Manual for Design and Detailings of Reinforced Concrete to Code of ...

Version 2.3 May 200831400×450For the 1400× 450 rectangle T2 = 200×= 131kNm111.94775×1062T22×131×10v t 1=== 1.035N/mm 22⎛hmin⎞ 2⎛450 ⎞hmin⎜hmax− ⎟ 450 ⎜1400− ⎟⎝ 3 ⎠ ⎝ 3 ⎠The total shear stress is 1 .035 + 0.82 = 1. 855 N/mm 2 < v = 4. 73 MPatu>cuAs 1 .035 0.067 f = 0. 396N/mm 2 , torsional shear reinforcement isrequired.x = 450 − 40×2 − 6×2 358 mm; y = 1400 − 40×2 − 6×2 1308mm1=1=AssvvT=0.8xy11( 0.87 f )6131×10=0.8×358×1308×0.87×4602=yv0.87 mmAsvAdding that for vertical shear, total = 0 .87 + 0.3 = 1. 17svUse T12 – 175 C.L. x 358 , y / 2 = 1308/ 2 654 ; use s =175 ≤ 200;x 11 =1=≤ and ≤ / 2 as per Cl. 6.3.7 of the Code.y 1It should be noted that the torsional shear link should be closed links of shapeas indicated in Figure 9.3 of the Code.As=Asvfyvs( x1+ y1) 0.87×460×( 358 + 1308)vfy=460= 1449 mm 2 . Use 13T12Incorporating the bottom 3T12 into the required flexural steel, the bottom steelarea required is 2865 + 113.1×3 = 3205 mm 2 . So use 4T32 at bottom and10T12 at sides.vThe sectional details is shown in Figure 3.33.1500T12 – 200 C.L.4001000T12T16T324T32450Figure 3.33 – Arrangement of torsional reinforcements47