Question

Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and permit a 1.58 rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. knowing that g 5 77.2 gpa, determine the maximum shearing stress in each shaft when a torque of t of magnitude 500 n?m is applied to the flange indicated. 3.57 the torque t is applied to flange
b.

Answer 1

Answer:

The maximum shear stress in shaft AB, $T_{ABmax}$ is 15 MPa

The maximum shear stress in shaft CD,  $T_{CDmax}$ is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = $\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}$

Therefore, we have

$J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}$

Where:

D$_{AB}$ = Diameter of shaft AB = 30 mm = 0.03 m

r$_{AB}$ = Radius of shaft AB = 15 mm = 0.015 m

∴ $J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}$ = 7.95 × 10⁻⁸ m⁴

and

$J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}$

Where:

D$_{CD}$ = Diameter of shaft CD = 36 mm = 0.036 m

r$_{CD}$ = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

$J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}$ = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= $1.58 \times \frac{2\pi }{360}$ rad = 2.76 × 10⁻² rad.

That is $\phi_r$ = 2.76 × 10⁻² rad.

However  $\phi_r = \phi_{C/D} - \phi_{B/A}$  

Where:

$\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$ and

$\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}$

$T_{AB}$ and $T_{CD}$= Torque on shaft AB and CD respectively

$T_{AB}$  = Required

$T_{CD}$= 500 N·m

$L_{AB}$ and $L_{CD}$ = Length of shafts AB an CD respectively

$L_{AB}$  = 600 mm = 0.6 m

$L_{CD}$ = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

$\phi_r = \phi_{C/D} - \phi_{B/A}$  =$\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

2.76 × 10⁻² rad =$\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

=$\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}$

Therefore;

T$_{AB}$ =  79.54 N.m

Where T = $T_{AB}$ + T$_{CD}$ =

Therefore T$_{CD total }$ = 500 - 79.54 = 420.46 N·m

τ$_{max}$ = $\frac{T\times R}{J}$

$\tau_{ABmax}$ = $\frac{T_{AB}\times R_{AB}}{J_{AB}}$ =  $\frac{79.54\times 0.015}{7.95\times 10^{-8}}$ = 15 MPa

$\tau_{CDmax}$ = $\frac{T_{CD}\times R_{CD}}{J_{CD}}$ = $\frac{420.46\times 0.018}{1.65\times 10^{-7}}$ = 45.9 MPa