IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Rigid Bodies: Equivalent System of Forces 13.12.2015 Dr. Engin Aktaş 1 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 In this chapter we will study • the effects of forces on a rigid body. • to replace the system of forces with a simpler equivalent system. 13.12.2015 Dr. Engin Aktaş 2 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 External and Internal Forces • The External Forces represent the action of other bodies on the rigid body under consideration. They control the external behavior of the body. F R1 W R2 • The Internal Forces Are the forces which hold together the particles forming the rigid body. 13.12.2015 Dr. Engin Aktaş 3 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Principle of Transmissibility Equivalent Forces The conditions of equilibrium or motion o a rigid body will remain unchanged if a force F acting at a given point of the rigid body is replaced by a force F’, F provided on the same line of action = 13.12.2015 W R2 F=F’ B acting at a different point F R1 same magnitude and direction F’ A = F R1 Dr. Engin Aktaş W R2 4 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Limitation P1 B A (a) P2 P2 (d) = P1=-P2 P1 B P’2 P1=-P2 B A A P1 = (c) (b) A = B A B P’2 (e) P1 B A = (f) While the principle of transmissibility may be used freely to determine the conditions of motion or equilibrium of rigid bodies and to compute the external forces acting on these bodies, it should be avoided, or at least used with care, in determining internal forces and deformations. 13.12.2015 Dr. Engin Aktaş 5 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Vector Product of Vectors V=PxQ V Q q Also referred as Cross Product of P and Q. Line of action of V P The magnitude of V V = P Q sin q The sense of V V Right Hand Rule Right hand’s fingers show the direction of P and when you curl your fingers toward Q the direction of your thumb is the sense of V. 13.12.2015 Dr. Engin Aktaş 6 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Properties V The magnitude of V is the area of the parallelogram. Q Q’ V = P x Q = P x Q’ P The vector products are not commutative The distributive property holds but Q x P = -(P x Q) P x (Q1 + Q2) = P x Q1 + P x Q2 The associative property does not apply 13.12.2015 QxPPxQ P x (Q x S) (P x Q) x S Dr. Engin Aktaş 7 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Vector Products Expressed in Terms of Rectangular Components y j i z x k ixi= 0 jxi= -k kxi= j ixj= k jxj= 0 kxj= -i ixk= -j jxk= i kxk= 0 j k 13.12.2015 i Dr. Engin Aktaş 8 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 V=PxQ V = P x Q = (Px i + Py j + Pz k) x (Qx i + Qy j + Qz k) V = (Py Qz - Pz Qy ) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx ) k i j k i j k i j V Px Py Pz Px Py Pz Px Py Qx Qy Qz Qx Qy Qz Qx Qy Repeat first and second columns to the right. The sum of the products obtained along the red line is then subtracted from the sum of the product obtained along the black lines 13.12.2015 Dr. Engin Aktaş 9 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Moment of a Force about a Point Mo Moment of F about O F q r O d Mo = r x F Mo A The magnitude of moment of F about O. Mo = r F sin q = F d The magnitude of Mo measures the tendency of the force F to make the rigid body rotate about a fixed axis directed along Mo. 13.12.2015 Dr. Engin Aktaş 10 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Problems Involving Only Two Dimensions F F Mo 13.12.2015 O O Mo Counterclockwise Clockwise Moment Mo = + F d Mo = - F d points out of screen points into the screen Dr. Engin Aktaş 11 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Varignon’s Theorem The moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O. (Varignon (1654-1722)) y F4 F5 F3 r O z 13.12.2015 x F1 F2 r x ( F1+F2+ … ) = r x F1 + r x F2 + … Dr. Engin Aktaş 12 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Rectangular Components of the Moment of a Force y r=xi+yj+zk Fy j F = Fx i + Fy j + Fz k yj Mo= r x F Fx i r O zk Mo = Mx i + My j + Mz k xi x Fz k z Mx = y Fz – z Fy 13.12.2015 My = z Fx – x Fz Dr. Engin Aktaş Mz= x Fy – y Fx 13 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 y (yA-yB) j MB= rA/B x F = (rA-rB) x F Fy j rA/B A B i M B xA/B Fx j k yA/B zA/B Fy Fz Fx i (xA-xB) i Fz k x O z xA/B = xA - xB 13.12.2015 yA/B = yA - yB Dr. Engin Aktaş zA/B = zA - zB 14 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture 2-D problems y Fy j F A (x, y, 0) z A Fx i yj xi Fy j B x O MB = MB k F Fx i (yA-yB) j y O AR231 Fall12/13 (xA-xB) i x Mo = M z k z MB = (xA-xB) Fy - (yA-yB) Fx Mo = (x Fy - y Fx) k Mo = Mz = x F y - y F x 13.12.2015 Dr. Engin Aktaş 15 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Sample Problems 13.12.2015 Dr. Engin Aktaş 16 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 A 60o O A 300-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine (a) the moment of the 300-N force about O 300 N (b) the magnitude of the horizontal force applied at A which creates the same moment about O. (c) the smallest force applied at A which creates the same moment about O. (d) how far from he shaft a 750-N vertical force must act to create the same moment about O. (Beer and Johnston) A a) Moment about O 300 N r d = (0.5 m) cos 60o = 0.25 m MO = F d =(300 N) (0.25 m) = 75.0 Nm 60o O d MO 13.12.2015 or r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m F = - 300 j N MO = r x F = (0.25 i +0.433 j) x (-300 N) j = -75.0 k Nm Dr. Engin Aktaş 17 IZMIR INSTITUTE OF TECHNOLOGY A AR231 Fall12/13 b) Horizontal Force d = (0.5 m) sin 60o = 0.433 m r d F Department of Architecture MO = F d =(F) (0.433 m) = 75.0 Nm O MO 60o F = 75.0 Nm / 0.433 m = 173.2 N or r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m F= FiN MO = r x F = (0.25 i +0.433 j) x (F ) i = -75.0 k Nm - 0.433 F k Nm = -75.0 k Nm F = 173.2 N 13.12.2015 Dr. Engin Aktaş 18 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 A F c) Smallest Force Since Mo = F d, when d is max F is the smallest. d = (0.5 m) MO = F d 60o O 75.0 Nm = F (0.5 m) F = 75.0 Nm / 0.5 m = 150.0 N MO 30o F A d) 750 N vertical force B 75 Nm = (750 N) d 750 N 60o O d d = 0.1 m OB cos 60o = d OB = 0.1 m / cos 60o = 200 mm MO 13.12.2015 Dr. Engin Aktaş 19 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 800 N 60o 160 mm A A force of 800 N acts on a bracket as shown. Determine the moment of the force about B. (Beer and Johnston) B 200 mm Fx = (693 N) j A MB = rA/B x F 800 N rA/B = (- 0.2 i +0.16 j ) m Fx = (400 N) i rA/B B F = (800 N) cos 60o i + (800 N) sin 60o j = 400 i + 693 j MB = (-0.2 i + 0.16 j) x 400i + 693 j = (- 138.6 Nm) k - 64.0 Nm) k = (- 202.6 Nm) k 13.12.2015 Dr. Engin Aktaş MB = 203 Nm 20 IZMIR INSTITUTE OF TECHNOLOGY y 60 mm C B z 30o Department of Architecture AR231 Fall12/13 200 N 60o x A A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. (Beer and Johnston) A(0,-50,0) C(60,25,0) rC/A = (0.06 i +0.075 j ) m MA = rC/A x F F = -(200 N) cos 30o j + (200 N) cos 60o k = -173.2 j + 100 k MA = (0.06 i + 0.075 j) x (-173.2 j + 100 k) = -10.39 k - 6 j + 7.5 i y MA = 7.5 i – 6 j –10.39 k MAz MA = 14.15 Nm MAx x z 13.12.2015 MAy MA Dr. Engin Aktaş 21 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Scalar Product of Two Vectors Q scalar product P•Q= q or P Q cos q dot product P The result is a scalar Properties Commutative Distributive X Associative 13.12.2015 P•Q=Q•P P • (Q1 + Q2) = P • Q1 + P • Q2 ? P • (Q • S) = (P • Q) • S P • (Q • S) and (P • Q) • S does not have a meaning Dr. Engin Aktaş 22 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Moment of a Force about a Given Axis MO y Let OL be an axis through O; we define the moment MOL of F about OL as the projection OC of the moment on the axis OL. F MOL = l • MO = l • (r x F) L C • l z 13.12.2015 O x A the moment MOL of F about OL measures the tendency of the force F to impart to the rigid body a motion of rotation about a fixed axis OL. Dr. Engin Aktaş 23 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Sample Problem A cube of side a is acted upon by a force P as C D A a B P G y E shown. Determine the moment of P a) About A b) About the edge AB c) About the diagonal AG of the cube F a)Moment about A C D A E j O k z 13.12.2015 rF/A = ai – aj B i G F x P P j 2 M A rF/A P a i j Dr. Engin Aktaş P k 2 P 2 j k aP i j k 2 24 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 b) Moment about AB M D C l B P A AB iM A i λ 13.12.2015 AG λ M A 2 AG 2 ai aj ak AG F M i j k aP c) Moment about diagonal AG. G E aP 1 3 a 3 i j k aP 1 3 i j k 2 Dr. Engin Aktaş i j k aP 6 1 1 1 aP 6 25 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Moment of a Couple Two forces F and –F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. rA x F + rB x (- F) = (rA – rB ) F B y -F r q M rB A rA O z rA – rB = r d F M=rxF The vector is called the moment of the couple; it is a vector perpendicular to the plane containing the forces and its magnitude x M = r F sin q d = r sin q M=Fd Since the r is independent of the choice of origin O, taking moment about another point would not change the result. Thus, the M of a couple is a free vector which may be applied at any point. 13.12.2015 Dr. Engin Aktaş 26 IZMIR INSTITUTE OF TECHNOLOGY y M y x 200 N y 150 mm O M 300 N x M z 200 N 300 N This property indicates that when a couple acts on a rigid body, it does not matter 300 N where the two forces forming the couple act, or what magnitude and x direction they have. The only thing which counts is the moment of the couple. Couples with the same moment will have the same effect on the rigid body. 300 N Two couples having the same moment M are equivalent, whether they are contained in the same plane or in parallel planes O z 13.12.2015 AR231 Fall12/13 Equivalent Couples O z Department of Architecture Dr. Engin Aktaş 27 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Addition of Couples -R - F1 Consider two intersecting planes P1 and P2. - F2 M = r x R = r x (F1 + F2) P2 B r P1 A F1 R M = r x F1 + r x F2 F2 M1 M M = M1 + M2 P2 P1 13.12.2015 M2 Dr. Engin Aktaş 28 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Couples May Be Represented by Vectors M y M=Fd y y M -F O z d F x = O x z = O x z y My O = Mx x Mz z 13.12.2015 Dr. Engin Aktaş 29 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Resolution of a Given Force into a Force at a Different Point and a Couple F F F A O F A r = MO r O A = O -F A force-couple system The force-couple system obtained by transferring a force F from a point A to a point O consists of F and a couple vector MO perpendicular to F. Conversely, any force-couple system consisting of a force F and a couple MO which are mutually perpendicular may be replaced by a single equivalent force. 13.12.2015 Dr. Engin Aktaş 30 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture y Sample Problem (Beer and Johnston) B C AR231 Fall12/13 Determine the components of the single couple equivalent to x the two couples shown 150 N 100 N E 100 N A 100 N Mx = - (150 N) (0.360 m) = - 54.0 N.m My = + (100 N) (0.240 m) = 24.0 N.m z D 100 N Mz = + (100 N) (0.180 m) = 18.00 N.m 150 N M = - (54.0 N.m) i +(24.0 N.m) j + (18.0 N.m) k 13.12.2015 Dr. Engin Aktaş 31 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Sample Problem (Beer and Johnston) B 60 mm 200 N 60o 400 N 200 N Replace the couple and force shown by an equivalent single force applied to the lever. Determine the distance from the shaft to the point of application of this equivalent force. O 150 mm F = -(400 N) j = -(24 N . m) k O -(24 N . m) k 150 mm O -(60 N . m) k -(400 N) j 13.12.2015 Dr. Engin Aktaş 32 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 C -(400 N) j -(84 N . m) k = 60o O O -(400 N) j -(84 N . m) k = OC x F = ( OC cos 60o i + OC sin 60o j ) x ( -400 N) j -(84 N . m) k = - OC cos 60o (400 N) k 84 N . m OC = = 0.42 m = 420 mm cos 60o (400 N) 13.12.2015 Dr. Engin Aktaş 33 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Reduction of a System of Forces to One Force and a Couple F2 M3 F2 M2 A2 F1 A1 r r1 O 2 r3 MO R F3 F1 = O R = F3 O A3 M1 R = SF 13.12.2015 MOR = SMO = S(r x F) Dr. Engin Aktaş 34 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Equivalent System of Forces Two systems of forces F1, F2, F3 , etc., and F’1, F’2, F’3, etc., are equivalent if, and only if, the sums of the forces and the sums of the moments about a given point O of the forces of the two systems are, respectively, equal. SF = SF’ 13.12.2015 SMO = SM’O and SFx = SF’x SFy = SF’y SMx = SM’x SMy = SM’y SFz = SF’z SMz = SM’z Dr. Engin Aktaş 35 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 Sample Problem (Beer and Johnston) 250 N 150 N 100 N 600 N A B 1.6 m 1.2 m 2m A 4.8 m beam is subjected to the forces shown. Reduce the given system of forces to a) An equivalent force couple system at A b) An equivalent force couple system at B c) A single force or resultant a) An equivalent force couple system at A R = SF = (150 N)j – (600 N)j + (100 N)j –(250 N)j = -(600 N)j MRA = S(r x F) = (1.6i) x (-600j) + (2.8i) x (100j) + (4.8i) x (-250j) = - (1880 N.m)k - (600 N)j - (1880 N)k B A 13.12.2015 Dr. Engin Aktaş 36 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 b) An equivalent force couple system at B MRB = MRA + BA x R = - (1880 N.m)k + (- 4.8m)i x (-600 N)j = - (1880 N.m)k + (2880 N.m)k = (1000 N.m)k - (600 N)j - (1880 Nm)k B A (2880 Nm)k - (600 N)j A 13.12.2015 B Dr. Engin Aktaş (1000 Nm)k 37 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13 c) A single force or resultant x - (600 N)j B A r x R = MRA (x)i x (-600 N)j = - (1880 N.m)k - x(600 N)k = - (1880 N.m)k x = 3.13 m 13.12.2015 Dr. Engin Aktaş 38 IZMIR INSTITUTE OF TECHNOLOGY 60 kN 200 kN 40 kN 100 kN A 3m y -(700 kNm)k x A square foundation mat supports the four columns shown. Determine the magnitude 2.5 m and point of application of the resultant of the four loads. 2.5 m C O 2m B R = SF -(400 kN)j O (600 kNm)i x MRO = S(r x F) r, m F, kN r x F, kN•m 0 -200j 0 5i -60j -300k 5i + 2.5k 2i + 5k z 13.12.2015 AR231 Fall12/13 Sample Problem (Beer and Johnston) y z Department of Architecture Dr. Engin Aktaş -40j -100j 100i – 200k 500i – 200k R = -400j MRO = 600i –700k 39 IZMIR INSTITUTE OF TECHNOLOGY y O r x R = MO R xi x (xi + zk) x (-400j) = 600i – 700k - 400xk + 400zi = 600i – 700k - 400x = – 700 x = 1.750 m 13.12.2015 AR231 Fall12/13 -(400 kN)j zk z Department of Architecture Dr. Engin Aktaş 400z = 600 z = 1. 500 m 40

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# Equivalent System of Forces