A compound bar is made of steel rod 19 mm in diameter surrounded by the closely fitting brass tube of 32mm outside diameter and two are securely fixed together at ends. Calculate the value of 'G' for brass if the angle of the twist over a length of 1 m is 7.2 degree when compound bar is subjected to twisting couple of 520 Nm.

Also calculate the maximum shear stress in the two material. Gs=80GPa (steel).

For the given composite shaft

the diameter of the steel rod is 19 mm (d)

torque is T= $(520\times 10^3)$

the rigidity modulus of steel is given by $8\times 10^9 {N\over mm^2}$

rigidity modulus of brass tube is to be found out

outside dia of brass tube is 32

length if the composite bar is given by L=1m=1000mm

common angle of twist is $ \theta = 7.2^0=0.125 rad $

$\tau \let\sp=^ [s]$ =shear stress in steel $\tau \let\sp=^ [b]$ = shear stress in brass

polar moment of inertia of steel rod and the Brass rod is given by

$J \let\sp=^ [s] $ = $\pi\times d^4\over 32 $ =12794.234

$J \let\sp=^ [b]$ = $\pi\times(D^4-d^4)\over 32 $= 90149.47

T = T'+T'' (1)

where T' =torque transmitted by Steel rod T'' = torque transmitted by the brass rod as the angle of twist is the same for both the rods using torsion equation we can calculate the torque transmitted in each rod

$T\over J$=$C\times\theta\over L$ (2)

T ' = 127942.3 by substuting T' in (1) T'' = 872057.7 and rigidity modulus of brass is given by $7.7018\times10^4 $ shear stress is calculated by $2\tau\over D$ = $ T\over J$ (3) by substituting the value of T' and T'', polar moment of steel and brass rods and diameter of steel rod we get the shear stress on the steel and brass rods are $\tau \let\sp=^ [s]$ =shear stress in steel = 95 $N\over mm^2$

$\tau \let\sp=^ [b]$ = shear stress in brass = 154.77 $N\over mm^2$

radius of brass=32/2=16 mm and radius of steel= 0.5x19=9.5 mm