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

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

3. 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 36

4. RESULTANT OF COPLANAR NON CONCURRENT
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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
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5. 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.
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6. MOMENT OF A FORCE ABOUT AN AXIS
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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.
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7. 7/23/2018 4 A B d F
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8. Consider the example of a pipe wrench. The force applied which is
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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’.
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9. It is computed as the product of the
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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
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10. The sense is obtained by the ‘Right Hand Thumb’ rule.
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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.
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11. VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
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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
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12. R Sin = P Sin + Q Sin  ----(1) From le AOB, p/AO = Sin 
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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
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13. VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION
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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.
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14. APPLICATIONS OF VARIGNON’S THEOREM
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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.
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15. in direction form a ‘couple’. The algebraic summation of the
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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
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16. of the forces. The sum of moments
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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
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17. 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.
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18. Moment of this couple, M = F × d.
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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).
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