Columns MCQ Quiz - Objective Question with Answer for Columns - Download Free PDF

Explanation:

The effective length factor K depends on the boundary conditions of the column. For a column with one end fixed and the other end hinged, the K value is 0.7.

Additional InformationEffective Length Factors (K) for Different Boundary Conditions:

1. Both ends fixed (K = 0.5): The column is well supported at both ends, so the effective length is reduced.

2. Both ends pinned (K = 1.0): The column can rotate freely at both ends, leading to no reduction in effective length.

3. One end fixed and the other free (K = 2.0): A free end increases the effective length, making the column more prone to buckling.

4. One end fixed and the other hinged (K = 0.7): This combination provides some resistance at one end, while allowing rotation at the other, so the effective length is reduced compared to a fully pinned column but not as much as a fully fixed column.

Explanation:

• When both ends of a column are fixed, it resists rotation at both ends, providing the highest resistance to buckling. This results in the lowest effective length factor (K), which is typically 0.5 for both ends fixed.

• A lower value of K indicates a column is more resistant to buckling.

Additional Information

One end fixed and the other free

• Effective Length Factor (K = 2.0): This setup results in the highest effective length factor because the free end does not resist rotation, leading to greater flexibility and susceptibility to buckling.

• Behavior: The column behaves like a cantilever, where the free end can deflect more easily compared to a fixed end.

• Common Use: Typically found in situations where one end of the column is fixed (like in a wall or frame) and the other is exposed (like a cantilevered beam or structural element).

One end fixed and the other pinned

• Effective Length Factor (K = 0.707): The column is restrained at one end and allowed to rotate at the other. This setup offers moderate buckling resistance.

• Behavior: The fixed end resists rotation, while the pinned end allows free rotation. The column has some flexibility but is still somewhat controlled.

• Common Use: This condition is common in columns where one end is embedded in the structure (e.g., a frame) and the other is pinned, allowing for some rotation but still providing support.

Both ends pinned

• Effective Length Factor (K = 1.0): Both ends are pinned, meaning the column can rotate freely at both ends but cannot translate horizontally. This provides a moderate level of buckling resistance.

• Behavior: Since both ends can rotate, the column is not as stable as one with fixed ends but is still better than if one end were completely free.

• Common Use: Found in simple structural systems where columns are connected at both ends to beams or slabs that do not fix the column against rotation.

Explanation:

Buckling/Critical load:

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

\({P_b} = \frac{{{\pi ^2}EI}}{{L_e^2}}\)
where E = Young's modulus of elasticity, Imin = Minimum moment of inertia, and Le = Effective length

Concept:

The crippling load for a strut with one end fixed and the other hinged is calculated using Euler's formula: \( P = \frac{\pi^2 E I}{L_e^2} \)

Where, \( L_e = \frac{L}{\sqrt{2}} \) for this end condition.

Given:

L = 4 m = 4000 mm, d = 80 mm, E = 2 × 10⁵ N/mm², π³ = 31

Calculation:

Effective length, \( L_e = \frac{4000}{\sqrt{2}} = 2828.4 \, \text{mm} \)

Moment of Inertia, \( I = \frac{\pi d^4}{64} = \frac{3.14 \times 80^4}{64} = 2010613 \, \text{mm}^4 \)

Crippling Load, \( P = \frac{\pi^2 \times 2 \times 10^5 \times 2010613}{(2828.4)^2} = 495142.4 \, \text{N} \approx 496 \, \text{kN} \)

Concept:

Euler buckling load for a column is given by,

\({{\rm{P}}_{\rm{E}}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where, Le is the effective length of the column that depends on the end support conditions, and EI is the flexural rigidity of the column.

Now, effective length (Le) for different end conditions in terms of actual length (L) are listed in the following table:

 ∴ For both ends fixed Effective length Le = 0.5 L

Concept:

Crippling load, \({{\rm{P}}_{{\rm{cr}}}} = \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where,

L_e = Effective length, and I = Moment of Inertia about bending axis

For a given E and I:

\({{\rm{P}}_{{\rm{cr}}}} \propto \frac{1}{{{\rm{L}}_{\rm{e}}^2}}\)

OR

\({\left( {{{\rm{P}}_{{\rm{cr}}}}} \right)_1}{\rm{\;}}{\left( {{\rm{L}}_{\rm{e}}^2} \right)_1} = {\left( {{{\rm{P}}_{{\rm{cr}}}}} \right)_2}{\left( {{\rm{L}}_{\rm{e}}^2} \right)_2}\)       ---(1)

Now

When both ends are fixed: \({L_{e1}} = \frac{L}{2}\)

When one end is free and another end is fixed:

Le₂ = 2L and (P_cr)₂ = P₂

Using equation (1), we get

\(\left( {\rm{F}} \right) \times {\left( {\frac{{\rm{L}}}{2}} \right)^2} = \left( {{{\rm{P}}_2}} \right) \times {\left( {2{\rm{L}}} \right)^2}\)

\(\Rightarrow {{\rm{P}}_2} = \frac{{\rm{F}}}{{16}}\)

Explanation:

We know that-

\(Slenderness\, ratio\, (\lambda) ={L_{eff}\over r_{min}}\)

Where Leff = Effective length of the column

rmin = Least radius of gyration

\(r_{min}=\sqrt{I_{min}\over A}\)

Calculation:

Given data:

Effective length of the column about the X-axis (L_eff-XX) = 3 m or 3000 mm

Effective length of the column about the Y-axis (Leff-YY) = 2.75 m or 2750 mm

Cross-section of the column = 400 mm × 600 mm

Width of column (b) = 400 mm

Depth of column (d) = 600 mm

\(I_{min}={db^3\over 12}={{600× 400^3}\over 12}=32× 10^8\, mm^4\)

\(Area\,(A)=bd=400\times 600=24\times 10^4\, mm^2\)

\(r_{min}=\sqrt{32\times 10^8\over 24\times 10^4} =115.470\, mm\)

Slenderness ratio about X-axis:

\(Slenderness\, ratio\, about\, X-axis\, (\lambda) ={L_{eff-XX}\over r_{min}}\)

\(Slenderness\, ratio\, about\, X-axis\, (\lambda_{X}) ={3000\over 115.47}=25.98\)

\(\lambda_{Y}=25.98 <32\)

Slenderness ratio about Y-axis:

\(Slenderness\, ratio\, about\, Y-axis\, (\lambda_{Y}) ={L_{eff-YY}\over r_{min}}\)

\(Slenderness\, ratio\, about\, Y-axis\, (\lambda_{Y}) ={2750\over 115.47}=23.815\)

\(\lambda_{Y}=23.815 <32\)

1. Short columns have a slenderness ratio of less than 32.
2. The medium column slenderness ratio is between 32 to 120.
3. The slenderness ratio of the long columns is greater than 120.

Hence the column is a short column.

Concept:

Effective length is defined as the distance between two adjacent points of zero bending moment or contra flexure.

Clamped free column means one end fixed and other ends free.

The value of the effective length varies according to the support condition.

Concept:

For column, the buckling load is taken as,

\({\rm{P}} = \frac{{{{\rm{\pi }}^2}{\rm{E\;}}{{\rm{I}}_{{\rm{min}}}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

E = Modulus of Elasticity, I_min = Minimum moment of inertia, and L_e = effective length of column.

Calculation:

In first case: Height = h, and cross-sectional area = a × 2a

\({\rm{I}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{12}} = {\rm{min}}\left\{ {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}},\frac{{{\rm{a}} \times {{\left( {2{\rm{a}}} \right)}^3}}}{{12}}} \right. = \frac{{2{{\rm{a}}^4}}}{{12}}\)

\({\rm{P}} = \frac{{{{\rm{\pi }}^2}{\rm{E\;}}\left( {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}}} \right){\rm{\;}}}}{{{{\rm{h}}^2}}} = \frac{{{{\rm{\pi }}^2}{\rm{E}}{{\rm{a}}^4}}}{{6{{\rm{h}}^2}}}\)

\({\rm{I}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{12}} = {\rm{min}}\left\{ {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}},\frac{{{\rm{a}} \times {{\left( {2{\rm{a}}} \right)}^3}}}{{12}}} \right. = \frac{{2{{\rm{a}}^4}}}{{12}}\)

In second case: Height = 1.5 h, and cross-sectional area = 3a × 0.5a

Concept:

Buckling load given by Euler is:

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

where,

E = Modulus of elasticity of the material

I = Moment of Inertia

\(I = \frac{\pi }{{64}}{D^4}\)

Leff = Effective length of the column

Calculation:

Let P1 be the buckling load of the column when the diameter of the column is "D"

\({P_1} \propto D^4\)

Let P2 be the buckling load of the column when the diameter is reduced by 20%, it becomes 0.8D. 

\({P_2} \propto {\left( {0.80D} \right)^4}\)

\(\% \;reduction = \;\frac{{{P_1} - {P_2}}}{{{P_1}}} \times 100\)

\(\% \;reduction = \;\frac{{{D^4} - {{\left( {0.80D} \right)}^4}}}{{{D^4}}} \times 100\)

% reduction = 59.04 %

Concept:

Slenderness ratio (λ):

Slenderness ratio of a compression member is defined as the ratio of its effective length to the radius of gyration

\({\rm{\lambda }} = \frac{{{{\rm{L}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}}\)

Where,

L_eff = Effective length of the compression member

r_min = Least radius of gyration

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{{{\rm{I}}_{{\rm{min}}}}}}{{\rm{A}}}} \)

I_min = Minimum moment of inertia of area

A = Area of the shape

Calculation:

Given,

For circular column

D = 40 cm, L = 4 m = 400 cm

\({\rm{A}} = \frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4}\), \({{\rm{I}}_{{\rm{min}}}} = \frac{{{\rm{\pi }}{{\rm{D}}^4}}}{{64}}\)

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{\frac{{{\rm{\pi }}{{\rm{D}}^4}}}{{64}}}}{{\frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4}}}} = \sqrt {\frac{{{D^2}}}{{16}}} = \frac{D}{4}\)

\({\rm{\lambda }} = \frac{{{{\rm{L}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}} = \frac{{400}}{{\frac{D}{4}}} = \;40\)

Hence the slenderness ratio of the circular column is 40

Concept:

Slenderness ratio is defined as the ratio of effective length of column to the least radius of gyration.

\({\rm{\lambda }} = \frac{{{{\rm{L}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}} \)

where,

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{\rm{I}}}{{\rm{A}}}} \)

\({\rm{I}} = \frac{{{\rm{\pi }} \times {{\rm{D}}^4}}}{{64}} \)

\({\rm{A}} = \frac{{\rm{\pi }}}{4} \times {{\rm{D}}^2} \)

Where,

L_eff = Effective length of column

r_min = Minimum radius of gyration

I = Moment of inertia of the centroidal axis.

A = Area of the cross section of the column

Calculation:

For column hinged at both ends, L_eff = Length of the column itself = 5 m = 500 cm

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{{{\rm{D}}^2}}}{{16}}} = \frac{{\rm{D}}}{4} \)

D = 16 cm

r_min = 4 cm = 0.04 m

Concept:

\(\frac{1}{{{P_R}}} = \frac{1}{{P_c}} + \frac{1}{{P_e}}\)

Where, PR = Rankine load, Pc = Crushing load, and Pe = Euler buckling load.

Calculation:

Given:

P_e = 1000 KN, P_c = 1500 KN

\(\frac{1}{{{P_R}}} = \frac{1}{{1000}} + \frac{1}{{1500}}\)

⇒ PR = 600 kN

Additional Information

The Euler buckling load (P_cr)of a steel column is

\({P_{cr}} = \frac{{{\pi ^2}EI}}{{L_{eff}^2}}\)

where EI = flexural rigidity, Leff = effective length of the column

Euler’s theory:

• This theory is valid only for long columns only.
• This theory is valid only when the slenderness ratio is greater or equal to the critical slenderness ratio.
• For any slenderness ratio above critical slenderness ratio, the column fails by buckling and for any value of slenderness ratio less than this value, the column fails in crushing not in buckling.

Euler’s critical load formula is,

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

Euler’s formula is applicable when,

Crushing stress ≥ Buckling stress

\({σ _{cr}} ≥ \frac{{{\pi ^2}E}}{{{λ _c^2}}}({λ _c}\;is\;critical\;slenderness\;ratio)\)

\({λ _{min}^2} = {λ _c^2} ≥ \frac{{{\pi ^2}E}}{{{σ _{cr}}}}\)    

For mild steel,

E = 2 × 105 N/mm2

σcr = 330 N/mm2

∴ λ ≥ 80 N/mm2

∴ When the slenderness ratio for mild steel columns is less than 80, Euler’s theory is not applicable.

Explanation:

According to Euler's column theory, the critical load (P) on the column for different types of end conditions is as follows:

\({{\rm{P}}_{\rm{e}}} = \frac{{{\rm{\pi E}}{{\rm{I}}_{{\rm{min}}}}}}{{{\rm{L}}_{\rm{e}}^2}} = \frac{{{\rm{n}}{{\rm{\pi }}^2}{\rm{E}}{{\rm{I}}_{{\rm{min}}}}}}{{{{\rm{L}}^2}}}\)

Where Pe = Buckling load, Imin = Minimum of [Ixx & Iyy], L = Actual length of the column, α = Length fixity coefficient, n = End fixity coefficient 

Le = αL and \({\bf{n}} = \frac{1}{{{\alpha ^2}}}\)

Euler's formula holds good only for long columns.

The Length fixity coefficient and End fixity coefficient for the given end conditions are given in the following table: