Sakalchand Patel College 7/23/2018 Sakalchand Patel College of Engineering,Visnagar 1

#Mechenical – 1 #Batch – t1 7/23/2018 Mechanics Of Solid #Mechenical – 1 #Batch – t1 Sr No:- Name:- Enrollment No:- Roll no:- 1. Patel sumitkumar S 130400106100 36

Structure Analysis II Force System 1. Patel sumitkumar S 130400106100 7/23/2018 Structure Analysis II Force System Sr No:- Name:- Enrollment No:- Roll no:- 1. Patel sumitkumar S 130400106100 36

RESULTANT OF COPLANAR NON CONCURRENT 7/23/2018 RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM   Coplanar Non-concurrent Force System:   This is the force system in which lines of action of individual forces lie in the same plane but act at different points of application. F2 F2 F1 F1 F5 F3 F3 F4 Fig. 1 Fig. 2 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 4 4

7/23/2018 2 TYPES Parallel Force System – Lines of action of individual forces are parallel to each other. Non-Parallel Force System – Lines of action of the forces are not parallel to each other. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 5 5

MOMENT OF A FORCE ABOUT AN AXIS 7/23/2018 3 MOMENT OF A FORCE ABOUT AN AXIS The applied force can also tend to rotate the body about an axis, in addition to causing translatory motion. This rotational tendency is known as moment. Definition: Moment is the tendency of a force to make a rigid body rotate about an axis. This is a vector quantity. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 6 6

7/23/2018 4 A B d F www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 7 7

Consider the example of a pipe wrench. The force applied which is 7/23/2018 5 EXAMPLE FOR MOMENT Consider the example of a pipe wrench. The force applied which is perpendicular to the handle of the wrench tends to rotate the pipe about its vertical axis. The magnitude of this tendency depends both on the magnitude of the force and the moment arm ‘d’. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 8 8

It is computed as the product of the 7/23/2018 MAGNITUDE OF MOMENT 6 It is computed as the product of the the perpendicular distance to the line of action of the force from the moment center and the magnitude of the force. MA = d×F Unit – Unit of Force × Unit of distance kN-m, N-mm etc. B A d F www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 9 9

The sense is obtained by the ‘Right Hand Thumb’ rule. 7/23/2018 7 SENSE OF THE MOMENT The sense is obtained by the ‘Right Hand Thumb’ rule. ‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the +ve sense of the moment vector’. M M For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or vice- versa. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 10 10

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS) 7/23/2018 8 Statement: The moment of a force about a moment center (axis) is equal to the algebraic sum of the moments of the component forces about the same moment center (axis). Proof (by Scalar Formulation): Let ‘R’ be the given force. ‘P’ & ‘Q’ be the component forces of ‘R’. ‘O’ be the moment center. p, r, and q be the moment arms of P, R, and Q respectively from ‘O’. , , and  be the inclinations of ‘P’, ‘R’, and ‘Q’ respectively w.r.t. the X – axis. Y Ry Q R Qy q P r Py  p   X A O www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 11 11

R Sin = P Sin + Q Sin  ----(1) From le AOB, p/AO = Sin  7/23/2018 9 Y We have, Ry = Py + Qy R Sin = P Sin + Q Sin  ----(1) From le AOB, p/AO = Sin  From le AOC, r/AO = Sin  From le AOD, q/AO = Sin  From (1), R ×(r/AO) = P ×(p/AO) + Q ×(q/AO) i.e., R × r = P × p + Q × q Moment of the resultant R about ‘O’ = algebraic sum of moments of the component forces P & Q about same moment center ‘O’. Ry Q R Qy r  q P Py  p  O X A www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 12 12

VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION 7/23/2018 VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION 10 Consider three forces F1, F2, and F3 concurrent at point ‘A’ as shown in fig. Let r be the position vector of ‘A’ w.r.t ‘O’. The sum of the moments about ‘O’ for these three forces by cross-product is, Mo = ∑(r×F) = (r×F1) + (r×F2) + (r×F3). By the property of cross product, Mo = r × (F1+F2+F3) = r × R where, R is the resultant of the three forces. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 13 13

APPLICATIONS OF VARIGNON’S THEOREM 7/23/2018 11 It simplifies the computation of moments by judiciously selecting the moment center. The moment can be determined by resolving a force into X & Y components, because finding x & y distances in many circumstances may be easier than finding the perpendicular distance (d) from the moment center to the line of action. 2. Location of resultant - location of line of action of the resultant in the case of non-concurrent force systems, is an additional information required, which can be worked out using Varignon’s theorem. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 14 14

in direction form a ‘couple’. The algebraic summation of the 7/23/2018 12 Two parallel, non collinear forces (separated by a certain distance) that are equal in magnitude and opposite in direction form a ‘couple’. The algebraic summation of the two forces forming the couple is zero. Hence, a couple does not produce any translation, but produces only rotation. F d F www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 15 15

of the forces. The sum of moments 7/23/2018 13 Moment of a Couple: Consider two equal and opposite forces separated by a distance ‘d’. Let ‘O’ be the moment center at a distance ‘a’ from one of the forces. The sum of moments of two forces about the point ‘O’ is, + ∑ Mo = -F × ( a + d) + F × a = -F× d Thus, the moment of the couple about ‘O’ is independent of the location, as it is independent of the distance ‘a’. The moment of a couple about any point is a constant and is equal to the product of one of the forces and the perpendicular distance between them. F d a F O www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 16 16

7/23/2018 14 F F F F F d Q Q M=F × d = = P P F Fig. (b) Fig. (a) Fig. (c) A given force F applied at a point can be replaced by an equal force applied at another point Q, together with a couple which is equivalent to the original system. Two equal and opposite forces of same magnitude F and parallel to the force F at P are introduced at Q. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 17 17

Moment of this couple, M = F × d. 7/23/2018 15 F F F F M=F × d d Q Q = = P P Fig. (a) F Fig. (b) Fig. (c) Of these three forces, two forces i.e., one at P and the other oppositely directed at Q form a couple. Moment of this couple, M = F × d. The third force at Q is acting in the same direction as that at P. The system in Fig. (c) is equivalent to the system in Fig. (a). www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 18 18