Consider a circular shaft that is attached to a fixed support at one end. If a torque T is applied to the other end, the shaft will twist, with its free end rotating through an angle called the angle of twist $θ$ . Since the internal torque is constant throughout, the line AB on the surface of the shaft is twisted into the helix AB throughout an angle $ϕ$ .

Image source : Mechanics of Materials , by Ferdinand Beer, Jr., E. Russell Johnston, John DeWolf , David Mazurek
Shearing strain $\tan ϕ \approx ϕ \approx \frac{A A^{'}}{B A} \frac{r θ}{L}$

If  is the intensity of shear stress at the surface of the shaft Shear strain = $\frac{s h e a r s t r e s s}{m o d u l a s o f r e g i d i t y} = \frac{r θ}{L} = \frac{τ_{s}}{C}$

Observations & sample calculations for torsion test on solid circular mild steel shaft:

Average diameter of the sample d =11.17mm (sent to user via email)
Length of the sample L=190mm (sent to user via email)
Polar moment of Inertia J= $\frac{π}{32} (d^{4}) = 1528.31 m m^{4}$
Typical torque(Nm)vs twist angle(degrees):

Typical plot of Torque vs Angle of twist

Yield torque $t_{y i e l d}$ = 62.78Nm (observed from plot)

Yield stress= $\frac{t_{y i e l d} \times d}{2 \times j} = \frac{62.78 \times 11.17 \times 10^{- 3}}{2 \times 1528.31 \times 10^{- 2}}$ =229.435Mpa

Torque at Rupture $t_{r u p t u r e}$ = 149.11 Nm (observed from plot)

Modulus of rupture = $\frac{t_{r u p t u r e} \times d}{2 \times j} = \frac{149.11 \times 11.17 \times 10^{- 3}}{2 \times 1528.31 \times 10^{- 2}}$ =544.90MPa

Plot to extract the slope(slope at linear portion of the graph: within elastic limit)

Slope of torque vs twist plot = 443.17 Nm/rad

Modulus of rigidity $G = (\frac{T}{θ}) \times \frac{L}{J} \frac{443.17 \times 11.17 \times 10^{- 3}}{1528.31 \times 10^{- 2}}$