Question

m₁ = 0.8kn·m, m₂ = 2.3kn·m, m₃ = 1.5kn·m, ab: d₁ = 4cm, bc: d₂ = 7cm。g = 80gpa, calculate φₐ₈ and φₐ₆。

Explanation:

Step1: Recall the formula for angle of twist

The formula for the angle of twist $\phi=\frac{TL}{JG}$, where $T$ is the torque, $L$ is the length of the shaft segment, $J = \frac{\pi d^{4}}{32}$ is the polar - moment of inertia of the circular cross - section, and $G$ is the shear modulus.

Step2: Calculate the polar - moment of inertia for segment AB

For segment AB with $d_1 = 4cm=0.04m$, $J_1=\frac{\pi d_1^{4}}{32}=\frac{\pi\times(0.04)^{4}}{32}m^{4}$.

Step3: Calculate the angle of twist for segment AB

The torque in segment AB is $T_{AB}=M_1 = 0.8\times10^{3}N\cdot m$, and $L_{AB}=0.8m$, $G = 80\times10^{9}Pa$. Using $\phi_{AB}=\frac{T_{AB}L_{AB}}{J_1G}$, we substitute the values:

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Step4: Calculate the polar - moment of inertia for segment BC

For segment BC with $d_2 = 7cm = 0.07m$, $J_2=\frac{\pi d_2^{4}}{32}=\frac{\pi\times(0.07)^{4}}{32}m^{4}$.

Step5: Calculate the torque in segment BC

The torque in segment BC is $T_{BC}=M_1 - M_2=0.8\times10^{3}-2.3\times 10^{3}=- 1.5\times10^{3}N\cdot m$.

Step6: Calculate the angle of twist for segment BC

$L_{BC}=1.0m$, using $\phi_{BC}=\frac{T_{BC}L_{BC}}{J_2G}$.

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Step7: Calculate the angle of twist for segment AC

$\phi_{AC}=\phi_{AB}+\phi_{BC}$.

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Answer:

$\phi_{AB}\approx0.556\text{ rad}$, $\phi_{AC}\approx0.363\text{ rad}$