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# Test: Struts & Columns - 3

## 10 Questions MCQ Test Topicwise Question Bank for Mechanical Engineering | Test: Struts & Columns - 3

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This mock test of Test: Struts & Columns - 3 for Mechanical Engineering helps you for every Mechanical Engineering entrance exam. This contains 10 Multiple Choice Questions for Mechanical Engineering Test: Struts & Columns - 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Struts & Columns - 3 quiz give you a good mix of easy questions and tough questions. Mechanical Engineering students definitely take this Test: Struts & Columns - 3 exercise for a better result in the exam. You can find other Test: Struts & Columns - 3 extra questions, long questions & short questions for Mechanical Engineering on EduRev as well by searching above.
QUESTION: 1

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, Imini = Minimum area moment of inertia of column, Le = 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 direct stress is 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. π2 EI/4L2 2. 4π2 EI/4L2 3. 2π2 EI/4L2 4. π2 EI/L2 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
PE = the crippling load given by Euler
Pc = the load at failure due to direct compression
PR = the load in accordance with the Rankine’s criterion of failure
Then PR is given by

Solution:

QUESTION: 5

If the crushing stress in the material of a mild steel column is 3300 kg/cm2. 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/cm2
= 330 N/mm2
But stress at proportional limit in mild steel
= 250 M/mm2
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 Leff 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 (σcritical) square : (σcritical)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:

when one end of hinged column is fixed and other free. New Lθ = 2L