Page 204 - Civil Engineering Formulas
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140                     CHAPTER FIVE

           ULTIMATE-STRENGTH DESIGN FOR TORSION

           When the ultimate torsion T is less than the value calculated from the T equa-
                                                               u
                               u
           tion that follows, the area A of shear reinforcement should be at least
                               v
                                          b w s
                                   A v   50                     (5.90)
                                          f y
             However, when the ultimate torsion exceeds T calculated from the T equa-
                                                               u
                                               u
           tion that follows, and where web reinforcement is required, either nominally or
           by calculation, the minimum area of closed stirrups required is
                                          50b w s
                                A v   2A t                      (5.91)
                                            f y
           where A is the area of one leg of a closed stirrup resisting torsion within a
                 t
           distance s.
             Torsion effects should be considered whenever the ultimate torsion exceeds
                                T u     0.5 f c  
x y           (5.92)
                                              2
           where     capacity reduction factor   0.85
                  T   ultimate design torsional moment
                   u
                  2
                
x y   sum for component rectangles of section of product of square of
                      shorter side and longer side of each rectangle (where T section
                      applies, overhanging flange width used in design should not exceed
                      three times flange thickness)
           The torsion T carried by the concrete alone should not exceed
                     c
                                             2
                                         f
                                     0.8 2 c   
x y
                             T c                                (5.93)
                                  2 1   (0.4V u /C t T u ) 2
                         2
           where C   b d/
x y.
                 t
                    w
             Spacing of closed stirrups for torsion should be computed from
                                     A t   f y   t x 1 y 1
                                 s                              (5.94)
                                     (T u   T c )
           where A   area of one leg of closed stirrup
                 t
                    0.66   0.33y /x but not more than 1.50
                             1
                 t
                               1
                f   yield strength of torsion reinforcement
                 y
                x   shorter dimension c to c of legs of closed stirrup
                 1
                y   longer dimension c to c of legs of closed stirrup
                 1
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