Question #124168

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


Expert's answer

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

length of shaft= 1.2 m , outside diameter=do=120d_o=120 mm, di=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

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


I=M1(K12)+M2(K22)I= M_1(K_1^2) + M_2(K_2^2)

I=150(1.22)+330(1.52)I= 150(1.2^2) + 330(1.5^2)

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

766.8×1000(π32)(12041154)=80×1000×θ1200\frac{766.8\times1000}{(\frac{\pi}{32})(120^4 - 115^4)}=\frac{80 \times 1000 \times \theta}{1200}


θ=0.0036\theta= 0.0036

(b)

And for the shear stress calculation

τR=Gθl\frac{\tau}{R}= \frac{G\theta}{l}


τ60=80×1000×0.00361200\frac{\tau}{60}= \frac{80\times 1000\times 0.0036}{1200}


τ=14.4MPa\tau = 14.4 MPa




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