RCC Structures Design The maximum shear stress (q) in concrete of a reinforced cement concrete beam is Lever arm/(Shear force × Width) Width/(Lever arm × Shear force) Shear force/(Lever arm × Width) (Shear force × Width)/Lever arm Lever arm/(Shear force × Width) Width/(Lever arm × Shear force) Shear force/(Lever arm × Width) (Shear force × Width)/Lever arm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A reinforced concrete cantilever beam is 3.6 m long, 25 cm wide and has its lever arm 40 cm. It carries a load of 1200 kg at its free end and vertical stirrups can carry 1800 kg. Assuming concrete to carry one-third of the diagonal tension and ignoring the weight of the beam, the number of shear stirrups required, is 45 40 30 35 45 40 30 35 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If ‘W’ is weight of a retaining wall and ‘P’ is the horizontal earth pressure, the factor of safety against sliding, is 1.0 1.5 3.0 2.0 1.0 1.5 3.0 2.0 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a doubly-reinforced beam if ‘c’ and ‘t’ are stresses in concrete and tension reinforcement, ‘d’ is the effective depth and ‘n’ is depth of critical neutral axis, the following relationship holds good (m + c)/t = n/(d + n) mc/t = (d - n)/t (t + c)/n = (d + n)/n mc/t = n/(d - n) (m + c)/t = n/(d + n) mc/t = (d - n)/t (t + c)/n = (d + n)/n mc/t = n/(d - n) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If ‘A’ is the sectional area of a pre-stressed rectangular beam provided with a tendon pre-stressed by a force ‘P’ through its centroidal longitudinal axis, the compressive stress in concrete, is 2A/P A/P P/2A P/A 2A/P A/P P/2A P/A ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Design of R.C.C. cantilever beams, is based on the resultant force at Mid span Fixed end Free end Mid span and fixed support Mid span Fixed end Free end Mid span and fixed support ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS