RCC Structures Design In a pre-stressed beam carrying an external load W with a bent tendon is having angle of inclination ? and pre-stressed load P. The net downward load at the centre is W - 2P sin θ W - P cos θ W - P sin θ W - 2P cos θ W - 2P sin θ W - P cos θ W - P sin θ W - 2P cos θ ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Based on punching shear consideration, the overall depth of a combined footing under a column A, is (Area of the column A × Safe punching stress)/Load on column A None of these (Perimeter of column A × Safe punching stress)/(Load on column A + Upward pressure × Area of the column) (Perimeter of column A × Safe punching stress)/(Load on column A × Upward pressure × Area of the column) (Area of the column A × Safe punching stress)/Load on column A None of these (Perimeter of column A × Safe punching stress)/(Load on column A + Upward pressure × Area of the column) (Perimeter of column A × Safe punching stress)/(Load on column A × Upward pressure × Area of the column) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design An intermediate T-beam reinforced with two layers of tensile steel with clear cover 13 cm encased with the floor of a hall 12 meters by 7 meters, is spaced at 3 meters from adjoining beams and if the width of the beam is 20 cm, the breadth of the flange is 176 cm 236 cm 300 cm 233 cm 176 cm 236 cm 300 cm 233 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In favourable circumstances a 15 cm concrete cube after 28 days, attains a maximum crushing strength 400 kg/cm² 200 kg/cm² 300 kg/cm² 100 kg/cm² 400 kg/cm² 200 kg/cm² 300 kg/cm² 100 kg/cm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the length of an intermediate span of a continuous slab is 5 m, the length of the end span is kept 4.1 m 4.7 m 4.5 m 4.0 m 4.1 m 4.7 m 4.5 m 4.0 m ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If P kg/m² is the upward pressure on the slab of a plain concrete footing whose projection on either side of the wall is a cm, the depth of foundation D is given by D = 0.00775 aP D = 0.07775 aP D = 0.0775 aP D = 0.775 Pa D = 0.00775 aP D = 0.07775 aP D = 0.0775 aP D = 0.775 Pa ANSWER DOWNLOAD EXAMIANS APP
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