RETAINING WALL ANALYSIS

Transcription

1 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: App'd by RETAINING WALL ANALYSIS In accordance with EN997:2004 incorporating Corrigendum dated February 2009 and the recommended values Retaining wall details Stem type; Propped cantilever Stem height; h stem = 5500 mm Prop height; h prop = 4000 mm Stem thickness; t stem = 500 mm Angle to rear face of stem; α = 90 deg Stem density; γ stem = 25 kn/m 3 Toe length; l toe = 2500 mm Heel length; l heel = 2500 mm Base thickness; t base = 500 mm Key position; p key = 550 mm Key depth; d key = 000 mm Key thickness; t key = 350 mm Base density; γ base = 25 kn/m 3 Height of retained soil; h ret = 5000 mm Angle of soil surface; β = 0 deg Depth of cover; d cover = 500 mm Depth of excavation; d exc = 200 mm Height of water; h water = 2500 mm Water density; γ w = 9.8 kn/m 3

2 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: App'd by Surcharge Prop 88 kn/m 2 88 kn/m Retained soil properties Soil type; Soft clay Moist density; γ mr = 7 kn/m 3 Saturated density; γ sr = 7 kn/m 3 Characteristic effective shear resistance angle; φ' r.k = 2 deg Characteristic wall friction angle; δ r.k = 8 deg

3 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Base soil properties Soil type; Stiff clay Moist density; γ mb = 20 kn/m 3 Characteristic cohesion; c' b.k = 25 kn/m 2 Characteristic adhesion; a b.k = 20 kn/m 2 Characteristic effective shear resistance angle; Characteristic wall friction angle; Characteristic base friction angle; Loading details φ' b.k = 8 deg δ b.k = 2 deg δ bb.k = 4 deg Permanent surcharge load; Surcharge G = 2 kn/m 2 Variable surcharge load; Surcharge Q = 5 kn/m 2 Calculate retaining wall geometry Base length; Base height; Saturated soil height; Moist soil height; Length of surcharge load; l base = l toe + t stem + l heel = 5500 mm h base = t base + d key = 500 mm h sat = h water + d cover = 3000 mm h moist = h ret h water = 2500 mm l sur = l heel = 2500 mm Distance to vertical component; x sur_v = l base l heel / 2 = 4250 mm Effective height of wall; h eff = h base + d cover + h ret = 7000 mm Distance to horizontal component; x sur_h = h eff / 2 d key = 2500 mm Area of wall stem; A stem = h stem t stem = 2.75 m 2 Distance to vertical component; x stem = l toe + t stem / 2 = 2750 mm Area of wall base; A base = l base t base + d key t key = 3. m 2 App'd by Distance to vertical component; x base = (l base 2 t base / 2 + d key t key (p key + t key / 2)) / A base = 304 mm Area of saturated soil; A sat = h sat l heel = 7.5 m 2 Distance to vertical component; x sat_v = l base (h sat l heel 2 / 2) / A sat = 4250 mm Distance to horizontal component; x sat_h = (h sat + h base) / 3 d key = 500 mm Area of water; A water = h sat l heel = 7.5 m 2 Distance to vertical component; x water_v = l base (h sat l heel 2 / 2) / A sat = 4250 mm Distance to horizontal component; x water_h = (h sat + h base) / 3 d key = 500 mm Area of moist soil; A moist = h moist l heel = 6.25 m 2 Distance to vertical component; x moist_v = l base (h moist l heel 2 / 2) / A moist = 4250 mm Distance to horizontal component; x moist_h = (h moist (t base + h sat + h moist / 3) / 2 + (h sat + h base) ((h sat + h base)/2 d key)) / (h sat + h base + h moist / 2) = 920 mm Area of base soil; A pass = d cover l toe =.25 m 2 Distance to vertical component; x pass_v = l base (d cover l toe (l base l toe / 2)) / A pass = 250 mm Distance to horizontal component; x pass_h = (d cover + h base) / 3 d key = 333 mm Area of excavated base soil; A exc = h pass l toe = 0.75 m 2 Distance to vertical component; x exc_v = l base (h pass l toe (l base l toe / 2)) / A exc = 250 mm Distance to horizontal component; x exc_h = (h pass + h base) / 3 d key = 400 mm

4 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Partial factors on actions Table A.3 Combination Permanent unfavourable action; γ G =.35 Permanent favourable action; γ Gf =.00 Variable unfavourable action; γ Q =.50 Variable favourable action; γ Qf = 0.00 Partial factors for soil parameters Table A.4 Combination Angle of shearing resistance; γ φ' =.00 Effective cohesion; γ c' =.00 Weight density; γ γ =.00 App'd by Retained soil properties Design effective shear resistance angle; Design wall friction angle; φ' r.d = atan(tan(φ' r.k) / γ φ') = 2 deg δ r.d = atan(tan(δ r.k) / γ φ') = 8 deg Base soil properties Design effective shear resistance angle; φ' b.d = atan(tan(φ' b.k) / γ φ') = 8 deg Design wall friction angle; δ b.d = atan(tan(δ b.k) / γ φ') = 2 deg Design base friction angle; δ bb.d = atan(tan(δ bb.k) / γ φ') = 4 deg Design effective cohesion; c' b.d = c' b.k / γ c' = 25 kn/m 2 Design adhesion; a b.d = a b.k / γ c' = 20 kn/m 2 Using Coulomb theory Active pressure coefficient; K A = sin(α + φ' r.d) 2 / (sin(α) 2 sin(α δ r.d) [ + [sin(φ' r.d + δ r.d) sin(φ' r.d β) / (sin(α δ r.d) sin(α + β))]] 2 ) = 0.60 Passive pressure coefficient; K P = sin(90 φ' b.d) 2 / (sin(90 + δ b.d) [ [sin(φ' b.d + δ b.d) sin(φ' b.d) / (sin(90 + δ b.d))]] 2 ) = Bearing pressure check Vertical forces on wall Wall stem; F stem = γ G A stem γ stem = 92.8 kn/m Wall base; F base = γ G A base γ base = 04.6 kn/m Surcharge load; F sur_v = (γ G Surcharge G + γ Q Surcharge Q) l heel = 96.8 kn/m Saturated retained soil; F sat_v = γ G A sat (γ sr γ w) = 72.8 kn/m Water; F water_v = γ G A water γ w = 99.3 kn/m Moist retained soil; F moist_v = γ G A moist γ mr = 43.4 kn/m Base soil; F pass_v = γ G A pass γ mb = 33.8 kn/m Total; F total_v = F stem + F base + F sat_v + F moist_v + F pass_v + F water_v + F sur_v = kn/m Horizontal forces on wall Surcharge load; F sur_h = K A cos(δ r.d) (γ G Surcharge G + γ Q Surcharge Q) h eff = 6.2 kn/m Saturated retained soil; F sat_h = γ G K A cos(δ r.d) (γ sr γ w) (h sat + h base) 2 / 2 = 58.5 kn/m Water; F water_h = γ G γ w (h water + d cover + h base) 2 / 2 = 34. kn/m

5 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: App'd by Moist retained soil; F moist_h = γ G K A cos(δ r.d) γ mr ((h eff h sat h base) 2 / 2 + (h eff h sat Total; Moments on wall Wall stem; Wall base; Surcharge load; Saturated retained soil; Water; Moist retained soil; Base soil; h base) (h sat + h base)) = 96.3 kn/m F total_h = F sat_h + F moist_h + F water_h + F sur_h = 550. kn/m M stem = F stem x stem = knm/m M base = F base x base = 38. knm/m M sur = F sur_v x sur_v F sur_h x sur_h = 8.2 knm/m M sat = F sat_v x sat_v F sat_h x sat_h = knm/m M water = F water_v x water_v F water_h x water_h = 355. knm/m M moist = F moist_v x moist_v F moist_h x moist_h = knm/m M pass = F pass_v x pass_v = 42.2 knm/m Total; M total = M stem + M base + M sat + M moist + M pass + M water + M sur = 49.6 Check bearing pressure knm/m Maximum friction force; Maximum base soil resistance; F friction_max = F total_v tan(δ bb.d) = 60.4 kn/m F pass_h_max = γ Gf K P cos(δ b.d) γ mb (d cover + h base) 2 / 2 = 99.6 kn/m Base soil resistance; F pass_h = min(max((m total + F total_h (h prop + t base) + F friction_max (h prop + t base) F total_v l base / 2) / (x pass_h h prop t base), 0 kn/m), F pass_h_max) = 0 kn/m Propping force; F prop_stem = min((f total_v l base / 2 M total) / (h prop + t base), F total_h) = 6.8 kn/m Friction force; F friction = F total_h F pass_h F prop_stem = kn/m Moment from propping force; M prop = F prop_stem (h prop + t base) = 278. knm/m Distance to reaction; x = (M total + M prop) / F total_v = 2750 mm Eccentricity of reaction; e = x l base / 2 = 0 mm Loaded length of base; l load = l base = 5500 mm Bearing pressure at toe; q toe = F total_v / l base = 7 kn/m 2 Bearing pressure at heel; q heel = F total_v / l base = 7 kn/m 2 Effective overburden pressure; q = max((t base + d cover) γ mb (t base + d cover + h water) γ w, 0 kn/m 2 ) = 0 kn/m 2 Design effective overburden pressure; q' = q / γ γ = 0 kn/m 2 Bearing resistance factors; N q = Exp(π tan(φ' b.d)) (tan(45 deg + φ' b.d / 2)) 2 = N c = (N q ) cot(φ' b.d) = 3.04 N γ = 2 (N q ) tan(φ' b.d) = Foundation shape factors; s q = s γ = s c = Load inclination factors; H = F total_h F prop_stem F friction = 0 kn/m V = F total_v = kn/m m = 2 i q = [ H / (V + l load c' b.d cot(φ' b.d))] m =

6 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Net ultimate bearing capacity; App'd by i γ = [ H / (V + l load c' b.d cot(φ' b.d))] (m + ) = i c = i q ( i q) / (N c tan(φ' b.d)) = n f = c' b.d N c s c i c + q' N q s q i q (γ mb γ w) l load N γ s γ i γ = 405. kn/m 2 Factor of safety; FoS bp = n f / max(q toe, q heel) = Partial factors on actions Table A.3 Combination 2 Permanent unfavourable action; γ G =.00 Permanent favourable action; γ Gf =.00 Variable unfavourable action; γ Q =.30 Variable favourable action; γ Qf = 0.00 Partial factors for soil parameters Table A.4 Combination 2 Angle of shearing resistance; γ φ' =.25 Effective cohesion; γ c' =.25 Weight density; γ γ =.00 PASS Allowable bearing pressure exceeds maximum applied bearing pressure Retained soil properties Design effective shear resistance angle; Design wall friction angle; φ' r.d = atan(tan(φ' r.k) / γ φ') = 9.7 deg δ r.d = atan(tan(δ r.k) / γ φ') = 6.4 deg Base soil properties Design effective shear resistance angle; φ' b.d = atan(tan(φ' b.k) / γ φ') = 4.6 deg Design wall friction angle; δ b.d = atan(tan(δ b.k) / γ φ') = 9.7 deg Design base friction angle; δ bb.d = atan(tan(δ bb.k) / γ φ') =.3 deg Design effective cohesion; c' b.d = c' b.k / γ c' = 20 kn/m 2 Design adhesion; a b.d = a b.k / γ c' = 6 kn/m 2 Using Coulomb theory Active pressure coefficient; K A = sin(α + φ' r.d) 2 / (sin(α) 2 sin(α δ r.d) [ + [sin(φ' r.d + δ r.d) sin(φ' r.d β) / (sin(α δ r.d) sin(α + β))]] 2 ) = 0.66 Passive pressure coefficient; K P = sin(90 φ' b.d) 2 / (sin(90 + δ b.d) [ [sin(φ' b.d + δ b.d) sin(φ' b.d) / (sin(90 + δ b.d))]] 2 ) = Bearing pressure check Vertical forces on wall Wall stem; Wall base; Surcharge load; Saturated retained soil; Water; Moist retained soil; Base soil; F stem = γ G A stem γ stem = 68.8 kn/m F base = γ G A base γ base = 77.5 kn/m F sur_v = (γ G Surcharge G + γ Q Surcharge Q) l heel = 78.8 kn/m F sat_v = γ G A sat (γ sr γ w) = 53.9 kn/m F water_v = γ G A water γ w = 73.6 kn/m F moist_v = γ G A moist γ mr = 06.3 kn/m F pass_v = γ G A pass γ mb = 25 kn/m

7 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: App'd by Total; F total_v = F stem + F base + F sat_v + F moist_v + F pass_v + F water_v + F sur_v = Horizontal forces on wall kn/m Surcharge load; F sur_h = K A cos(δ r.d) (γ G Surcharge G + γ Q Surcharge Q) h eff = Saturated retained soil; Water; 44.9 kn/m F sat_h = γ G K A cos(δ r.d) (γ sr γ w) (h sat + h base) 2 / 2 = 47.8 kn/m F water_h = γ G γ w (h water + d cover + h base) 2 / 2 = 99.3 kn/m Moist retained soil; F moist_h = γ G K A cos(δ r.d) γ mr ((h eff h sat h base) 2 / 2 + (h eff h sat Total; Moments on wall Wall stem; Wall base; Surcharge load; Saturated retained soil; Water; Moist retained soil; Base soil; h base) (h sat + h base)) = 60.6 kn/m F total_h = F sat_h + F moist_h + F water_h + F sur_h = kn/m M stem = F stem x stem = 89. knm/m M base = F base x base = knm/m M sur = F sur_v x sur_v F sur_h x sur_h = 27.6 knm/m M sat = F sat_v x sat_v F sat_h x sat_h = knm/m M water = F water_v x water_v F water_h x water_h = 263 knm/m M moist = F moist_v x moist_v F moist_h x moist_h = 43.2 knm/m M pass = F pass_v x pass_v = 3.3 knm/m Total; M total = M stem + M base + M sat + M moist + M pass + M water + M sur = Check bearing pressure Maximum friction force; Maximum base soil resistance; knm/m F friction_max = F total_v tan(δ bb.d) = 96.5 kn/m F pass_h_max = γ Gf K P cos(δ b.d) γ mb (d cover + h base) 2 / 2 = 8.9 kn/m Base soil resistance; F pass_h = min(max((m total + F total_h (h prop + t base) + F friction_max (h prop + t base) F total_v l base / 2) / (x pass_h h prop t base), 0 kn/m), F pass_h_max) = 0 kn/m Propping force; F prop_stem = min((f total_v l base / 2 M total) / (h prop + t base), F total_h) = 64.6 Friction force; Moment from propping force; Distance to reaction; Eccentricity of reaction; kn/m F friction = F total_h F pass_h F prop_stem = 388. kn/m M prop = F prop_stem (h prop + t base) = knm/m x = (M total + M prop) / F total_v = 2750 mm e = x l base / 2 = 0 mm Loaded length of base; l load = l base = 5500 mm Bearing pressure at toe; q toe = F total_v / l base = 88 kn/m 2 Bearing pressure at heel; q heel = F total_v / l base = 88 kn/m 2 Effective overburden pressure; q = max((t base + d cover) γ mb (t base + d cover + h water) γ w, 0 kn/m 2 ) = 0 kn/m 2 Design effective overburden pressure; q' = q / γ γ = 0 kn/m 2 Bearing resistance factors; N q = Exp(π tan(φ' b.d)) (tan(45 deg + φ' b.d / 2)) 2 = N c = (N q ) cot(φ' b.d) = 0.7 N γ = 2 (N q ) tan(φ' b.d) =.447

8 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Foundation shape factors; s q = Load inclination factors; Net ultimate bearing capacity; s γ = s c = H = F total_h F prop_stem F friction = 0 kn/m V = F total_v = kn/m m = 2 i q = [ H / (V + l load c' b.d cot(φ' b.d))] m = App'd by i γ = [ H / (V + l load c' b.d cot(φ' b.d))] (m + ) = i c = i q ( i q) / (N c tan(φ' b.d)) = n f = c' b.d N c s c i c + q' N q s q i q (γ mb γ w) l load N γ s γ i γ = kn/m 2 Factor of safety; FoS bp = n f / max(q toe, q heel) = PASS Allowable bearing pressure exceeds maximum applied bearing pressure RETAINING WALL DESIGN In accordance with EN992:2004 incorporating Corrigendum dated January 2008 and the recommended values Concrete details Table 3. Strength and deformation characteristics for concrete Concrete strength class; C40/50 Characteristic compressive cylinder strength; f ck = 40 N/mm 2 Characteristic compressive cube strength; f ck,cube = 50 N/mm 2 Mean value of compressive cylinder strength; f cm = f ck + 8 N/mm 2 = 48 N/mm 2 Mean value of axial tensile strength; f ctm = 0.3 N/mm 2 (f ck / N/mm 2 ) 2/3 = 3.5 N/mm 2 5% fractile of axial tensile strength; f ctk,0.05 = 0.7 f ctm = 2.5 N/mm 2 Secant modulus of elasticity of concrete; E cm = 22 kn/mm 2 (f cm / 0 N/mm 2 ) 0.3 = N/mm 2 Partial factor for concrete Table 2.N; γ C =.50 Compressive strength coefficient cl.3..6(); α cc =.00 Design compressive concrete strength exp.3.5; f cd = α cc f ck / γ C = 26.7 N/mm 2 Maximum aggregate size; Reinforcement details h agg = 20 mm Characteristic yield strength of reinforcement; f yk = 500 N/mm 2 Modulus of elasticity of reinforcement; E s = N/mm 2 Partial factor for reinforcing steel Table 2.N; γ S =.5 Design yield strength of reinforcement; f yd = f yk / γ S = 435 N/mm 2 Cover to reinforcement Front face of stem; Rear face of stem; Top face of base; Bottom face of base; Check stem design for maximum moment Depth of section; c sf = 40 mm c sr = 50 mm c bt = 50 mm c bb = 75 mm h = 500 mm

9 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Rectangular section in flexure 6. Design bending moment; Depth to tension reinforcement; Lever arm; Depth of neutral axis; M = 04.8 knm/m d = h c sr φ sr / 2 = 442 mm K = M / (d 2 f ck) = 0.03 K' = 0.96 App'd by K' > K No compression reinforcement is required z = min( ( 3.53 K) 0.5, 0.95) d = 420 mm x = 2.5 (d z) = 55 mm Area of tension reinforcement required; A sr.req = M / (f yd z) = 574 mm 2 /m Tension reinforcement provided; c/c Area of tension reinforcement provided; A sr.prov = π φ sr 2 / (4 s sr) = 005 mm 2 /m Minimum area of reinforcement exp.9.n; A sr.min = max(0.26 f ctm / f yk, 0.003) d = 806 mm 2 /m Maximum area of reinforcement cl.9.2..(3); A sr.max = 0.04 h = mm 2 /m Crack control 7.3 Limiting crack width; max(a sr.req, A sr.min) / A sr.prov = PASS Area of reinforcement provided is greater than area of reinforcement required w max = 0.3 mm Variable load factor EN990 Table A.; ψ 2 = 0.6 Serviceability bending moment; M sls = 7.7 knm/m Tensile stress in reinforcement; σ s = M sls / (A sr.prov z) = 69.8 N/mm 2 Load duration; Long term Load duration factor; k t = 0.4 Effective area of concrete in tension; A c.eff = min(2.5 (h d), (h x) / 3, h / 2) = mm 2 /m Mean value of concrete tensile strength; f ct.eff = f ctm = 3.5 N/mm 2 Reinforcement ratio; ρ p.eff = A sr.prov / A c.eff = Modular ratio; α e = E s / E cm = Bond property coefficient; k = 0.8 Strain distribution coefficient; k 2 = 0.5 Maximum crack spacing exp.7.; Maximum crack width exp.7.8; Rectangular section in shear 6.2 Design shear force; k 3 = 3.4 k 4 = s r.max = k 3 c sr + k k 2 k 4 φ sr / ρ p.eff = 562 mm w k = s r.max max(σ s k t (f ct.eff / ρ p.eff) ( + α e ρ p.eff), 0.6 σ s) / E s w k = mm w k / w max = PASS Maximum crack width is less than limiting crack width V = 95.7 kn/m C Rd,c = 0.8 / γ C = 0.20 k = min( + (200 mm / d), 2) =.673 Longitudinal reinforcement ratio; ρ l = min(a sr.prov / d, 0.02) = v min = N /2 /mm k 3/2 f ck 0.5 = N/mm 2

10 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Design shear resistance exp.6.2a & 6.2b; Rectangular section in flexure 6. Design bending moment; Depth to tension reinforcement; Lever arm; Depth of neutral axis; App'd by V Rd.c = max(c Rd.c k (00 N 2 /mm 4 ρ l f ck) /3, v min) d V Rd.c = 2.7 kn/m V / V Rd.c = PASS Design shear resistance exceeds design shear force M = 72.6 knm/m d = h c sf φ sx φ sf / 2 = 442 mm K = M / (d 2 f ck) = K' = 0.96 K' > K No compression reinforcement is required z = min( ( 3.53 K) 0.5, 0.95) d = 420 mm x = 2.5 (d z) = 55 mm Area of tension reinforcement required; A sf.req = M / (f yd z) = 398 mm 2 /m Tension reinforcement provided; c/c Area of tension reinforcement provided; A sf.prov = π φ sf 2 / (4 s sf) = 894 mm 2 /m Minimum area of reinforcement exp.9.n; A sf.min = max(0.26 f ctm / f yk, 0.003) d = 806 mm 2 /m Maximum area of reinforcement cl.9.2..(3); A sf.max = 0.04 h = mm 2 /m Crack control 7.3 Limiting crack width; max(a sf.req, A sf.min) / A sf.prov = PASS Area of reinforcement provided is greater than area of reinforcement required w max = 0.3 mm Variable load factor EN990 Table A.; ψ 2 = 0.6 Serviceability bending moment; M sls = 49.5 knm/m Tensile stress in reinforcement; σ s = M sls / (A sf.prov z) = 3.8 N/mm 2 Load duration; Long term Load duration factor; k t = 0.4 Effective area of concrete in tension; A c.eff = min(2.5 (h d), (h x) / 3, h / 2) = mm 2 /m Mean value of concrete tensile strength; f ct.eff = f ctm = 3.5 N/mm 2 Reinforcement ratio; ρ p.eff = A sf.prov / A c.eff = Modular ratio; α e = E s / E cm = Bond property coefficient; k = 0.8 Strain distribution coefficient; k 2 = 0.5 Maximum crack spacing exp.7.; Maximum crack width exp.7.8; k 3 = 3.4 k 4 = s r.max = k 3 c sf + k k 2 k 4 φ sf / ρ p.eff = 577 mm w k = s r.max max(σ s k t (f ct.eff / ρ p.eff) ( + α e ρ p.eff), 0.6 σ s) / E s w k = 0.27 mm w k / w max = PASS Maximum crack width is less than limiting crack width Rectangular section in shear 6.2 Design shear force; V = 95.7 kn/m

11 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: C Rd,c = 0.8 / γ C = 0.20 k = min( + (200 mm / d), 2) =.673 Longitudinal reinforcement ratio; ρ l = min(a sf.prov / d, 0.02) = Design shear resistance exp.6.2a & 6.2b; App'd by v min = N /2 /mm k 3/2 f ck 0.5 = N/mm 2 V Rd.c = max(c Rd.c k (00 N 2 /mm 4 ρ l f ck) /3, v min) d V Rd.c = 2.7 kn/m V / V Rd.c = Horizontal reinforcement parallel to face of stem 9.6 PASS Design shear resistance exceeds design shear force Minimum area of reinforcement cl.9.6.3(); A sx.req = max(0.25 A sr.prov, 0.00 t stem) = 500 mm 2 /m Maximum spacing of reinforcement cl.9.6.3(2); Transverse reinforcement provided; s sx_max = 400 mm c/c Area of transverse reinforcement provided; A sx.prov = π φ sx 2 / (4 s sx) = 393 mm 2 /m FAIL Area of reinforcement provided is less than area of reinforcement required Check base design Depth of section; h = 500 mm Rectangular section in flexure 6. Design bending moment at toe; Depth to tension reinforcement; Lever arm; Depth of neutral axis; M = knm/m d = h c bb φ bb / 2 = 45 mm K = M / (d 2 f ck) = K' = 0.96 K' > K No compression reinforcement is required z = min( ( 3.53 K) 0.5, 0.95) d = 394 mm x = 2.5 (d z) = 52 mm Area of tension reinforcement required; A bb.req = M / (f yd z) = 579 mm 2 /m Tension reinforcement provided; c/c Area of tension reinforcement provided; A bb.prov = π φ bb 2 / (4 s bb) = 253 mm 2 /m Minimum area of reinforcement exp.9.n; A bb.min = max(0.26 f ctm / f yk, 0.003) d = 757 mm 2 /m Maximum area of reinforcement cl.9.2..(3); A bb.max = 0.04 h = mm 2 /m Crack control 7.3 Limiting crack width; max(a bb.req, A bb.min) / A bb.prov = PASS Area of reinforcement provided is greater than area of reinforcement required w max = 0.3 mm Variable load factor EN990 Table A.; ψ 2 = 0.6 Serviceability bending moment; M sls = 89.6 knm/m Tensile stress in reinforcement; σ s = M sls / (A bb.prov z) = 9.4 N/mm 2 Load duration; Long term Load duration factor; k t = 0.4 Effective area of concrete in tension; A c.eff = min(2.5 (h d), (h x) / 3, h / 2) = mm 2 /m Mean value of concrete tensile strength; f ct.eff = f ctm = 3.5 N/mm 2 Reinforcement ratio; ρ p.eff = A bb.prov / A c.eff = 0.07

12 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Modular ratio; α e = E s / E cm = Bond property coefficient; k = 0.8 Strain distribution coefficient; k 2 = 0.5 Maximum crack spacing exp.7.; Maximum crack width exp.7.8; Rectangular section in shear 6.2 Design shear force; k 3 = 3.4 k 4 = App'd by s r.max = k 3 c bb + k k 2 k 4 φ bb / ρ p.eff = 457 mm w k = s r.max max(σ s k t (f ct.eff / ρ p.eff) ( + α e ρ p.eff), 0.6 σ s) / E s w k = 0.25 mm w k / w max = PASS Maximum crack width is less than limiting crack width V = 26.6 kn/m C Rd,c = 0.8 / γ C = 0.20 k = min( + (200 mm / d), 2) =.694 Longitudinal reinforcement ratio; ρ l = min(a bb.prov / d, 0.02) = Design shear resistance exp.6.2a & 6.2b; Rectangular section in flexure 6. Design bending moment at heel; Depth to tension reinforcement; Lever arm; Depth of neutral axis; v min = N /2 /mm k 3/2 f ck 0.5 = N/mm 2 V Rd.c = max(c Rd.c k (00 N 2 /mm 4 ρ l f ck) /3, v min) d V Rd.c = 244. kn/m V / V Rd.c = PASS Design shear resistance exceeds design shear force M = 70.6 knm/m d = h c bt φ bt / 2 = 440 mm K = M / (d 2 f ck) = K' = 0.96 K' > K No compression reinforcement is required z = min( ( 3.53 K) 0.5, 0.95) d = 48 mm x = 2.5 (d z) = 55 mm Area of tension reinforcement required; A bt.req = M / (f yd z) = 939 mm 2 /m Tension reinforcement provided; c/c Area of tension reinforcement provided; A bt.prov = π φ bt 2 / (4 s bt) = 795 mm 2 /m Minimum area of reinforcement exp.9.n; A bt.min = max(0.26 f ctm / f yk, 0.003) d = 803 mm 2 /m Maximum area of reinforcement cl.9.2..(3); A bt.max = 0.04 h = mm 2 /m Crack control 7.3 Limiting crack width; max(a bt.req, A bt.min) / A bt.prov = PASS Area of reinforcement provided is greater than area of reinforcement required w max = 0.3 mm Variable load factor EN990 Table A.; ψ 2 = 0.6 Serviceability bending moment; M sls = 56.3 knm/m Tensile stress in reinforcement; σ s = M sls / (A bt.prov z) = N/mm 2 Load duration; Long term

13 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Load duration factor; k t = 0.4 App'd by Effective area of concrete in tension; A c.eff = min(2.5 (h d), (h x) / 3, h / 2) = mm 2 /m Mean value of concrete tensile strength; f ct.eff = f ctm = 3.5 N/mm 2 Reinforcement ratio; ρ p.eff = A bt.prov / A c.eff = 0.02 Modular ratio; α e = E s / E cm = Bond property coefficient; k = 0.8 Strain distribution coefficient; k 2 = 0.5 Maximum crack spacing exp.7.; Maximum crack width exp.7.8; Rectangular section in shear 6.2 Design shear force; k 3 = 3.4 k 4 = s r.max = k 3 c bt + k k 2 k 4 φ bt / ρ p.eff = 45 mm w k = s r.max max(σ s k t (f ct.eff / ρ p.eff) ( + α e ρ p.eff), 0.6 σ s) / E s w k = mm w k / w max = PASS Maximum crack width is less than limiting crack width V = 73.8 kn/m C Rd,c = 0.8 / γ C = 0.20 k = min( + (200 mm / d), 2) =.674 Longitudinal reinforcement ratio; ρ l = min(a bt.prov / d, 0.02) = Design shear resistance exp.6.2a & 6.2b; v min = N /2 /mm k 3/2 f ck 0.5 = N/mm 2 V Rd.c = max(c Rd.c k (00 N 2 /mm 4 ρ l f ck) /3, v min) d V Rd.c = kn/m V / V Rd.c = Secondary transverse reinforcement to base 9.3 Minimum area of reinforcement cl.9.3..(2); A bx.req = 0.2 A bb.prov = 503 mm 2 /m Maximum spacing of reinforcement cl.9.3..(3); s bx_max = 450 mm Transverse reinforcement provided; PASS Design shear resistance exceeds design shear force c/c Area of transverse reinforcement provided; A bx.prov = π φ bx 2 / (4 s bx) = 565 mm 2 /m Check key design Depth of section; Rectangular section in flexure 6. Design bending moment at key; Depth to tension reinforcement; Lever arm; Depth of neutral axis; PASS Area of reinforcement provided is greater than area of reinforcement required h = 350 mm M = 7.7 knm/m d = h c bb φ k / 2 = 269 mm K = M / (d 2 f ck) = K' = 0.96 K' > K No compression reinforcement is required z = min( ( 3.53 K) 0.5, 0.95) d = 256 mm x = 2.5 (d z) = 34 mm Area of tension reinforcement required; A k.req = M / (f yd z) = 60 mm 2 /m

14 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: Tension reinforcement provided; c/c Area of tension reinforcement provided; A k.prov = π φ k 2 / (4 s k) = 565 mm 2 /m App'd by Minimum area of reinforcement exp.9.n; A k.min = max(0.26 f ctm / f yk, 0.003) d = 49 mm 2 /m Maximum area of reinforcement cl.9.2..(3); A k.max = 0.04 h = 4000 mm 2 /m Crack control 7.3 Limiting crack width; max(a k.req, A k.min) / A k.prov = PASS Area of reinforcement provided is greater than area of reinforcement required w max = 0.3 mm Variable load factor EN990 Table A.; ψ 2 = 0.6 Serviceability bending moment; M sls = 0 knm/m Tensile stress in reinforcement; σ s = M sls / (A k.prov z) = 0 N/mm 2 Load duration; Long term Load duration factor; k t = 0.4 Effective area of concrete in tension; A c.eff = min(2.5 (h d), (h x) / 3, h / 2) = mm 2 /m Mean value of concrete tensile strength; f ct.eff = f ctm = 3.5 N/mm 2 Reinforcement ratio; ρ p.eff = A k.prov / A c.eff = Modular ratio; α e = E s / E cm = Bond property coefficient; k = 0.8 Strain distribution coefficient; k 2 = 0.5 Maximum crack spacing exp.7.; k 3 = 3.4 k 4 = s r.max = k 3 c bb + k k 2 k 4 φ k / ρ p.eff = 635 mm Maximum crack width exp.7.8; w k = s r.max max(σ s k t (f ct.eff / ρ p.eff) ( + α e ρ p.eff), 0.6 σ s) / E s w k = 0 mm w k / w max = 0 PASS Maximum crack width is less than limiting crack width Rectangular section in shear 6.2 Design shear force; V = 40 kn/m C Rd,c = 0.8 / γ C = 0.20 k = min( + (200 mm / d), 2) =.862 Longitudinal reinforcement ratio; ρ l = min(a k.prov / d, 0.02) = v min = N /2 /mm k 3/2 f 0.5 ck = N/mm 2 Design shear resistance exp.6.2a & 6.2b; V Rd.c = max(c Rd.c k (00 N 2 /mm 4 ρ l f ck) /3, v min) d V Rd.c = 5.3 kn/m V / V Rd.c = PASS Design shear resistance exceeds design shear force

15 Retaining Wall Analysis Example (EN997:2004) GEODOMISI Ltd. Dr. Costas Sachpazis Consulting Company for Tel.: (+30) , Fax.: App'd by