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ENCI426 > Walls > Flashcards

Walls Flashcards

1
Q

ductility formula

A

ultimate / yield

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2
Q

lp

A

equivalent length of plastic hinge

- lower bound approximation: lp = h/2

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3
Q

yield strain

A

= fy / Es

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4
Q

yield curvature

A

= 2 ey / lw

  • function of the yield strain of the steel
  • not a function of section geometry
  • nothing to do with strength
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5
Q

estimate of wall neutral axis

A 
k = 0.25
c = klw = lw / 4
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6
Q

Determining structural ductility factor

A
  • longest wall yields and reaches max ductility first
  • max drift targeting is limited by ductility of longest wall
  • ductility demand on other walls decreases as length of wall decreases
  • structural response determined by summing wall responses
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7
Q

benefit of lower stiffness

A

smaller loads

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8
Q

benefit of bigger stiffness

A

smaller drifts

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9
Q

columns of different height

A
  • same hc = same yield curvature
  • taller column = larger yield displacement
  • shortest column = highest local ductility demand
  • shortest column sets structural ductility factor
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10
Q

columns of different depths

A
  • smaller columns = higher yield curvature and therefore displacement
  • column with largest depth = greatest local ductility demand
  • deepest column sets structural ductility factor
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11
Q

estimation of effective flange width

A
  • width of wall flanges in tension or compression is reduced because of the effects of shear lag
  • based on the number of rebars put into strain
  • moment applied - web in shear - transferred into C/T in flanges
  • as moment increases and so too the associated tension in the flange, the section of flange resisting increases
  • for nominal bending tension flange: hw + bw (26.6 degrees or 1:2)
  • reduced width in compression flange after loaded in tension - less contact further away from web
  • can’t close cracks unless very high axial load
  • compression: 0.3hw + bw
  • for overstrength this is conservative, so consider 45 degrees for tension
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12
Q

coupling effect

A

typically provides 40-60% of overturning capacity

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13
Q

two types of coupling beams

A
  • conventionally reinforced

- diagonally reinforced cage

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14
Q

advantage of diagonally reinforced cage

A

concrete not needed outside of diagonals -

can put service ducts there

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15
Q

basketing

A

to keep wedges of concrete formed in diagonal cage from falling out

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16
Q

simplifying assumption for wall design

A
  • assume all steel yielding in tension or compression

- percentage difference less than 1%

17
Q

ultimate curvature

A

0.003 / c

18
Q

flexural overstrength factor

A

= Mo / Me

19
Q

maximum credible earthquake

A
  • structures are designed for the DBE but should be capable of resisting a larger earthquake
  • NZ standards may design a maximum credible earthquake (MCE) in the future
  • for now MCE taken as 2.0 x ULS actions
  • not OK to fail after ULS
20
Q

Comments on Singly Reinforced Walls

A
  • should be avoided
  • 150mm thick walls are TOO THIN
    • cant get reinforcement in
    • anti-buckling reinforcement should also be provided over central region for yielding walls
21
Q

Thin walls with Drossbach

A
  • need to have transfer stirrups like columns
  • properly detailed end of wall that look like mini-columns
  • anchor floor starters and other embedments properly
  • anchor horizontal bars in boundary elements
22
Q

Comments on Precast Panel Slices

A
  • walls roughened to min of 5 mm to prevent slipping/sliding
  • failure of lap splices observed where drossbach ducts were not confined
  • drossbach ducts cause stress concentrations at panel joints
  • Drossbach ducts should be fully confined by stirrups
    • clamp bars to drossbach and ensure bond maintained
  • panel splices must allow for de-bonding of reinforcement
23
Q

Comments on concrete strength

A
  • higher than in lab testing
  • code equations based on lab testing (different behaviour to concrete in buildings)
  • min. reinforcement requirements may be insufficient
  • min steel is to ensure fan cracking: bar yields and elongates
  • if real life concrete&raquo_space;> stronger than lab testing : no fan cracking
  • case studies showed failure of bars in walls and inelastic capacity of bars exhausted
24
Q

order of most likely strained elements

A
  • walls
  • beams
  • columns
25
Q

Comments on distribution of reinforcement

A
  • large cracks where insufficient reinforcing to form fan crack pattern
  • results in high strain demands on reinforcing
  • walls with even distribution of steel develop extremely high strain demands on the bars at the extreme fibres
  • can lead to significant strain hardening or fracture of the bars under ‘elastic’ loads
    RECOMMEND:
  • lump reinforcing steel at the ends of the walls and distribute minimum steel through the central portion
26
Q

effect of neutral axis on ductility

A

longer neutral axis = significantly less ductility

27
Q

reason for boundary elements

A
  • ductile walls tend to buckle

- boundary elements (flanges) provide adequate flexural rigidity at end of wall section

28
Q

lapping in PPHZ

A

must stagger laps

  • ensure fan cracking still occurs and not horizontal crack at lap
  • no more than 1/3 steel lapped at same levels
29
Q

effect of size of penetrations in wall

A
  • smaller penetration = greater shear capacity between walls
  • therefore larger tension/compression forces in wall
  • therefore higher overturning capacity
30
Q

three failure modes for walls

A

flexure, shear, sliding

  • avoid sliding by roughening cold joints to 5 mm
  • likely combined flexural shear failure
31
Q

Effect of penetration size on coupled walls

A
  • larger penetrations means less depth in coupling beam and therefore reduced shear capacity
  • less load transferred into axial loads in walls
  • reduces coupling effect
  • also need to be careful of zipper effect
32
Q

Diagonally reinforced cages

A
  • cages designed as mini-columns, with 6db stirrups
  • need to extend 1.5 Ld into wall to anchor
  • basketing extending 100-150 mm into walls to prevent diagonal wedges of concrete falling out
  • advantage of services placement
33
Q

Boundary Elements

A
  • ends of walls: 0.15 lw
  • designed like columns
  • more steel required in BE (0.5 sqrt(f’c) / fy ) than in middle of wall (0.25 sqrt(f’c) / fy )
  • more steel in BE to ensure multiple cracks form
  • cracking steel runs full height of wall to prevent big horizontal crack forming at top and allow for higher modes
  • can use enlarged boundary elements (flanges) to provide flexural rigidity to ductile walls
  • steel lumped in ends of walls
  • lapping staggered
34
Q

Shear Requirements

A
  • horizontal rebar fully anchored at ends of walls
  • vertical spacing <= lw/5, 3t, or 450 mm
  • horizontal spacing <= lw/3, 3t or 450 mm
  • limit nominal shear stress to prevent diagonal compression failure in potential plastic hinge region
    <= (0.2f’c or 8MPa)
35
Q

Antibuckling

A
  • common issue observed in Canterbury earthquakes was localised failure of reinforcing bars
  • need cross ties to hold stirrups and rebars in to prevent buckling, cover spalling and contain concrete core
36
Q

double the strength halve the ductility - fallacy

A
  • traditional view based on force-based approach which uses equal displacement rule
  • suggests that by doubling the strength through using more rebars, the ductility would half
  • in reality, changing the number of bars has no impact on the yield displacement
  • the stiffness of the section increases, but the ductility remains the same
  • only things that affect ductility: yield strain of steel and length of wall

ENCI426 (2 decks)

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