764in- k
Mu
= 23.6 - in3 < 26.0-in3 = ZW10x22 (386.7X103 mm3 < 426.1X103 mm3)
=
=
Z req 'd
φ y 0.9(36ksi)
F
Use W10x22 (254mmX1.05KN/m) for transverse beams
Design longitudinal platform beams (see Figures J2-2, and J2-4);
P1u = (1/2)38.2k = 19.1k (89.96KN)
P2u = (1/2)P1u = 9.6k (42.70KN)
Therefore, reactions are; Ru = 19.1k (89.96KN)
The maximum moment occurs at mid span as;
Mu = [19.1k(5' 9.6k(2.5' ](12"/1' = 858in-k (96.95KN-m)
)
)
)
Figure J2-4. Loading on longitudinal beams
Note: Beam is laterally supported throughout its span
858in- k
Mu
= 26.5 - in3 > 26.0 = ZW10x22 (434.3X103 mm3 > 426.1X103 mm3)
Z req 'd =
=
φ y 0.9(36ksi)
F
(okay because of conservative assumptions)
Use W10x22 (254mmX1.05KN/m) for longitudinal beams
Design columns;
Using load combination U = 1.4D;
Pu / column = 1.4(Wp1 + Wp2 )/# columns = 1.4(54.6k + 5k ) / 4 = 20.9 k /column (93.0KN/column)
Try W10x22;
Relevant properties of a W10x22 are as follows;
A = 6.49-in2 (4.19X103 mm2)
ry = 1.33-in (33.8mm)
rx = 4.27-in (108.5mm)
Check capacity of W10x22;
Note: Elements are in a braced frame with K = 1.0
Since KLy = KLx;
KL y 1.0(10' )(12"/1' )
=
= 90.2
ry
1.33"
φc Fcr = 19.94ksi (137.5MPa)
From AISC LRFD Table 3-36;
(interpolated)
φc Pn = φc Fcr A g = 19.94ksi(6.49 - in2 ) = 129.4 k (575.6KN)
φc Pn = 129.4 k > 20.9 k = Pu (575.6KN > 93.0KN)
Use W10x22 (W254mmX1.05KN/m) for columns
Design transverse beam to longitudinal beam connection (see Figures J2-2, and J2-5);
Use; Fy = 36ksi (248.2MPa), and Fu = 58ksi (399.9MPa)
Note: It was previously determined that the middle transverse beam supports half of Wp1 with the
distribution as shown in Figure J2-3. Therefore, the reaction can be taken as Ru = (1/2)38.2k = 19.1k
(89.96KN). One design will be made for this beam and used throughout for all transverse beam
connections. Formulas are taken from Part 9 of AISC LRFD volume II, 2nd edition.
Relevant properties of a W10x22 are as follows;
tw = 0.240-in (6.1mm)
bf = 5.750-in (146.1mm)
tf = 0.360-in (9.1mm)
d = 10.17-in (258.3mm)
Determine coping dimensions c and dc;
5.750"- 0.240"
c=
= 2.76 - in (70.1mm)
Say c = 3.0-in (76.2mm)
2
dc = 0.360" + 0.500" = 0.86" (21.8mm)
Say dc = 1.0-in (25.4mm)
Check flexural yielding of coped section assuming two 7/8-in. (22.2mm) φbolts;
φM
Ru ≤ b n
where; φ= 0.9
e
Mn = FySnet
J2-4
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