Bearing Wall Design

Lecture 21 – Load-Bearing Walls The ACI recognizes the “Empirical “ Empirical Design Method” Method ” for the design of loadbearing concrete walls having the following limitations:

Load-bearing Wall

1. Wall must be solid, with a rectangular cross-section  Lu 25

2. Minimum wall thickness “t w” = larger of

4” ← Interior walls only 7½” ← Exterior walls

3. Applied load eccentricity “e” (including lateral load moments) < 4. Wall must be adequately anchored at top and bottom

Lecture 21 – Page 1 of 7 

t w 6

5. Must have horizontal & vertical reinforcement, with 2 rows of reinforcing for walls with t w > 10” Vert. bars A s = 0.0012(tw)(12”) #3, #4, #5 grade 60 bars →

Horz. bars As = 0.0020(tw)(12”) Vert. bars A s = 0.0015(t w)(12”)

#6 and bigger grade 60 bars →

Horz. bars As = 0.0025(t w)(12”) 3tw

Maximum bar spacing = smaller of

or 18”

6. Effective length of wall for beam reaction = L eff Center-to-center spacing of beams Leff = smaller of Width of bearing + 4t w 7. Design factored axial strength of wall = P n

⎡ ⎛  KLu  ⎞ 2 ⎤ ⎟⎟ ⎥ Pn = 0.55φ  f  ' c  Ag ⎢1 − ⎜⎜ t  32 ⎢⎣ ⎝  w  ⎠ ⎥⎦ where: φ = 0.70 f’c = specified concrete strength, PSI K = end fixity of walls = 1.0 for pinned-pinned = 0.70 for fixed-pinned Ag = Gross effective area of wall section, in 2 = Leff x tw Lu = unbraced height of wall, inches tw = thickness of wall, inches

Lecture 21 – Page 2 of 7 

Example GIVEN: A poured-in-place concrete wall supports W18x35 steel beams spaced 6’-0” apart and rests on 10” wide steel bearing plates. Use the following:

• Beam end reaction = 22 KIPS Service Dead Load (Not incl. wall wt.) = 15 KIPS Service Live Load

• • • •

Wind pressure on wall = 25 PSF (service load) Concrete f’c = 4000 PSI Use #5 grade 60 vertical and horizontal bars Assume “K” = 1.0 for wall end fixity

REQUIRED: Design the wall using the Empirical Design Method. PDEAD = 22 KIPS PLIVE = 15 KIPS

Steel beam 10” bearing plate Wall Slab on grade

Lecture 21 – Page 3 of 7 

   F    S    P    5    2   =    d   n    i    W

   ”    0      ’    3    1   =   u    L

Fin. Grade

Step 1 – Determine “Trial” thickness of wall, t w:  Lu 25

Minimum wall thickness “t w” = larger of

=

13'−0" (12" /  ft ) 25

= 6.24”

4” 7½”

Largest

TRY tw = 8” which is > 7½” Step 2 – Determine maximum SERVICE wind moment M wind on wall: Assume a 1’-0” wide “strip” of wall: 2

Mwind =

=

wL 8

( 25PSFx1'−0" )(13' )

2

8

Mwind = 528 Ft-Lb per 1’-0” length of wall = 528 Ft-Lb(6’-0”) for beam spacing = 3168 Ft-Lb per 6’-0” length of wall Step 3 – Determine the maximum SERVICE vertical load on wall P total: Ptotal = PDEAD + PLIVE = (PDEAD + wall weight) + PLIVE

⎛  8"  ⎞ 0.150 KCF ) ⎟⎟13'−0" )) + 15 KIPS ⎝ 12" /  ft   ⎠

= (22 KIPS + (6’-0” ⎜⎜

= 29.8 KIPS + 15 KIPS Ptotal = 44.8 KIPS

Lecture 21 – Page 4 of 7 

Step 4 – Determine if eccentricity is acceptable for Empirical Design: e = eccentricity =

=

 M 

Service wind load moment

P

3.168Kip − Ft (12" /  ft ) 44.8 KIPS

Service axial load (see above)

= 0.85 inches

The empirical design method dictates that e < t w 6

=

t w 6

8"

t w 6

Wall thickness

6

= 1.33” > 0.85”

OK to use Empirical Design Method

Step 5 – Determine applied factored axial load, P u: Pu = 1.2(PDEAD) + 1.6(PLIVE) = 1.2(29.8 KIPS) + 1.6(15 KIPS) = 59.8 KIPS Step 6 – Determine factored bearing strength of concrete: A1 = bearing contact area from beam, in 2 = (bearing plate width)(t w) = (10”)(8”) = 80 in2 Concrete bearing capacity = 0.9(0.85f’ cA1) = 0.9(0.85(4000 PSI)(80 in 2)) Concrete bearing capacity = 244,000 Lbs > 59.8 KIPS

Lecture 21 – Page 5 of 7 

OK

Step 7 – Determine design axial strength of wall P n: Leff = smaller of

Center-to-center spacing of beams = 6’(12”/ft) = 72” Width of bearing + 4t w = 10” + 4(8”) = 42”

USE

Ag = Gross effective area of wall section, in 2 = Leff x tw = (42”)(8”) = 336 in2

⎡ ⎛  KLu  ⎞ 2 ⎤ ⎟⎟ ⎥ Pn = 0.55φ  f  ' c  Ag ⎢1 − ⎜⎜ ⎢⎣ ⎝ 32t w  ⎠ ⎥⎦ 2 ⎡ ⎤ ⎛   ⎞  x ft  ( 1 . 0 )( 13 ' 12 "  /  ) ⎟⎟ ⎥ = 0.55(0.70)(4000 PSI )(336in 2 ) ⎢1 − ⎜⎜ 32(8" ) ⎢⎣ ⎝   ⎠ ⎥⎦

Pn = 325,300 Lbs. > 59.8 KIPS

wall is acceptable

Step 8 – Determine vertical and horizontal bars: #3, #4, #5 grade 60 bars →

Vert. bars A s = 0.0012(tw)(12”) Horz. bars As = 0.0020(tw)(12”)

a) Vert. bars → As = 0.0012(8”)(12”) = 0.115 in 2 per 1’-0” length of wall

⎛ 0.31in 2 _  per _#5 _ bar  ⎞ ⎟⎟ spacing = 12” ⎜⎜ 2 in 0 . 115 ⎝   ⎠ = 32.3”

32.3” or 3tw = 3(8”) = 24”

Maximum bar spacing = smaller of

or 18”

USE #5 vertical bars @ 18” o.c.

Lecture 21 – Page 6 of 7 

USE

b) Horizontal bars → As = 0.0020(tw)(12”) = 0.0020(8”)(12”) = 0.192 in 2

⎛ 0.31in 2 _  per _#5 _ bar  ⎞ ⎟⎟ spacing = 12” ⎜⎜ 2 0 . 192 in ⎝   ⎠ = 19.4” 19.4” or

Maximum bar spacing = smaller of

3tw = 3(8”) = 24” or 18”

USE

USE #5 horizontal bars @ 18” o.c.

Step 9 – Draw “Summary Sketch”:

W18x35 Steel beam

18”

10” bearing plate

8” thick concrete Wall

#5 @ 18” ea. way

   ”    0      ’    3    1   =   u    L

Slab on grade

NOTES: 1) Conc. f’ c = 4000 PSI. 2) All bars grade 60 3) Footing dowels and “keyway” not shown

Lecture 21 – Page 7 of 7 

Fin. Grade