R.C. FOUNDATION ANALYSIS, DETAILS & DETAILING

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1 1 R.C. FOUNDATION ANALYSIS, DETAILS & DETAILING By 0 Dec 016

2 PAD FOOTING ANALYSIS & DESIGN (BS8110) FOR PROPOSED PIGS BUILDING FOR PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997) Pad footing details Length of pad footing; L = 100 mm Width of pad footing; B = 100 mm Area of pad footing; A = L B = m Depth of pad footing; h = 300 mm Depth of soil over pad footing; h soil = 850 mm Density of concrete; ρ conc = 4.0 kn/m 3 Column details Column base length; Column base width; Column eccentricity in x; Column eccentricity in y; l A = 300 mm b A = 300 mm e PxA = 0 mm e PyA = 0 mm Soil details Density of soil; ρ soil = 19.0 kn/m 3 Design shear strength; φ = 0.0 deg Design base friction; δ = 15.0 deg

3 3 Allowable bearing pressure; P bearing = 110 kn/m Axial loading on column Dead axial load on column; Imposed axial load on column; Wind axial load on column; Total axial load on column; P GA = 37.0 kn P QA = 18.0 kn P WA = 0.5 kn P A = 55.5 kn Foundation loads Dead surcharge load; F Gsur = kn/m Imposed surcharge load; F Qsur = kn/m Pad footing self weight; F swt = h ρ conc = 7.00 kn/m Soil self weight; F soil = h soil ρ soil = kn/m Total foundation load; F = A (F Gsur + F Qsur + F swt + F soil) = 33.6 kn Horizontal loading on column base Dead horizontal load in x direction; Imposed horizontal load in x direction; Wind horizontal load in x direction; Total horizontal load in x direction; Dead horizontal load in y direction; Imposed horizontal load in y direction; Wind horizontal load in y direction; Total horizontal load in y direction; H GxA = 0.0 kn H QxA = 0.0 kn H WxA = 1.0 kn H xa = 1.0 kn H GyA = 0.0 kn H QyA = 0.0 kn H WyA = 1.0 kn H ya = 1.0 kn Check stability against sliding Resistance to sliding due to base friction H friction = max([p GA + (F Gsur + F swt + F soil) A], 0 kn) tan(δ) = 18.9 kn Passive pressure coefficient; K p = (1 + sin(φ )) /(1- sin(φ )) =.040 Stability against sliding in x direction Passive resistance of soil in x direction; Total resistance to sliding in x direction; H xpas = 0.5 K p (h + h h soil) B ρ soil = 14.0 kn H xres = H friction + H xpas = 3.9 kn PASS - Resistance to sliding is greater than horizontal load in x direction Stability against sliding in y direction Passive resistance of soil in y direction; Total resistance to sliding in y direction; H ypas = 0.5 K p (h + h h soil) L ρ soil = 14.0 kn H yres = H friction + H ypas = 3.9 kn PASS - Resistance to sliding is greater than horizontal load in y direction Check stability against overturning in x direction Total overturning moment; M xot = M xa + H xa h = knm Restoring moment in x direction Foundation loading; Axial loading on column; Total restoring moment; M xsur = A (F Gsur + F swt + F soil) L / = knm M xaxial = (P GA) (L/ - e PxA) =.170 knm M xres = M xsur + M xaxial = knm PASS - Restoring moment is greater than overturning moment in x direction

4 4 Check stability against overturning in y direction Total overturning moment; M yot = M ya + H ya h = knm Restoring moment in y direction Foundation loading; Axial loading on column; Total restoring moment; Calculate pad base reaction Total base reaction; Eccentricity of base reaction in x; Eccentricity of base reaction in y; Check pad base reaction eccentricity M ysur = A (F Gsur + F swt + F soil) B / = knm M yaxial = (P GA) (B/ - e PyA) =.170 knm M yres = M ysur + M yaxial = knm PASS - Restoring moment is greater than overturning moment in y direction T = F + P A = 89.1 kn e Tx = (P A e PxA + M xa + H xa h) / T = 3 mm e Ty = (P A e PyA + M ya + H ya h) / T = 3 mm abs(e Tx) / L + abs(e Ty) / B = Base reaction acts within combined middle third of base Calculate pad base pressures q 1 = T / A - 6 T e Tx /(L A) - 6 T e Ty /(B A) = kn/m q = T / A - 6 T e Tx /(L A) + 6 T e Ty /(B A) = kn/m q 3 = T / A + 6 T e Tx /(L A) - 6 T e Ty /(B A) = kn/m q 4 = T / A + 6 T e Tx /(L A) + 6 T e Ty /(B A) = kn/m Minimum base pressure; q min = min(q 1, q, q 3, q 4) = kn/m Maximum base pressure; q max = max(q 1, q, q 3, q 4) = kn/m PASS - Maximum base pressure is less than allowable bearing pressure 61.9 kn/m 63.9 kn/m 59.8 kn/m 61.9 kn/m

5 5 Partial safety factors for loads Partial safety factor for dead loads; γ fg = 1.40 Partial safety factor for imposed loads; γ fq = 1.60 Partial safety factor for wind loads; γ fw = 0.00 Ultimate axial loading on column Ultimate axial load on column; P ua = P GA γ fg + P QA γ fq + P WA γ fw = 80.5 kn Ultimate foundation loads Ultimate foundation load; F u = A [(F Gsur + F swt + F soil) γ fg + F Qsur γ fq] = 47.1 kn Ultimate horizontal loading on column Ultimate horizontal load in x direction; Ultimate horizontal load in y direction; H xua = H GxA γ fg + H QxA γ fq + H WxA γ fw = 0.0 kn H yua = H GyA γ fg + H QyA γ fq + H WyA γ fw = 0.0 kn Ultimate moment on column Ultimate moment on column in x direction; Ultimate moment on column in y direction; Calculate ultimate pad base reaction Ultimate base reaction; Eccentricity of ultimate base reaction in x; Eccentricity of ultimate base reaction in y; M xua = M GxA γ fg + M QxA γ fq + M WxA γ fw = knm M yua = M GyA γ fg + M QyA γ fq + M WyA γ fw = knm T u = F u + P ua = 17.6 kn e Txu = (P ua e PxA + M xua + H xua h) / T u = 0 mm e Tyu = (P ua e PyA + M yua + H yua h) / T u = 0 mm Calculate ultimate pad base pressures q 1u = T u/a - 6 T u e Txu/(L A) - 6 T u e Tyu/(B A) = kn/m q u = T u/a - 6 T u e Txu/(L A) + 6 T u e Tyu/(B A) = kn/m q 3u = T u/a + 6 T u e Txu/(L A) - 6 T u e Tyu/(B A) = kn/m q 4u = T u/a + 6 T u e Txu/(L A) + 6 T u e Tyu/(B A) = kn/m Minimum ultimate base pressure; q minu = min(q 1u, q u, q 3u, q 4u) = kn/m Maximum ultimate base pressure; q maxu = max(q 1u, q u, q 3u, q 4u) = kn/m Calculate rate of change of base pressure in x direction Left hand base reaction; Right hand base reaction; Length of base reaction; Rate of change of base pressure; Calculate pad lengths in x direction Left hand length; Right hand length; f ul = (q 1u + q u) B / = kn/m f ur = (q 3u + q 4u) B / = kn/m L x = L = 100 mm C x = (f ur - f ul) / L x = kn/m/m L L = L / + e PxA = 600 mm L R = L / - e PxA = 600 mm Calculate ultimate moments in x direction Ultimate moment in x direction; M x = f ul L L / + C x L L 3 / 6 - F u L L / ( L) = knm Calculate rate of change of base pressure in y direction Top edge base reaction; Bottom edge base reaction; Length of base reaction; Rate of change of base pressure; f ut = (q u + q 4u) L / = kn/m f ub = (q 1u + q 3u) L / = kn/m L y = B = 100 mm C y = (f ub - f ut) / L y = kn/m/m

6 6 Calculate pad lengths in y direction Top length; Bottom length; L T = B / - e PyA = 600 mm L B = B / + e PyA = 600 mm Calculate ultimate moments in y direction Ultimate moment in y direction; M y = f ut L T / + C y L T 3 / 6 - F u L T / ( B) = knm Material details Characteristic strength of concrete; f cu = 5 N/mm Characteristic strength of reinforcement; f y = 500 N/mm Characteristic strength of shear reinforcement; f yv = 500 N/mm Nominal cover to reinforcement; c nom = 50 mm Moment design in x direction Diameter of tension reinforcement; Depth of tension reinforcement; φ xb = 16 mm d x = h - c nom - φ xb / = 4 mm Design formula for rectangular beams (cl ) K x = M x /(B d x f cu) = K x = K x < K x' compression reinforcement is not required Lever arm; z x = d x min([0.5 + (0.5 - K x / 0.9)], 0.95) = 30 mm Area of tension reinforcement required; A s_x_req = M x /(0.87 f y z x) = 11 mm Minimum area of tension reinforcement; A s_x_min = B h = 468 mm Tension reinforcement provided; 6 No. 16 dia. bars bottom (5 centres) Area of tension reinforcement provided; A s_xb_prov = N xb π φ xb / 4 = 106 mm PASS - Tension reinforcement provided exceeds tension reinforcement required Moment design in y direction Diameter of tension reinforcement; Depth of tension reinforcement; φ yb = 16 mm d y = h - c nom - φ xb - φ yb / = 6 mm Design formula for rectangular beams (cl ) Lever arm; K y = M y / (L d y f cu) = K y = K y < K y' compression reinforcement is not required z y = d y min([0.5 + (0.5 - K y / 0.9)], 0.95) = 15 mm Area of tension reinforcement required; A s_y_req = M y / (0.87 f y z y) = 19 mm Minimum area of tension reinforcement; A s_y_min = L h = 468 mm Tension reinforcement provided; 6 No. 16 dia. bars bottom (5 centres) Area of tension reinforcement provided; A s_yb_prov = N yb π φ yb / 4 = 106 mm Calculate ultimate shear force at d from top face of column PASS - Tension reinforcement provided exceeds tension reinforcement required Ultimate pressure for shear; q su = (q 1u - C y (B / + e PyA + b A / + d y) / L + q 4u) / q su = kn/m Area loaded for shear; A s = L (B / - e PyA - b A / - d y) = 0.69 m Ultimate shear force; V su = A s (q su - F u / A) = kn

7 7 Shear stresses at d from top face of column (cl ) Design shear stress; v su = V su / (L d y) = N/mm From BS 8110:Part 1: Table 3.8 Design concrete shear stress; v c = 0.79 N/mm min(3, [100 A s_yb_prov / (L d y)] 1/3 ) max((400 mm / d y) 1/4, 0.67) (min(f cu / 1 N/mm, 40) / 5) 1/3 / 1.5 = N/mm Allowable design shear stress; v max = min(0.8 N/mm (f cu / 1 N/mm ), 5 N/mm ) = N/mm Calculate ultimate punching shear force at face of column PASS - v su < v c - No shear reinforcement required Ultimate pressure for punching shear; q pua = q 1u+[(L/+e PxA-l A/)+(l A)/] C x/b-[(b/+e PyA-b A/)+(b A)/] C y/l = kn/m Average effective depth of reinforcement; d = (d x + d y) / = 34 mm Area loaded for punching shear at column; A pa = (l A) (b A) = m Length of punching shear perimeter; Ultimate shear force at shear perimeter; Effective shear force at shear perimeter; Punching shear stresses at face of column (cl ) u pa = (l A)+ (b A) = 100 mm V pua = P ua + (F u / A - q pua) A pa = kn V puaeff = V pua = kn Design shear stress; v pua = V puaeff / (u pa d) = 0.69 N/mm Allowable design shear stress; v max = min(0.8n/mm (f cu / 1 N/mm ), 5 N/mm ) = N/mm Calculate ultimate punching shear force at perimeter of 1.5 d from face of column PASS - Design shear stress is less than allowable design shear stress Ultimate pressure for punching shear; q pua1.5d = q 1u+[L/] C x/b-[(b/+e PyA-b A/-1.5 d)+(b A+ 1.5 d)/] C y/l = kn/m Average effective depth of reinforcement; d = (d x + d y) / = 34 mm Area loaded for punching shear at column; A pa1.5d = L (b A+ 1.5 d) = 1.0 m Length of punching shear perimeter; Ultimate shear force at shear perimeter; Effective shear force at shear perimeter; u pa1.5d = L = 400 mm V pua1.5d = P ua + (F u / A - q pua1.5d) A pa1.5d = kn V pua1.5deff = V pua1.5d 1.5 = kn Punching shear stresses at perimeter of 1.5 d from face of column (cl ) Design shear stress; v pua1.5d = V pua1.5deff / (u pa1.5d d) = N/mm From BS 8110:Part 1: Table 3.8 Design concrete shear stress; v c = 0.79 N/mm min(3, [100 (A s_xb_prov / (B d x) + A s_yb_prov / (L d y)) / ] 1/3 ) max((800 mm / (d x + d y)) 1/4, 0.67) (min(f cu / 1 N/mm, 40) / 5) 1/3 / 1.5 = N/mm Allowable design shear stress; v max = min(0.8n/mm (f cu / 1 N/mm ), 5 N/mm ) = N/mm PASS - v pua1.5d < v c - No shear reinforcement required

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Annex - R C Design Formulae and Data

Annex - R C Design Formulae and Data The design formulae and data provided in this Annex are for education, training and assessment purposes only. They are based on the Hong Kong Code of Practice for Structural Use of Concrete 2013 (HKCP-2013).