For structural tees, bl should be taken as one-half the width of the flange.
bl = 7.12/2 = 3.56 in.
According to Article 6.7.4.3, transverse top and bottom bracing members (i.e. top and bottom struts of
internal cross frames) are required to ensure that the cross section shape is retained. Whenever
longitudinal flange stiffeners are present, the bottom transverse bracing members are to be attached to the
longitudinal stiffener(s) to better control the transverse distortion of the box flange. At other locations, the
bottom transverse member is to be attached directly to the box flange. The cross-sectional area and
stiffness of the top and bottom transverse bracing members is not to be less than the area and stiffness of
the diagonal members. At the pier section (the point of maximum compressive flexural stress in a box
flange in most cases), the bottom transverse bracing member, when properly attached to the longitudinal
flange stiffener, can be assumed to provide the required transverse stiffening of the box flange. Use a
W10x68 (I = 394 in4) for the bottom transverse bracing member.
The longitudinal flange stiffener should be attached to the internal diaphragm with a pair of clip angles as
shown in Figure D-2 (page D-81).
D-33
Girder Stress Check Section 5-5 G2 Node 36
Design of the Internal Diaphragm
Article 6.11.1 directs the designer to the provisions of Article 6.7.4 for general design considerations for
cross-frames and diaphragms.
Try a 1-inch thick A36 diaphragm plate.
Compute the maximum factored vertical shear in the diaphragm.
Load Shear Source
Steel 47 + |-46| = 93 k 3D Finite Element Analysis
Deck 185 + |-185| = 370 k (in critical web from Table C2)
SupImp 44 + |-41| = 85 k Unfactored results are shown
FWS 58 + |-55| = 113 k
LL + IM 160 + |-155| = 315 k
Vu = kips
1.25(93 + 370 + 85) + 1.5(113) + 1.75(315) = 1406
The internal diaphragm is subject to vertical bending over the bearing sole plates in addition to shear.
Therefore, Article 6.11.8.1.1 requires that the principal stresses in support diaphragms not exceed the
factored compressive resistance given by Eq (C6.11.8.1.1-1), which is a yield criterion for combined
stress. The example box is supported by two bearings, therefore, f in this equation is taken as 0.0 ksi
by
since it is typically relatively small.
Compute the maximum factored shear stress in the diaphragm web. First, separate out the shears due
to bending, Vb, and due to St. Venant torsion, VT. [ Pobierz całość w formacie PDF ]