# Statics (Engineering Mechanics-i) STATICS (ENGINEERING MECHANICS-I) Force Systems, Moment, and couple moment By Dr. Fahed Alrshoudi 02/09/2020 1 Objectives To define what a force is and its types To demonstrate through examples how to find rectangular and nonrectangular components of a give force To explain scalar and vector definitions of moment of a force about an axis To derive expressions for scalar and vector calculations of moment of a given force about a point To illustrate the use of above expressions through a simple example

To define couple and couple moments To derive scalar and vector expressions of couple moments To illustrate the concept of couple through examples 02/09/2020 2 Force Systems Force: Action of one body on another. Specification include: Magnitude Direction Point of application (or line of action) Effects of a force External: Reactions Internal: Stresses and strains In dealing with the mechanics of rigid bodies concern is only to the net external effects of forces. 02/09/2020 3

Two-Dimensional Force Systems (Addition of two Forces) Parallelogram law F2 A 02/09/2020 Triangle law R F1 A R

F2 F1 4 Two-Dimensional Force Systems (Force Components) y j Fy Rectangular Components Non-rectangular Components F F2

A x i Fx Fx F cos ; Fy F sin 2 x 2 y F F F F Fx Fy

F Fx i Fy j 02/09/2020 F F1 90 0 Fy and tan Fx 1 F F1 F2 5 Law of sines and cosines Law of sines

B c A b Law of cosines a C D a b c sin A sin B sin C 02/09/2020

c A B b a C D c 2 a 2 b 2 2ab cos C c 2 a 2 b 2 2ab cos D 6 Problem-1 The force F has a magnitude of 600 N. Express F as a vector in terms of the ' unit vecto rs i and j . Identify t he x and y scalar components of F .

y F 600 N 300 x 02/09/2020 7 Solution y F 600 N Fy 600 cos 300 30 0 Fx 600 sin 30 02/09/2020 0

x 0 Fx 600 sin 30 ( i ) 0 Fy 600 cos 30 j F Fx Fy F 300i 520 j N 8 CHAPTER 2 FORCE SYSTEMS EXAMPLE - 1

02/09/2020 9 CHAPTER 2 FORCE SYSTEMS EXAMPLE 1 (CONTINUED) Rectangular components Problem-2 Determine the magnitude of the resultant R of the two forces (shown below), and the angle which R makes with the positive xaxis. 02/09/2020 T1 7 kN T2 5 kN O 150

450 x 11 Solution This problem can be viewed how two non-rectangular force components can be replaced by a single resultant force R. 2 2 T2 5 kN R 2

Use cosine law : c a b 2ab cos C 2 1 2 2 0 0 R T T 2T1 T2 cos( 45 15 ) 2 2 0 150 450

150 5 kN 0 7 5 2 5 7 cos( 45 15 ) T1 7 kN 450 x T1 7 kN 6.24 kN T2 5 kN sin sin( 450 150 )

sin( 450 150 ) sin T2 T2 R R 0 O 150 450 x sin( 60 ) 5 0.6939 sin 1 (0.6939) 43.90 6.24 450 43.90 450 88.90 sin

02/09/2020 12 Alternative Solution This method can be also discussed in rectangular components. Rx Fx T1 cos 450 T2 cos150 7 cos 450 5 cos150 0.12 kN Ry Fy T1 sin 450 T2 sin 150 7 sin 450 5 sin 150 6.24 kN R Rx2 Ry2 0.122 6.242 6.24 kN R 6.24 0 tan y tan 1 88.9 0.12 Rx 1 02/09/2020 T1 7 kN

T2 5 kN O 150 450 x 13 CHAPTER 2 FORCE SYSTEMS EXAMPLE - 2 02/09/2020 14 CHAPTER 2 FORCE SYSTEMS EXAMPLE 2 (CONTINUED) 02/09/2020

15 Moment 02/09/2020 16 Moment In addition to the tendency to move a body in the direction of its application, a force can also tend to rotate a body about an axis. The axis may be any line which neither intersects nor is parallel to the line of action of the force. M=Fd 02/09/2020 17 Moment about a Point Point of application of a force and line of action of a force are as in Figure:

F line of action of a force point of application The magnitude of the moment or tendency of the force to rotate the body about the axis perpendicular to the plane of the body is proportional both to the magnitude of the force and to the moment arm d, which is the perpendicular distance from the axis to the line of action of the force. 02/09/2020 18 Moment: Vector Definition In some two-dimensional and many of the three-dimensional problems to follow, it is convenient to use a vector approach for moment calculations.

M r F O r Position v ector that runs from the moment reference point A to any po int the line of action of F MA r F d Direction : The right - hand rule is used to identify the direction. O

Thus the moment of F about O-O may be identified as a vector pointing in the direction of thumb, with the fingers curled in the direction of the tendency to rotate. 02/09/2020 19 Clockwise and Counterclockwise moments: If tendency of the force is to rotate its moment arm in a counter clockwise manner it is called counterclockwise moment (CCW). We will assume CCW moment as a positive moment (+). If tendency of the force is to rotate its moment arm in a clockwise manner it is called clockwise moment (CW). We will assume CW moment as a negative moment (-). 02/09/2020 20 Zero moments - If line of action of the force intersects the axis of rotation, there is no moment of the force about this axis of rotation - If line of action of the force is parallel to the axis of rotation, there is no moment of the force about this axis of rotation

02/09/2020 21 Varignons Theorem The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point. 02/09/2020 22 Problems For the figure shown below, find moment of 400 N force about point A using: 1.Scalar approach 2.vector approach. y 3m O

45 x 0 400 N 6m A 02/09/2020 23 Scalar Approach 3m y O

45 x 0 400 N 6m A M A 400 cos 450 6 400 sin 450 3 M A 2545.2 N.m (Countercl ock wise) Ans. 02/09/2020 24 Vector Approach

M A r F (-3,0) 3m y O x 450 400 N 6m (0,-6) A

F 400 cos 450 ( i) 400 sin 450 ( j ) 282.8i 282.8 j r ( x2 x1 )i ( y2 y1 ) j ( 3 0)i (0 ( 6)) j 3i 6 j j i k M A r F 3 6 0 2545.2k N.m 282.8 282.8 0 02/09/2020 Ans. 25 02/09/2020

26 Alternative Solution (Using Varignons Theorem): Assuming that the counterclockwise moment is positive; MB = - Fx.(0.16m) - Fy.(0.2m) where Fx= 400 N and Fy= 693 N MB = -400N(0.16m)-693N(0.2m) = -202.6 N.m We can say MB = -203 N.m Since we assumed that the counterclockwise moment is positive and we obtained a negative moment value, this implies that the direction of MB is actually clockwise. So; the result is exactly the same as before, i.e. 02/09/2020 27 Couple and Couple Moment 02/09/2020

28 Couple Moment The moment produced by two equal, opposite, and non-collinear forces is called a couple. Mo=F(a+d)-Fa Mo=Fd Its direction is counterclockwise 02/09/2020 29 Vector We may also express the moment of a couple by using vector algebra. M rA F rB ( F ) rA F rB F (rA rB ) F r F

M r F where rA and rB are position vectors which run from point O to arbitrary points A and B on the lines of action of F and F, respectively. 02/09/2020 30 Equivalent Couples The Figure shows four different configurations of the same couple M. 02/09/2020 31 Force-Couple Systems The replacement of a force by a force and a couple is illustrated in the Figure, where the given force F acting at point A is replaced by an equal force F at some point B and the counterclockwise couple M=Fd. The transfer is seen in the middle figure, where the equal and opposite forces F and F are added at point B

without introducing any net external effects on the body. We now see that the original force at A and the equal and opposite one at B constitute the couple M=Fd, which is counterclockwise for the sample chosen, as shown in the right-hand part of the figure. Thus, we have replaced the original force at A by the same force acting at a different point B and a couple, without altering the external effects of the original force on 02/09/2020 the body. The combination of the force and couple in the right-hand part of the Figure is referred to as a forcecouple system. 32 Counterclockwise and Clockwise couples 02/09/2020 33 Problem-1 Calculate the combined moment of the two 2-kN forces, shown in the Figure, about point O and A using : (i) Scalar approach

y (ii)Vector approach 2 kN 1.6 m A 1.6 m 0.8 m 2 kN 02/09/2020 O x 34 y 2 kN 1.6 m

i- scalar As 2 kN forces are forming a couple, moments will be the same about A and O as couple moment is independent of moment centres. A 1.6 m 0.8 m 2 kN O x M O M A Fd 2 (1.6 1.6 0.8) 8 kN.m 02/09/2020 35 y

ii- vector (0, 3.2) As 2 kN forces are forming a couple, moments will be the same about A and O as couple moment is independent of moment centres. M r F 2( i) 1.6 m A 1.6 m 0.8 m 2i

r O x (0,-0.8) r ( x2 x1 )i ( y2 y1 ) j r (0 0)i (3.2 ( 0.8)) j 4 j m Therefore M O M A r F 4 j 2( i) 8k kN.m 02/09/2020 36 Problem-2 The top view of a revolving entrance door is shown in the Figure. Two persons simultaneously approach the door and exert forces of equal magnitude as shown. If the resulting moment about point O is 25 N.m, determine the force magnitude F. 02/09/2020 37

The two components F cos15 and -F cos15 will form a couple. The moment of this couple would be M O Fcos15o 1.6 25 N.m F 16.18 N Note : The components F sin 15 and -F sin 15 will not cause any moment as their lines of action are passing through O. 02/09/2020 38 Problem-3 A square plate of 200 mm 200 mm is subjected to two forces, each of magnitude 50 N, as shown in the figure: 1. Calculate the moment of the forces about points O, A, C, and D. 2. Find the moment of the forces about y-axis. y 100 mm E D 50 N

450 C 50 N 450 200 mm B 100 mm O 02/09/2020 A 200 mm x 39

y 1- 100 mm E D 50 N 450 C 50 N d 450 200 mm B

100 mm O A 200 mm x The perpendicu lar distance d between two 50 N forces 1002 1002 141.4 mm Since two 50 N forces form a couple. The moment of this couple is M Fd 50 141.4 7070 N.mm As couple is independen t of moment centers, M O M A M C M D M 7070 N.mm Ans. 02/09/2020 40 y 100 mm 2-

E D 50 N 450 C 50 N d 450 200 mm B 100 mm O

A 200 mm x Since forces are acting in xy plane, they will either intersect the axes or they will be parallel to them. Therefore, moment about x- or y-axis of all the coplanar forces = 0 02/09/2020 41 Problem-4 Replace the 12-kN force acting at point A by a force-couple system acting at point O . 12 kN 30 0 A

02/09/2020 y O x 4m 42 12 kN 30 0 A 4m y y

12 kN O x 30 0 A O MO x M O 12 d 12 4 sin 300 24 kNm 02/09/2020 43 Problem

A force of 400 N is applied at A to the handle of control level which is attached to the fixed shaft OB. Replace this force by an equivalent force at O and a couple. Describe this couple as a vector . M 44 Solution F 400( i ) 400i N r ( x2 x1 )i ( y2 y1 ) j ( z2 z1 )k

0i (200 0) j (125 0) k r 200 j 125k mm r 0.2 j 0.125k m i M r F 0.2 j k

0 0.125 0i 50 j 80k 400 0 0 M 50 j 80k N.m 45

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