冢
Solution:
Rv ⳱ 0.75 ⫻ 0.6F y d c tw 1 Ⳮ
Calculate the flange forces and panel-zone shear force: From Equation 2.1-2, the force at each flange need not be taken greater than Puf ⳱ ⳱
3b f t 2f d b d c tw
冣
⳱ 0.75 ⫻ 0.6(50 ksi)(18.67 in.)(1.875 in.)
1.1 R y F y Z Ⳮ V u a d ⫺ tf
冢
⫻ 1Ⳮ
in.3 ) Ⳮ (150
1.1(1.1)(50 ksi)(356 kips)(22.5 in.) (35.85 in. ⫺ 0.940 in.)
3(16.695 in.)(3.035 in.)2 (35.85 in.)(18.67 in.)(1.875 in.)
⳱ 1,080 kips ⬎ V u ⳱ 714 kips
冣
o.k.
To prevent seismic shear buckling in the panel-zone, from Equation 2.2-7,
⳱ 714 kips
Neglecting the effects of story shear, the panel-zone web shear force is determined from Equation 2.1-5 as
tw min ⳱
V u ⳱ Puf ⳱ 714 kips
⳱
Determine the design panel-zone web shear strength: In a high-seismic application, either Equation 2.2-5 or Equation 2.2-6 is used.
d m Ⳮ d c ⫺ 2t f 90 (35.85 in. ⫺ 0.940 in.) Ⳮ 18.67 in. ⫺ 2(3.035 in.) 90
⳱ 0.528 in. ⬍ tw ⳱ 1.875 in.
P y ⳱ F y A ⳱ (50 ksi)(125 in. ) ⳱ 6,250 kips 2
o.k.
Therefore, the web of the W14⫻426 is adequate to resist the panel-zone web shear without reinforcement.
1,000 kips Pu ⳱ ⳱ 0.160 6,250 kips Py
Determine the transverse stiffener requirements: As indicated in Section 2.3, transverse stiffeners are required to match the configuration used in the qualifying
Since this ratio is less than 0.75, Equation 2.2-5 is applicable.
W14x426, Fy = 50 ksi W36x150, Fy = 50 ksi
Centerline of radiuscut RBS (location of plastic hinge)
22.5” Check if column stiffening is required Figure 6-13 Framing arrangement for Example 6-13 (problem statement). 60
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