# Question: SOİL MECHANİCS… For the RC retaining wall given in the figure below a) Calculate the required extent of the base slab forward the back of the wall such that a factor of safety of 2.0 against overturning is met. -Free Course Hero Question Answer.

Question Description:

SOİL MECHANİCS…

For the RC retaining wall given in the figure below

a) Calculate the required extent of the base slab forward the back of the wall such that a factor of safety of 2.0 against overturning is met.

b) Subsequently check if the required minimum factor of safety of 1.5 is met against base sliding. If not, design a base key. Calculate the passive pressure starting from the bottom level of the base slab. Also, use 2/3 of the shear strength parameters of the base soil in calculations and apply an FS 2.0 for the passive resistance.

Image transcription text

`For the RC retaining wall given in the figure below a) Calculate the required extent of the base slab forward the back of the wall such that a factor of safety of 2.0 against overturning is met. b) Subsequently check if the required minimum factor of safety of 1.5 is met against base sliding. Ifnot, design a base key. Calculate the passive pressure starting from the bottom level of the base slab. Also use 2H3 of the shear strength parameters of the base soil in calculations and apply a FS 2.0 for the passive resistance. ¢=25° 3 "3:1 7 kN/m 10 ' Base soil (silt clay) 3 c=65 kPa, ¢=35', 7:20 kNJ'm`

(1) x= 3.86m

Total B=5.36m

(2) FOS= 1.66

Q_t=346.29 KN/m2

Q_h= -63.9 KN/m2

I have attached the solution.

Step-by-step explanation

Image transcription text

`Given, 2= 35 KN/ M 2 so sucharge Backfill 7m 40 = 250 Hill = 17 KN/m2 rcone = 24 KN/m 3 Im 0.5ml im I x - B Base soil C = 65KPa D = 350 = 20 KN/M3 Simple diagram for this with acting forces, (2) TM V 2 0 -53 B Now Ka = tam2 (45- 25101- - Ka = 0.406 Pa = (2 +12 )ka - 2c /Kq`

Image transcription text

`2 * at botturn Pa = (0 . 40 6 x 35 ] +0 = 14. 21 KN / m 2 + at top Pa - ( 35 + 17 * 8 ) X 0.406 . = 69 43 KN /m 2 14.21 - 455. 22 Now calculating all forces, &amp; moment through table S . No Forces Distance moment V1 = 1X7X24 = 168 0 .5+0.5 166 2 V = 7XXX 17=119x ( 2 + 1.5 ) (* * 1:5 ) 119x 3 V3 = 1 x(1. 5+ 21 ) x 2 4 ( 1. 5+ 2 ] 1.5+ 2 (36 +24 2( ) = 36+24x ZY EV= 204+143x Em = m, + 12 +my = 715x2 + 214 5x + 195 4 HI = 14.21 X 8 4 454. 72 15 H 2 = 55 . 227 8 x 1 - 220.8 8/ 3 5.89 EH = 3345 8 m = 1043.7 a ) a = Emr EMOV = Fos ) overturning 7 0 = 71.5x2+ 214. 5x+ 195 1043. 73 2 71:5x2 + 214. 5x + 18 92. 46 1 = 3. 86m`

Image transcription text

`PAGE NO. : B = 0 .5 + 1 93- 86 = 5. 36 m 3 Fos ) sliding = Ev tand't = 201 2/3 BC ) D' = 20 = 2X35 = 23:33' 3 207 FOS = (175. 98 x tam 23,33) + ( 2/2x65 x5-36) (33 4:56 - 1. 668 Pressure on soil at toe &amp; heel Xp = 1 + 8in35 1- \$17135 = 3.69 * Passive pressure Pp = Kp ( Z xr ) + 2 c Jkp = 2 X65 X ( 3 6 9) .5 = 249.7 KPQ e = B _ (me - Emo) - 1.30ms I toe = EV [It be] I heel - 758.9 [ 1-/6*).30) B B 5. 36 = 755.9 1 1 + ( 6 X 130)7 9Peel - - 63- 9 3 KN/ mz 5. 3 6 5. 36 OR = 346 . 29 KN/ M2`