Question

Using LRFD, select a 30-ft-long W section of A992 steel with a nominal depth of 10 in (a W12) to support a tensile service dead load PD=130kand a tensile service live load PL=110k. In addition, the section is to comply with the slenderness ratio limit from AISC D1 to prevent sag of the member. As shown in the figure below, the member is to have two lines of bolts in each flange for 7/8-in bolts (at least three in a line 4-in on center).

Answer 1

Before Solve the problem know about#AISC LRFD Specifications are as follow:=
The design strength φt Pn is the lesser of
a) φt Fy Ag (1)
b) φt Fu Ae. (2)
c) The block shear strength, φt Rn . (3)

AISC LRFD Specification

The first expression (Eq. 1) is satisfied if
the minimum gross area is at least equal to
the following:

min Ag= Pu/φt Fy (4)

– The second expression (Eq. 2) is satisfied
if the minimum value of Ae is at least

Min Ae= Pu/φt Fu (5)

AISC LRFD Specifications
– And since Ae = U An for bolted members, the
minimum value of An is given by

min An = min Ae/U= Pu/φt FuU (6)

Then the minimum Ag for the second expression

(Eq. 2) must be at least equal the minimum

value of An plus the estimated hole area:

min Ag = Pu/φt FuU + estimated hole areas (7)

– The third expression (Eq. 3) can be
evaluated once a trial shape has been
selected, and the other parameters related
to the block shear strength are known.
– The designer can substitute into Eqs. 4
and 7, taking the larger value of Ag so
obtained for an initial size estimate

– The designer also has to check the
slenderness ratio that it would not exceed
a value of 300, that is
L/r=300

or . (8)

min r= L/300

– If no load involved other than the dead and
live loads, then the designer must check
the following load factor expressions and
take the larger:

Pu= 1.4D (9)

Pu=1.2D+1.6L. (10)

*Solution of Problem =

Considering the load factor expressions of
Eqs. 9 and 10:

Pu = 1.4 D = 1.4 (130) =182 k

Pu =1.2D +1.6L = 1.2 (130)+1.6(110)=332k
Computing the minimum Ag required using
Eqs. 4 and 7:

min Ag =Pu/φt Fy = 332/0.9(50)= 7.38in^2

Assume U = 0.9 and assume the flange
thickness is 0.380 in from the manual for W12.

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