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Lesson8

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Dan Abrams + Magenes Course on Masonry

Dan Abrams + Magenes Course on Masonry

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  • 1. Seismic design and assessment of Seismic design and assessment of Masonry Structures Masonry Structures Lesson 8 October 2004 Masonry Structures, lesson 8 slide 1 Strength design/assessment of urm walls Recent masonry codes are increasingly adopting strength design/assessment as a fundamental element within a Limit State or a Performance Based approach, for both unreinforced and reinforced masonry. Eurocodes, as an example, are entirely based on a Limit State approach, both for unreinforced and reinforced masonry and for non-seismic and seismic design/assessment. In the following, attention is focused on strength evaluation of urm walls subjected to in-plane forces (e.g. shear walls). Masonry Structures, lesson 8 slide 2
  • 2. Definitions • Consider a wall with free lateral edges, subjected to self weight and to forces applied at top and bottom sections. • For each section, by integrating the stresses it is possible to define a resultant normal (axial) force N, a resultant shear force V, a moment M (which can be defined as a product of N and its eccentricity e w. respect to the vertical axis of the wall). The resultant forces are contained within the middle plane of the wall. • The following equilibrium equations hold: Ninf = Nsup + P V ⋅ h = M sup + M inf = N sup esup + N inf einf Masonry Structures, lesson 8 slide 3 Flexural (tensile) cracking • Section of a wall with thickness t , length l subjected to axial force N and moment M M • Tensile strength of bedjoint: fjt N 2 ⎛N ⎞ l t M crack = N ⋅ e = ⎜ + f jt ⎟ ⋅ ⎝ lt ⎠ 6 N • If fjt = 0 the well known condition on fjt maximum eccentricity to avoid cracking for σ a no-tension material is found: e e =l/6 • Flexural cracking can be considered a damage or serviceability limit state, not an ultimate limit state. Masonry Structures, lesson 8 slide 4
  • 3. Flexural strength (ultimate) • Section of a wall with thickness t , length l subjected to axial force N and moment M • Compressive strength of masonry : fu • Average compression stress on the section: σ0 • Assume equivalent rectangular stress block N σ0 = l ⋅t κ (=k3 )= 0.75-1 a / x (= k1)= 0.67-0.85 N a= κf u t ⎛ l − a ⎞ Nl ⎛ N ⎞ σ 0 l 2t ⎛ σ 0 ⎞ Mu = N⎜ ⎟= ⎜1 − ⎟= ⋅ ⎜1 − ⎟ ⎝ 2 ⎠ 2 ⎜ κf u lt ⎟ ⎝ ⎠ 2 ⎜ κf u⎝ ⎟ ⎠ Masonry Structures, lesson 8 slide 5 Flexural strength (ultimate) • Note: it is easy to verify that if σ0 is very low compared to fu the equation reduces to the overturning resistance of a rigid block. • Example: Equilibrium of rigid block around O: N l l V Vu ⋅ h = ( N + P) ⋅ ⇒ Vu = ( N + P ) 2 2h P h Using flexural strength equation : ⎛ l − a ⎞ σ 0l t ⎛ σ 0 ⎞ σ 0l 2t 2 l Mu = N⎜ ⎟= ⋅ ⎜1 − ⎟≅ = N inf ⎝ 2 ⎠ 2 ⎜ κf u ⎝ ⎟ ⎠ 2 2 O l l l M u = Vu ⋅ h ⇒ Vu = N inf = ( N + P) 2h 2h Masonry Structures, lesson 8 slide 6
  • 4. Flexural strength (ultimate) • In-plane flexural/rocking cyclic response of a brick masonry wall 80 60 Note: for low 40 values of 20 force (kN) mean 0 compression, -20 flexural failure mode is -40 somewhat -60 “ductile” -80 -15 -10 -5 0 5 10 15 displacement (mm) (Anthoine, Magenes and Magonette, 1994) Masonry Structures, lesson 8 slide 7 Shear strength (ultimate) • The so-called shear failure can be associated to different cracking patterns, but is essentially generated by the effect of the combination of shear stresses, mostly due to horizontal loading, with the normal stresses. • Two main shear failure modes can be defined: diagonal cracking (a) or shear-sliding (b). • Mixed modes are also possible stepped crack w. w. shear/tensile joint failure cracking of units Masonry Structures, lesson 8 slide 8
  • 5. Shear strength (ultimate) Definition of shear strength criteria Problems: - experimental data are highly scattered (typical of brittle failure modes) - in a real wall the distribution of stresses is highly non uniform, and its evaluation is difficult (squat elements where beam theory is questionable, subjected to tensile cracking). In practical design/assessment, simplifications must be introduced, sometimes to the detriment of accuracy. Some common simplified criteria used in design/assessment: - Maximum principal tensile stress criterion - Coulomb-like criterion Masonry Structures, lesson 8 slide 9 Shear strength (ultimate) Maximum tensile stress criterion From experimental results of shear-compression tests on urm panels, where it was observed that the attainment of shear strength corresponds to the onset of diagonal cracks at the centre of the panel, it was postulated that shear failure takes place when the principal tensile stress attains a limit value ftu , which is assumed as the referential or conventionall tensile strength of masonry. 2 principal tensile stress: ⎛σz ⎞ σ σt = ⎜ ⎟ + τ zx − z = f tu 2 ⎝ 2 ⎠ 2 at the centre of the panel : 2 N V ⎛σ0 ⎞ σ σ =σ0 = ; τ = bτ 0 = b σt = ⎜ ⎟ + (bτ ) − 0 = f tu 2 lt lt ⎝ 2 ⎠ 2 f tu lt σ b varies with the aspect ratio h/l of the wall. A simple criterion to Vu = 1+ 0 evaluate b is (Benedetti & Tomaževič, 1984): b= 1.5 when h/l ≥ 1.5 b f tu (slender walls), b = 1 when h/l≤1.0, and b=h/l per 1 < b < 1.5. Masonry Structures, lesson 8 slide 10
  • 6. Shear strength (ultimate) • In-plane cyclic response of a brick masonry wall failing in shear 100 80 60 40 force (kN) 20 0 -20 -40 -60 -80 -100 -8 -6 -4 -2 0 2 4 6 8 displacement (mm) (Anthoine, Magenes and Magonette, 1994) Masonry Structures, lesson 8 slide 11 Shear strength (ultimate) Coulomb-like criterion: τ = c + µσ where the shear stress τ and the normal stress σ can have different meaning, depending on the criterion. The approach followed by e.g. Eurocode 6 and the Italian code is that both stresses must be considered average stresses on the compressed part of the wall, ignoring any part that is in tension. According to Eurocode 6, the characteristic shear strength of an urm wall is expressed via the shear strength per unit area fvk times the compressed area of the wall: V Rk = f vk ⋅ t ⋅ lc where lc is the length of the compressed zone, t is the thickness of the wall and the characteristic shear strength fvk is defined as : fvk = fvk0 + 0.4 σd with fvk ≤ fvk,lim σd : average normal stress on the compressed area fvk0 : characteristic shear strength under zero compressive stress fvk,lim : limit value of fvk depending on the type of units (e.g. 0.065fb where fb is the normalized compressive strength of the units). Masonry Structures, lesson 8 slide 12
  • 7. Shear strength (ultimate) According to Eurocode 6, the length of the compression zone and the corresponding average stresses can be evaluated as follows: M= N . e lc /3 l lc l e if e> = −e 6 3 2 N therefore: ⎛1 e⎞ lc = β ⋅ l = 3 ⋅ ⎜ − ⎟ ⋅ l = ⎝2 l ⎠ V ⎛1 M ⎞ ⎛1 V ⎞ = 3 ⋅ ⎜ − ⎟ ⋅ l = 3 ⋅ ⎜ − αV ⎟ ⋅ l lc= x ⎝ 2 Nl ⎠ ⎝2 N ⎠ l/2 l/2 M H0 where: αV = = Vl l is the shear ratio of the section and H0 is the shear span of the section Masonry Structures, lesson 8 slide 13 Shear strength (ultimate) In this latter case, a closed form expression can be written for the shear strength based on a Coulomb-like criterion as: ⎛ ⎞ ⎜ ⎟ ⎛ N ⎞ 1.5c + µσ 0 ⎟ Vu = βlt ⋅ ⎜ c + µ ⎜ ⎟ = lt ⋅ ⎜ = lt ⋅τ u ⎝ βlt ⎟ ⎠ ⎜ cαV ⎟ ⎜ 1+ 3 σ ⎟ ⎝ 0 ⎠ Note: when this sort of criterion is applied to a wall, failure is always predicted at the section with the highest moment, i.e. the section with the smallest compressed area (scketch on blackboard). This criterion seems more suitable to describe a sliding shear failure, rather than a diagonal cracking failure. Masonry Structures, lesson 8 slide 14
  • 8. Shear strength (ultimate) When walls are subjected to low vertical forces and high horizontal cyclic forces, horizontal cracks may develop all along the whole section of the wall, typically in bedjoints, and the resistance to sliding would then be expressed as: ⎛ N ⎞ Vu = β lt ⋅ ⎜ µ sl ⎜ ⎟ = µ sl ⋅ N ⎝ βlt ⎟ ⎠ Note: the application of such strength formula has proven to give too conservative and unrealistically low strength predictions. This is especially the case for the upper stories of a building. In reality: - the low tensile strength of bedjoints, usually neglected in computations, can prevent the formation of fully cracked sections at upper stories; - when a flexural cracks closes in compression it is arguable to assume that the residual strength is based on friction only; a more realistic approach should assume a residual cohesion. Masonry Structures, lesson 8 slide 15 Flanged sections (urm walls) See Tomaževič, pages 151-154. Masonry Structures, lesson 8 slide 16
  • 9. Response of Building Systems Masonry Structures, lesson 8 slide 17 Response of building systems Response of an elementary cell to horizontal loading: role of diaphragms and ring beams Out-of-plane collapse Out-of-plane collapse Out-of-plane collapse mechanism prevented, presence of ring prevented, rigid beam (tie-beam), flexible diaphragm diaphragm Masonry Structures, lesson 8 slide 18
  • 10. Response of building systems Response to horizontal loading : role of diaphragms and ring beams • Ring beams and diaphragms contribute to restrain the out-of- plane deflections of walls and to avoid out-of-plane collapse (i.e they “hold the box together”). • To exert an effective restraint ring beams and diaphragm must be able to transmit tensile forces and tensile stresses and must be effectively connected to the walls. • A ring beam connecting walls in the same plane confers robustness and redundancy to the system, allowing force redistribution among walls. (Note: a similar role, in part, is played by tie-rods, used also in ancient buildings) • Rigidity of diaphragms affects horizontal load distribution among shear walls Masonry Structures, lesson 8 slide 19 Response of building systems Response to horizontal loading: role of tie-rods to prevent out-of-plane collapse Masonry Structures, lesson 8 slide 20
  • 11. Response of building systems Response to horizontal loading: role of tie-rods to improve in-plane response urm with tie-rods tie-rods to restrain overturning and mobilitate in- plane resistance of walls. urm without tie-rods (Giuffré, 1993) Masonry Structures, lesson 8 slide 21 Response of building systems Example of reinforced concrete ring beam: PLAN VIEW at corners longitudinal reinforcement, min. 6-8 cm2 stirrups 6-8 mm dia. min., max. spacing 25-30 cm ⎧ 2 Prescriptions suitable for general (non- ⎪ t ⎧h b0 ≥ ⎨ 3 seismic) design, low-rise buildings. h0 ≥ ⎨ More stringent requirements are ⎪12 cm ⎩ ⎩t/2 suggested for seismic design. Masonry Structures, lesson 8 slide 22
  • 12. Flexible Diaphragms Distribution of Lateral Force to Shear Walls L1 L2 F3 wall “1” F1 wall “3” F2 flexible wall “2” diaphragm wind or earthquake forces = w For fully flexible diaphragms, walls attract lateral forces based on tributary areas F1 = wL1/2 F2 = wL1/2 + wL2/2 F3 = wL2/2 I.e. horizontal forces are resisted by the shear walls are taken from the floors to which they are directly connected, unless a “semi-rigid” analysis is carried out. Masonry Structures, lesson 8 slide 23 Flexible Diaphragms Diaphragm Deflections A A w Em ∆ L1 floor or roof d 6t Ei L2 d brick or A block 5 wL4 wall ∆= 2 t 384 E m I bL3 Ei L1 Section A-A I =n 1 + ∑ Ad 2 n= d= 12 Em 2 Masonry Structures, lesson 8 slide 24
  • 13. Flexible Diaphragms Imposed Deflections on Out-of-Plane Walls ∆ = imposed from flexible diaphragm H Hh 3 ∆= 3 EI h 3 E m I∆ M = Hh = h2 3 E m I∆ t 3 Em ∆ fb = M = h2 = 2 =F + f 2 t a S S h t P M ⎛ h⎞ 2 A ( Ft + f a )⎜ ⎟ ∆ ⎝t⎠ allowable = t 1.5 E m tensile stress Masonry Structures, lesson 8 slide 25 Flexible Diaphragms Imposed Deflections on Out-of-Plane Walls Values of allowable diaphragm deflection, ∆, divided by wall thickness, t* h/t Ft + fa (psi) 10 15 20 25 30 20 0.00133 0.00300 0.00533 0.00833 0.01200 40 0.00267 0.00600 0.01067 0.01666 0.02400 60 0.00400 0.00900 0.0160 0.02500 0.03600 80 0.00533 0.01200 0.02133 0.03333 0.04800 100 0.00667 0.01500 0.02667 0.04166 0.06000 120 0.00800 0.01800 0.03200 0.05000 0.07200 140 0.00933 0.02100 0.03730 0.05833 0.08400 *based on Em = 1,000 ksi Masonry Structures, lesson 8 slide 26
  • 14. Example: Flexible Diaphragms Determine deflections of the diaphragm in the north-south direction north 80’-0” two-wythe URM brick walls 10’-0” story height 30’-0” timber floor f’m = 2000 psi 7.63” Type N mortar with masonry cement fa dead = 47 psi 3 w = 400 lb/ft nbl 1 5 wl24 I= + ∑ Ad 2 ∆= 12 384 EI 6 ( 7.63quot; )2 A = 6t 2 = = 2.43 ft 2 5 0.400 kip / ft ( 80' )4 x 12 144 ∆= = 0.0108quot; 384 1500 ksi x 144 ( 1094 ft 4 ) E m = 750 f 'm = 750 × 2.0 ksi = 1500 ksi 30' d= = 15' ; ∑ Ad 2 = 2( 2.43 x 15 2 ) = 1094 ft 4 2 bl13 neglecting for simplicity I = 1094 ft 4 12 Masonry Structures, lesson 8 slide 27 Example: Flexible Diaphragms Check Cracking of the Out-of-Plane Walls as per UBC h 120quot; allowable deflection per Table for = = 15.7 t 7.63quot; Ft = 15 psi, f a + Ft = 62 psi ∆ = 0.00900 for E m = 1000 ksi t for E m = 1500 ksi ∆ 1000 ksi = 0.00900 = 0.0060 ; ∆ = 0.006 x 7.63 = 0.046quot; > 0.0108quot; ok t 1500 ksi Masonry Structures, lesson 8 slide 28
  • 15. Flexible Diaphragms Reducing Flexibility with Drag Struts without strut with strut B B A A drag strut C C w w Masonry Structures, lesson 8 slide 29 Flexible Diaphragms Determine the lateral force attracted to each wall (case 1). A w = 1.0 kips/ft lateral force, kips 24’-0” wall w/o strut w/strut drag strut A 33.0 12.0 B B 0.0 33.0 C 33.0 21.0 42’-0” total 66.0 66.0 C 30’-0” 32’-0” Masonry Structures, lesson 8 slide 30
  • 16. Flexible Diaphragms Determine the lateral force attracted to each wall (case 2). w = 1.0 kips/ft A lateral force, kips 24’-0” wall w/o strut w/strut drag strut A 25.4 12.0 B 12.0 33.0 B C 28.6 21.0 42’-0” total 66.0 66.0 C 30’-0” 32’-0” Masonry Structures, lesson 8 slide 31 Rigid Diaphragms Translation Without Rotation symmetrical about centerline analogy: springs in series F1 F2 F1 ∆ wall “2” wall “l” wall “l” k1 k2 k1 ∆ H wind or earthquake forces equilibrium k i = lateral stiffness of wall H = F1 + F2 + F1 = ∑ Fi bE m ki = for solid, cantilever ed wall = k1∆ + k2 ∆ + k1∆ = ∑ ki ∆ ⎛ h ⎞⎡ ⎛ h ⎞ ⎤ 2 ⎜ ⎟ ⎢ 4⎜ ⎟ + 3⎥ Η ki ⎝ L ⎠⎢ ⎝ L ⎠ ∆= Fi = ki ∆ = H ⎣ ⎥ ⎦ ∑ ki ∑ ki or, summation of pier stiffnesses for perforated shear wall Masonry Structures, lesson 8 slide 32
  • 17. Rigid Diaphragms Translation with Rotation y xr xr1 center of stiffness wall “1” find center of stiffness : yr1 ∑ k xi yi ∑ k yi xi yr = xr = ∑ k xi ∑ k yi ey yr2 where : yr Px k xi = lateral stiffness of wall quot; iquot; parallel to x direction wall “2” k yi = lateral stiffness of wall quot; iquot; xr2 x parallel to y direction Py ex xi , yi = distance from centroid of wall quot; iquot; to some datum Plan View of Roof or Floor System Masonry Structures, lesson 8 slide 33 Rigid Diaphragms Translation with Rotation ∑ Mo = 0 M ext = Py e x + Px e y = M int y r 1Θ considering in - plane stiffness only : M int = [ Fx 1 y rl + F y 2 x r 2 ] Fx 1 = k x 1 y r 1Θ M int = [( k x 1 y rl Θ ) y rl + ( k y 2 x r 2Θ ) x r 2 ] M int = J rΘ Θ where J r = [ k x 1 y rl 2 + k y 2 x r 2 2 ] ey o M Θ= Jr Px M Fx 1 = k x 1 y rlΘ = k x 1 y rl ( ) Jr x r 2Θ M F y 2 = k y 2 x r 2Θ = k y 2 x r 2 ( ) Py Jr ex k xi yri k yi xri Fy 2 = k y 2 x r 2 Θ Fxi = M ext Fyi = Jr M ext Jr Masonry Structures, lesson 8 slide 34
  • 18. Rigid Diaphragms Translation with Rotation Forces attracted to shear walls: translatio n rotation k xi k y Fxi = ( ) Px + ( xi ri ) ( Py e x + Px e y ) ∑ k xi Jr k yi k yi x ri Fyi = ( ) Py + ( ) ( Py e x + Px e y ) ∑ k yi Jr 2 2 J r = ∑ k xi y ri + ∑ k yi x ri Masonry Structures, lesson 8 slide 35 Example: Rigid Diaphragms Determine which wall is the most vulnerable to a NS wind load of 20 psf A B north 4” brick 70’-0” collar joint filled a with mortar 30’-0” 1 9.25” 10’ a Section a-a double wythe brick wall 8” 50’-0” CMU 8” concrete block 30’-0” fully grouted type N mortar b b 7.63” 2 Section b-b floor plan 10’ Walls of the one-story building are 15’ tall and the roof system has been strengthened to be rigid. Assume f’m equal to 5333 psi for the brick, and 2667 psi for the block. Masonry Structures, lesson 8 slide 36
  • 19. Example: Rigid Diaphragms Determine center of stiffness. wall Em b h/L kxi kyi yi xi kxyi kyxi ksi inches kip/in. kip/in. feet feet kip-feet/in. kip-feet/in. A 3000 9.25 1.5 - 1542 - 0 - 0 B 2000 7.63 0.5 - 7630 - 70 - 534,100 1 3000 9.25 0.5 13,875 - 50 - 693,750 - 2 2000 7.63 1.5 848 - 0 - 0 - y ∑ kxi = 14,723 ∑ kyi = 9172 ∑ k xi yi = 693,750 ∑ k yi xi = 534,100 A B 70’ 1 ∑ k yi xi 534,100 ki = bEm xr = = = 58.2' ∑ k yi 9172 ⎛ h ⎞⎡ ⎛ h ⎞ ⎤ 2 ⎜ ⎟ ⎢ 4 ⎜ ⎟ + 3⎥ 50’ ⎝ L ⎠⎢ ⎝ L ⎠ ⎣ ⎥ ⎦ ∑ k xi yi 693,750 yr yr = = = 47.1' xr x Em = 750 f 'm < 3000 ksi ∑ k xi 14,723 2 Masonry Structures, lesson 8 slide 37 Example: Rigid Diaphragms Determine torsional constant. 2 2 torsional stiffness = J r = ∑ k xi y ri + ∑ k yi x ri wall kxi yri kyi xri Jr kip/in feet kip/in feet kip.ft2/in A - - 1542 58.2 5,223,124 B - - 7630 11.8 1,062,401 1 13,875 2.9 - - 116,689 2 848 47.1 - - 1,881,212 Jr = 8,283,426 Masonry Structures, lesson 8 slide 38
  • 20. Example: Rigid Diaphragms Determine wall shear forces. ki k y Fxi = ( )Px + ( xi ri )( Py e x + Px e y ) ∑ ki Jr ki k x ri Fyi = ( ) Py + ( yi )( Py e x + Px e y ) ∑ ki Jr Wind in north-south direction will result in larger wall shears than for wind in east-west direction because of larger wind surface area. Px = 0 15' Py = ( )( 70' )w = 525 w where w = wind pressure in psf = 20 psf 2 Py = 10.5 kips e x = 58.2' − 35.0' = 23.2' Py e x ( 10.5 kips )( 23.2' ) 1 = = in/ft Jr ( 8 ,283 ,426 kip − ft 2 / in .) 34 ,004 Masonry Structures, lesson 8 slide 39 Example: Rigid Diaphragms Determine wall shear forces. wall direct shear torsional shear total shear Ae fv=V/Ae Fv(1) Fv/fv kips kips kips inches psi psi 1542 1542( 58.2) A (10.5) = 1.76 = 2.64 4.41 1152 3.8 29.1 7.7 9172 34,004 governs 7630 − 7630(11.9) B (10.5) = 8.73 = − 2.64 6.09 2776 2.2 30.6 13.6 9172 34,004 13,875( 2.9) 1 0 = 1.18 1.18 3372 0.35 29.1 83.1 34,004 − 848( 47.1) 2 0 = − 1.18 1.18 945 1.25 30.6 24.5 34,004 (1)F = 1.33 x 0.3 f’m0.5 = 29.1 psi < 80 psi for brick wall v Fv = 1.33 x 23 psi = 30.6 psi for concrete block with Type N mortar Masonry Structures, lesson 8 slide 40
  • 21. Example: Rigid Diaphragms Determine wall flexural tensile stresses. wall Sg shear fbi = Hih/Sg fa dead -fa + fb inches3 kips psi psi psi A 23,944 4.41 33.2 30.0 3.2 B 168,320 6.09 6.5 30.0 -23.5 1 204,967 1.18 1.0 30.0 -29.0 2 19,495 1.18 10.9 30.0 -19.1 worst case is flexural tension for wall A: -30.0 + (w/20)(33.2) = Ft = 1.33 x 15.0 psi maximum wind load, w= 30 psf Masonry Structures, lesson 8 slide 41 Rigid diaphragms: summary (elastic analysis) External torque due to eccentricity of shear force Vtot w. respect to center of stiffness: M tot = Vtot , y ⋅ ( xC − xR ) − Vtot , x ⋅ ( yC − y R ) = Vtot , y ⋅ eV , x − Vtot , x ⋅ eV , y Shear force in wall i : K xi K ⋅ ( yi − y R ) Vix = ⋅ Vtot , x − xi ⋅ M tot ; K x,tot J p,tot K yi K yi ⋅ ( xi − x R ) Viy = ⋅ Vtot , y + ⋅ M tot ; K y ,tot J p ,tot Kθi Ti = ⋅ M tot J p ,tot where: K x,tot = ∑ K xi ; K y ,tot = ∑ K yi ; are respectively: total translational stiffnesses i i J p,tot = ∑ K xi ⋅ ( yi − y R ) 2 + ∑ K yi ⋅ ( xi − x R ) 2 + ∑ Kθi total torsional stiffness i i i Masonry Structures, lesson 8 slide 42
  • 22. Vertical structures: degree of coupling Role of coupling provided by floors and/or spandrel beams deflected shape and shears and moments deflected shape shears and moments crack pattern and crack pattern (b) (a) deflected shape and shears and moments crack pattern (c) Masonry Structures, lesson 8 slide 43 Vertical structures: modeling Some possible modelling approaches for multistorey masonry walls a) cantilever model b) equivalent frame c) equivalent frame d) 2-D or 3-D finite with rigid offsets element modelling Masonry Structures, lesson 8 slide 44