Question

# In tensile test the sample with 172 mm length is subjected to load of $35 KN$ and the deformation is $0.5 mm$, knowing that the specimen diameter is $1.2 cm$, determine The modulus of elasticity:- A. 212 Gpa B. $106 Mpa$ C. $212 Mpa$ D. $106 Gpa$

Solution

Verified

The expression for uniaxial deflection is:

$\Delta L = \dfrac{PL}{AE}$

From here we can express the modulus of elasticity as:

$E=\dfrac{PL}{A\Delta L}$

Where:

$P=35 \text{ kN}$ - applied force $L=172\text{ mm}$-length of the rod $\Delta L=0.5\text{ mm}$-deflection of the rod $A=6^2\pi\text{ mm}^2$-area of cross-section

Put all the values in expression for $E$ to get:

$E=\dfrac{35\cdot172}{6^2\pi\cdot0.5}=106.46\text{ }\dfrac{\text{kN}}{\text{mm}^2}=106.46\text{ MPa}$

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