Columns MCQ Quiz in తెలుగు - Objective Question with Answer for Columns - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Explanation:

The buckling load of a given material depends on Slenderness ratio, Area of a cross-section, and Modulus of elasticity.

Buckling load:

Analysis of long column is done using Euler’s formula:

Elastic Critical stress (fcr)

\({{\rm{f}}_{{\rm{cr}}}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{E}}}}{{{{\rm{\lambda }}^2}}}\)

where E = Modulus of elasticity of the material, and λ = slenderness Ratio

\({\rm{\lambda \;}} = {\rm{\;}}\frac{{{\rm{l}}{{\rm{e}}_{{\rm{ff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}}\)  & Imin = A × rmin2

where rmin = Minimum radius of gyration of the section, Imin = Minor principle Moment of Inertia, A = Cross-sectional Area, and Leff­ = It depends on the end conditions of the column section.

Eulers Load is given as

\({\rm{P\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{E}}}}{{{\rm{l}}_{{\rm{eff}}}^2}}{\rm{r}}_{{\rm{min}}}^2 \times {\rm{A\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{E\;Imin}}}}{{{\rm{l}}_{{\rm{eff}}}^2}}\)

Thus by Euler’s load formula,

\({\rm{Load\;}} \propto {\rm{\;}}{{\rm{I}}_{{\rm{min}}}}{\rm{\;and\;Load}} \propto \frac{1}{{{{\left( {{{\rm{l}}_{{\rm{eff}}}}} \right)}^2}}}\)

The load at which column buckle is termed as buckling load. Buckling load is given by:

\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{L_e^2}}\)

where E = Young’s modulus of elasticity, Imin = Minimum moment of inertia, and Le = Effective length

We know that MOI for a circular section is \(I = \frac{\pi d^4}{64}\)

Concept:

Crippling load:

\(P = \frac{{{\pi ^2}EI}}{{L_e^2}}\)

Calculation:

\({P_c} = \frac{{{\pi ^2}EI}}{{L_e^2}}\), where L_e = effective length of the column.

For one end fixed and one end free L_e = 2L:

Crippling load:

\(P_{Initial} = \frac{{{\pi ^2}EI}}{{4L^2}}=60\;kN \\ \therefore \frac{{{\pi ^2}EI}}{{L^2}}=240\;kN\)

For one end fixed and one end free  Le = \(\frac{L}{{\sqrt 2 }}\):

Crippling load:

\(P_{Final} = \frac{{{2\pi ^2}EI}}{{L^2}}= 2\times240=480~ kN\)

\(P_{Final} =8\times P_{Initial}\)

According to Euler's column theory, the crippling load for a column of length L,

\({P_{cr}}= \;\frac{{{\pi ^2}EI}}{{{{\left( {L_e} \right)}^2}}}\)

Where Leq is the effective length of the column.

∴ From the above conditions mentioned we can clearly see that while applying Euler’s theory of columns to calculate the crippling load, equivalent calculations are made by referencing column with both ends hinged

Explanation:

The Rankine formula is valid for both short and long columns. Hence failure by, both crushing and buckling should be considered.

If P = Failure load

1. PC = Crushing load = σC A
2. PE = Buckling load = \( = \frac{{{\pi ^2}E}}{{{l^2}}}\)

\(P = \frac{{{\sigma _C}A}}{{1 + \propto {\lambda ^2}}}\)

where ∝ = Rankine constant = \( = \frac{{{\sigma _c}}}{{{\pi ^2}E}}\), λ = Slenderness ratio,

σC = Crushing stress, A = cross-sectional area

Additional Information Rankine’s constant for the type of materials are:

Concept:

A radius of gyration is given by:

\(r = \sqrt {\frac{I}{A}} \)

I = Moment of inertia about C.G

A = Area of the x-section

\(I = \frac{{B{D^3}}}{{12}}\)

A = BD

\(r = \sqrt {\frac{{B{D^3}}}{{12 \times BD}}}\)

\(r = \frac{D}{{2\sqrt 3 }}\)

Explanation:

Short Column:

Short columns are those whose slenderness ratio is less than 32 or length to diameter ratio is less than 8. Such columns are always subjected to direct compressive stress only.

Long Column:

 Long columns have a slenderness ratio of more than 120 or a length to diameter ratio of more than 30.

Modes of failure of the column:

• In the case of short columns, the failure occurs by crushing the material under the compression yield stress.
• In the case of a long column, the failure occurs by buckling. i.e By lateral deflection of the bar
• The intermediate column fails in combined Buckling and Crushing.

Note:

Euler's theory is only applicable for long column

Rankine theory considers combining crushing and buckling modes of failure.

Concept:

Crippling load for the column:

\(\rm{\begin{array}{l} P = \frac{{{\pi ^2}EI}}{{L_e^2}} \Rightarrow P \propto I \propto {d^4}\\\end{array}}\)

Calculation:

Given:

d₂ = 0.9d₁

\(\rm{ \frac{{{P_2}}}{{{P_1}}} = \frac{{d_2^4}}{{d_1^4}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4} = {\left( {\frac{{0.9d}}{d}} \right)^4} =0.6561}\)

\({P_2} = 0.6561\;{P_1}\)

Concept:

The diameter of the core of a circular short column is d/4.

The core of a solid circular section:-

\(\begin{array}{l} {\sigma _{min}} = 0 = \frac{P}{A} + \frac{{{M_x}}}{{{I_x}}}\left( { - y} \right)\\ O = \frac{P}{{\frac{\pi }{4}{d^2}}} + \frac{{{P_e}}}{{\frac{\pi }{{64}}{d^4}}}\left( { - \frac{d}{2}} \right)\\ {e_y} = \frac{d}{8} \end{array}\)

I.e. for a maximum eccentricity of d/8, there will be no tensile stress in the section

∴ So the diameter of the core will be \(= \frac{d}{8} + \frac{d}{8}\)

\(= \frac{d}{4}\)

Effective length of columns under different end conditions are:

1. For both ends hinged: L_e = L

2. One end fixed and another end free: L_e = 2L

3. Both ends fixed = \({L_e} = \frac{L}{2}\)

4. One end fixed and other is hinges: \({L_e} = \frac{L}{{\sqrt 2 }}\)