Question

In tensile test the sample with 172 mm length is subjected to load of 35KN35 KN and the deformation is 0.5mm0.5 mm, knowing that the specimen diameter is 1.2cm1.2 cm, determine The modulus of elasticity:- A. 212 Gpa B. 106Mpa106 Mpa C. 212Mpa212 Mpa D. 106Gpa106 Gpa

Solution

Verified

The expression for uniaxial deflection is:

ΔL=PLAE\Delta L = \dfrac{PL}{AE}

From here we can express the modulus of elasticity as:

E=PLAΔLE=\dfrac{PL}{A\Delta L}

Where:

P=35 kNP=35 \text{ kN} - applied force L=172 mmL=172\text{ mm}-length of the rod ΔL=0.5 mm\Delta L=0.5\text{ mm}-deflection of the rod A=62π mm2A=6^2\pi\text{ mm}^2-area of cross-section

Put all the values in expression for EE to get:

E=3517262π0.5=106.46 kNmm2=106.46 MPaE=\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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