Answer:
the maximum shear stress in the bar = 9.55 MPa
the twisting angle = [tex]0.146^0[/tex]
Explanation:
From the diagram below;
Torque (T) = 2 kN × 5 mm + 2 kN × 5 mm
= [tex]2*10^3 *0.05 + 2*10^3 *0.05[/tex]
= 200 N.m
Now; If we divide the shaft into two parts AC & BC
Then :
[tex]T = T_1 + T_2 = 200[/tex] ------------ equation (1)
At the junction :
[tex]\phi _{AC} = \phi _{BC}[/tex]
[tex]\phi = \frac{Tl}{GJ}[/tex]
⇒ [tex]\frac{T_1 \ l_{AC}}{GJ_1}= \frac{T_2 \ l_{BC}}{GJ_2}[/tex]
[tex]T_1l_{AC} = T_2 l_{BC}[/tex]
[tex]T_1*400 = T_2 * 600\\T_1 = \frac {600T_2}{400}\\T_1 = 1.5 T_2[/tex]
replace [tex]T_1 = 1.5 T_2[/tex] into above equation (1)
[tex]1.5 T_2 +T_2 = 200\\2.5 T_2 = 200\\T_2 = \frac{200}{2.5}\\T_2 = 80 N.m[/tex]
Again:
[tex]T_1 +T_2 = 200\\T_1 + 80 = 200\\T_1 = 200 - 80\\T_1 = 120 N.m[/tex]
Now ; we can deduce that the maximum shear comes from [tex]\phi_{AC}[/tex] since [tex]T_1 = 120 \ N.m[/tex]
So;
[tex]\gamma_{max} = \frac{T_1 *R}{ \frac {\pi D^4}{32}}[/tex]
where;
R = 20 mm
D = 40 mm = 0.04 m
Then;
[tex]\gamma_{max} = \frac{120 *20*10^{-3}}{ \frac {\pi (0.04)^4}{32}}[/tex]
[tex]\gamma_{max} = 9.55 \ MPa[/tex]
Thus ; the maximum shear stress in the bar = 9.55 MPa
b)
Again: [tex]\phi_c = \frac{T_1l_{AC}}{JG} =\frac{T_2l_{AC}}{JG}[/tex]
[tex]\phi_c = \frac{120*0.4}{75*10^9*\pi *\frac{(0.04)^2}{32}}[/tex]
[tex]\phi_c = 2.55 *10^{-3} \ rads[/tex]
[tex]\phi_c = 0.146^0[/tex]
Thus, the twisting angle = [tex]0.146^0[/tex]
