Balancing rotating masses pdf
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA. Home current Explore. Swaying couple. The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure on the rails, which results in hammering action on the rails.
Kumar perpendicular to the line of stroke is known as a. Hence the couple is known as swaying couple. Problem No. The reciprocating mass per cylinder is 1. Open navigation menu. Close suggestions Search Search. User Settings. Skip carousel. Carousel Previous. Carousel Next. What is Scribd? Explore Ebooks. Bestsellers Editors' Picks All Ebooks. Explore Audiobooks. Bestsellers Editors' Picks All audiobooks. Explore Magazines. Editors' Picks All magazines. Explore Podcasts All podcasts.
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Mechanics of Machines 1 For: 2nd Year Mechanical Technology By: Taimoor Asim Centrifugal Force When a mass rotates in a circle, centripetal force acts on it, which is towards the center of the circle According to Newtons 3rd Law, centrifugal force acts on the mass in the opposite direction i. Kiran Varma. Jaren Gan. V Dhinakaran. Jose Prado. Muhammed Shameem N Edavannappara. Nitin Kukreja. Chong Jie Mee.
Muhummad Zeeshan. BalRam Dhiman. In this case, the mass m lies in the plane A and the balancing masses lie in the planes L and M, as shown in Fig. As discussed above, the following conditions must be satisfied in order to balance the system, i. Balancing of Several Masses Rotating in the Same Plane Consider any number of masses say four of magnitude m1, m2, m3 and m4 at distances of r1, r2, r3 and r4 from the axis of the rotating shaft.
Balancing of several masses rotating in the same plane. Analytical method The magnitude and direction of the balancing mass may be obtained, analytically, as discussed below : 1. A car assembly line. Note : This picture is given as additional information and is not a direct example of the current chapter. Chapter 21 : Balancing of Rotating Masses l 2. Resolve the centrifugal forces horizontally and vertically and find their sums, i.
The balancing force is then equal to the resultant force, but in opposite direction. Graphical method The magnitude and position of the balancing mass may also be obtained graphically as discussed below : 1. First of all, draw the space diagram with the positions of the several masses, as shown in Fig. Find out the centrifugal force or product of the mass and radius of rotation exerted by each mass on the rotating shaft. Now draw the vector diagram with the obtained centrifugal forces or the product of the masses and their radii of rotation , such that ab represents the centrifugal force exerted by the mass m1 or m1.
Similarly, draw bc, cd and de to represent centrifugal forces of other masses m2, m3 and m4 or m2. Now, as per polygon law of forces, the closing side ae represents the resultant force in magnitude and direction, as shown in Fig. The balancing force is, then, equal to the resultant force, but in opposite direction.
Four masses m1, m2, m3 and m4 are kg, kg, kg and kg respectively. The corresponding radii of rotation are 0. Find the position and magnitude of the balance mass required, if its radius of rotation is 0. But we shall solve the problem by both the methods one by one.
Analytical method Fig. Resolving m1. Graphical method The magnitude and the position of the balancing mass may also be found graphically as discussed below : 1.
First of all, draw the space diagram showing the positions of all the given masses as shown in Fig Since the centrifugal force of each mass is proportional to the product of the mass and radius, therefore m1. Now draw the vector diagram with the above values, to some suitable scale, as shown in Fig. The closing side of the polygon ae represents the resultant force. The balancing force is equal to the resultant force, but opposite in direction as shown in Fig.
Since the balancing force is proportional to m. Balancing of Several Masses Rotating in Different Planes When several masses revolve in different planes, they may be transferred to a reference plane briefly written as R. The effect of transferring a revolving mass in one plane to a reference plane is to cause a force of magnitude equal to the centrifugal force of the revolving mass to act in the reference plane, together with a couple of magnitude equal to the product of the force and the distance between the plane of rotation and the reference plane.
In order to have a complete balance of the several revolving masses in different planes, the following two conditions must be satisfied : 1. The forces in the reference plane must balance, i. The couples about the reference plane must balance, i. Let us now consider four masses m1, m2, m3 and m4 Diesel engine.
The relative angular positions of these masses are shown in the end view [Fig. The magnitude of the balancing masses mL and mM in planes L and M may be obtained as discussed below : 1. Take one of the planes, say L as the reference plane R. The distances of all the other planes to the left of the reference plane may be regarded as negative, and those to the right as positive. Tabulate the data as shown in Table The planes are tabulated in the same order in which they occur, reading from left to right.
Table Balancing of several masses rotating in different planes. A couple may be represented by a vector drawn perpendicular to the plane of the couple. The vector representing this couple is drawn in the plane of the paper and perpendicular to Om1 as shown by OC1 in Fig. The couple vectors as discussed above, are turned counter clockwise through a right angle for convenience of drawing as shown in Fig. We see that their relative positions remains unaffected.
Hence the couple vectors are drawn radially outwards for the masses on one side of the reference plane and radially inward for the masses on the other side of the reference plane. Now draw the couple polygon as shown in Fig. Since the balanced couple CM is proportional to mM. Now draw the force polygon as shown in Fig. The vector eo in the direction from e to o represents the balanced force. Since the balanced force is proportional to mL. Example A shaft carries four masses A, B, C and D of magnitude kg, kg, kg and kg respectively and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm in planes measured from A at mm, mm and mm.
The balancing masses are to be placed in planes X and Y. If the balancing masses revolve at a radius of mm, find their magnitudes and angular positions. The position of planes and angular position of the masses assuming the mass A as horizontal are shown in Fig. Assume the plane X as the reference plane R.
The data may be tabulated as shown in Table
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