4. If the formula of sigma =500varepsilon ^0.5 exists between the true stress and true strain for a steel. Obtain amount of elastic strain at a strain equivalent to 0.25 ? (E=200GPa) 5.During a tensile test on a specimen of 1cm^2 cross-section,maximum load of 80 kN and area of cross-section at neck was 0.5cm^2 Calculate ultimate tensile strength UTS) of specimen.

Question

Answer 1

4. To find the elastic strain at a strain equivalent to 0.25, we can use the given formula

. We need to rearrange the formula to solve for

.Given:

\varepsilon

\varepsilon ^{0.5} = \frac{\sigma}{500}

\sigma

\varepsilon ^{0.5} = \frac{E \cdot \varepsilon}{500}

E

\varepsilon ^{0.5} = \frac{\varepsilon}{500/E}

\varepsilon = \left(\frac{\varepsilon}{500/E}\right)^2

\varepsilon = \frac{\varepsilon^2}{500^2/E^2}

E^2

E^2 \cdot \varepsilon = \frac{\varepsilon^2}{500^2}

\varepsilon

E \cdot \varepsilon = \frac{\varepsilon}{500^2}

E \cdot \varepsilon = \frac{\varepsilon}{250000}

E

\varepsilon = \frac{\varepsilon}{250000E}

\varepsilon = \frac{1}{250000}

\frac{1}{250000}

UTS = \frac{Force}{Area}

UTS = \frac{80000}{0.5 \times 10^{-4}}

UTS = 80000 \times 10^4

UTS = 800000000$Therefore, the ultimate tensile strength (UTS) of the specimen is 800 MPa.

Soru