For all these problems, select sizes with LRFD expressions and check the selecte

Question

For all these problems, select sizes with LRFD expressions and check the selected sections with both the LRFD and the ASD expressions. Select sections for the conditions described, using Fy=50 ksi and Fu-65 ksi,unless otherwise noted, and neglecting block shear. Select the lightest W10 section available to support working tensile loads of PD 120 k and PE 160 k. The member is to be 18 ft long and is assumed to have two lines of holes for 3/4-in bolts in each flange. There will be at least three holes in each line 3 in on center. (Ans. W10 × 33 LRFD and ASD) 4-1 to 4-8. 4-1. 4-2. Repeat Prob. 4-1 selecting a W12 section. 4-3. Select the lightest WT5 available to support a factored tensile load of P-200 kor Pa = 135 k. Assume there are two lines of 3/4-in& bolts in the flange (at least three bolts in each line 3.5 in on center). The member is to be 24 ft long. (Ans. WT5 × 19,5 LRFD and ASD) select the lightest L4 × 4 section that will safely support the service tensile loads Po 45 k and PL-25 k. The member is to be 15 ft long and is assumed to have one line of holes for 3/4-in Ø bolts in one leg. Assume that there are at least three holes in each line 4 in on center. Use A36 steel 44,

Explanation / Answer

4.4) Given:

Dead load, Pd = 45 k and

Live load, Pl = 25 k

For A36 Steel , Fy = 36 ksi, and Fu =58 ksi

Load combinations:

1: 1.2 D + 1.6 L = 1.2 * 45 + 1.6 * 25 = 94 k

2:   1.4 D = 1.4 * 45 = 63 k (controls)

Hence applied factored load, P = 94 k

Required area:

for Gross section yielding: Ag = P/(0.9Fy) = 94/(0.9*36) = 2.601 in^2

for Net section rupture , effective area Ae = P/(0.75Fu) = 94/(0.75*58) = 2.161 in^2

For at least three bolts in a single line and single leg connection, U = 0.85

Net Area, An = Ae/U = 2.161/0.85 = 2.54 in^2

Now since angle is connected with only one line of bolts.

(Ag) = An + 1*3/4*t

(Ag) = 2.54 + 3/4*t ----------------------------(1)

where t = thickness of leg

Check for Slenderness

Rmin = L/300 = 15*12/300 = 0.6 in

Now from Rmin value select a appropriate thickness which satisfies Gross area requirements as per equation 1.

Section as per required Gross area = 2.61 in^2

Gorss Area as per

Ag = An +3/4*t

Hence Provide L4*4*7/16 with R = 0.785 in which also satisfies slenderness criteria.

Section Gross Area, Ag

Gorss Area as per

Ag = An +3/4*t

Remark L4*4*3/8 2.86 2.82 Net Area not Available L4*4*7/16 3.31 2.868 Net Area Available OK

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