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QUESTION: 1

Compression members always tend to buckle in the direction of the

Solution:

QUESTION: 2

Euler’s formula gives 5 to 10% error in crippling load as compared to experimental results in practice because

Solution:

**Euler's formula to calculate crippling load in the column**

\({P_{cr}} = \frac{{{\pi ^2}EI_{mini}}}{{L_e^2}}\)

Where P = Crippling load, E = young's modulus, I_{mini} = Minimum area moment of inertia of column, L_{e} = Effective length of a column

The **assumption made** while developing **Euler Formula** for the crippling load of a column **does not meet** in the real-life practice due to which therefore Euler's formula gives 5 to 10 % error compared to experimental results or true value.

**The following assumptions are made in Euler's column theory :**

- The column is initially
**perfectly straight**and the load is applied axially. - The cross-section Of the column is
**uniform**throughout its length. - The column material is
**perfectly elastic, homogeneous, and isotropic**and obeys Hooke's law. - The length of the column is
**very large**as compared to its lateral dimensions. - The
**very small**as compared to the bending stress. - The self-weight of the column is negligible.
- The column will
**fail by buckling**alone. - Pin joints are free from
**friction.**

QUESTION: 3

Match List-I (Long column) with List-ll (Critical Load) and select the correct answer using the codes given below the lists:

List-I

A. Both ends hinged

B. One end fixed and other end free

C. Both ends fixed

D. One end fixed and other end hinged

List-II

1. π^{2} EI/4L^{2}

2. 4π^{2} EI/4L^{2}

3. 2π^{2} EI/4L^{2}

4. π^{2} EI/L^{2}

Codes:

A B C D

(a) 2 1 4 3

(b) 4 1 2 3

(c) 2 3 4 1

(d) 4 3 2 1

Solution:

(i) Both ends hinged

(ii) One end fixed and other end free

(iii) Both ends fixed

(iv) One end fixed and other end hinged

QUESTION: 4

Given that

P_{E} = the crippling load given by Euler

P_{c }= the load at failure due to direct compression

P_{R} = the load in accordance with the Rankine’s criterion of failure

Then P_{R} is given by

Solution:

QUESTION: 5

If the crushing stress in the material of a mild steel column is 3300 kg/cm^{2}. Euler’s formula for crippling load is applicable for slenderness ratio equal to/greater than

Solution:

For Euler’s formula to be applicable the critical stress must not exceed the proportional limit.

Now crushing stress in mild steel

= 3300 kg/cm^{2}

= 330 N/mm^{2}

But stress at proportional limit in mild steel

= 250 M/mm^{2}

Euler’s buckling stress

λ = 88.55 ≈ 89

Thus slenderness ratio should be more than or equal to 89 ideally. Option (d) is the most close one.

QUESTION: 6

For a circular column having its ends hinged, the slenderness ratio is 160. The L/d ratio of the column is

Solution:

Slenderness ratio,

where r is the least radius of gyration and L_{eff} is effective length

∴

QUESTION: 7

When a column is subjected to an eccentric load, the stress induced in the column will be

Solution:

QUESTION: 8

An aluminium column of square cross-section (10 mm x 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler ’s critical load (σ_{critical}) square : (σ_{critical})circular is

Solution:

For same material, same end condition and same length

QUESTION: 9

If one end of a hinged column is made fixed and the other free, how much is the critical load compared to the original value?

Solution:

Original load =

when one end of hinged column is fixed and other free. New L_{θ} = 2L

∴ New load

QUESTION: 10

A short column of external diameter D and internal diameter d carries and eccentric load W. The greatest eccentricity which the load can be applied without producing tension on the cross-section of the column would be

Solution:

Direct stress

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