At its ends, the girder will bear on masonry buttresses. The total depth of the girder is re-
stricted to approximately 70 in (1778.0 mm). Select the cross section, establish the spac-
ing of the transverse stiffeners, and design both the intermediate stiffeners and the bearing
stiffeners at the supports.
Calculation Procedure:
- Construct the shear and bending-moment diagrams
These diagrams are shown in Fig. 8.
- Choose the web-plate dimensions
Since the total depth is limited to about 70 in (1778.0 mm), use a 68-in (1727.2-mm) deep
web plate. Determine the plate thickness, using the Specification limits, which are a slen-
derness ratio hltw of 320. However, if an allowable bending stress of 22,000 lb/in^2
(151,690.0 kPa) is to be maintained, the Specification imposes an upper limit of
24,0001(22,000)°^5 = 162. Then tw = h/162 = 68/162 = 0.42 in (10.668 mm); use a^7 /i 6 -in
(11.112-mm) plate. Hence, the area of the web Aw = 29.75 in^2 (191.947 cm^2 ).
- Select the flange plates
Apply the approximation Af = McI(IJy^1 ) - AJ6, where y = distance from the neutral axis
to the centroidal axis of the flange, in (mm).
Assume 1-in (25.4-mm) flange plates. Then Af = 4053(12)(35)/[2(22)(34.5)^2 ] -
29.75/6 = 27.54 in^2 (177.688 cm^2 ). Try 22 x P/ 4 in (558.8 x 31.75 mm) plates with Af =
27.5 in^2 (177.43 cm^2 ). The width-thickness ratio of projection = 11/1.25 = 8.8 < 16. This
is acceptable.
Thus, the trial section will be one web plate 68 x^7 / 16 in (1727 x 11.11 mm); two flange
plates 22 x \% in (558.8 x 31.75 mm).
- Test the adequacy of the trial section
For this test, compute the maximum flexural and shearing stresses. Thus, / =
(1/12)(0.438)(68)^3 + 2(27.5)(34.63)^2 = 77,440 in^3 (1,269,241.6 cm^3 ); / - McII =
4053(12)(35.25)/77,440 = 22.1 kips/in^2 (152.38 MPa). This is acceptable. Also, v =
207/29.75 = 6.96 < 14.5 kips/in^2 (99.98 MPa). This is acceptable. Hence, the trial section
is satisfactory.
- Determine the distance of the stiffeners from the girder ends
Refer to Fig. 8 d for the spacing of the intermediate stiffeners. Establish the length of the
end panel AE. The Specification stipulates that the smaller dimension of the end panel
shall not exceed 11,000(0.438)/(6960)^05 = 57.8 < 68 in (1727.2 mm). Therefore, provide
stiffeners at 56 in (1422.4 mm) from the ends.
- Ascertain whether additional intermediate stiffeners
are required
See whether stiffeners are required in the interval EB by applying the Specification crite-
ria.
Stiffeners are not required when h/tw < 260 and the shearing stress within the panel is
below the value given by either of two equations in the Specification, whichever equation
applies. Thus EB = 396 - (56 + 96) = 244 in (6197.6 mm); h/tw = 68/0.438 = 155 < 260;
this is acceptable. Also, a/h = 244/68 = 3.59.
In lieu of solving either of the equations given in the Specification, enter the table of
a/h, h/tw values given in the AISC Manual to obtain the allowable shear stress. Thus, with
a/h > 3 and h/tw = 155, ^allow = 3.45 kips/in
2
(23.787 MPa) from the table.
