DESIGN OF CANTILEVER RETAINING WALL WITH M HEIGHT WITH INDIAN STANDARD

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1 DESIGN OF CANTILEVER RETAINING WALL WITH M HEIGHT WITH INDIAN STANDARD Nisarg Modi 1 1 (RKDF Institute of Science and Technology, SRK University. Bhopal (M.P), India) Abstract Retaining structures hold back soil or other loose material where an abrupt change in ground elevation occurs. The retained material or backfill material is push on the vertical member and that reason of wall is overturn or in some case slide it or both. Different type of Retaining wall is constructed like Gravity Retaining wall, Semi Gravity Retaining wall, Cantilevered retaining wall and counterfort retaining wall. The cantilever is the most common type of retaining wall is constructed in site. Here in this article presents Analysis and Design Cantilevered Retaining wall which is made from an internal stem of steel-reinforced, cast-in-place concrete. Detailed analyses and design for this type Retaining of walls which include estimation of primary basic dimensions of the Retaining wall, as per the adopt this dimension through Design of Retaining wall and must be calculated the different checks for design like stability, overturning and many other in this paper. The factor of safety against sliding, overturning and bearing were calculated. The shear resistance for the base, the tension stresses in the stem and the tension stresses for the base were checked. Here in the design of cantilever Retaining wall in the only Vertical member which is called Stem is only design and as per analyze the member Calculation of reinforcement for stem of the wall were done. Keywords Retaining wall, Cantilever, Vertical Stability, shear, Crack and Stress 1. INTRODUCTION Retaining wall is a such type of structure used for maintaining the ground surfaces at different heights on either side for it. Retaining wall provide lateral support to vertical slopes to soil. They are many types of materials that can be used to create retaining walls like concrete blocks. Poured concrete, treated concrete rocks or boulders. The material is retained or supported by wall so this type of material called backfill. this may have its top surface Retaining wall to Horizontal or in some case inclined. The Backfill material is lying above the horizontal plane at the of top of retaining wall is a called surcharge & its inclination to the horizontal is called as Surcharge angle. Retaining walls have primary function of retaining soil at an angle of in excess of the soils nature angle of repose. The retained soil is referred to as a backfill. Walls within the design height range are designed to provide the necessary resistance by either their own mass or by the principle of leverage. Retaining walls are such a type of structures to be designed the restrain soil to a slope that it would not naturally keep to (typically a steep, near-vertical or vertical slope). A basement wall is called a cantilever retaining wall, which is a freestanding vertical Member without lateral support at its top. The Vertical member must be resist the lateral pressures which is mostly generated by loose soils. Always Retaining wall is supports a wedge of soil. The soil which is above a limit of the failure plane of the soil type present at a Backward and Forward portion of wall and calculated once the soil friction angle is known as a wedge of retaining wall. The most important consideration in retaining walls is a retained material to move downslope due to gravity force. In behind of Retaining wall in Backfill material is a develop lateral or Horizontal earth pressure which is depends on the cohesive strength of soil and angle of internal friction of the Backfill or retained material, the direction and magnitude of movement the retaining structure moves undergoes. Earth pressures behind the wall will push the wall forward or overturn it if not properly set. Also, if ground water behind the wall or underside of wall that is not dissipated by a drainage system causes hydrostatic pressure on the wall. In design of retaining wall the taking some consideration is overturning of the wall does not occur, forward sliding does not occur, materials used are suitable and subsoil is not overloaded. 2. TYPE OF RETAINING WALL 3. OBJECTIVE THE STUDY This paper shows the analysis and design of the cantilever retaining wall. In this Paper the different steps of Analysis of Bending Moment, Lateral & Vertical Force for Active Earth Pressure and Live Load Surcharge and Self-weight of abutment & load due to Backfill at base of stem. In this type of Vertical member check the vertical stability and Verification of Structural Strength at abutment shaft bottom for design. After all this procedure carried out the Design of Volume: 04 Issue:

2 Abutment shaft and different design check like Shear, stress and Cracking in this type of member. 4. DATA OF RETAINING WALL Width of the Bridge (m) Skew angle with vertical, θ (degrees) Angle of Repose of Soil, f (degrees) Bulk Density of Soil (dry), g b (t/m3) Inclination of Backfill with horizontal, b (degrees) Equiv. Ht. of backfill earth for Live load surcharge (m) Coefficient of Friction at base between soil & concrete Unit weight of concrete (in t/m3) s Concrete Grade (MPa) Grade of Steel Allowable Bearing Capacity (t/m 2 ) Passive Earth Pressure to be considered (1 for Yes, 2 for No) Angle of Repose of River Soil, f (degrees) Clear Cover to reinforcement in foundation (mm) Dia. of main reinforcement in foundation (mm) Clear cover to reinf. in stem (mm) Dia. of main reinforcement in stem (mm) 18 Top R.L. of Retaining wall (m) G.L. of abutment (m) Foundation R.L. of abutment (m) Live Load Surcharge ( 1 for YES, 0 for NO) Fig I Section of Retaining wall 1.0 Lateral and Vertical Loads due to Active Earth Pressur Horizontal component of active earth pressure = 0.5*ka*g*h2*W*cos(a+d) Vertical component of active earth pressure = 0.5*ka*g*h2*W*sin(a+d) According to Clause , IRC : , ka = 0.794= x x Active earth pressure in dry condition 1 Active earth pressure at Bottom of Stem 2.4 m H= Horizontal component Pah2 = 0.5x0.32x2.00x5.73^2xcos( ) = t acting at 0.42h = m Vertical component Pav2 = 0.5x0.32x2.00x5.73^2xsin( ) = t acting at x = m Net Moment = (9.659x2.407) - (4.099x0.300) = 22 t-m 2 Active earth pressure at Base of Foundation m H = m Horizontal component Pah3 = 0.5x0.32x2.00x6.63^2xcos( ) = t acting at 0.42h = m Vertical component Pav3 = 0.5x0.32x2.00x6.63^2xsin( ) = t acting at x = m Net Moment = (12.931x2.785) - (5.488x-0.100) = t-m 2.0 Lateral and Vertical Loads due to Live Load Surcharge Live Load surcharge in dry condition 1 Live load surcharge at Base of Stem m h = m Horizontal component Ph1 = 0.32x2.00x1.20x5.73x cos( ) = 4.04 t acting at h/2 = m Vertical component Pv1 = 0.32x2.00x1.20x5.73x sin( ) = 1.72 t acting at x = m 2 Live load surcharge at Base of Foundation m h = Horizontal component Ph1 = 0.32x2.00x1.20x6.63x cos( ) = t acting at h/2 = m Vertical component Pv1 = 0.32x2.00x1.20x6.63x sin( ) = 1.99 t acting at x = m Volume: 04 Issue:

3 4.0 Verification of Stability Fig II Part of Retaining wall 3.0 Self-Weight of abutment & Load due to Backfill At Base of Stem 5.0 Verification of Structural Strength at Abutment Shaft Bottom 5.1 Design of Abutment Shaft Concrete compressive strength At Base of Foundation where, α = f ck = 30 MPa γ m =1.500 Design yield strength of reinforcement Load Combinations as per Annex B of IRC: Notations for Various Loads EP Active Earth Pressure LLS Live Load Surcharge SW Self Weight BW Back fill Weight where, f yk = 500 MPa γ s = Now, from Fig. A2-4, IRC: Effective depth d = mm where, Volume: 04 Issue:

4 Width b = mm Factor, η = 1.0 (Cl. A2.9 (2), IRC : ) Factor, λ =0.8 (Cl. A2.9 (2), IRC : ) Rearranging the above equation, Rearranging the above equation, Substituting for the values given and solving for (x/d), 0.40 (x/d)2 - (x/d) + (x/d) = 0 Therefore, x =95.998mm<x u max=235.8mm Under Reinforced Also, β cc (t)= exp(s(1-(28/t/t 1 )^0.5)) (Cl (1), IRC ) t = age of concrete in days=28 days t1 = 1 days β cc (t) = 1.00 α = 0.7 for t >= 28 days = (β cc (t)) α.f ctm =2.5 =2.9 or 2.5 whichever is greter (Cl , Page 124, IRC ) = 2.9 kc = 0.40 <= 1 k = 0.65 for h > 800 mm Act = mm 2 As min = mm 2 For Crack Control Therefore, Asmin = mm2 Therefore, As,reqd. = mm2 Hence, Thus, A s = mm 2 Width of Stem at Base =0.600 m From Table 6.5, IRC : , f ctm = 2.5 MPa For R.C. Walls, f ctm.fl (Wall) = MAX( )*2.5 or 2.5 = 2.5 (Eq. 6.6, IRC : ) Also, diameter shall not be less than 8 mm and minimum As should not be less than *Ac but not less than 12mm spacing Φ & 200 shall mm not spacing be more than 300mm As,min = 720 mm2(cl , IRC : ) 5.2Check for Shear (Clause and Of IRC : ) For wall, VEd = t takes account of stress distribution within section just prior to cracking NEd and = 0.00 t of the change of the lever arm =0.4(1-( distance deff from face of abutment wall. (Eq. 12.2, c/ k1*(h/h*)*fct.eff)) IRC ) <= d 1 =513 mm =NEd/bh (Eq. 12.4, IRC ) =(0.12*K*(80* > 1*fck)^ * Ned=0.000kN (vmin+0.15* cp)*bw.d b=1000.0mm h=600mm Therefore, c=0.000 N/mm2 K=1+(200/d)^0.5 or 2mm whichever is less = 1.62 or 2 = 1.62 mm k1=coefficient considering the effects of axial forces and stress distribution cp = NEd/Ac or 0.2*fcd whichever is less =1.5 Now h*=h for h< 1.0 m fct.eff=mean value of tensile strength of the concrete effective at the time when the cracks may first be expected to occur fct.eff=fctm or lower, fctm(t) f ctm (t) = (β cc (t)) α.f ctm (Cl (5), IRC ) Required 25 mm dia bars at 200mm c/c. Provide 25 mm dia bars at 175 mm c/c Earth face Provide12mm dia bars at150mm c/c on Outer face As,prov. =754.0mm2 Distribution Reinforcement should be max of 25% of total horizontal steel or 0.001Ac. Distribution Reinforcement should be max of Distribution Reinforcement = mm2 Therefore, At each face of wall = / 2 = mm2 Required 10mm dia bars at 200mm c/c. Provide 10 mm dia bars at 190mm c/c. Provided = mm2 bw =1000.0mm Ac = mm2 cp = 0.000< = Asl/bw.d Asl = 2805 mm2 1 = = vrdc = t Volume: 04 Issue:

5 vmin = 0.031*k1.5*fck0.5 = vrdc = t Ved = t < t Hence Shear Reinforcement is not required Limit State of Serveciability Short term Modulus of Elasticity of Concrete, Ecm = 22(fcm/12 (Sr no 88, Errata to IRC: ) In general, the effects of long term loading (due to creep) shall be obtained seperately and added to those obtained from short term analysis. The value of Ecm can be modified by a factor (1/1+ (Cl (4) (iii), IRC : ) For long term creep effects at age of 70 years, Notional size ho, 2Ac/u (Table 6.9, IRC-112 : 2011) Ac = Cross sectional area =600000mm2 u = Perimeter in contact with atmosphere =12264mm Therefore, Notional size, 2Ac/u =97.85mm Final creep coefficient of concrete for 70 yrs at 28 days age of loading β H = Therefore, β H = β (25550,28) = φ (25550,28) =β (25550,28) X = 0.996x1.923 = Therefore, modification factor=1/(1+1.92) x (45/(30+10))^ (Amendment No. 2, IRC: ) Therefore, long term Modulus of Elasticity of Concrete Ecm Modulus of Elasticity of Steel Es= Modular Ratio (Es/Ecm) =200000/11,344= Verification of Serviceability Limit State at Abutment Shaft Bottom ) accounting for long term creep effects. 6.1Stress Check = = (Table 6.9, IRC-112 : 2011) Modulus of Elasticity of Concrete Ec = MPa The development of creep with time Modulus of Elasticity of Steel Es = Mpa Modular Ratio (m = Es/Ec) = Width of Section, b (Eq. 6.16, IRC-112 = : 2011) 1000 mm where Depth of Section, D = 600 mm Effective Depth of Section, d = mm Moment of Inertia, I (Eq. 6.17, IRC-112 = : 1.80x ) mm 4 where Depth of neutral axis x = A s *f yd / η f cd* λ*b X =113.8mm<x u max=235.8mm Under Reinforced t is the age of concrete in days at the time considered Moment, t M=32.53 = t-m days ( 100 years) Actual Max. Tensile stress in concrete=8.79mpa > 2.50 MPa t 0 is the age of concrete in days at time of loading Section is t 0 Cracked = 28 days (t-t 0 ) is the actual duration of loading in days (t-t 0 ) =36472 Cracked Moment of Inertia Days Icr = b*x3/3 + m*ast*(d-x)2 Ast =2805 mm2 Icr =8.35x109mm4 Cl and of IRC : for f cm <= 45 (Eq. 6.18, IRC-112 : Max. Compressive Stress in Concrete for Rare Combinations 2011) f c =M*x/Icr<0.48*fck f c =4.430 MPa<14.4MPa OK where Max. Compressive Stress in Concrete for Quasi-Permanent Combination RH = Relative humidity expressed as percent f c = =M*x/Icr<0.36*fck 80 % RH 0 = 100 (i.e. 100 percent) f c = = MPa<10.8MPa % OK h 0 = notional member size Max. = Tensile 98 mm Stress in Steel for Rare Combinations f y =m*m*(d-x)/icr<0.80*fyk, 300 Whichever is lesser f y = MPa< 300 MPa OK Max. Tensile Stress in Steel for Quasi-Permanent Combinations f y =m*mxx*(d-x)/icr<0.80*fyk, 300 Whichever is lesser f y = MPa<300MPaOK 6.2 Check for Cracking (Clause of IRC 112) Crack width Wk S r,max (ε sm -ε cm ) = (ε sm -ε cm ) = σ sc -(k t *f ct,eff *(1+α e* ρ peff )/ρ peff )/Es Volume: 04 Issue:

6 Or 0.6*σsc/Es whichever is more Structures like Residential, industrial, water structure and Bridge Structure Retaining wall is used in a small type of Bridge structure which is called Culverts. Here in this paper the Retaining Wall is design for Bridge Structure. In Bridge Structure the Retaining wall is provided backward side of Retaining wall which is 90o perpendicular to abutment. The Retaining wall is Resist the earth pressure of behind the abutment material and another side of Retaining wall is the resist the pressure of protection material of Bridge. Generally, in Bridge Structure the Cantilever Retaining wall. In this paper, the Design of Retaining wall in the first of take the analysis of Bending moment and calculation of Lateral and Vertical Force of Retaining wall. In this the Check two type of Design one of its Vertical Stability and another one is the Structural Strength at abutment shaft at bottom portion. As per this all of result take Step of Design of Retaining wall, and as per configuration of Retaining Wall Shear Check, Cracking check and stress Check. Here show all of Results for Design of Retaining wall for stem and detailing. REFERENCES 5*(c+ /2) = > 175 Hence, Sr,max = 501 mm Sr no 56, Errata to IRC: Wk = < 0.3 OK (Table 12.1, IRC : ) [1] IRC: Standard Specifications and Code of Practice for Road Bridges SECTION II Load and Stresses. [2] IRC: Standard Specifications and Code of Practice for Road Bridges. [3] IRC: Standard Specifications and Code of Practice for Road Bridges [4] Yash chaliawala, gunvant solanki and Anuj.K.Chandiwala, Comparative Study of Cantilever and Counter fort Retaining wall, International Journal of Advance Engineering and Research Development, Dec [5] Ivan Nešović, Stepa Paunović,Milan Petrović,Ninoslav Ćirić, The Stability of Gravity Retaining Structures Contemporary achievements in civil engineering 24. April 2015 [6] Design of Reinforced Concrete Design Part-2 by H.J. SHAH Table I : Schedule of Reinforcement of Retaining wall SR. NO. TYPE OF BARS DIA OF BARS (mm) SPACING 1 a b c d CONCLUSION Retaining wall provide as a stability member to resist the soil in a site of retaining wall. Retaining wall is mostly used member vertical for 3 to 8 m height. If a small height requires retaining wall then using a PCC Retaining wall, but Required Strength of member is above 25 n/mm2 and height is more than 3 m so that reason RCC Retaining wall used in structure. Retaining wall is used in different type of Volume: 04 Issue: