Building

The area A₁, in²(mm²), 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

A₁=R /0.70f^‘_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=(A₁)^½+0.5(0.95d–0.8b)
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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/0.35f^‘_c
Where
R=beam reaction, kip (kN),
f ^‘_c= specified compressive strength of the concrete, ksi (MPa).
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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:

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:

Normal shearing stresses may be computed on the assumption of a parabolic stress distribution over the shell thickness:
where V=unit shear force normal to middle surface.

The AISC specification for ASD specifies the following allowable shear stresses F_v, ksi ( ksiX6.894=MPa)
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For a compact section bent about the major axis, the unbraced length L_b of the compression flange, where plastic hinges may form at failure, may not exceed L_pd, given by Eqs. given in post. For beams bent about the minor axis and square and circular beams, L_b is not restricted for plastic analysis.

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The maximum fiber stress in bending for laterally supported beams and girders is F_b = 0.66F_y if they are compact, except for hybrid girders and members with yield points exceeding 65 ksi (448.1 MPa). F_b = 0.60F_y for non-compact sections. F_y is the minimum specified yield strength of the steel, ksi (MPa). Table lists values of F_b for two grades of steel.
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Plastic analysis of prismatic compression members in buildings is permitted if (F_y)^½(l/r) does not exceed 800 and F_uless than equal to 65 ksi (448 MPa). For axially loaded members with b/t less than equal to (lambda)_r , the maximum load P _u , ksi (MPa= 6.894 X ksi), may be computed from

P_u = 0.85A_gF_cr

Where A_g=gross cross-sectional area of the member
F _cr =0.658 ^lambda F_y for lambda <= 2.25

=0.877 F_y/lambda for lambda  > 2.25

lambda =(Kl/r)²(F_y/286,220)

The AISC specification for LRFD presents formulas for designing members with slender elements.

The AISC specification for allowable-stress design (ASD) for buildings provides two formulas for computing allowable compressive stress Fa, ksi (MPa), for main members. The formula to use depends on the relationship of the largest effective slenderness ratio Kl/r of the cross section of any unbraced segment to a factor C _c defined by the following equation and table below:
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Based on the American Institute of Steel Construction (AISC) specifications for load-and-resistance factor design (LRFD) for buildings, the shear capacity V _u , kip (kN = 4.448 X kip), of flexural members may be computed from the following:
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