Question

EGM3520L Materials Testing Laboratory 2020 Summer 6 75 Observations and Calculations Diameter of specimen D- Length of specimen L Rotation Rotation Torsion reading, R: angle, reading, R2 (Degrees) (N-M) (radians) 0 0 o Imm) (mm) Shear strain, Y (0.001radians) Shear Torsion, T (N-M) stress, (MPa) 6.92 3.24 4.99 10.02 0.12 3-2 4X10 4.8x10 3 76.39 14.90 x 103 7x10 s 117.66 0.332 15. 45x10 13.28x107 128.50 19.01 5.45 19.99 6.42 21.00 7.39 22.04 7.87 23.03 7.92 24.00 7.95 2L Dx Ф Y = TXD TED 2) 32 47 Name: Submission Date:

Answer 1

Since the torque reading is directly in N.m unit, we don't need to convert this.

Strain of the specimen is required in the unit 0.001 radian. This means a strain of 4.8 x 10^-3 radian will be written as 4.8.

With this...the chart will be

Diameter = 6 mm = 6 x 10-3 m Length = 75 mm = 75 x 10-3 m Polar moment of innertia (J) = ( 11.D4 )/32 = 1.272345 x 10-1° m4 Rotation reading, Torsion reading, Rotation angle, Torsion, T (N-Shear strain, y Sear stress, T R1 (Degrees) R2 (N-M) 0 (radians) M) (0.001 radians) (Mpa) 0 0 0.00 0 0.00 0.00 6.92 3.24 0.12 3.24 +4.83 76.39 10.02 4.99 0.17 4.99 7.00 117.66 19.01 5.45 0.33 5.45 13.27 128.50 19.99 6.42 0.35 6.42 13.96 151.37 21 7.39 0.37 7.39 14.66 174.25 22.04 7.87 0.38 7.87 15.39 185.56 23.03 7.92 0.40 7.92 16.08 186.74 24 7.95 0.42 7.95 16.76 187.45

For this data, if we plot stress strain curve, we get...

STRESS - STRAIN CURVE (SHEAR) 200.00 190.00 180.00 170.00 160.00 150.00 140.00 130.00 o 120.00 o 110.00 Sear stress, t (Mpa) 100.00 90.00 80.00 o 70.00 60.00 50.00 40.00 30.00 20.00 10.00 0.00 0 0.00 2.00 4.00 6.00 14.00 16.00 18.00 20.00 8.00 10.00 12.00 Shear strain, y (0.001 radians)

The modulus of rigidity, G is equal to the slope of the initial linear part. If we calculate the linear section to be between stress 0 MPa to 76.39 MPa, the modulus of rigidity will be equal to...

Modulus of rigidity

$G=\frac{(76.39 - 0)}{(4.83-0)} \frac{(MPa)}{(0.001 radians)}$

$\therefore G=15.816 GPa$