2011.05.11 - Design of Tilt-Up Wall Systems

Tilt-Up Construction Par artt I - Plan ann ning Greg Riley Steven Schaefer Associates, Inc.

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Introduction 



Two part presentation: 

Part Part I – Plann Plannin ing g



Part Part II – Desi Design gn

Target audience: 

“General” structural engineer that designs a wide variety var iety of structure types. Engineer may or may not have tilt-up experience.

2

Introduction 

Definition: 

ACI 116R: “Tilt-up construction is a technique for casting concrete elements in a horizontal position at the jobsite and tiltin tiltin them them to their their final final ositio osition n in the struct structure” ure”

3

Introduction 

Introduction to Tilt-up construction process: 1.

.

Concrete slab on grade and exterior wall foundations (for the tilt-up walls) are placed. (exterior face face down on the slab). The formwork and rustication strips (architectural reveals) are connected to the slab on grade.

4

Introduction 

Formwork and reveals:

5

Introduction 


Bondbreaker Bondbreaker is sprayed sprayed inside inside the forms on the slab slab on grade

4.

Chairs, reinforcement, and embedded items (lifting/bracing , , . placed.

5.

Concrete placed, finished, and cured.

6.

Forms removed

7.

Crane connects to lifting inserts cast into interior face of wall panel and tilts panel up and places it on foundation

6

Introduction 

Panels with reinforcing:

7

Introduction 


Temporary braces are installed to support the wall panel prior to the crane releasing the panel

. 10.

Structural connections are made to diaphragm(s)

11.

Braces are removed once LFRS complete

12.

Non-structural work may occur at any time after erection ere ction

8

Introduction 

Lifting sequence:

9

Introduction 

Lifting sequence:

10

Introduction 

Lifting sequence:

11

Introduction 

Lifting sequence:

12

Introduction 

Lifting sequence:

13

Introduction 

History: 

Tilt-up has been used in the United States since the early 1900’s, but most of the development and momentum momen tum with using tilt tilt-u -u as a cons constr truc ucti tion on tech techni ni ue is ostost- Worl World d War War II. II.

14

Introduction 

History: 



Traditionally, tilt-up is thought of for large, warehouse structures

The proper mindset: 

Don’t limit yourself and the owner ow ner based on traditional ways to construct something. Consider tilt-up and other alternatives!



Tilt-up can be economically and effectively used for a wide variety of structures: office buildings, theaters, churches, retail, etc. Also in non-building structures such as retaining walls, screen walls, signs, tanks, etc.

15

Introduction 

Industrial:

16

Introduction 

Industrial/Warehouse:

17

Introduction 

Office:

18

Introduction 

Office:

19

Introduction 

Office:

20

Introduction 

Retail:

21

Introduction 

The proper mindset (cont.): 

You are part of a design team.



If at all possible, meet and discuss d iscuss construction type possibilities process so a sound decision is made.



Design-build obviously lends itself well to this or you need an experienced team for design-bid-build.

22

Step Step 1 – Is tilttilt-up up a viabl viable e alterna alternative tive? ? 

Crane access: Will there be plenty of room for the crane to move in and around the site? Truck crane costs can be ~ $3000/day, so efficiently using the crane is essential to the econom of a tilt-u buildin .



Building Building square footage: footage: 5000-7000 5000-7000 sf is a decent rule of thumb for a minimum square footage size to economically consider tilt-up.

23


Building wall height-to-floor slab width ratio is key. Ideally the panel height is less than ½ the building depth. Not a deal-breaker but definitely adds cost.

24


Building height: 



Don’t automatically disregard tilt-up if the building is multistory. A multistory panel isn’t any more difficult to design than a singlestor anel. 3-story buildings are reasonable without upsizing the crane that would potentially potentially be used. 4-stories would require a bigger bigger crane than usual. For above 4 stories you will potentially need to stack the wall panels.

25


Exterior elevations 

Will the exterior wall surface essentially be flat with some rustication joints? , should be at least 60% solid for tilt-up to be economical.

26


Exterior elevations (cont.): 

Will curved panels be required?



Can most of the panels be supported on foundations and not



Modularity is important.

27

Tilt-up Comparisons 

Precast 

Is there a qualified qualified precaster precaster in the general general area? Transportation costs aren’t cheap (especially these days). dimensions that can be supplied. Precast panels are typically 8’ or 12’ wide, which can limit the openings and opportunities for architectural expression.



The narrow panel width of precast panels can be a disadvantage compared to tilt-up if it is possible to use drilled piers and span the tilt-up panels pier-to-pier (no grade beam).

28


Precast (cont.) 

More panel joints = more caulk and maintenance. Caulk joint life is ~ 5-7 years. . precast, you want the steel erected first and waiting for the precast. In tilt-up, the walls come first and the steel brought to the walls.

29


Precast (cont.): 

Dimensional stability: Precast panels are frequently insulated. The result of the insulation between the exterior and interior anel face is a thermal radient between the faces. The resulting gradient can cause a bow in the panel. This bow results in problems where interior finishes are connected to the panels.



Future flexibility: flexibility: Tilt-up panels are typically somewhat easier to modify in the future for new or enlarged windows, doors, louvers, etc.

30


Precast thermal bow:

31


Precast thermal bow:

32


Masonry: 

Dependent on local material costs and labor costs for masons.



Building footprint and wall square footage is a big determinant. ~ , - , more economical choice.



The taller the wall, the more mor e economical tilt-up becomes relative to CMU.



In general, a tilt-up building can be erected faster than a CMU building.

33


Pre-engineered metal buildings (PEMB): 

PEMB’s are not really an apples-to-apples apples-to- apples comparison with conventional structures. framed structures structures in terms of durabilit durability y (future maintenan maintenance), ce), future flexibility, and fire resistance.



PEMB a great selection for owners whose primary pr imary desire is to have the lowest initial cost possible. Costs increase rapidly when trying to build in the features above into a PEMB.

34


PEMB (cont.): 

It is possible to clad a PEMB with tilt-up. Requires additional coordination with PEMB manufacturer.

35


Wood structures: 

Similar to PEMB, not really an apples-to-apples apples-to- apples comparison.



If it can be done with traditional wood construction it is cheapest .

36

Tilt-up Considerations 

Assume the design team has selected tilt-up for the construction of the project. Now what needs to be discussed and considered?

37


SEOR design vs. delegated design 

Lifting and bracing insert design. Most commonly done by insert manufacturer. . designed bracing inserts. They will specify a brace reaction to the slab on grade or potentially design the deadman/anchor if the brace is to the exterior.



Is the slab on grade a structurally designed item or non-structural specified/chosen item?

38


Foundations 

Strip foundations (trench formed) are the easiest (as compared comp ared to formed wall). – potential option if deep foundations are required

-

39


Slab on grade (SOG) construction 





Where are the panels going to be cast? Building SOG, casting bed, stack casting? Contractor’s call based on available slab square footage and owners SOG desires. Reference ACI publications for appropriate SOG design and construction techniques. 6” minimum

40


Slab on grade (SOG) construction (cont.) 

Construction loads on the SOG will typically far exceed exce ed the inuse loadings for the slab on grade. Options:   



Design thickened strips for crane travel Mandatory means/methods specifications for dunnage/cribbing under outriggers, keep outriggers off control joints (especially intersections), etc. You break it you bought it strategy. Sometimes you just have to repair or replace some slab. 41



SOG is cast prior to building being enclosed. Pay special attention to slab and panel protection pr otection from the elements (wind, tem erature, humidit . Plastic shrinka e crackin is es eciall problematic (hot and windy days with high evaporation rates).

42


Plastic shrinkage cracking on panels:

43



Bondbreaker Bondbreaker is important important between between the the panel and the the SOG



Make sure the bond breaker is compatible with the curing



Combination bond breakers/curing compounds are available. Popular with contractors.



Both bond breakers and curing compounds come in 2 general classes: Non-membrane forming vs. membrane forming.

44



Joints are important as all joints/slab imperfections are reflected r eflected in the tilt-up panels ,

,



Lots of decent skim coats, etc. available on the market now. This is preferable to grinding imperfections



Column locations: Covers with a thin layer layer of concrete on top, plywood/plastic covers, place slab over footing and core co re a hole for a column at a later date.

45



The bottom line is that the SEOR on a tilt-up building needs to pay more attention to the SOG and its details on than the would would on on most most oth other er str struct uctur ures es..

46


Panels  



How should the panel information be conveyed? Elevation format with reinforcing keys is conventional “structural “structur al detailing” Tri-elevation format is closer to shop drawing level of detailing

47


Tri-Elevation format:

48


Panels (cont.) 

Panel layout layout has many considerat considerations ions - Set a maximum maximum panel weight by balancing crane considerations with architectural considerations. 

Truck crane: Common panel weight limit limit is ~ 20T.



Crawlers: Common panel weight limit is ~ 60T.



You can upsize the size of the truck crane to increase the flexibility of the picks and minimize the number of set-ups.

49


Panels (cont.) 

Quick panel thickness estimate in inches is vertical vertica l span in ft/4. Ideally keep h/t <50. . ., pocket locations, reveal strip locations, etc.)

,



Ideally joists (and especially girders) don’t bear at panel joint locations



Joist and girder bearing detail: Pockets vs. vs. face mounted with seat angle

50


Panels (cont.) 

Attempt to maintain ~ 2’ jamb widths (dimension from edge of opening to edge of panel) strongbacks

-

’



If possible, limit the reveal depths to ¾”. Make sure the reveals are chamfered to facilitate formwork removal and reduce tendency for crack formation



Corner joint options: Lapped vs. mitered

51


Panels (cont.) 

 

Chamfer corners of panels and openings for less spalls and cleaner look ” ” for shorter panels. Concrete slump ~ 4”-5”. Use non-AE. Be extra careful with dimensions

52


Panels (cont.): 

Many panel finish options Paint 



Form liners



Thinset Thinset brick/v brick/vene eneer er



Exposed aggregate

53


LGMF/Foam board feature:

54


Masonry feature:

55


Panels (cont.) 

Insulation  Industrial applications leave the lower 8’ left uninsulated  Interior walls furred out and insulated in office/retail applications

56


Panels (cont.) 

Sandwich panels 





Constructed Constructed with with 2 wythes sandwiching sandwiching an interior interior insulation insulation May increase in popularity with the trend for increased energy efficiency Most popular now in extremely cold climates

57


Panels (cont.) 

Sandwich panels 

Can be composite or non-composite



In non-composit non-composite, e, the the inner wythe wythe is structural structural





In composite, a thermal gradient concern is present pres ent (similar to precast) If you use composite, make sure you are specifying a reliable connector product and have appropriate quality control measures in the field because it is big money to retrofit a failed system. 58

Questions prior to Part II?

59

Tilt-Up Construction Par artt II - De Desi sig gn John Ashbaugh Steven Schaefer Associates, Inc.

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Tilt-Up Design Presentation Overview

 

Code Evolution ACI 318, 14.8 - “Alternative Design of Slender Walls”

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Desi n Exam les

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Misc. Design Topics



Design Tips

61

Code Evolution



Before Slender Wall Design Provisions 

ACI h/t limits resulted in uneconomical designs 





Example: Max h/t = 25 resulted in 14½” thick panel for 30’ tall bearing wall

1979 SEAOSC “Recommended Tilt-Up Wall Design” (Yellow Book) 

Max h/t: 36 unstiffened bearing walls, 42 for stiffened bearing walls



Included second-order effects

1982 SEAOSC/ACI “Test Report on Slender Walls” (Green Book) 

Full scale testing testing of tilt-up tilt-up panels – showed stability stability under large large deflections deflections



Report stated “no validity for fixed h/t limits”



Report stated need for deflection limits (h/100) 62

Code Evolution



1988 UBC - “Alternate “Alternate Design Design Slender Slender Walls” Walls” 

Considered eccentric gravity load effects



Considered P-delta effects



Included service load deflection limit (h/150)



Basis of current ACI design procedure procedur e for slender walls

63

Code Evolution



ACI 318-99, 14.8 - “Alternative Design for Slender Walls” 



ACI’s first “slender wall” wall” design procedure



Similar to UBC 97, but strength streng th design



No h/t limits

ACI 318-08, 14.8 - “Alternative Design for Slender Walls” 

Current “slender wall” design procedure

64

Code Evolution



ACI 551.2R-10 – Design Guide for Tilt-Up Construction 

Expands on slender wall provisions of ACI 318 Section 14.8



Provides a “comprehensive procedure for the design” of tilt-up



Provides recommendations for various conditions not specifically covered in ACI 318

65

ACI 318, 14.8 14.8 – “Alternative Design of Slender Slender Walls”



Assumptions / Requirements – Section 14.8.2 

Simply supported axially loaded member subjected to out-ofplane lateral load, with max. moment & deflections at midspan



Tension-contro Tension-controlled lled (c / d < 0.375 – refer to ACI ACI 318 R9.3.2.2) R9.3.2.2)



Vertical stress Pu/Ag at midheight ≤ 0.06f c’

66




Assumptions / Requirements – Section 14.8.2 (cont’d) 

φMn ≥ Mcr

Eq. (14-2)

where Mcr = f r S where f r = 7.5 λ 

f c '

Concentrated gravity load distribution 

Bearing width + 2 vert / 1 horiz slope down to design section



Not greater than spacing of concentrated loads



Not extending beyond edges of wall panel

67




Assumptions / Requirements – Section 14.8.2 (cont’d) 

Concentrated gravity load distribution: ACI 551.2R-11, Fig. 4.2

68




Design Design Momen Momentt Strength Strength – Section Section 14.8.3 14.8.3 

φMn ≥ Mu



φMn is determined per ACI 318 Ch. 10

Eq. (14-3)

φMn = φ Ase f y (d – a/2) 

Effective area of steel (Ase) accounts for increased bending moment resistance due to axial load Ase = As + (Pu / f y) (h / 2d)

69




Design Design Moment Moment Strength Strength – Section Section 14.8.3 (cont’d) (cont’d) 

Mu includes moment due to applied loads & due to P∆



Fig. 3.1 from ACI 551.2R:

70





Mu can be determined using Iteration Method to account for P∆  



Mu = Mua + Pu∆u

Eq.(14-4)

Mua = max. factored factored M at midheight midheight due to lateral lateral loads loads & eccentric vertical loads (does not include P∆) ∆u =

5 M u l c 2

Eq. (14-5)

(0.75)48 E c I cr

71





Or, Mu can be determined using Moment Magnification Magn ification Method M ua

Μu = 1−

where Icr =

u c

Eq. (14-6)

2

(0.75) 48 E c I cr

E s E c

( A s

+

P u h

)(d − c) f y 2d

2

+

l wc 3 3

Eq. (14-7)

and Es / Ec ≥ 6

72




Minimu Minimum m Reinforc Reinforceme ement nt - ACI 318, 318, 14.3 14.3 



14.3.2 - Min vertical vertical reinforceme reinforcement nt ratio 

(a) 0.0012 for #5 bars or smaller, f y ≥ 60 ksi



(b) 0.0015 for other deformed bars



(c) 0.0012 for WWR not larger than W31 or D31

14.3.3 – Min horizonta horizontall reinforcement reinforcement ratio 

(a) 0.0020 for #5 bars or smaller, f y ≥ 60 ksi



(b) 0.0025 for other deformed bars



(c) 0.0020 for WWR not larger than W31 or D31 73




Minimum Minimum Reinforc Reinforcement ement - ACI 318, 14.3 14.3 (cont’d) (cont’d) 

14.3.7 – In addition addition to min reinforcemen reinforcement, t, bars are are required around windows, doors, and similar openings. Bars shall be anchored to develo f at corners of o enin . 

Panel with 2 layers of reinforcement: not less than (2) #5’s



Panel with 1 layer of reinforcement: not less than than (1) #5

74




Maximum Maximum Out-of-Plane Out-of-Plane Deflection Deflection – Section Section 14.8.4 14.8.4 

Max. out-of-plane deflection due to service loads (including (includ ing P-D effects), ∆s, shall not exceed lc / 150 a

cr

∆s = ( 2 / 3)∆ cr + 

If Ma ≤ 2/3 Mcr : ∆s =

( M a )

( M cr )

∆ cr

( M a

− ( 2 / 3) M cr )

( M n

− ( 2 / 3) M cr )

(∆ n

− ( 2 / 3) ∆ cr )

Eq. (14-8)

Eq. (14-9)

75




Max. Out-of-Plane Out-of-Plane Deflecti Deflection on – Section Section 14.8.4 (cont’d) (cont’d) 

Where… ∆cr =

5 M cr l c

2

Eq. (14-10)

c g

∆n =

5 M nl c 2 48 E c I cr

Eq. (14-11)

And Icr is per Eq. (14-7)

76




Max. Out-of-Plane Out-of-Plane Deflecti Deflection on – Section Section 14.8.4 (cont’d) (cont’d) 

Eq. (14-8) accounts for a rapid increase in out-of-plane deflections when Ma > 2/3 Mcr combinations for calculating service level deflections: 

D + 0.5L + 0.7W



D + 0.5L + 0.7E

77

Reinforcing Steel Location



One layer 

Vertical bars located at or near center cen ter of panel thickness



Typical for solid panels

78

Reinforcing Steel Location



Two layers 

Vertical bars typically located minimum clear from each face



Typical for panels with openings, and economical for some solid panels .

79

Design Design Exampl Example e 1 – Solid Solid Pane Panell

80




Concrete & reinforcing steel properties: 

f c’ = 4000 psi



γc = 150 pcf



f r = 7.5 λ



f y = 60,000 psi



Es = 29,000 ksi



Ec = 57



Es / Ec = 8.044

f c ' =

f c '

474 psi

= 3605 ksi

81




Panel wt. above design section: (7.25 in/12)(150 pcf)(14 ft)(16.0 ft) / 1000 = 20.3 20 .3 k



For sim licit in this exam le we we’ll on onl consider wind suction and only one design load case: 1.2D + 1.6W + 0.5 Lr



Factored applied axial load at top of wall Pua = 1.2 (0.45 (0.45 klf klf x 14 ft) + 0.5 0.5 (0.6 (0.6 klf klf x 14 ft) ft) = 11.8 11.8 k

82




Factore Factored d axial axial load at midhei midheight ght of wall wall Pum = 11.8 k + 1.2 (20.3 k) = 36.1 k



Check vert. rt. stres ress at midhei hei ht < 0.06 .06f ’ = 240 si Pum/Ag = 36,100 lb / (7.25in x 14ft x 12 in/ft) = 29.6 psi < 240 



OK

Trial reinforcing: (18) #6 vertical bars, (1) layer centered in panel As = 7.92 in2 & d = 3.625 in 83




Check the design moment strength Ase = As + (Pum / f y) (h / 2d) A = 7.92 in2 + 36.1 k / 60 ksi 7.25 in/ 2 x 3.625 in = 8.52 in2 a=

A se f y 0.85 f c ' b

=

8.52in 2 (60ksi )

= 0.895 in

0.85(4ksi)(14 ft )(12in / ft )

c = a / 0.85 = 0.895 in/ 0.85 = 1.053 in c / d = 1.053 / 3.625 = 0.291 < 0.375 OK (tension controlled)

84




Check the design moment strength (continued) Icr =

E s E c

( A s

+

P um h

)(d − c) f y 2d

2

+

l wc 3 3

Icr = 8.044(8.52 in2)(3.6 )(3.625 25 – 1.053 1.053))2 + (14 ft x 12 in/ft)(1.053)3 / 3 = 519 in4 φMn = φ Ase f y (d–a/2) = 0.9(8.52 in2)(60 ksi)(3.625-0.895/2) φMn = 1462 in-k = 122 ft-k

85




Check min. reinforcement per section 14.8.2.4 Mcr = f r S = 0.474 0.474 ksi ksi [1/6 (14 (14 ft x 12 in/ft)(7.2 in/ft)(7.25) 5)2 = 698 in-k φMn = 1462 in-k > Mcr



OK

Check min. reinforcement per section 14.3.2 ρ = As / (bh) = 7.92 in2 / [(14 ft x 12 in/ft)(7.25 in)] = 0.0065 ρ = 0.0065 > 0.0015

OK

86




Check Mu using moment magnification (Eq. 14-6) wu = 1.6 x 18 psf x 14 ft ft /1000 = 0.403 klf Axial load applied to top of wall panel (previously calc’d) Pua = 11.8 k Factored moment, excluding P∆ effects: 2 Mua = wu l c + P ua ecc

8

2

Mua = 0.403 0.403 klf klf (32 ft) ft)2 /8 + 11.8 k (0.33 ft) / 2 = 53.5 ft-k

87




Check Mu using moment magnification (Eq. 14-6) Axial load at midheight of wall, Pum = 36.1k (previously calc’d) Factored moment, including P∆ effects: ua

Mu = 1−

5 P uml c 2 (0.75) 48 E c I cr 53.5 ft − k

Mu =

1−

5(36.1k )(32 ft ) 2 0.75(48)(3605ksi )(519in 4 ) / 144

= 88.5 ft-k < φMn =122 ft-k 

OK 88




Check service load deflection with 1.0D + 0.5Lr + 0.7W ∆allowable = lc / 150 = (32 ft x 12 in/ft) / 150 = 2.56 in

Ig = (1/12) (14 ft x 12) ( 7.25 in)3 = 5335 in4 ∆cr =

cr c

48 E c I g

= 5(698in − k )(32 ftx12in / ft )

= 0.558 in

48(3605ksi )(5335in 4 )

Initial iteration service load moment (without P∆): Msa = wl2/8 + (Paxecc)/2 Pa = 0.45 0.45 klf klf x 14 ft ft + 0.5 0.5 (0.6 (0.6 klf klf x 14 ft) = 10.5 10.5 k

89




Check service load deflection (continued) Msa =

0.7(0.018 x14 ft )(32 ft ) 8

2 +

10.5k (0.33 ft ) 2

Msa = 24.3 ft-k < 2/3 Mcr = 38.8 ft-k … use Eq. (14-9) ∆s = (Msa / Mcr ) x ∆cr = (24.3 / 58.2) x 0.558 in = 0.233 in

90




Check service load deflection (continued) Now including P∆ effects: Ma = Msa + Psm ∆s Psm = 10.5 k + 20.3 k = 30.8 k Ma = 24.3 ft-k + 30.8 k(0.233 in/12) = 24.9 ft-k < 2/3 Mcr ∆s =(Ma / Mcr ) x ∆cr = (24.9 / 58.2) x 0.558 in = 0.239 in

Final ∆s = 0.24 in < 2.56 in OK

91

Design Example Example 2 – Panel with Opening Opening

92





f c’ = 4000 psi



γc = 150 pcf



f r = 7.5 λ



f y = 60,000 psi



Es = 29,000 ksi



Ec = 57



Es / Ec = 8.044

f c ' =

f c '

474 psi

= 3605 ksi

93




Panel wt. above design section acting on design strip: (7.25 in/12)(150 pcf)(7 ft)(16.0 ft) / 1000 = 10.2 k



For sim licit in this exam le we we’ll on onl consider wind suction and only one design load case: 1.2D + 1.6W + 0.5 Lr



Factored applied axial load at top of wall Pua = 1.2 (0.45 (0.45 klf klf x 7 ft) ft) + 0.5 (0.6 (0.6 klf klf x 7 ft) ft) = 5.9 k

94







Check vert. rt. stres ress at midhei hei ht < 0.06 .06f ’ = 240 si Pum/Ag = 18,100 lb / (7.25in x 30 in) = 83 psi < 240  OK



Trial reinforcing: (5) #6 vertical bars each face, 1 ½” clr. As = 2.2 in2 & d = 7.25 in -1.5 in – 0.75 in / 2 = 5.375 5.375 in

95





A se f y 0.85 f c ' b

=

2.40in 2 (60ksi )

= 1.41 in

0.85( 4ksi)(30in)


96





E s E c

( A s

+

P um h

)(d − c) f y 2d

2

+

l wc 3 3

Icr = 8.044(2.4 in2)(5. )(5.37 375 5 – 1.66 1.66))2 + (30 in)(1.66)3 / 3 = 312 in4 φMn = φ Ase f y (d–a/2) = 0.9(2.4 in2)(60 ksi)(5.375-1.41/2) φMn = 605 in-k = 50.4 ft-k

97




Check min. reinforcement per section 14.8.2.4 Mcr = f r S = 0.474 0.474 ksi ksi [1/6 [1/6 (30 in)(7. in)(7.25) 25)2 = 125 in-k φMn = 605 in-k > Mcr



OK

Check min. reinforcement per section 14.3.2 ρ = As / (bh) = 2.2 in2 / [(30 in)(7.25 in)] = 0.0101 ρ = 0.0101 > 0.0015

OK

98




Check Mu using moment magnification (Eq. 14-6) wu = 1.6 x 18 psf x 7 ft /1000 = 0.202 klf Axial load applied to top of wall panel (previously calc’d) Pua = 5.9 k Factored moment, excluding P∆ effects: 2 Mua = wu l c + P ua ecc

8

2


99





Mu = 1−

5 P uml c 2 (0.75) 48 E c I cr 26.8 ft − k

Mu =

1−

5(18.1k )(32 ft ) 2 0.75(48)(3605ksi )(312in 4 ) / 144

= 40.0 ft-k < φMn =50.4 ft-k 

OK 100




Check service load deflection with 1.0D + 0.5Lr + 0.7W ∆allowable = lc / 150 = (32 ft x 12 in/ft) / 150 = 2.56 in

Ig = (1/12) (30 in) ( 7.25 in)3 = 953 in4 ∆cr =

cr c

48 E c I g

= 5(125in − k )(32 ftx12in / ft )

= 0.559 in

48(3605ksi)(953in 4 )

Initial iteration service load moment (without P∆): Msa = wl2/8 + (Pa x ecc)/2 Pa = 0.45 0.45 klf klf x 7 ft + 0.5 0.5 (0.6 (0.6 klf klf x 7 ft) ft) = 5.3 5.3 k

101




Check service load deflection (continued) Msa =

0.7(0.018 x7 ft )(32 ft ) 8

2 +

5.3k (0.33 ft ) 2

Msa = 12.2 ft-k = 146 in-k > 2/3 Mcr = 83 in-k … use Eq. (14-8) ∆s = ( 2 / 3)∆ cr +

Where ∆n =

( M sa

− ( 2 / 3) M cr )

( M n

− ( 2 / 3) M cr )

5 M nl c

2

48 E c I cr

=

(∆ n

− ( 2 / 3) ∆ cr )

5(672in − k )(32 ftx12) 2 4

= 9.18 in

48(3605ksi)(312in )

102




Check service load deflection (continued) ∆s = ( 2 / 3)(0.559in) + ∆s = 1.31 in

(146 − (2 / 3)125) (672 − (2 / 3)125)

(9.18in − (2 / 3)(0.559in)

Now including P∆ effects: Ma = Msa + Psm ∆s Psm = 5.3 k + 10.2 k = 15.5 k Ma = 12.2 ft-k + 15.5 k(1.31 in/12) Ma = 13.9 ft-k = 167 in-k > 2/3 Mcr

103




Check service load deflection (continued) ∆s = ( 2 / 3)(0.559in) + ∆s = 1.62 in

(167 − (2 / 3)125) (672 − (2 / 3)125)

(9.18in − (2 / 3)(0.559in)

Continue to iterate using Eq. (14-8)… Final ∆s = 1.7 in < 2.56 in OK

104

Design Example Example 3 – Concentrat Concentrated ed Gravity Gravity Load

105





f c’ = 4000 psi



γc = 150 pcf



f r = 7.5 λ



f y = 60,000 psi



Es = 29,000 ksi



Ec = 57



Es / Ec = 8.044

f c ' =

f c '

474 psi

= 3605 ksi

106




12 ft wide “design strip” for concentrated gravity load



Panel wt. above design section acting on design strip: 7.25 in/12 150 cf 12 ft 16.0 ft / 1000 = 17.4 k



For simplicity in this example we’ll only consider wind suction and only one design load case: 1.2D + 0.8W + 1.6 Lr



Factored applied axial load at top of wall Pua = 1.2 (16 k) + 1.6 (22 k) = 54.4 k 107







Check vert. rt. stres ress at midhei hei ht < 0.06 .06f ’ = 240 si Pum/Ag = 75,300 lb / (7.25in x 144 in) = 72 psi < 240  OK



Trial reinforcing: (12) #6 vertical bars each each face, 1½” clr. As = 5.28 in2 & d = 7.25 in -1.5 -1.5 in – 0.75 in / 2 = 5.375 in

108





A se f y 0.85 f c ' b

=

6.13in 2 (60ksi)

= 0.75 in

0.85(4ksi )(144in)


109





E s E c

( A s

+

P um h

)(d − c) f y 2d

2

+

l wc 3 3

Icr = 8.044(6.13in2)(5. )(5.37 375 5 – 0.88 0.88))2 + (144 in)(0.88)3 / 3 = 1029 in4 φMn = φ Ase f y (d–a/2) = 0.9(6.13 in2)(60 ksi)(5.375-0.75/2) φMn = 1655 in-k = 138 ft-k

110




Check min. reinforcement per section 14.8.2.4 Mcr = f r S = 0.474 0.474 ksi ksi [1/6 [1/6 (144 in)(7 in)(7.25 .25))2 = 598 in-k φMn = 1655 in-k > Mcr



OK

Check min. reinforcement per section 14.3.2 ρ = As / (bh) = 5.28 in2 / [(144 in)(7.25 in)] = 0.0051 ρ = 0.0051 > 0.0015

OK

111




Check Mu using moment magnification (Eq. 14-6) wu = 0.8 x 18 psf x 12 ft ft /1000 = 0.173 klf Axial load applied to top of wall panel (previously calc’d) Pua = 54.4 k Factored moment, excluding P∆ effects: 2 Mua = wu l c + P ua ecc

8

2


112





Mu = 1−

5 P uml c 2 (0.75) 48 E c I cr 31.1 ft − k

Mu =

1−

5(75.3k )(32 ft ) 2

= 53.2 ft-k

0.75(48)(3605ksi)(1029in 4 ) / 144 < φMn =138 ft-k 

OK 113

Panel Base Connections



Code language 





ACI 318, 14.2.8: “Transfer of force to footing footing at base of wall shall be in accordance with 15.8” , . using reinforcement, dowels, or mechanical connectors. ACI 318, Chapter 16 – Precast Construction 

R16.1.1: “Tilt-up concrete concrete construction construction is a form of precast construction”



16.5.1.3 – Vertical Tension Ties: (b) “Precast wall panels shall have a minimum of two ties per panel, with a nominal tensile strength not less than 10,000 lb per tie” 114




Code language (continued): 

16.5.1.3 – Vertical Tension Ties: (c) When design forces result in no tension at the base, the ties required by 16.5.1.3(b) shall be permitted to be anchored into an appropriately reinforced concrete oor s a -on-groun .



R16.5.1.3: R16.5.1.3 : Base connections at shear walls walls “are designed to transfer all all design design forces and and moments. The minimum minimum tie requirements of 16.5.1.3 are not additive to these design requirements.”



16.5.1.4: “Connection details that rely solely on friction caused by gravity loads shall not be used”

115




Code language (continued): 

ACI 551.2R Section Section 8.2 - In-Plane Shear / Resistance Resistance to sliding: sliding: “Resistance to sliding forces can be obtained by a combination of friction friction between between the bottom bottom of anel and the footin footin and connecti connections ons to the floor slab or foundation (refer to ACI 318, Section 16.5.1.3) “Where panels are subjected to seismic forces, the contribution of friction resistance may not be permitted by some buildings codes. In addition, connections between the panel and floor slab or footing is a compulsory requirement in many building codes, particularly for seismic forces.”

116




Conclusions 

If calculated uplift, base connection shall be designed for uplift forces



If no calculated uplift, minimum vertical tension tie connections shall be rovided at anel base er ACI 318 Section 16.5.1.3



ACI 318 Section Section 16.5.1.3(c): If wall panels panels anchored to slab-on-ground, slab-on-ground, is the slab-on-ground required to resist two 10,000 lb nominal tensile forces per panel? panel? What is an “appropriately “appropriately reinforced concrete concrete slabon-ground”?



In-plane shear forces at base of footing can be resisted by a combination of friction and base connections, but not friction only

117




Sampling of details from ACI 551.1R:

118





119

Panel & Floor / Roof Connections



In addition to calculated gravity, uplift, in-plane lateral, and outof-plane lateral forces, check minimum seismic anchorage forces (ASCE 7, Section 12.11 “Structural Walls and Their Anchorage”) Anchor age”)



Design connections according to provisions of ACI 318, App. D



Ductility for seismic connections

120





121





122

Panel to Panel Connections





Connections restrain shrinkage and thermal expansion/contraction… try to avoid ACI 551.1R recommendations: “ shrinkage and thermal expansion and contraction” 

Reinforcing bar anchors preferred over short headed studs



Delay welding as long as possible to allow majority of panel shrinkage to occur

123

Panel to Panel Connections



ACI 551.1R, Fig 7.14:

124

Design for Lifting & Bracing

  



Analysis is often by supplier of embedded lift & bracing inserts EOR reviews submittal by embedded insert supplier Additional reinforcement &/or &/or “stiffbacks” are sometimes required by 

Multi-story, large openings with small jambs, and single layer reinforcement



Review early in design to avoid costly measures later

ACI 551.2R-10 states that further development of design procedures will be included in future editions

125

In-Plane Shear



Typically resisted by tilt-up “shear walls”



If high % of panel openings, may need to design as frames rather than solid shear wall elements



Resistance to overturning (OT) 

Resistance to OT typically provided by panel wt., roof loads, and floor loads

126

In-Plane Shear



Resistance to OT (continued) 

If panels alone do not provide OT capacity:   



Increase panel width &/or thickness Anchor panel to foundation with tension tension tie connections, or Connect 2 or more panels together to create a larger shear wall

Resistance to Sliding 

ACI 551.2R states that resistance can be through combination of friction & connections to floor slab or foundation, but cautions that friction may not be allowed to resist seismic forces

127

Dock Walls / Retained Soil



Lateral support is commonly provided by slab-on-ground, creating continuity as depicted in Fig 7.5 from ACI 551.2R:

128

Dock Walls / Retained Soil





For large slab to roof span relative to slab to foundation span, continuity has often been neglected and panel is designed as simply supported from slab to roof. More rigorous analysis that considers continuity is presented pr esented in ACI 551.2R – Example B.6



Large horizontal forces in slab and at foundation must be fully developed



Consider larger span from roof to footing that may occur temporarily during construction

129

Multi-Story Panels



Example Example problem problem is presented presented in in ACI 551.2R 551.2R – Example Example B.5 

Moment diagram for continuous span is determined without considering P∆ effects. Section 14.8.3 and applied to both max. positive and max. negative moments.



Consider lifting stresses for larger spans that may occur oc cur temporarily during construction.

130

Large Concentrated Gravity Loads



What are options if wall is compression controlled? 

Increase panel thickness



Move panel joints &/or openings to provide more “b”



Design & detail as “column” per ACI 318 Ch. 10



Add pilaster at concentrated load



Add steel column at concentrated load

131

Design Tips



Repetition/modularity is key to efficiency & economy



Look out for… 

Hanging/spandrel panels



Panel to panel connections



Panel hold-down connections at shear walls



“L” and and “T” “T” – shaped shaped panel panels s



Small jambs adjacent to large openings



Heavy girder bearing on skinny jambs or panel joints

132

Design Tips



Look out for… (continued) 

Lifting stresses for multi-story panels



Dock wall wall & retained retained soil – details details & load path at slab & footing footing



Frequent openings may require moment frames (vs. shear walls) wa lls)



Tall & narrow shear wall panels (check OT)



Reveal joints joints – may affects affects bar location location for exterior exterior face face bars



Out-of-plane deflection at small jambs adjacent to large openings



Embedded plates near the ends of panels 133

Design Tips



Look out for… (continued) 

Restraint created by panel to foundation (or slab) connections



Corrosion projection may be required for panel to foundation



Corner joint details at “big box” buildings (thermal expansion/contraction in roof)



Too-small scupper size at parapets in “big box” buildings



QC dimensions!!!!



Site visit prior to first panel pour is recommended

134

Questions?

135

2011.05.11 - Design of Tilt-Up Wall Systems

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