Plate Girder In Buildings

For greatest resistance to bending, as much of a plate girder cross section as practicable should be concentrated in the flanges, at the greatest distance from the neutral axis. This
might require, however, a web so thin that the girder would fail by web buckling before it reached its bending capacity.

To preclude this, the AISC specification limits h/t.

For an unstiffened web, this ratio should not exceed.
h / t = 14,000 / (F Y (F y + 16.5) ) ½
where F y = yield strength of compression flange, ksi (MPa).

Larger values of h/t may be used, however, if the web is stiffened at appropriate intervals.

For this purpose, vertical angles may be fastened to the web or vertical plates welded to it. These transverse stiffeners are not required, though, when h/t is less than the value
computed from the preceding equation or Table 9.4.

Critical h/t for Plate Girders in Buildings should be taken from the codes

With transverse stiffeners spaced not more than 1.5 times the girder depth apart, the web clear-depth/thickness ratio may be as large as
h / t = 2000 / ( F y ) ½
If, however, the web depth/thickness ratio h/t exceeds 760 / (F b ) ½ , where F b , ksi (MPa), is the allowable bending stress in the compression flange that would ordinarily apply, this stress should be reduced to F b , given by the following equations:

F b = R P G R e F ¬b
R P G = [ 1 – 0.0005 ( A w / A f ) ( h/t – 760 / ( F b )) ½ ] ? 1.0

R e = [ ( 12 + ( A w /A f ) (3a – a 3 ) ) / ( 12 + 2 (A w / A f ) ] ? 1.0

Where A w = web area , in 2 (mm 2 )

A f = area of compression flange, in 2 (mm 2 )

a = 0.6 F y w / F b ? 1.0

F y w = minimum specified yield stress, ksi, (MPa), of web steel

In a hybrid girder, where the flange steel has a higher yield strength than the web, the preceding equation protects against excessive yielding of the lower strength web in the vicinity of the higher strength flanges. For nonhybrid girders, R e = 1.0.

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Bearing Of Milled Surfaces

In building construction, allowable bearing stress for milled surfaces, including bearing stiffeners, and pins in reamed, drilled, or bored holes, is F p = 0.90F y , where F y is the yield strength of the steel, ksi (MPa)
For expansion rollers and rockers, the allowable bearing stress, kip/linear in (kN/mm), is
F p = ( ( F y – 13 ) / 20 ) 0.66d
where d is the diameter, in (mm), of the roller or rocker. When parts in contact have different yield strengths, F y is the smaller value.

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Column Base Plates

The area A 1 , in 2 (mm 2 ), required for a base plate under a column supported by concrete should be taken as the larger of the values calculated from the equation cited earlier, with R taken as the total column load, kip (kN), or

A 1 = R / 0.70 f c

Unless the projections of the plate beyond the column are small, the plate may be designed as a cantilever assumed to be fixed at the edges of a rectangle with sides equal to 0.80b and 0.95d, where b is the column flange width, in (mm), and d is the column depth, in (mm).

To minimize material requirements, the plate projections should be nearly equal. For this purpose, the plate length N, in (mm) (in the direction of d), may be taken as
N = ( A 1 ) ½ + 0.5 ( 0.95d – 0.80b )
The width B, in (mm), of the plate then may be calculated by dividing A1 by N. Both B and N may be selected in full inches (millimeters) so that BN ³ A 1 . In that case, the bearing pressure f p , ksi (MPa), may be determined from the preceding equation. Thickness of plate, determined by cantilever bending, is given by
t = 2p (f p / F y ) ½
where F y = minimum specified yield strength, ksi (MPa), of plate; and p = larger of 0.5 (N - 0.95d) and 0.5 ( B - 0.80b).
When the plate projections are small, the area A 2 should be taken as the maximum area of the portion of the supporting surface that is geometrically similar to and concentric with the loaded area. Thus, for an H-shaped column, the column load may be assumed distributed to the concrete over an H-shaped area with flange thickness L, in (mm), and web thickness 2L:
L = ( 1 / 4 ) (d + b) – 1 / 4 ( (d + b) 2 - 4R / F P ) ½
where F p = allowable bearing pressure, ksi (MPa), on support. (If L is an imaginary number, the loaded portion of the supporting surface may be assumed rectangular as discussed earlier.) Thickness of the base plate should be taken as the larger
of the values calculated from the preceding equation and
t = L (¬3 f p / F b ) ½

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Bearing Plates

To resist a beam reaction, the minimum bearing length N in the direction of the beam span for a bearing plate is deter- mined by equations for prevention of local web yielding and web crippling. A larger N is generally desirable but is limited by the available wall thickness.
When the plate covers the full area of a concrete support, the area, in 2 (mm 2 ), required by the bearing plate is

A 1 = R / 0.35 f c

Where R = beam reaction, kip (kN), f c = specified compressive strength of the concrete, ksi (MPa). When the plate covers less than the full area of the concrete support, then, as determined from Table

 Full area of concrete support  0.30 f 'c
 Sandstone and limestone  0.40
 Brick in cement mortar  0.25


Units in MPa = 6.895 X ksi

where A 2 = full cross-sectional area of concrete support, in 2 (mm 2 ).
With N established, usually rounded to full inches (millimeters), the minimum width of plate B, in (mm), may be calculated by dividing A 1 by N and then rounded off to full inches (millimeters), so that BN ³ A 1 . Actual bearing pres- sure f p , ksi (MPa), under the plate then is
F p = R / BN
The plate thickness usually is determined with the assumption of cantilever bending of the plate:
t = ( ( 1 / 2) B - k ) ( 3 f p / F b ) ½

where t = minimum plate thickness, in (mm)
k = distance, in (mm), from beam bottom to top of web fillet
F b = allowable bending stress of plate, ksi (MPa)

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Stresses In Thin Shells

Results of membrane and bending theories are expressed in terms of unit forces and unit moments, acting per unit of length over the thickness of the shell. To compute the unit stresses from these forces and moments, usual practice is to assume normal forces and shears to be uniformly distributed over the shell thickness and bending stresses to be linearly distributed.
Then, normal stresses can be computed from equations of the form:

f x = N x / t + ( M x / t 3 /12) z

where z = distance from middle surface

t = shell thickness

M x = unit bending moment about an axis parallel to direction of unit normal force N x .

Similarly, shearing stresses produced by central shears T and twisting moments D may be calculated from equations of the form:

V xy = T / t ± D / ( t 3 / 12 ) z
Normal shearing stresses may be computed on the assumption of a parabolic stress distribution over the shell thickness:
V xy = V / t 3 / 6 ( t 2 / 4 - z 2 )

where V = unit shear force normal to middle surface.

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