An R.C.C. column of 30 cm diameter is reinforced with 6 bars 12 mm φ placed symmetrically along the circumference. If it carries a load of 40,000 kg axially, the stress is
A. 49.9 kg/cm2
B. 100 kg/cm2
C. 250 kg/cm2
D. 175 kg/cm2
Answer: Option A
Solution (By Examveda Team)
Given: Diameter of R.C.C. column D = 30 cm; 6 bars of 12 mm (1.2 cm) diameter; Axial load P = 40,000 kgGross area of column (Ag): (π/4)D2 = (π/4)(30)2 = 706.86 cm2
Area of one bar: (π/4)d2 = (π/4)(1.2)2 = 1.131 cm2 (approx.)
Total steel area (As): 6 × 1.131 = 6.786 cm2
Net concrete area (Ac): Ag − As = 706.86 − 6.786 = 700.07 cm2
Assume modular ratio (m): 15 (typical for RCC working stress method)
Transformed area carrying load: Ac + mAs = 700.07 + 15×6.786 = 801.86 cm2
Concrete stress (working): σc = P / (Ac + mAs) = 40,000 / 801.86 = 49.9 kg/cm2
Therefore, the stress is 49.9 kg/cm2 ⇒ Option A.
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Comments (2)
Distribution of shear intensity over a rectangular section of a beam, follows:
A. A circular curve
B. A straight line
C. A parabolic curve
D. An elliptical curve
If the shear stress in a R.C.C. beam is
A. Equal or less than 5 kg/cm2, no shear reinforcement is provided
B. Greater than 4 kg/cm2, but less than 20 kg/cm2, shear reinforcement is provided
C. Greater than 20 kg/cm2, the size of the section is changed
D. All the above
In a pre-stressed member it is advisable to use
A. Low strength concrete only
B. High strength concrete only
C. Low strength concrete but high tensile steel
D. High strength concrete and high tensile steel
In a simply supported slab, alternate bars are curtailed at
A. $${\frac{1}{4}^{{\text{th}}}}$$ of the span
B. $${\frac{1}{5}^{{\text{th}}}}$$ of the span
C. $${\frac{1}{6}^{{\text{th}}}}$$ of the span
D. $${\frac{1}{7}^{{\text{th}}}}$$ of the span
Stress =Load/Ac+mAs
Ac= 3.14 d2/4 =706.5
Ast= 6x 3.14xd2/4 =7.065
modular ratio m=13
Putting Values Stress= 40000/706.5+ 13x7.065 = 50 apx =49.9
Kese solve Kiya