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Structural Design of Footings Prof. Jie Han, Ph.D., PE The University of Kansas Outline of Presentation • Introduction • Design for Square Footings • Design for Continuous Footings • Design for Rectangular Footings • Connections with Superstructures Factored Normal Load Pu = 1.2 D + 1.6 L (ACI) Pu = 1.25 D + 1.75 L (AASHTO) D = dead load L = live load Minimal Requirements d T db 3in. 3in. Flexural steel Effective depth d = T − 3in. − d b ≥ 6in. Minimal thickness T ≥ 12in. Common Selection of Materials Concrete Compressive strength fc' = 2,000 to 3,000psi (15 to 20MPa) Steel Grade 40 steel fy = 40,000psi (300MPa) Grade 60 steel fy = 60,000psi (420MPa) Design Criterion for Shear Design Criterion Vuc ≤ φVnc Vuc = factored shear force on critical surface φ = resistance factor for shear = 0.75 Vnc = nominal shear capacity on critical surface Nominal Shear Capacity Vnc = Vc + Vs Vnc = nominal shear capacity on critical surface Vc = nominal shear capacity of concrete Vs = nominal shear capacity of steel (ignored) Design for Square Footings Design of Square Footings for One-Way Shear Pu B c d T d Factored shear force on one critical surface Pu B − (c + 2d ) Vuc = 2 B No need to check one-way shear if there is no applied moment or shear loads Design of Square Footings for One-Way Shear Pu Mu c Vu d d Most critical face Factored shear force on one critical surface 2 2 ⎛ B − c − 2d ⎞ ⎛ Pu 3Mu ⎞ ⎛ Vu ⎞ Vuc =⎜ ⎟ ⎜ + ⎟ +⎜ ⎟ ⎝ B ⎠ ⎝ 2 B ⎠ ⎝ 2⎠ Design of Square Footings for Two-Way Shearing B c Pu d/2 Mu d/2 T d d c+d Most critical face Factored shear force on one critical surface ⎛ Pu Mu ⎞⎛ Base area of outer block ⎞ ⎛ Pu Mu ⎞⎛ B 2 − (c + d )2 ⎞ Vuc =⎜ + ⎟⎜ ⎟=⎜ + ⎟⎜ ⎜ ⎟ ⎟ ⎝ 4 c + d ⎠⎝ ⎠ ⎝ 4 c + d ⎠⎝ 2 total base area B ⎠ Design of Square Footings for Two-Way Shearing Pu Vu Most critical faces Most critical faces c+d Factored shear force on one critical surface ⎛ Pu ⎞ ⎛ Vu ⎞ ⎛ B − (c + d ) ⎞ 2 2 2 2 2 ⎛ Pu ⎞ ⎛ Vu ⎞ ⎛ base area of outer block ⎞ 2 Vuc = ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎜ ⎜ ⎟ ⎟ ⎝ 4 ⎠ ⎝ 2⎠ ⎝ ⎠ ⎝ 4 ⎠ ⎝ 2⎠ ⎝ 2 total base area B ⎠ Nominal One-Way Shear Capacity Vnc = Vc = 2bw d fc' English Units 1 Vnc = Vc = bw d fc' SI Units 6 Vnc = nominal shear capacity on one critical surface (lb or N) Vc = nominal shear capacity of concrete (lb or N) bw = length of one critical surface = B (in or mm) d = effective depth (in or mm) fc’ = 28-day compressive strength of concrete (psi or MPa) Nominal Two-Way Shear Capacity – English Units Smallest of Vnc = Vc = 4b0d fc' ⎛ 4⎞ ⎜ 2 + ⎟b 0d f c' Vnc = Vc = ⎜ ⎝ βc ⎟ ⎠ ⎛ αsd ⎞ Vnc = Vc = ⎜ 2 + ⎜ ⎟b 0d f c' ⎝ 4b 0 ⎟ ⎠ Vnc = nominal shear capacity on one critical surface (lb) βc = long side length cl/short side length cs of columns αs = 40, 30, and 20 for interior, edge, and corner columns b0 = length of one critical surface = c + d (in) c = column width (in or mm) d = effective depth (in) Nominal Two-Way Shear Capacity – SI Units 1 Smallest of Vnc = Vc = b 0 d f c' 3 1⎛ 2⎞ Vnc = Vc = ⎜1 + ⎟b 0d f c' 6 ⎜ βc ⎟ ⎝ ⎠ 1⎛ αsd ⎞ Vnc = Vc = ⎜ 2 + ⎟b 0d f c' 12 ⎜⎝ 4b 0 ⎟⎠ Vnc = nominal shear capacity on one critical surface (N) βc = long side length cl/short side length cs of columns αs = 40, 30, and 20 for interior, edge, and corner columns b0 = length of one critical surface = c + d (mm) c = column width (mm) d = effective depth (mm) Critical Locations for Flexure Critical sections Flexural Design Principles b Neutral axis 0.85fc’ c C a/2 a d d-a/2 fy T=Asfy Nominal Moment Capacity ⎛ a⎞ M n = As fy ⎜ d − ⎟ ⎝ 2⎠ ρdfy As a= ρ= 0.85fc' bd ⎛ 0.59 As f y ⎞ M n = As f y ⎜ d − ⎜ ' ⎟ ⎟ ⎝ bf c ⎠ As = cross-sectional area of reinforcing steel ρ= steel ratio b = width of flexural member Required Cross-Sectional Area of Reinforcing Steel Design Criterion Muc = φM n Required Area of Reinforcing Steel ⎛ f c'b ⎞⎛ ⎞ As = ⎜ ⎟⎜ d − d 2 − 2.355M uc ⎟ ⎜ 1.176 f ⎟⎜ φf c'b ⎟ ⎝ y ⎠⎝ ⎠ Muc = factored bending moment φ = 0.9 for flexure in reinforced concrete b = width of flexural member Factored Bending Moment at Critical Section Pu Mu c Pu l 2 2Mu l Muc = + 2B B b=? l B Location of Critical Section for Bending c c c c/4 cp l l l B B B l = (B − c ) / 2 l = (B − c / 2) / 2 [ ] l = 2B − (c + c p ) / 4 Concrete Masonry Steel Minimum Steel Area Requirement For grade 40 (metric grade 300) steel As > 0.0020 Ag For grade 60 (metric grade 420) steel As > 0.0018 Ag Ag = gross cross-sectional area Design Data for Steel Reinforcing Bars Nominal dimensions Bar size Available Cross-sectional designation grades Diameter, db area, As English SI English SI (in) (mm) (in2) (mm2) #3 #10 40,60 300,420 0.375 9.5 0.11 71 #4 #13 40,60 300,420 0.500 12.7 0.20 129 #5 #16 40,60 300,420 0.625 15.9 0.31 199 #6 #19 40,60 300,420 0.750 19.1 0.44 284 #7 #22 60 420 0.875 22.2 0.60 387 #8 #25 60 420 1.000 25.4 0.79 510 #9 #29 60 420 1.128 28.7 1.00 645 #10 #32 60 420 1.270 32.3 1.27 819 #11 #36 60 420 1.410 35.8 1.56 1006 #14 #43 60 420 1.693 43.0 2.25 1452 #18 #57 60 420 2.257 57.3 4.00 2581 Development Length - Develop proper anchorage c ld l B ld = l − 3in. (or 70mm) Development of Bars in Tension ld 15 f y αβγλ Atr fyt = K tr = d b 16 f c' ⎛ cd + K tr ⎞ 260s ⋅ n SI ⎜ ⎜ d ⎟ ⎟ ⎝ b ⎠ Use Ktr=0 for conservative α = reinforcement location factor β = coating factor γ = reinforcement factor λ = lightweight concrete factor ld = minimum required development length (mm) db = nominal bar diameter (mm) cd = half-spacing of bars or concrete cover dimension (mm) s = max. c-c spacing of transverse reinf. within ld (mm) n = number of bars Atr = total cross-sectional area of all transverse reinf. (mm2) fyt = yield strength of transverse reinforcement (MPa) Development of Bars in Tension ld 3 fy αβγλ Atr fyt = K tr = d b 40 f c' ⎛ cd + K tr ⎞ 1500s ⋅ n ⎜ ⎜ d ⎟ ⎟ English ⎝ b ⎠ Use Ktr=0 for conservative α = reinforcement location factor β = coating factor γ = reinforcement factor λ = lightweight concrete factor ld = minimum required development length (in) db = nominal bar diameter (in) cd = half-spacing of bars or concrete cover dimension (in) s = max. c-c spacing of transverse reinf. within ld (in) n = number of bars Atr = total cross-sectional area of all transverse reinf. (in2) fyt = yield strength of transverse reinforcement (psi) Definition of Transverse Reinforcement Atr Atr = 2Ab Atr = 4Ab Modifiers for Development Length Factor Condition Modifier α Top reinf. (bars w/ 12in. concrete below) 1.3 Bottom bars 1.0 β Epoxy-coated bars w/ cover < 3db or 1.5 Clear spacing < 6db All other epoxy-coated bars 1.2 Uncoated reinforcement 1.0 γ No. 6 and smaller bars & deformed wire 0.8 No. 7 and larger bars 1.0 λ Lightweight-aggregate concrete 1.3 Normal weight concrete 1.0 Development of Bars in Compression ld = λ s lab English db f y lab = 0.02 ≥ 0.0003d b f y f c' ld = minimum required development length (in) lab = basic development length (in) db = nominal bar diameter (in) λs = modifying multiplier Excess reinforcement: λs = required As/provided As Spirally enclosed reinforcement: λs = 0.75 fc’ = compressive strength of concrete (psi) Design for Continuous Footings Continuous Footings c d T d B d Transverse Reinforcement - One-Way Shear c ⎛ B − c − 2d ⎞ Vuc / b = Pu / b⎜ ⎟ ⎝ B ⎠ d d Vuc / b = φVnc / b φ = 0.75 d Pu / b(B − c ) d= English B 48φB f + 2 Pu / b c ' 1500 Pu / b(B − c ) d= SI 500φB f c' + 3Pu / b Transverse Reinforcement - Flexure 450 Zone of compression No transverse steel is needed if the entire base is within a 450 frustum Transverse Reinforcement - Flexure c Pu / bl 2 2M u / bl M uc / b = + 2B B l M uc / b = φM n / b d Required Area of Reinforcing Steel B ⎛ f c'b ⎞⎛ ⎞ As = ⎜ ⎟⎜ d − d 2 − 2.355M uc ⎟ ⎜ 1.176 f ⎟⎜ φf c'b ⎟ ⎝ y ⎠⎝ ⎠ Design for Rectangular Footings Design Philosophy • Similar to those for square footings • Check one-way and two-way to determine minimal d & T • Design long steel bars and evenly distribute them • Design short steel bars & distribute the portion (E) of the total short steel area within the inner zone 2 E= L / B +1 Critical Shear Surfaces One-way shear surface Two-way shear surface Long Steel and Short Steel Long steel Short steel Distribution of Short Steel B Outer zone Inner zone Outer zone B 2 E= L / B +1 L Connections with Superstructures Use of Dowels for Connection Wall steel Concrete or masonry Lap joint column Dowel Design for Compressive Loads Nominal column bearing capacity Pnb = 0.85 f c' A1 on column (use column f c' ) Pnb = 0.85 f c' A1s on footing (use footing f c' ) Design criterion A1 Pu ≤ φPnb Pu = factored column load A2 A1 = cross-sectional area of the column = c2 s = (A2/A1)0.5 < 2 if c + 4d < B; otherwise, s = 1 A2 = (c + 4d)2 φ = 0.7 Concrete or Masonry Columns Use at least four dowels with a total area of steel, As, at least equal to that of the column steel or 0.005 times the cross- sectional area of the column, whichever is greater Dowels may not be larger than #11 bars and must have a 90o hook at the bottom Normally, the number of dowels is equal to the number of vertical bars in the column Design for Moment Loads 1200d b Development length ldh = f c' ldh 12db Design for Shear Loads The minimum required dowel steel area Vu As = Vu ≤ 0.2φf c' Ac φf y μ μ = 0.6 if the cold joint not intentionally roughened or 1.0 if the cold joint is roughened by heavy raking or grooving