RE WALL Design as Per Irc 102 | Friction | Structural Load

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   Calculations for Reinforced TitleHand Calculations for modular block wall of height 10.75 mReferenceBS 00! 1#$010%ate #%esigned b&#Checked b& #'((ro)ed b& #%esign *n(ut +arameters Reinforced Soil Data 'ngle of *nternal friction,$ ° -nit wt 1 1.5 k/cu.m Retained Backfill Soil Data 'ngle of *nternal friction $ ᶲ ,0 ° -nit wt1.5 k/cu.m Foundation Soil Data Cohesion0 k+a'ngle of *nternal riction , ᶲ ,02-nit wt1.5 k/cu.m Crash Barrier Data Stri( load due to crash barrier 315.45 k+ai)e oad 3l$, k+ai)e oad should be considered as (er (ro)isions of *RC#7 $0146ater table is considered below the influence one.8eneral Shear ailure is considered. COMPUTATO! OF #T R  10.75 m H*8H 6' B ST'T*C '/'S*S Coefficient of acti$e earth %ressure (k a)& or reinforced soil#ka 9 :l sin;:l<sin; ᶲ ᶲ kal90.,071 ᶲ 9  32 6all batter4.$,or retained backfill soil# ᶲ 1   ᶿ=  ka$90.,,,=echanical wall height H 910.75 m ength of reinforcement 97.!0 m oundation (ro(erties #=inimum embedment de(th1 m -nit weight of foundation soil >f1 kN/m3 Su''ar of %artial factors to e used +artial factorsSoil material factors #to be a((lied tan0?to be a((lied Cto be a((lied CuSoilreinforcement interaction factorsSliding across surface of reinforcement :fs;+ullout resistance of reinforcement :f(;+artial factors of safet&oundation bearing ca(acit& # to be a((liedSliding along base of the structure or an& horiontal surface where there is soil soil contact Partial load factors for load co'inations associated *ith *alls @ffects    A =ass of the reinforced soil bod&:ffs;=ass of the backfill on to( of the reinforced soil wall :ffs;@arth (ressure behind the structu:ffs;Traffic load# An reinforced soil b:f;Behind reinforced soil block:f;1.5 51 *RC#S+#10$ $014 C'[email protected] #'A'%SSelf weight of Reinforced Soil 6alle)er >9 iDHDDff5'rm9 1.5D10.75D7.!D1.5k/mE*9,.9$$!7.1 k/mStri( load due to Crash Barrier >$9 3 F bDfts  width of the Stri( oadb 91.!>$9 15.5D1.!D1.5k/mE$90.9,7.0 k/m>ertical load due to i)e oad>,9 3l F Dffs9 $,D7.!D1.5k/mE,9,. $!$.$ k/mResultant >ertical oadR) 9 >1<>$<>,9$5!!.45 k/mHoriontal [email protected] (ressure behind reinforced soil block 9 0.5D0.,,,D1.5D10.75D10.75D1.595,4.$$ k/[email protected] +ressure due to i)e oad #+$ 9 ka$ D3lD H D ffs9 0.,,,D$,D10.75D1.5 1$,.5! k/mCheck for Sliding along the baseC'[email protected]or long term stabilit& where there is soil to soil contact at the base of the structurefsRh R):tan0?(fms;<:CDfms;Rhis the horiontal factored disturbing forceR)is the )ertical factored resultant force0?(is the (eak angle of shearing resistance under effecti)e stress cfmsis the (artial materials factor a((lied to tan0?(I C?IC-fsis the (artial factor against base slidingis the effecti)e base width for slidingSliding force :Rh;9:+i<+$;9!57.7 k/mResisting force9:R) D tanA?(;fms9!.J1 k/m7J.,4!.J1Hence structure is safe in sliding stabilit& 5$*RCGS+G 10$ $014 Check for Bearing ailureC'[email protected] #'A)erturning =oment=o 9 :+* D H, < +$ D H$;+i 1$ D ka$ D)$ D H$Dffs  9$57.4 k/ mmResisting =oment=r 9 :>* D $ <c$ F >$<>,D$;9J!41., K/ m[email protected]& :e; of resultant load R) about the centre line of thebase of width :=r =0;e 9$ *:>1<>$<>,9,.70!$.$5!!.4!9,.0 $.75$91.04 mBearing (ressure r due to =e&erhof distributionr9R) $e is the reinforcement length at the base of the wallR)is the resultant of all factored )ertical loads9$5!!.4!5.50494!!.$J k/mruit fms< D %mfms is (artial material factor a((lied to uIt-ltimate bearing ca(acit& of foundation soiluit 9 Lcf for c( 9 ,02/c9 ,0.14 /9 1.4 /&9 $$.40 $e 9 5.50uit 9 /< 0.5 : $e; &f /&9 1440.0k/m$r1047.14k/m$45!.$J 1047.14Hence foundation is safe against bearing ca(acit& failure 5,*RCiS+G 10$ $014 CA=+-T'T*A= A */[email protected]/' ST'B**T AR 10.75 m H*8H 6' B ST'T*C '/'S*SCheck for Ru(tureor reinforced soil#kal90.,07
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