Answered: A round steel alloy bar with a diameter…

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 round steel alloy bar with a diameter of 19mm and a gauge length of 76mm was subjected to tension, with the results shown in Table.

Using a computer spreadsheet program, plot the stress–strain relationship. From the graph, determine the Young’s modulus of the steel alloy and the deformation corresponding to a 37kN load.

Deformation,
Load, kN
mm
6.
0.0286
18
0.0572
27
0.0859
36
0.1145
45
0.1431
54
0.1718

Expert Solution

Step 1 To determine the stresses for all given loading:

Given data:

Diameter of round steel alloy bar = D = 19 mm

Gauge length = L = 76 mm

Area of steel bar is calculated as:

$A = \frac{π}{4} d^{2} = \frac{π}{4} \times 19^{2} = 283.53 m m^{2}$

Now, stress in the bar for a load of 9 kN is calculated as:'

$σ_{1} = \frac{F o r c e}{A r e a} = \frac{9 \times 1000 N}{283.53 m m^{2}} = 31.74 M P a$

Stress for loads $(18, 27, 36, 45, 54) k N$ are calculated as follows:

$σ_{2} = \frac{F o r c e}{A r e a} = \frac{18 \times 1000 N}{283.53 m m^{2}} = 63.48 M P a$

$σ_{3} = \frac{F o r c e}{A r e a} = \frac{27 \times 1000 N}{283.53 m m^{2}} = 95.23 M P a$

$σ_{4} = \frac{F o r c e}{A r e a} = \frac{36 \times 1000 N}{283.53 m m^{2}} = 126.97 M P a$

$σ_{5} = \frac{F o r c e}{A r e a} = \frac{45 \times 1000 N}{283.53 m m^{2}} = 158.71 M P a$

$σ_{6} = \frac{F o r c e}{A r e a} = \frac{54 \times 1000 N}{283.53 m m^{2}} = 190.46 M P a$

Step 2 To determine the strain corresponding the given load:

Now, the strain is calculated as:

$ε = \frac{δ}{L} w h e r e, ε = s t r a i n i n t h e b a r c o r r e s p o n d i n g t o t h e s t r e s s δ = d e f o r m a t i o n i n t h e b a r c o r r e s p o n d i n g t o t h e l o a d (v a l u e s g i v e n i n t h e t a b l e)$

As per the values of deformation given in the table, the strain corresponding to different deformations is calculated as:

$ε_{1} = \frac{0.0286 m m}{76 m m} = 3.763 \times 10^{- 4} ε_{2} = \frac{0.0572 m m}{76 m m} = 7.562 \times 10^{- 4} ε_{3} = \frac{0.0859 m m}{76 m m} = 1.130 \times 10^{- 3} ε_{4} = \frac{0.1145 m m}{76 m m} = 1.5066 \times 10^{- 3} ε_{5} = \frac{0.1431 m m}{76 m m} = 1.8828 \times 10^{- 3} ε_{6} = \frac{0.1718 m m}{76 m m} = 2.2505 \times 10^{- 3}$

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