If the effective length of a 32 cm diameter R.C.C. column is 4.40 m, its slenderness ratio, is
A. 40
B. 45
C. 50
D. 55
Answer: Option D
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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
slenderness ratio = Le/k
where Le = effective length
k= radius of gyration
k= square root of ( I xx /A)
where I = moment of inertia
A = Area
Ixx= (pie D^4)/64 = (3.14 × 0.32^4)/64 =0.000514
A= (pie D^2)/4 = (3.14 × 0.32^2)/4 =0.080
now, radius of gyration k = square root of (0.000514/0.08)
=0.08
slenderness ratio = Le / k = 4.4÷0.08 = 55
so the answer is option D.
Slenderness Ratio=l/r(min) -------------(1)
Here ,
l=4.4m=440cm
And ,
r(min)=[I/A]^1/2 --------------(2)
Here ,
I=pi*d^4/64
=3.24*(32)^4/64
=51445.76cm^4
Now,
A=pi*d^2/4
=3.14*(32)^2/4
=803.84cm^2
Now putting values in eqn (2)
r(min)=[51445.76cm^4/803.84cm^2]^2
=8cm
Now, putting all the values in eqn (1)
Slenderness Ratio=440cm/8cm
=55.
explain plz?
L/r
R = . 25 d= 8 cm
So, 4.4/.08= 55