Question

A shaft is supported in bearings 1.8 m apart and projects 0.45 m beyond bearings at each end. The shaft carries three pulleys one at each end and one at the middle of its length. The mass of end pulleys is 48 kg and 20 kg and their centre of gravity are 15 mm and 12.5 mm respectively from the shaft axis. The centre pulley has a mass of 56 kg and its centre of gravity is 15 mm from the shaft axis. If the pulleys are arranged so as to give static balance, determine :

1. relative angular positions of the pulleys, and 

2. dynamic forces produced on the bearings when the shaft rotates at 300 r.p.m.

Answer 1

m_A = 48 kg ; m_C = 20 kg ; r_A = 15 mm = 0.015 m ; r_C = 12.5 mm = 0.0125 m ; m_B = 56 kg; r_B = 15 mm = 0.015 m ; N = 300 r.p.m. or 0) = 2 rt x 300/60 = 31.42 rad/s.

1.  Relative angular position of the pulleys The position of the shaft and pulleys is shown in Fig.(a) Let m_L and m_M = Mass at the bearings Land M, and r_L and r_M = Radius of rotation of the masses at Land M respectively. Assuming the plane of bearing L as reference plane, the data may be tabulated as below:

Plane (1)	Mass (m) kg (2)	Radius (r) m (3)	Cent force/ω2(m.r) kg-m(4)	Distance from plane L(l)m(5)	Couple/ω2(m.r.l) kg-m2(6)
A L(R.P) B M C	48 mL 56 mM 20	.015 rL 0.015 rM 0.0125	0.72 mL.rL 0.84 mM.rM 0.25	-0.45 0 0.9 1.8 2.25 0.72 mL.rM 0.25	-0.324 0 0.756 1.8 mMrM 0.5625

First of all, draw the force polygon to some suitable scale, as shown in Fig. (c), from the data given in Table (column 4). It is assumed that the mass of pulley B acts in a vertical direction. We know that for the static balance of the pulleys, the centre of gravity of the system must lie on the axis of rotation. Therefore a force polygon must be a closed figure. Now in Fig.(b), draw OA parallel to vector bc and OC' parallel to vector co. By measurement, we find that Angle between pulleys B and A 

= 1610 Ans. Angle between pulleys A and C = 760 and Angle between pulleys C and B = 1230.

Fig (a) Position of shaft of pulleys

Fig (b) Angular position of pulleys.

2. Dynamic forces at the two bearings In order to find the dynamic forces (or reactions) at the two bearings Land M, let us first calculate the values of m L. r _L and m M.r_M as discussed below: 

Fig (c) Force polygon

Fig (d) Couple polygon

Fig (e)  Force polygon

1. Draw the couple polygon to some suitable scale, as shown in Fig.  (dL from the data given in Table (column 6). 

The closing side of the polygon (vector co) represents the balanced couple and is proportional to 1.8 m M.r_M. By measurement, we find that 1.8 m M. r_M = vector c₀ = 0.97 kg-m₂ or m M. r_M = 0.54 kg-m

Dynamic force at the bearing M = m_M.r_M.co₂ = 0.54 (31.42) = 533 N

2.  Now draw the force polygon, as shown in Fig.(e), from the data given in Table (column 4) and taking 

m_M.r_M = 0.54 kg-m. The closing side of the polygon (vector do) represents the balanced force and is proportional to m L.r_L. By measurement, we find that m_L.r_L = 0.54 kg-m

Dynamic force at the bearing L = m_L.r_L.CO₂ =  0.54 (31.42)2 = 533 N