Answer to Question #124168 in Mechanical Engineering for Nikitha Dharana

Question

The playground roundabout, shown below, has a central shaft attached to the ground with a bearing at the top that supports the roundabout. As part of a preliminary design, you have been asked to check the effect of an impact resulting from the top bearing suddenly seizing.

Use the following data:

Mass of roundabout 150 kg and radius of gyration 1.2 m

Mass of people (6 at 55 kg each) = 330 kg and radius of gyration 1.5 m

Speed of rotation 0.8 turns/sec

The central shaft extends 1.2 m from the ground and is made from a steel tube, 120 mm outside diameter and 5 mm

thick. The modulus of rigidity G = 80 GPa and the elastic modulus E = 200 GPa.

a) Determine the maximum angular displacement of the shaft due to the impact.

b) Determine the maximum shear stress in the shaft due to the impact

Answer 1

As here figure is missing so i m asuming my own condition and solving it

length of shaft= 1.2 m , outside diameter=$d_o=120$ mm, $d_i=$ 115 mm,Modulus of rigidity= G=80 GPa, Modulus of elasticity=E=200 GPa,

Mass of round about = 150 kg, and radius of gyration 1.2 m

Mass of people (6 at 55 kg)=330 kg and radius of gyration=1.5 m

and speed of rotation =0.8 turns/sec

(a) For maximum angular of the shaft due to impacts

$\frac{T}{J}= \frac{G \theta}{l}$ , here J= polar moment of inertia

$I= M_1(K_1^2) + M_2(K_2^2)$

$I= 150(1.2^2) + 330(1.5^2)$

$I= 958.5,T= 958.5 \times 0.8 = 766.8$

$\frac{766.8\times1000}{(\frac{\pi}{32})(120^4 - 115^4)}=\frac{80 \times 1000 \times \theta}{1200}$

$\theta= 0.0036$

(b)

And for the shear stress calculation

$\frac{\tau}{R}= \frac{G\theta}{l}$

$\frac{\tau}{60}= \frac{80\times 1000\times 0.0036}{1200}$

$\tau = 14.4 MPa$