Walls

Question 1

ductility formula

Answer 1

ultimate / yield

Question 2

lp

Answer 2

equivalent length of plastic hinge

- lower bound approximation: lp = h/2

Question 3

yield strain

Answer 3

= fy / Es

Question 4

yield curvature

Answer 4

= 2 ey / lw

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

Question 5

estimate of wall neutral axis

Answer 5

k = 0.25
c = klw = lw / 4

Question 6

Determining structural ductility factor

Answer 6

• 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

Question 7

benefit of lower stiffness

Answer 7

smaller loads

Question 8

benefit of bigger stiffness

Answer 8

smaller drifts

Question 9

columns of different height

Answer 9

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

Question 10

columns of different depths

Answer 10

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

Question 11

estimation of effective flange width

Answer 11

• 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

Question 12

coupling effect

Answer 12

typically provides 40-60% of overturning capacity

Question 13

two types of coupling beams

Answer 13

• conventionally reinforced

- diagonally reinforced cage

Question 14

advantage of diagonally reinforced cage

Answer 14

concrete not needed outside of diagonals -

can put service ducts there

Question 15

basketing

Answer 15

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

Question 16

simplifying assumption for wall design

Answer 16

• assume all steel yielding in tension or compression

- percentage difference less than 1%

Question 17

ultimate curvature

Answer 17

0.003 / c

Question 18

flexural overstrength factor

Answer 18

= Mo / Me

Question 19

maximum credible earthquake

Answer 19

• 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

Question 20

Comments on Singly Reinforced Walls

Answer 20

• 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

Question 21

Thin walls with Drossbach

Answer 21

• 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

Question 22

Comments on Precast Panel Slices

Answer 22

• 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

Question 23

Comments on concrete strength

Answer 23

• 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»_space;> stronger than lab testing : no fan cracking
• case studies showed failure of bars in walls and inelastic capacity of bars exhausted

Question 24

order of most likely strained elements

Answer 24

• walls
• beams
• columns

Question 25

Comments on distribution of reinforcement

Answer 25

- 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

Question 26

effect of neutral axis on ductility

Answer 26

longer neutral axis = significantly less ductility

Question 27

reason for boundary elements

Answer 27

- ductile walls tend to buckle | - boundary elements (flanges) provide adequate flexural rigidity at end of wall section

Question 28

lapping in PPHZ

Answer 28

must stagger laps - ensure fan cracking still occurs and not horizontal crack at lap - no more than 1/3 steel lapped at same levels

Question 29

effect of size of penetrations in wall

Answer 29

- smaller penetration = greater shear capacity between walls - therefore larger tension/compression forces in wall - therefore higher overturning capacity

Question 30

three failure modes for walls

Answer 30

flexure, shear, sliding - avoid sliding by roughening cold joints to 5 mm - likely combined flexural shear failure

Question 31

Effect of penetration size on coupled walls

Answer 31

- 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

Question 32

Diagonally reinforced cages

Answer 32

- 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

Question 33

Boundary Elements

Answer 33

- 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

Question 34

Shear Requirements

Answer 34

- 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)

Question 35

Antibuckling

Answer 35

- 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

Question 36

double the strength halve the ductility - fallacy

Answer 36

- 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