Respuesta :
Answer:
The maximum shear stress in shaft AB, [tex]T_{ABmax}[/tex] is 15 MPa
The maximum shear stress in shaft CD, [tex]T_{CDmax}[/tex] is 45.9 MPa
Explanation:
The formula for a shaft polar moment of inertia, J is given by
J = [tex]\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}[/tex]
Therefore, we have
[tex]J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}[/tex]
Where:
D[tex]_{AB}[/tex] = Diameter of shaft AB = 30 mm = 0.03 m
r[tex]_{AB}[/tex] = Radius of shaft AB = 15 mm = 0.015 m
∴ [tex]J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}[/tex] = 7.95 × 10⁻⁸ m⁴
and
[tex]J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}[/tex]
Where:
D[tex]_{CD}[/tex] = Diameter of shaft CD = 36 mm = 0.036 m
r[tex]_{CD}[/tex] = Radius of shaft CD = 18 mm = 0.018 m
Therefore,
[tex]J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}[/tex] = 1.65 × 10⁻⁷ m⁴
Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;
1.58 °= [tex]1.58 \times \frac{2\pi }{360}[/tex] rad = 2.76 × 10⁻² rad.
That is [tex]\phi_r[/tex] = 2.76 × 10⁻² rad.
However [tex]\phi_r = \phi_{C/D} - \phi_{B/A}[/tex]
Where:
[tex]\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}[/tex] and
[tex]\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}[/tex]
[tex]T_{AB}[/tex] and [tex]T_{CD}[/tex]= Torque on shaft AB and CD respectively
[tex]T_{AB}[/tex] = Required
[tex]T_{CD}[/tex]= 500 N·m
[tex]L_{AB}[/tex] and [tex]L_{CD}[/tex] = Length of shafts AB an CD respectively
[tex]L_{AB}[/tex] = 600 mm = 0.6 m
[tex]L_{CD}[/tex] = 900 mm = 0.9 m
G = Shear modulus of the material = 77.2 GPa
Therefore;
[tex]\phi_r = \phi_{C/D} - \phi_{B/A}[/tex] =[tex]\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}[/tex]
2.76 × 10⁻² rad =[tex]\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}[/tex]
=[tex]\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}[/tex]
Therefore;
T[tex]_{AB}[/tex] = 79.54 N.m
Where T = [tex]T_{AB}[/tex] + T[tex]_{CD}[/tex] =
Therefore T[tex]_{CD total }[/tex] = 500 - 79.54 = 420.46 N·m
τ[tex]_{max}[/tex] = [tex]\frac{T\times R}{J}[/tex]
[tex]\tau_{ABmax}[/tex] = [tex]\frac{T_{AB}\times R_{AB}}{J_{AB}}[/tex] = [tex]\frac{79.54\times 0.015}{7.95\times 10^{-8}}[/tex] = 15 MPa
[tex]\tau_{CDmax}[/tex] = [tex]\frac{T_{CD}\times R_{CD}}{J_{CD}}[/tex] = [tex]\frac{420.46\times 0.018}{1.65\times 10^{-7}}[/tex] = 45.9 MPa
