## 4

Therefore, the section is compact.

Determine the allowable flexural stress, Fb: Since the section is compact and the lateral unbraced length, Lb = 60 in., is less than Lc = 83.4 in., the allowable bending stress from Equation 4.28 is 0.66Fy = 24 ksi.

Determine the section modulus of the beam with cover plates:

q _ ^combination section

Sx,combination section

'23.57s

Determine flexural capacity of the beam with cover plates:

^x,combination section Sx,combination sectionFb (192)(24) 4608 k in.

Since the flexural capacity of the beam without cover plates is

the increase in flexural capacity is 68.4%.

Determine diameter and longitudinal spacing of bolts: From Mechanics of Materials, the relationship between the shear flow, q, the number of bolts per shear plane, n, the allowable bolt shear stress, Fv, the cross-sectional bolt area, Ab, and the longitudinal bolt spacing, s, at the interface of two component elements of a combination section is given by nFvAb — = q

Substituting n = 2, q = VQ/I = (100)[(7)(±)(12.035)]/2364 = 1.78 k/in. into the above equation, we have

FvAb

If 2 in. diameter A325-N bolts are used, we have Ab = p(i)2/4 = 0.196 in. and Fv = 21 ksi (from Table 4.12), from which s can be solved from the above equation to be 4.57 in. However, for ease of installation, use s = 4.5 in.

In calculating the section properties of the combination section, no deduction is made for the bolt holes in the beam flanges nor the cover plates, which is allowed provided that the following condition is satisfied:

0.5FuAfn > 0.6FyAfg where Fy and Fu are the minimum specified yield strength and tensile strength, respectively. Afn is the net flange area and Afg is the gross flange area. For this problem

Beam flanges [0.5FuAfn = 0.5(58)(7.005 - 2 x 1/2)(0.505) = 87.9 kip] >[0.6Fy Afg = 0.6(36)(7.005)(0.505) = 76.4 kip]

Cover plates [0.5FuAfn = 0.5(58)(7 - 2 - 1/2)(1/2) = 87kip] >[0.6Fy Afg = 0.6(36)(7)(l/2) = 75.6 kip]

so the use of gross cross-sectional area to compute section properties is justified. In the event that the condition is violated, cross-sectional properties should be evaluated using an effective tension flange area Afe given by

5 Fu

6 Fy

So, use 1-in. diameter A325-N bolts spaced 4.5 in. apart longitudinally in two lines 4 in. apart to connect the cover plates to the beam flanges.

4.5.1.2 Load and Resistance Factor Design 4.5.1.2.1 Flexural Strength Criterion

Flexural members must be designed to satisfy the flexural strength criterion of fbMn > Mu

where f bMn is the design flexural strength and Mu is the required strength. The design flexural strength is determined as given below.

4.5.1.2.1.1 Compact Section Members Bent about Their Major Axes — For Lb < Lp (plastic hinge formation)

For Lp < Lb < Lr (inelastic lateral torsional buckling)

For Lb > Lr (elastic lateral torsional buckling) For I-shaped members and channels:

1 IyCw

For solid rectangular bars and symmetric box sections:

The variables used in the above equations are defined in the following:

Lb = lateral unsupported length of the member Lp, Lr = limiting lateral unsupported lengths given in the following table:

Structural shape

I-shaped sections, channels

1.76ry /V (E/Ff where ry = radius of gyration about minor axis E = modulus of elasticity Fyf = flange yield strength

[rrX1/FL]V[1 + V (1 + X2Fl2)]} where ry = radius of gyration about minor axis, in. Xi = (p/Sx)p(EGJA/2) X2 = (4Cw/1y)(Sx/ GJ)2 Fl = smaller of (Fyf - Fr) or Fyw Fyf = flange yield stress, ksi Fyw = web yield stress, ksi Fr = 10 ksi for rolled shapes, 16.5 ksi for welded shapes Sx = elastic section modulus about the major axis, in.3 (use Sxc, the elastic section modulus about the major axis with respect to the compression flange if the compression flange is larger than the tension flange)

Iy = moment of inertia about the , 4

minor axis, in. = torsional constant, in.4 = warping constant, in.6 = modulus of elasticity, ksi shear modulus, ksi

Solid rectangular bars, symmetric box

[0.13ryV (JA)]/Mp where ry = radius of gyration about minor axis E = modulus of elasticity J = torsional constant A = cross-sectional area Mp = plastic moment capacity = FyZx

Fy = yield stress Zx = plastic section modulus about the major axis

[2ryV (JA)]/Mr where ry = radius of gyration about minor axis J = torsional constant A = cross-sectional area Mr = Fyf Sx Fy = yield stress Fyf=flange yield strength Sx = elastic section modulus about the major axis

Note: Lp given in this table are valid only if the bending coefficient Cb is equal to unity. If Cb > 1, the value of Lp can be increased. However, using the Lp expressions given above for Cb > 1 will give conservative value for the flexural design strength.

Mr = FLSx for I-shaped sections and channels, FfSx for solid rectangular bars and box sections Fl = smaller of (Fyf — Fr) or F^ Fyf = flange yield stress, ksi Fy„ = web yield stress, ksi

Fr = 10 ksi for rolled sections, 16.5 ksi for welded sections Fy = specified minimum yield stress sections

Sx = elastic section modulus about the major axis Zx = plastic section modulus about the major axis Iy = moment of inertia about the minor axis J = torsional constant Cw = warping constant E = modulus of elasticity G = shear modulus

Mmax, Ma, Mb, Mc = absolute value of maximum moment, quarter-point moment, midpoint moment, and three-quarter point moment along the unbraced length of the member, respectively.

Cb is a factor that accounts for the effect of moment gradient on the lateral torsional buckling strength of the beam. Lateral torsional buckling strength increases for a steep moment gradient. The worst loading case as far as lateral torsional buckling is concerned is when the beam is subjected to a uniform moment resulting in single curvature bending. For this case Cb = 1. Therefore, the use of Cb = 1 is conservative for the design of beams.

4.5.1.2.1.2 Compact Section Members Bent about Their Minor Axes — Regardless of Lb, the limit state will be plastic hinge formation fbMn = 0.90Mpy = 0.90Fy Zy (4.43)

4.5.1.2.1.3 Noncompact Section Members Bent about Their Major Axes — For Lb < Lp (flange or web local buckling)

Lp, Lr, Mp, Mr are defined as before for compact section members, and

For flange local buckling:

l = bf/2tf for I-shaped members, bf/tf for channels lp, 1r are defined in Table 4.8 For web local buckling:

in which bf is the flange width, tf is the flange thickness, hc is twice the distance from the neutral axis to the inside face of the compression flange less the fillet or corner radius, and tw is the web thickness.

For Lp < Lb < Lr (inelastic lateral torsional buckling), fbMn is given by Equation 4.40 except that the limit 0.90Mp is to be replaced by the limit 0.90M^.

For Lb > Lr (elastic lateral torsional buckling), fbMn is the same as for compact section members as given in Equation 4.41 or 4.42.

4.5.1.2.1.4 Noncompact Section Members Bent about Their Minor Axes — Regardless of the value of Lb, the limit state will be either flange or web local buckling, and f bMn is given by Equation 4.42.

4.5.1.2.1.5 Slender Element Sections — Refer to Section 4.10.

4.5.1.2.1.6 Tees and Double Angle Bent about Their Major Axes — The design flexural strength for tees and double-angle beams with flange and web slenderness ratios less than the corresponding limiting slenderness ratios 1r shown in Table 4.8 is given by fbMn = 0.90

1^/EIyGJ

where

Use the plus sign for B if the entire length of the stem along the unbraced length of the member is in tension. Otherwise, use the minus sign. b equals 1.5 for stems in tension and equals 1.0 for stems in compression. The other variables in Equation 4.46 are defined as before in Equation 4.41.

4.5.1.2.2 Shear Strength Criterion

For a satisfactory design, the design shear strength of the webs must exceed the factored shear acting on the cross-section, that is f vVn > Vu

Depending on the slenderness ratios of the webs, three limit states can be identified: shear yielding, inelastic shear buckling, and elastic shear buckling. The design shear strength that corresponds to each of these limit states are given as follows:

fvVn = 0.90[0.60FywAw ] For 2.45y/(E/F^) < h/tw < 3.07^(E/Fy») (inelastic shear buckling of web)

0.60Fyw A,

0 0