 NEPAL ENGINEERING SERVICES

## Tuesday, July 9, 2019

### Design of Basement Wall

Introduction
Basement wall is constructed to retain the earth and to prevent moisture from seeping into the building. Since the basement wall is supported by the mat foundation, the stability is ensured and the design of the basement wall is limited to the safe design of vertical stem.

Basement walls are exterior walls of underground structures (tunnels and other earth sheltered buildings), or retaining walls, which must resist lateral earth pressure as well as additional pressure due to other type of loading. Basement walls carry lateral earth pressure generally as vertical slabs supported by floor framing at the basement level and upper floor level. The axial forces in the floor structures are, in turn, either resisted by shear walls or balanced by the lateral earth pressure coming from the opposite side of the building.

Although basement walls act as vertical slabs supported by the horizontal floor framing, keep in mind that during the early construction stage when the upper floor has not yet been built, the wall may have to be designed as a cantilever.

Design of Vertical System
The basement wall is designed as the cantilever wall with the fixity provided by the mat foundation.

Design of Basement Wall
 Reference Step Calculation Output IS 456:2000, Cl.32.5.1 IS 456:2000, Cl.26.5.2.1 IS 456:2000, 26.5.2.2 IS 456:2000, Cl.32.5.b IS 456:2000, Cl.31.6.2.1 IS 456:2000, Table-19 IS456:2000, Cl.23.2.a IS 456:2000, Cl.32.5.d IS 456:2000, Cl.26.2.1 1. 2. 3. 4. 5. 6. 7. 8. Design Constants Floor to floor heights, Basement, h = 2.844 m Unit weight of soil, = 17 kN/m3 Angle of internal friction of the soil, = 300 Surcharge produced due to vehicular movement,  Ws = 10 kN/m2 Safe bearing capacity of the soil, qs = 125 kN/m2 Height of soil from base to ground level (h)= 2.48 m Moment calculation Here, Coefficient of Earth Pressure, Lateral load due to soil pressure, Pa = = = 22.92 kN/m Lateral load due to surcharge load, Ps = Ka Ws h                               = 0.333 10 2.844     = 9.48 kN/m Characteristic bending moment at the base of wall is calculated below. Since the weight of the wall gives insignificant moment, so this can be neglected in the design. Mc = = = 35.21 kN-m/m Design moment, M = 1.5Mc = 1.5 35.21                                               = 52.81 kN-m/m Approximate design of section Let effective depth of wall = d BM =0.138 ƒck bd2 (For Fe415) or, 52.81 106 = 0.138 20 1000 d2 or, d = 138.36 mm Let clear cover is 40mm & bar size is 20mm-Ф Overall depth of  wall, D = 138.36 + 40 + 10                                          = 188.36 mm Take D = 200 mm So, d  = 200-40-10 = 150 mm D = 200mm 200mm So, two curtains of reinforcement need to be provided. Calculation of Main Steel Reinforcement  or, Ast = 1161.83 mm2 Minimum Ast = 0.12% b D = 0.0012 1000 200                       = 240 mm2< Ast Maximum Diameter of bar = = 25 mm Providing 16mm-Ф bar, Spacing of bars (s)  on the front as well as back face: or, s = = 346.11 mm/m             Provide 16 mm-Ф bar @275 mm c/c So, Provided = 1462.3 mm2   > Ast min Max. Spacing = 3d = 3x300 = 450 mm or 450 mm                                                      OK. Check for Shear The critical section for shear strength is taken at a distance, ‘d’ from the face of the support .Thus, the critical section is at d =(200-40-8)= 0.152 m from the top of mat foundation i.e. at (2.844- 0.152) = 2.692 m below the top edge of wall. Shear force at critical section is,      Vu =  = 44.242 KN Nominal shear stress , = 0.29 N/mm2 Permissible shear stress , = 0.614 N/mm2   Here, > Hence, it is safe in shear. Check for Deflection m Allowable deflection = =11.68 mm Actual Deflection = = = 6.85 mm < 11.68 mm                                     which is less than the allowable deflection, hence it is safe. Calculation of Horizontal Reinforcement Steel bar Area of Hz. Reinforcement = 0.002Dh            = 0.002 200 2844 = 1137.6 mm2 As the temperature change occurs at front face of basement wall, 2/3 of horizontal reinforcement is provided at front face and 1/3 of horizontal reinforcement is provided in inner face. Front face Horizontal Reinforcement steel,                           = = 758.4 mm2 Providing  10mm-Ф bars, No. of bars required, N = . Spacing = = = 275.4 mm Provide 10mm-Ф bar @ 250 mm c/c Inner  face Horizontal Reinforcement steel,                           = mm2 Providing  10mm-Ф bars, No. of bars required, Spacing = = = 688.5 mm Max. spacing = 3d = 3x152 = 456 mm or 450 mm So, provide 10mm-Ф bar @ 450 mm c/c Curtailment of Reinforcement No bars can be curtailed in less than Ld distance from the bottom of stem.       Ld = = = 752.19 mm 755 mm      M1  = 0.87 fy * Ast * d * (1 - )        = 0.87 * 415 * 1462.3 * 152 * (1 - )        =  64.230* 106 N-mm        = 64.230 KN-m Vu =44.242 KN 1.3 M1/Vu + Lo = 1.3*64.230*106/(44.242*103) +                               12*16                           = 2081.38 mm > Ld (ok)               The curtailment of bars can be done in two layers, at 1/3 and 2/3 heights of the stem above the base. Let us curtail bars at 1/3 distance i.e. 948 mm from the base. Pa = 22.92            kN/m Ps = 9.48 kN/m                                                                                M  = 52.81         kN-m/m D = 200 mm d  = 150 mm Ast = 1161.83            mm2 s = 275 mm on front and back face = 0.975% Vu = 44.242 kN = 0.29          N/mm2 = 0.413         N/mm2 Front face: Provide 10mm-Ф bar @ 250 mm c/c Inner face: Provide 10mm-Ф bar @ 450 mm c/c