Answered: A brass specimen of the circular…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

A brass specimen of the circular cross-section is fractured at 151 kN force and the final length of the specimen at fracture is 49 mm. The fracture strength of the specimen is found to be 74 kN/mm². The percentage of elongation of the specimen is 42 %.

Determine the following

(i) Diameter of the specimen

ii) Initial length of the specimen

iii) Stress under an elastic load of 16 kN

iv) Young's Modulus if the elongation is 1.6 mm at 16 kN

(v) Final diameter if the percentage of reduction in area is 20 %

solve:

Initial Cross-sectional Area (in mm²) =

The Diameter of the Specimen (in mm) =

Initial Length of the Specimen (in mm) =

Expert Solution

Step 1 Initial Cross-sectional Area and diameter of the specimen

We know,

$f r a c t u r e s t r e n g t h = \frac{f r a c t u r e l o a d}{c r o s s s e c t i o n a l a r e a} A = \frac{151}{74} A = 2.04 m m^{2}$

Hence, the initial cross-sectional area is 2.04 mm²

Also,

$A = \frac{π}{4} \times d^{2} d = \sqrt{\frac{4 \times 2.04}{π}} d = 1.61 m m$

Hence, the diameter of the specimen is 1.61 mm

Step 2 Initial length of the specimen

We have,

$% e l o n g a t i o n = \frac{f i n a l l e n g t h - i n i t i a l l e n g t h}{i n i t i a l l e n g t h} \times 100 42 = \frac{49 - L_{i}}{L_{i}} \times 100 L_{i} = 34.51 m m$

Hence, the initial length of the specimen is 34.51 mm

Step 3 Stress under an elastic load of 16 kN

The stress is given by,

$σ = \frac{P}{A} σ = \frac{16}{2.04} σ = 7.84 k N / m^{2}$

Hence, the stress under an elastic load of 16 kN is 7.84 kN/m²

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