RCC Structures Design The toe projection of foundation slabs is taken Equal to heel slab Below ground surface As one sixth of overall height of the wall As one third of the base Equal to heel slab Below ground surface As one sixth of overall height of the wall As one third of the base ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If K is a constant depending upon the ratio of the width of the slab to its effective span l, x is the distance of the concentrated load from the nearer support, bw is the width of the area of contact of the concentrated load measured parallel to the supported edge, the effective width of the slab be is Kx (1 + x/l) + bw All listed here K/x (1 + x/d) + bw Kx (1 - x/l) + bw Kx (1 + x/l) + bw All listed here K/x (1 + x/d) + bw Kx (1 - x/l) + bw ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design According to I.S. : 456, slabs which span in two directions with corners held down, are assumed to be divided in each direction into middle strips and edge strips such that the width of the middle strip, is Two-third of the width of the slab Half of the width of the slab Four-fifth of the width of the slab Three-fourth of the width of the slab Two-third of the width of the slab Half of the width of the slab Four-fifth of the width of the slab Three-fourth of the width of the slab ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Top bars are extended to the projecting parts of the combined footing of two columns L distance apart for a distance of 0.1 L from the outer edge of column One-fourth the distance of projection Half the distance of projection 0.1 L from the centre edge of column 0.1 L from the outer edge of column One-fourth the distance of projection Half the distance of projection 0.1 L from the centre edge of column ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The Young's modulus of elasticity of steel, is 275 KN/mm² 250 KN/mm² 150 KN/mm² 200 KN/mm² 275 KN/mm² 250 KN/mm² 150 KN/mm² 200 KN/mm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If d is the diameter of a bar, ft is allowable tensile stress and fb, is allowable bond stress, the bond length is given by π ft .d²/fb (π/4). (ft .d3/fb) (π/4). (ft .d/fb) ft .d/4fb π ft .d²/fb (π/4). (ft .d3/fb) (π/4). (ft .d/fb) ft .d/4fb ANSWER DOWNLOAD EXAMIANS APP
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