(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Calculating torque on a portable cement mixer which is to be designed as a hollow cylinder with a closed base that rotates on a shaft (bearing). Dimentions being 0.5m diameter (D) and a height of 0.9m (h); the thickness of the cylinder is 0.006m (t). the drum is tilted at an angle of 15 degrees (@) to horizontal. it has aggregate of mass 363 kg (Ma) inside it. the drum operational speed is 10 rpm (sp). the drum is made of mild steel.

2. Relevant equations

Torque = moment of inertia * angular acceleration

moment of inertia = mass * radius of gyration^2 (for hollow cylinder)

F = mg

density of mild steel (ds) = 7860 kg/m3

final angular velocity = initial angular velocity + (angular acceleration * time)

3. The attempt at a solution

r = D/2

volume of steel used on cylinder (Vc) = [(pi * r^2 * t)] + pi * [ {(r + t)^2} - (r^2)] * h

mass of steel used on cylinder (Mc) = ds * Vc

I = (Mc + Ma) * r^2

assume it takes 2 second to reach speed of 10 rpm on start up

angular acceleration (aa) = {[(sp * 2 * pi) /60] / 2}

T = I * aa

but i know the material in the cylinder will occupy a certain volume and have a centroid where the weigth will act, and where does the effect of the drum being tilted come into affect

OR

T = Fr

F = (Mc + Ma) g

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# Homework Help: Design problem on a cement mixer

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