Question

: The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the torque, determine the maximum shear stress in the bar. If the shear modulus of the steel is 75 GPa, determine the twisting angle at C.

Answer 1

Answer:

the maximum shear stress in the bar = 9.55 MPa

the twisting angle = $0.146^0$

Explanation:

From the diagram below;

Torque (T) = 2 kN × 5 mm +  2 kN × 5 mm

= $2*10^3 *0.05 + 2*10^3 *0.05$

= 200 N.m

Now; If we divide the shaft into two parts AC & BC

Then :

$T = T_1 + T_2 = 200$          ------------ equation (1)

At the junction :

$\phi _{AC} = \phi _{BC}$

$\phi = \frac{Tl}{GJ}$

⇒ $\frac{T_1 \ l_{AC}}{GJ_1}= \frac{T_2 \ l_{BC}}{GJ_2}$

$T_1l_{AC} = T_2 l_{BC}$

$T_1*400 = T_2 * 600\\T_1 = \frac {600T_2}{400}\\T_1 = 1.5 T_2$

replace $T_1 = 1.5 T_2$ into above equation (1)

$1.5 T_2 +T_2 = 200\\2.5 T_2 = 200\\T_2 = \frac{200}{2.5}\\T_2 = 80 N.m$

Again:

$T_1 +T_2 = 200\\T_1 + 80 = 200\\T_1 = 200 - 80\\T_1 = 120 N.m$

Now ; we can deduce that the maximum shear comes from $\phi_{AC}$ since $T_1 = 120 \ N.m$

So;

$\gamma_{max} = \frac{T_1 *R}{ \frac {\pi D^4}{32}}$

where;

R = 20 mm

D = 40 mm = 0.04 m

Then;

$\gamma_{max} = \frac{120 *20*10^{-3}}{ \frac {\pi (0.04)^4}{32}}$

$\gamma_{max} = 9.55 \ MPa$

Thus ; the maximum shear stress in the bar = 9.55 MPa

b)

Again: $\phi_c = \frac{T_1l_{AC}}{JG} =\frac{T_2l_{AC}}{JG}$

$\phi_c = \frac{120*0.4}{75*10^9*\pi *\frac{(0.04)^2}{32}}$

$\phi_c = 2.55 *10^{-3} \ rads$

$\phi_c = 0.146^0$

Thus, the twisting angle = $0.146^0$