Strength of Materials Test 2

Result

Strength of Materials Test 2

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Question 1
1 / -0

The volume of a tensile specimen remains constant during plastic deformation under the influence of an axial load and its diameter changes from 2d to 1.5d. The true strain induced in the specimen is

A
1.78

B
-1.78

C
0.575

D
-0.575

Solution

Concept:
True Strain: It is given as
\({\varepsilon _t} =\ln\frac{\ell }{{{\ell _0}}}\)
Where, ℓ = final length, ℓ₀ = initial length
Calculation:
Let the initial length be ‘ℓ’ and final length be ℓ’.
Since volume remains constant for the specimen.
∴ Initial volume = Final volume
\(\Rightarrow \frac{\pi }{4}{\left( {2d} \right)^2} \times \ell = \frac{\pi }{4} \times {\left( {1.5\;d} \right)^2} \times \ell {\rm{'}}\)
\(\Rightarrow \frac{{\ell '}}{\ell } = \frac{4}{{2.25}}\)
From the definition of true strain
\({\varepsilon _t} = \ln\;\left( {\frac{{\ell '}}{\ell }} \right) = \ln\;\left( {\frac{4}{{2.25}}} \right) = 0.575\)
Question 2
1 / -0

The ratio of bulk modulus to the elastic modulus of a material having Poisson's ratio 0.35 will be ______ . (up to two decimal place)

A
1.1-1.15

B
1.14-1.151

C
1.17-1.156

D
1.19-1.152

Solution

Concept:
The relationship between elastic modulus, bulk modulus and Poisson's ratio is given by
\(E=3K(1-2\mu)\)
The relationship between elastic modulus, modulus of rigidity and Poisson's ratio is given by
\(E=2G(1+\mu)\)
The relationship between elastic modulus, modulus of rigidity and bulk modulus is given by
\(E={9KG\over{G+3K}}\)
Hence, combining all these equations and solving it together we can get,
\(\begin{array}{l} \mu = \frac{{3K - 2G}}{{6K + 2G}} \end{array}\)
Calculation:
\(E=3K(1-2\mu)\)
\(E=3K(1-2\times 0.35)\)
\(E=3K(0.3)\)
\({E\over K}=3\times 0.3\)
\({K\over E}={1\over0.9}=1.11\)
Question 3
1 / -0

A steel rod 3 m long with a cross sectional area of 250 mm² is stretched between two fixed points. The tensile force is 12000 N at 40°C. Using E = 200 × 10³ N/mm² and α = 11.7 × 10^-6 mm/mm°C.
Calculate the temperature at which the tensile stress in the bar will be 100 N/mm².

A
22.2°C

B
62.2°C

C
17.8°C

D
34.5°C

Solution

Explanation:
The rod is stretched so there will be initial tensile stress.
Initial stress \(= \frac{\sigma }{A} = \frac{{12000}}{{250}} = 48\;N/m{m^2}\)
Final stress = 100 N/mm²
Additional stress required due to temperature change:
σ_th = σ_fin = σ_ini = 100 – 48 = 52 N/mm²
Thermal stress:
σ_th = αΔTE
52 = 11.7 × 10^-6 × ΔT × 200 × 10³
ΔT = T₂ – T_i = 22.22 °C
Note: - (Temperature will be reduced as tensile stress needs to be generated)
40°– T2 = 22.22 ⇒ T₂ = 17.8°C