Ishik University / Sulaimani Architecture Department 1 Structure ARCH 214 Chapter -4- Force System Resultant 2 1
CHAPTER OBJECTIVES To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. To provide a method for finding the moment of a force about a specified axis. To define the moment of a couple. To present methods for determining the resultants of nonconcurrent force systems. To indicate how to reduce a simple distributed loading to a resultant force having a specified location. 3 CHAPTER OUTLINE Moment of a Force Scalar Formation Principle of Moments Moment of a Force about a Specified Axis Moment of a Couple Equivalent System Resultants of a Force and Couple System Reduction of a Simple Distributed Loading 4 2
4.1 MOMENT OF A FORCE SCALAR FORMATION Moment of a force about a point or axis a measure of the tendency of the force to cause a body to rotate about the point or axis. Case 1 Consider horizontal force F x, which acts perpendicular to the handle of the wrench and is located d y from the point O. 5 F x tends to turn the pipe about the z axis. The larger the force or the distance d y, the greater the turning effect. Torque tendency of rotation caused by F x or simple moment (M o ) z Moment axis (z) is perpendicular to shaded plane (x-y) F x and d y lies on the shaded plane (x-y) Moment axis (z) intersects the plane at point O 6 3
Case 2 Apply force F z to the wrench. Pipe does not rotate about z axis. Tendency to rotate about x axis. The pipe may not actually rotate F z creates tendency for rotation so moment (M o ) x is produced. Moment axis (x) is perpendicular to shaded plane (y-z). Fz and dy lies on the shaded plane (y-z). 7 Case 3 Apply force F y to the wrench. No moment is produced about point O. Lack of tendency to rotate as line of action passes through O. 8 4
In General Consider the force F and the point O which lies in the shaded plane. The moment M O about point O, or about an axis passing through O and perpendicular to the plane, is a vector quantity. Moment M O has its specified magnitude and direction. 9 Magnitude For magnitude of M O, M O = Fd where d = moment arm or perpendicular distance from the axis at point O to its line of action of the force. Units for moment is N.m. Direction Direction of MO is specified by using right hand rule. - fingers of the right hand are curled to follow the sense of rotation when force rotates about point O. - Thumb points along the moment axis to give the direction and sense of the moment vector. - Moment vector is upwards and perpendicular to the shaded plane. 10 5
Direction M O is shown by a vector arrow with a curl to distinguish it from force vector. Example (Fig b) M O is represented by the counterclockwise curl, which indicates the action of F. 11 Direction Arrowhead shows the sense of rotation caused by F Using the right hand rule, the direction and sense of the moment vector points out of the page. In 2D problems, moment of the force is found about a point O. Moment acts about an axis perpendicular to the plane containing F and d. Moment axis intersects the plane at point O. 12 6
Resultant Moment of a System of Coplanar Forces Resultant moment, M Ro = addition of the moments of all the forces algebraically since all moment forces are collinear. M Ro = Fd taking clockwise to be negative. o A clockwise curl is written along the equation to indicate that a positive moment if directed along the + z axis and negative along the z axis 13 14 7
Moment of a force does not always cause rotation. Force F tends to rotate the beam clockwise about A with moment. M A = Fd A Force F tends to rotate the beam counterclockwise about B with moment. M B = Fd B Hence support at A prevents the rotation. 15 16 8
17 Example 4.1 For each case, determine the moment of the force about point O. 18 9
Solution; Line of action is extended as a dashed line to establish moment arm d. Tendency to rotate is indicated and the orbit is shown as a colored curl. ( a) M ( b) M o o = (100N)(2m) = 200N. m( CW ) = (50N)(0.75m) = 37.5N. m( CW ) 19 Solution; ( c) M ( d) M ( e) M o o o = (40N)(4m + 2cos30 m) = 229N. m( CW ) = (60N)(1sin 45 m) = 42.4N. m( CCW ) = (7kN)(4m 1m) = 21.0kN. m( CCW ) 20 10
Example 4.2 Determine the moments of the 800N force acting on the frame about points A, B, C and D. 21 Solution; Scalar Analysis M = (800N)(2.5m) = 2000N.m(CW) M M A B C = (800N)(1.5m) = 1200N.m(CW) = (800N)(0m) = 0kN.m Line of action of F passes through C M = (800N)(0.5m) = 400N.m(CCW) D 22 11
Example 4.3 Determine the resultant moment of the four forces acting on the rod shown in figure about point O. 23 24 12
4.2 PRINCIPLES OF MOMENTS The guy cable exerts a force F on the pole and creates a moment about the base ata, M A = Fd If the force is replaced by F x and F y at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment. 25 F y create zero moment about A M A = F x h Apply principle of transmissibility and slide the force where line of action intersects the ground at C, F x create zero moment about A M A = F y b 26 13
Example 4.3 The force F acts at the end of the angle bracket. Determine the moment of the force about point O. 27 Solution Method 1: M O = 400sin30 N(0.2m)-400cos30 N(0.4m) = -98.6N.m = 98.6N.m (CCW) As a Cartesian vector, M O = {-98.6k}N.m 28 14
Solution Method 2: Express as Cartesian vector r = {0.4i 0.2j}N F = {400sin30 i 400cos30 j}n = {200.0i 346.4j}N For moment, M O = = rxf = 98.6k N. m i 0.4 200.0 j 0.2 346.4 k 0 0 29 Example 4.4 The member is subjected to a force of F=6 kn. If θ= 45º, determine the moment produced by F about point A. Ans; 30 15
Example 4.5 The two boys push on the gate with forces of FA=30 Ib and FB=50 Ib as shown. Determine the moment of each force about C. Which way will the gate rotate, clockwise or counterclockwise? Neglect the thickness of the gate. Ans; The gate will rotate C.C.W. 31 4.3 MOMENT OF A COUPLE Couple - two parallel forces. - same magnitude but opposite direction. - separated by perpendicular distance d. Resultant force = 0 Tendency to rotate in specified direction. Couple moment = sum of moments of both couple forces about any arbitrary point. Example; Position vectors r A and r B are directed from O to A and B, lying on the line of action of F and F. Couple moment about O M = r A X (-F) + r B X (F) Couple moment about A M = r X F since moment of F about A = 0 32 16
4.3 MOMENT OF A COUPLE A couple moment is a free vector - It can act at any point since M depends only on the position vector r directed between forces and not position vectors r A and r B, directed from O to the forces. - Unlike moment of force, it do not require a definite point or axis. Scalar Formulation Magnitude of couple moment M = Fd Direction and sense are determined by right hand rule. In all cases, M acts perpendicular to plane containing the forces. 33 4.3 MOMENT OF A COUPLE Vector Formulation For couple moment, M = r X F If moments are taken about point A, moment of F is zero about this point. r is crossed with the force to which it is directed. Resultant Couple Moment Couple moments are free vectors and may be applied to any point P and added vectorially. For resultant moment of two couples at point P, M R = M 1 + M 2 For more than 2 moments, M R = (r X F) 34 17
Equivalent Couples Two couples are equivalent if they produce the same moment. Since moment produced by the couple is always perpendicular to the plane containing the forces, forces of equal couples either lie on the same plane or plane parallel to one another. 35 Example 4.6 : Force Couple 36 18
Example 4.7 : Force Couple 37 38 19
Example 4.8 Determine tile resultant couple moment acting on the beam. Ans: 39 Example 4.9 Determine the resultant couple moment acting on the triangular plate. Ans: 40 20
Example 4.10 If F=200 Ib, determine the resultant couple moment. 41 Solution; 42 21
43 Example 4.11 Determine the required magnitude of force F if the resultant couple moment on the frame is 200 Ib.ft, clockwise. 44 22
Solution; 45 4.4 Equivalent System A force has the effect of both translating and rotating a body. The extent of the effect depends on how and where the force is applied. We can simplify a system of forces and moments into a single resultant and moment acting at a specified point O. A system of forces and moments is then equivalent to the single resultant force and moment acting at a specified point O. 46 23
4.4 Equivalent System Point O is on the Line of Action Consider body subjected to force F applied to point A. Apply force to point O without altering external effects on body. - Apply equal but opposite forces F and F at O. 47 4.4 Equivalent System Point O is on the Line of Action - Two forces indicated by the slash across them can be cancelled, leaving force at point O. - An equivalent system has be maintained between each of the diagrams, shown by the equal signs. 48 24
4.4 Equivalent System Point O is on the Line of Action - Force has been simply transmitted along its line of action from point A to point O. - External effects remain unchanged after force is moved. - Internal effects depend on location of F. 49 4.4 Simplification of a Force and Couple System Simplifying any force and couple system, F R = F M R = M C + M O If the force system lies on the x-y plane and any couple moments are perpendicular to this plane, F Rx = F x F Ry = F y M Ro = M C + M O 50 25
4.4 Simplification of a Force and Couple System MOVING A FORCE OFF OF ITS LINE OF ACTION 51 Example 4.12 52 26
53 54 27
Example 4.13 55 56 28
57 Example 4.14 Replace the force system acting on the beam by an equivalent force and couple moment at point A. 58 29
solution: 59 Example 4.15 Replace the force system acting on the beam by an equivalent force and couple moment at point A. 60 30
solution: 61 Homework -4- Replace the force and couple moment system acting on the overhang beam by a resultant force, and specify its location along AB measured from point A. 62 31