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Dan Abrams + Magenes Course on Masonry

- 1. Seismic design and assessment of Seismic design and assessment of Masonry Structures Masonry Structures Lesson 10 October 2004 Masonry Structures, lesson 10 slide 1 Out-of-plane seismic response of urm walls Parapet failure (Newcastle Earthquake Study, The Institution of Engineers, Australia, 1990) Out-of-plane damage (from the 1997 Umbria-Marche earthquake, Italy, Blasi et al., 1999). Masonry Structures, lesson 10 slide 2
- 2. Out-of-plane seismic response of urm walls Global and partial overturning mechanisms of façade walls Masonry Structures, lesson 10 slide 3 Out-of-plane collapse followed by global collapse Tests performed at Ismes, Bergamo, (Benedetti et al. 1996) Masonry Structures, lesson 10 slide 4
- 3. Out-of-plane seismic response of urm walls Seismic Load Path for Unreinforced Masonry Buildings Floor diaphragm response Shaking amplifies accelerations and motion transmits load to out-of- plane walls Parapet wall In-plane shear walls response filters the ground motion and transmits to floor Earthquake Excitation at diaphragms footings Out-of-plane wall (Doherty 2000,adapted after Priestley) Masonry Structures, lesson 10 slide 5 Out-of-plane seismic response of urm walls Important issues: -Strength of wall against out-of-plane forces and relevant mechanisms of resistance -Out-of-plane displacement capacity of walls -Evaluation of out-of-plane dynamic response -Definition of seismic demand on walls considering filtering effect of building and diaphragms Masonry Structures, lesson 10 slide 6
- 4. Out-of-plane strength of urm walls Wall subjected to horizontal distributed loading (wind or inertia forces) (A) (B) Overburden Force = O Self weight above crack = W/2 ∆c One-way bending, One-way bending, before cracking after cracking Base reaction = O+W Masonry Structures, lesson 10 slide 7 Out-of-plane strength of urm walls, low vertical stress Relastic Force Control/Dynamic Loading (Explosive) Un-cracked Elastic Capacity Displacement Control Loading Applied lateral Force, w (kN/m) Semi-rigid Non-linear F-∆ Res(1) Relationship Un-cracked Linear Elastic Behaviour Mid-height Displacement (∆) ∆instability (Doherty 2000) Masonry Structures, lesson 10 slide 8
- 5. Out-of-plane strength of urm walls, high vertical stress Res(1) Semi-rigid nonlinear F-∆ Applied lateral Force, w (kN/m) relationship Relastic Force/Dynamic Control Loading Displacement Control Loading Un-cracked Linear Elastic Behaviour Mid-height Displacement (∆) ∆instability (Doherty 2000) Masonry Structures, lesson 10 slide 9 Out-of-plane flexural strength of nonloadbering walls Two fundamental resistance mechanisms: tensile strength of masonry to resist one-way and two-way arching action bending Masonry Structures, lesson 10 slide 10
- 6. Flexural strength of nonloadbering walls Vertical one-way flexure: already treated Clearly, when zero or very low vertical compression is present, tensile strength of bedjoints assume an importan role and should not be neglected. Horizontal flexure: Masonry Structures, lesson 10 slide 11 Horizontal flexure Empirical relations for brick masonry (single-wythe): ⎛ σ ⎞ f tp ' = C f jt ⎜1 + n ⎟ ⎜ f jt ⎟ ⎝ ⎠ for toothed failure, where fjt = tensile stregth normal to bedjoints, C=2 for stress in MPa; f tp ' = 0.45 f tb + 0.55 f jt for failure in units, where ftb = tensile strength of unit (can be assumed as 0.1 times the compressive strength) Masonry Structures, lesson 10 slide 12
- 7. Two-way flexure (single-wythe) Most walls are supported on three or four sides. Evaluation is difficult because the walls are statically indeterminate and the material is anisotropic. Wall tend to show fracture lines whose pattern depends also on geometry of wall, degree of rotational restraint, presence of vertical compression, besides properties of materials. Common design/assessment methods make use of •simplistic conservative approaches such as the crossed strip method or •yield line and fracture line approaches Masonry Structures, lesson 10 slide 13 Arching action Masonry Structures, lesson 10 slide 14
- 8. Arching action C = fc (1−γ )t Resisting moment: M = C(γt − ∆0 ) 8C p= (γt − ∆0 ) h2 According to EC6 and BS: (1-γ)=0.1 2 ⎛t ⎞ From which, neglecting deflection (ok for h/t < 25): p = 0.72fc ⎜ ⎟ ⎝ h⎠ Masonry Structures, lesson 10 slide 15 Arching action with gap Arch action can be 4(γt)2 g≤ activated only if L From geometry of similar triangles: g 2∆g = 2γt L − g g(L − g) gL ∆g = ≅ 4γt 4γt 8C This can be inserted in the equation of the previous slide: p= (γt − ∆g ) h2 Masonry Structures, lesson 10 slide 16
- 9. Influence of movement of supports on arching action The thrust force C can produce some displacement of the supports. The displacement can be treated as an equivalent gap. If the movement is a function of the thrust force, a number of trials or iterations may be required to determine the optimum stress level. Axial shortening due to compression of the wall and corresponding deflection can be treated similarly as an equivalent gap. Masonry Structures, lesson 10 slide 17 Post-cracking behaviour loadbearing walls in single bending Priestley, 1985 From moment equilibrium about O: From moment W ⋅∆ w ⋅ h2 W ∆ W ⋅ ∆ h w ⋅ h2 equilibrium about R: H = R⋅x = + ⋅ + P⋅∆ + ⋅ = + R⋅∆ 2⋅h 8 2 2 2⋅h 2 8 8 P P w= ⋅ R ⋅ (x − ∆ ) h2 H=∆W/2h H=∆W/2h ∆/2 h/2 W/2 h/2 ∆ w w = ma h W/2 Static instability ∆ W/2 occurs when ∆=x o x H=∆W/2h P+W R Masonry Structures, lesson 10 slide 18
- 10. Post-cracking behaviour in single bending: statics Calculation of force-deflection curve: first, calculate curvature at midheight section and the associated moment, then evaluate displacement at midheight assuming a given curvature distribution. Mcrack=Rt/6 fcrack fcrack=2R/t R φcrack=fcrack/Et x x=t/6 t w ⋅ h2 8 ⋅ M crack 5 ⋅ wcrack ⋅ h 4 M crack = crack ⇒ wcrack = ∆ crack = 8 h2 384 ⋅ E ⋅ J Masonry Structures, lesson 10 slide 19 Post-cracking behaviour in single bending: statics After cracking, calculation behaviour is non linear and the following relationship are used for a given value of deformation/stress at the right edge of the section, from which the corresponding value of x is found: M =Rx Conservatively, it can be assumed that displacement si f f =2R/(3(t/2-x)) proportional to curvature at midheight : R φ =f/(3E(t/2-x)) φ x ∆= ⋅ ∆ crack x φcrack t At ultimate, a rectangular stress block is assumed and the corresponding curvature is evaluated, from which the displacement at midheight. Masonry Structures, lesson 10 slide 20
- 11. Post-cracking behaviour in single bending: statics Having determined the displacement ∆ at midheight corresponding to a given value of x, from the equation: 8 w= ⋅ R ⋅ (x − ∆ ) h2 the corresponding value of distributed horizontal force is determined, and the nonlinear force-displacement curve is determined point by point. Note: static instability may occur before the attainment of the ultimate stress. The behaviour is elastic nonlinear, i.e. the wall will load and unload along the same curve. Masonry Structures, lesson 10 slide 21 Post-cracking behaviour in single bending: rigid block behaviour Note: if the two halves of the cracked wall were considered as rigid blocks, the force-displacement curve would be given by the blue line, which can be obtained by simple equilibrium of rigid blocks.: F0 Bi-linear F- ∆ 'Semi-rigid threshold Relationship resistance' Real semi-rigid Non-linear F- ∆ Relationship Applied Lateral Force K0 ∆u=∆instability Masonry Structures, lesson 10 slide 22
- 12. Post-cracking behaviour in single bending: rigid block behaviour ∆ ∆ Masonry Structures, lesson 10 slide 23 Out-of-plane dynamic experimental behaviour Wall specimen 5. ACCEL (TA) Timber wall 1. LVP catch (TWD) 7. ACCEL (TWA) 2 LVP 8. ACCEL (MWD) (MWA) 9. ACCEL (BWA) 3. LVDT Stationary (BWD) 4. INSTRON 6. ACCEL reference frame: (TTA) Rigid connection to strong floor Tests performed at University of Adelaide, Australia (Doherty et al. 2000) Masonry Structures, lesson 10 slide 24
- 13. Braced steel Cornice frame support (Representin g URM shear wall action) Wall Specimen Table 10mm stiff rubber spacer bearing support rails (after Doherty, 2000) Masonry Structures, lesson 10 slide 25 Out-of-plane dynamic experimental behaviour 300% NAHANNI DISPLACEMENTS 15 DISPLACEMENT (mm) 10 5 PGD 14mm 0 0.01 0.62 1.23 1.84 2.45 3.06 3.67 4.28 4.89 5.5 6.11 6.72 7.33 7.94 8.55 9.16 9.77 10.4 -5 -10 TIME (secs) 300% NAHANNI VELOCITIES 0.25 0.2 300% PGV VELOCITY (m/S) 0.15 0.1 207mm/sec 0.05 Nahanni 0 0.01 0.57 1.13 1.69 2.25 2.81 3.37 3.93 4.49 5.05 5.61 6.17 6.73 7.29 7.85 8.41 8.97 9.53 10.1 10.6 -0.05 -0.1 -0.15 TIME (secs) 300% NAHANNI ACCELERATIONS 0.8 0.6 ACCELERATION (g) PGA 0.4 0.2 0.65g 0 0 0.56 1.11 1.67 2.22 2.78 3.33 3.89 4.44 5 5.55 6.11 6.66 7.22 7.77 8.33 8.88 9.44 9.99 10.5 -0.2 -0.4 -0.6 TIME (secs) Masonry Structures, lesson 10 slide 26
- 14. Out-of-plane dynamic experimental behaviour 80% PACOIMA DAM DISPLACEMENTS 40 30 20 PGD DISPLACEMENT (mm) 10 43.2mm 0 0.02 0.6 1.18 1.76 2.34 2.92 3.5 4.08 4.66 5.24 5.82 6.4 6.98 7.56 8.14 8.72 9.3 9.88 10.5 11 11.6 12.2 12.8 13.4 -10 -20 -30 -40 -50 TIME (s ecs ) 80% PACOIMA DAM VELOCITIES 0.3 80% 0.25 PGV 0.2 VELOCITY (m/S) 0.15 0.1 Pacoima 245mm/sec 0.05 0 Dam 0.02 0.62 1.22 1.82 2.42 3.02 3.62 4.22 4.82 5.42 6.02 6.62 7.22 7.82 8.42 9.02 9.62 10.2 10.8 11.4 12 12.6 13.2 -0.05 -0.1 -0.15 TIME (s ecs ) 80% PACOIMA DAM ACCELERATIONS 0.4 0.3 PGA 0.2 ACCELERATION (g) 0.1 0 0.35g 0 0.58 1.16 1.74 2.32 2.9 3.48 4.06 4.64 5.22 5.8 6.38 6.96 7.54 8.12 8.7 9.28 9.86 10.4 11 11.6 12.2 12.8 13.3 -0.1 -0.2 -0.3 -0.4 TIME (s ecs ) Masonry Structures, lesson 10 slide 27 Safety wih respect to collapse: force or displacement? Clearly, any methodology should take into consideration the nonlinear nature of the response. Earlier proposal by Priestley (1985): force based assessment based on equal energy equivalence; Equivalent “resistance” of a linear response Masonry Structures, lesson 10 slide 28
- 15. Design accelerations to be compared with “resistance” Masonry Structures, lesson 10 slide 29 A more rigorous evaluation of the flexibility of diaphragms would be made by an adequate dynamic global model. (from Tena-Colunga & Abrams) Masonry Structures, lesson 10 slide 30
- 16. Force-based assessment with q-factor Eurocode 8 & Italian code approach The effect of the seismic action may be evaluated considering a force system proportional to the masses (concentrated or distributed) of the element; the resultant force (Fa) on the element, is computed as: Fa = Wa Sa / qa Wa is the weight of the element. qa is the behaviour factor, to be considered equal to 1 for cantilever elements (for example chimneys and parapets fixed only at the base) equal to 2 for non- structural walls, equal to 3 (Italian code) for structural walls which do not exceed specified slenderness limits. Sa is the seismic coefficient to apply at the structural or non-structural wall, that considers dynamic interaction with the building. Masonry Structures, lesson 10 slide 31 Force-based assessment with q-factor a g S ⎡ 3(1 + Z/H) ⎤ a gS Sa = ⎢ − 0.5⎥ ≥ g ⎣1 + (1 − Ta /T1 ) 2 ⎦ g agS is the design ground acceleration Z is the height from the foundation of the centre of the mass of the element H is the height of the structure g is the gravity acceleration Ta is the first period of vibration of the wall element in the considered direction (out-of-plane), estimated also in an approximate way. T1 is the first period of vibration of the structure in the considered direction. Masonry Structures, lesson 10 slide 32
- 17. Recent trends: displacement based assessment Griffith et al., 2000-today F0 F0 Rigid body Applied Lateral Force Calculated curve Fu FU K1 K2 KS K0 ∆2 ∆u ∆1 ∆2 ∆MAX ∆U Mid-height Displacement Dynamic response is estimated via an equivalent secant stiffness (K2) and an equivalent s.d.o.f. system with suitable effective damping (5% suitable for one-way vertical bending). Masonry Structures, lesson 10 slide 33 Recent trends: displacement based assessment Displacement demand must be evaluated using an elastic spectrum which takes into account the filtering effect of the structure, e.g.(Italian code) Ts2 ⎛ 3 (1 + Z H ) ⎞ Ts < 1.5T1 ∆ d (Ts ) = a gS ⎜ − 0.5 ⎟ 4π ⎜ 1 + (1 − Ts T1 ) 2 2 ⎟ ⎝ ⎠ 1.5T1Ts ⎛ Z⎞ 1.5T1 ≤ Ts < TD ∆ d (Ts ) =a g S ⎜1.9 + 2.4 ⎟ 4π 2 ⎝ H⎠ 1.5T1TD ⎛ Z⎞ TD ≤ Ts ∆ d (Ts ) =a gS ⎜ 1.9 + 2.4 ⎟ 4π 2 ⎝ H⎠ agS is the design ground acceleration Z is the height from the foundation of the centre of the mass of the element H is the height of the structure g is the gravity acceleration Ta is the first period of vibration of the element in the considered direction, estimated also in an approximate way. T1 is the first period of vibration of the structure in the considered direction. Masonry Structures, lesson 10 slide 34

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