Strength of Materials Test 7

Result

Strength of Materials Test 7

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Question 1
1 / -0

A tensile test was conducted on a mild steel bar. The following data was obtained from the test:
(i) The diameter of the steel bar: 2 cm
(ii) The gauge length of the bar: 20 cm
(iii) Load of elastic limit: 200 kN
(iv) Extension at a load of 150 kN: 0.2 mm
(v) Maximum load: 400 kN
(vi) Total extension: 60 mm
(vii) The diameter of the rod at failure: 1.2 cm
Calculate the stress at the elastic limit (MPa)

A
635.5-637.2

B
635.54-637.21

C
635.57-637.26

D
635.59-637.22

Solution

Concept:
Stress at the elastic limit is calculated as:
\({σ _e} = \frac{{Load\;at\;elastic\;limit}}{{Area}}\)
Calculation:
Given:
The load of elastic limit = 200 kN = 200000 N
Diameter (d) = 2 cm = 20 mm
\(A = \frac{\pi }{4}{d^2} = \frac{\pi }{4}{\left( {20} \right)^2} = 314.16 \:m{m^2}\)
∴ Stress at the elastic limit:
\({σ _e} = \frac{{Load\;at\;elastic\;limit}}{{Area}} = \frac{{200 \times {{10}^3}}}{{314.16}}\)
∴ σ_e = 636.62 MPa
Question 2
1 / -0

A steel rod of 50 mm diameter is 4m long. In a test, a pull of 80 kN is suddenly applied to it. Take E = 200 GPa. Elongation due to suddenly applied load is _____mm.

A
1-2

B
14-21

C
17-26

D
19-22

Solution

Concept:
As load is suddenly applied, the maximum stress produced is double the stress due to gradual loading.
\(Maximum\;stress = \;\sigma = \frac{{2P}}{A}\)
Calculation:
\(Area = \;\frac{\pi }{4} \times {50^2} = 1963.5\;m{m^2}\)
\(Maximum\;stress = \;\sigma = \frac{{2 \times 80000\left( N \right)}}{{1963.5\;(m{m^2})}} = 81.487\;\left( {\frac{N}{{m{m^2}}}} \right)\)
Maximum instantaneous elongation is given by,
\(\delta L = \;\frac{\sigma }{E} \times L\)
\(\delta L = \;\frac{{81.487\left( {\frac{N}{{m{m^2}}}} \right)}}{{200 \times 1000\left( {\frac{N}{{m{m^2}}}} \right)}} \times 4000\left( {mm} \right)\)
∴ δL = 1.63 mm