Answer:
a) the maximum shear stress Ļ[tex]_{max}[/tex] the bar is 16T[tex]_{max}[/tex] /ĻdĀ³
b) the angle of twist between the ends of the bar is 16tLĀ² / ĻGdā“ Ā
Explanation:
Given the data in the question, as illustrated in the image below;
d is the diameter of the prismatic bar of length AB
t is the intensity of distributed torque
(a) Determine the maximum shear stress tmax in the bar
Maximum Applied torque Ā T_max = tL
we know that;
shear stress Ļ = 16T/ĻdĀ³
where d is the diameter
so
Ļ[tex]_{max}[/tex] = 16T[tex]_{max}[/tex] /ĻdĀ³
Therefore, the maximum shear stress Ļ[tex]_{max}[/tex] the bar is 16T[tex]_{max}[/tex] /ĻdĀ³
(b) Determine the angle of twist between the ends of the bar.
let theta ([tex]\theta[/tex]) be the angle of twist
polar moment of inertia [tex]I_p}[/tex] = Ļdā“/32
now from the second image;
lets length dx which is at distance of "x" from "B"
Torque distance x
T(x) = tx
Elemental angle twist = d[tex]\theta[/tex] = T(x)dx / G[tex]I_{p}[/tex]
so
d[tex]\theta[/tex] = tx.dx / G(Ļdā“/32)
d[tex]\theta[/tex] = 32tx.dx / ĻGdā“
so total angle of twist [tex]\theta[/tex] will be;
[tex]\theta[/tex] = Ā [tex]\int\limits^L_0 \, d\theta[/tex]
[tex]\theta[/tex] = Ā [tex]\int\limits^L_0 \,[/tex] 32tx.dx / ĻGdā“
[tex]\theta[/tex] = 32t / ĻGdā“ Ā [tex]\int\limits^L_0 \, xdx[/tex]
[tex]\theta[/tex] = 32t / ĻGdā“ [ LĀ²/2]
[tex]\theta[/tex] = 16tLĀ² / ĻGdā“ Ā
Therefore, Ā the angle of twist between the ends of the bar is 16tLĀ² / ĻGdā“ Ā
