PSI - Issue 44

6

Fabio Di Trapani et al. / Procedia Structural Integrity 44 (2023) 496–503 Di Tr pani et al./ Structural Integrity Procedia 00 (2022) 000–000

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10 20 30 40 50 60 70 80 90 100 Shear force [kN]

10 20 30 40 50 60 70 80 90 100 Shear force [kN]

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100

S1B - Leeward column

S1A - Windward column

S1A - Leeward column

S1B - Windward column

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80

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Shear force [kN]

Shear force [kN]

Cut 4 Cut 5 Cut 6 Average

Cut 1 Cut 2 Cut 3 Average

Cut 4 Cut 5 Cut 6 Average

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20

0 5 10 15 20 25 30 35 40 Displacement [mm] 0

0 5 10 15 20 25 30 35 40 Displacement [mm] 0

0 5 10 15 20 25 30 35 40 Displacement [mm] 0

0 5 10 15 20 25 30 35 40 Displacement [mm] 0

(a)

(b)

100 120 140 160 180 200

100 120 140 160 180 200

100 120 140 160

100 120 140 160

S1C - Leeward column

S1C - Windward column

Spec. 5 - Windward column

Spec. 5 - Leeward column

20 40 60 80

20 40 60 80

20 40 60 80

20 40 60 80

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Shear force [kN]

Shear force [kN]

Shear force [kN]

Shear force [kN]

Cut 4 Cut 5 Cut 6 Average

Cut 1 Cut 2 Cut 3 Average

Cut 4 Cut 5 Cut 6 Average

0 5 10 15 20 25 30 35 Displacement [mm] 0

0 5 10 15 20 25 30 35 Displacement [mm] 0

0 5 10 15 20 25 30 35 40 Displacement [mm] 0

0 5 10 15 20 25 30 35 40 Displacement [mm] 0

(c)

(d)

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120

100 120 140 160 180

100 120 140 160 180

Spec. 9 - Leeward column

Spec. 8 - Windward column

Spec. 9 - Windward column

Spec. 8 - Leeward column

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100

80

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20 40 60 80

20 40 60 80

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Cut 1 Cut 2 Cut 3 Average

Shear force [kN]

Shear force [kN]

Shear force [kN]

Shear force [kN]

Cut 4 Cut 5 Cut 6 Average

Cut 4 Cut 5 Cut 6 Average

Cut 1 Cut 2 Cut 3 Average

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20

0 5 10 15 20 25 30 35 Displacement [mm] 0

0 5 10 15 20 25 30 35 Displacement [mm] 0

0 5 10 15 20 25 30 35 Displacement [mm] 0

0 5 10 15 20 25 30 35 Displacement [mm] 0

(f)

(e)

Fig. 7. Total shear demand at the windward and leeward column ends for Cavaleri & Di Trapani (2014) specimens: (a) S1A; (b) S1B; (c) S1C and Mehrabi & Shing (1996) specimens: (d) 5; (e) 8; (f) 9. 4. Prediction of the additional shear demand using macro-modelling approach Equivalent strut macro-models do not allow assessing local shear demand due to frame-infill interaction. Nevertheless, considering Fig. 8a, it is possible to imagine that the total shear demand at the end of a column adjacent to the infill ( V d,tot ) can be decomposed as the sum of the drift-related shear on the frame ( V d,frame ) and of the additional shear demand due to frame-infill interaction ( V d,inf ), that is: (1) While V d,frame is already available as shear internal force from the frame, the term V d,inf is unknown. However it can be reasonably assumed the shear force V d,inf is a rate of the axial force acting on the equivalent strut. In fact, considering the forces acting on a portion of infill at the end of a column, the translational equilibrium equation provides: V N cos T d ,inf − = θ (2) meaning that the additional shear demand is the difference between the horizontal component of the axial force on the equivalent strut and the tangential friction force at the interface ( T ). The latter is related to the vertical component ( σ v ) of the normal stress acting on the strut ( σ v ) through the friction coefficient ( µ ) and acts on a contact length ( α l ), that is a portion of the total length of the infill ( α l , with α ≤1) . The tangential force at the interface is therefore as: t l T v α µσ ⋅ ⋅ = (3) where θ σ σ sin n v = and N / w t n ⋅ = σ (4) d ,inf d , frame d ,tot V V V + =