IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Rigid Bodies: Equivalent System of Forces 25. 10. 2020 Dr. Engin Aktaş 1

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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. 25. 10. 2020 Dr. Engin Aktaş 2

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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 R 1 • The Internal Forces 25. 10. 2020 Dr. Engin Aktaş W R 2 3

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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 A = 25. 10. 2020 W R 2 F=F’ B F R 1 same magnitude and direction F’ acting at a different point = Dr. Engin Aktaş F R 1 W R 2 4

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Limitation P 1 B A (a) P 2 (d) = P 1=-P 2 P 1 B P’ 2 P 1=-P 2 B P 2 A A P 1 B A = (c) (b) A = B P’ 2 (e) P 1 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. 25. 10. 2020 Dr. Engin Aktaş 5

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Vector Product of Vectors V=Px. Q 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. 25. 10. 2020 Dr. Engin Aktaş 6

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Properties V The magnitude of V is the area of the parallelogram. Q Q’ V = P x Q’ P The vector products are not commutative Q x P P x Q The distributive property holds P x (Q 1 + Q 2) = P x Q 1 + P x Q 2 The associative property does not apply 25. 10. 2020 but Q x P = -(P x Q) Dr. Engin Aktaş P x (Q x S) (P x Q) x S 7

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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 25. 10. 2020 Dr. Engin Aktaş i 8

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 V=Px. Q 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 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 25. 10. 2020 Dr. Engin Aktaş 9

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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. 25. 10. 2020 Dr. Engin Aktaş 10

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Problems Involving Only Two Dimensions F F Mo 25. 10. 2020 d d 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 AR 231 Fall 12/13 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)) F 4 F 5 y F 3 r O x F 1 F 2 z 25. 10. 2020 r x ( F 1+F 2+ … ) = r x F 1 + r x F 2 + … Dr. Engin Aktaş 12

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Rectangular Components of the Moment of a Force y r=xi+yj+zk Fy j F = Fx i + Fy j + Fz k Mo = r x F yj O zk Mo = Mx i + My j + Mz k Fx i r xi x Fz k z Mx = y F z – z F y 25. 10. 2020 My = z F x – x F z Dr. Engin Aktaş Mz = x F y – y F x 13

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 y (y. A-y. B) j MB= r. A/B x F = (r. A-r. B) x F Fy j r. A/B A )k B -z B A z ( Fz k Fx i (x. A-x. B) i x O z x. A/B = x. A - x. B 25. 10. 2020 y. A/B = y. A - y. B Dr. Engin Aktaş z. A/B = z. A - z. B 14

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture 2 -D problems Fy j y Fy j F A (x, y, 0) A (y. A-y. B) j y yj AR 231 Fall 12/13 Fx i r z xi x O Fx i r A/B B O F MB = MB k (x. A-x. B) i x Mo = Mz k z MB = (x. A-x. B) Fy - (y. A-y. B) Fx Mo = (x Fy - y Fx) k Mo = Mz = x F y - y F x 25. 10. 2020 Dr. Engin Aktaş 15

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Sample Problems 25. 10. 2020 Dr. Engin Aktaş 16

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 50 0 m m A 60 o 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) 50 0 m m A a) Moment about O 300 N r d = (0. 5 m) cos 60 o = 0. 25 m MO = F d =(300 N) (0. 25 m) = 75. 0 Nm 60 o O d MO 25. 10. 2020 or r = (0. 5 m cos 60 o) i + (0. 5 m sin 60 o) 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 F AR 231 Fall 12/13 b) Horizontal Force 50 0 m d m A Department of Architecture d = (0. 5 m) sin 60 o = 0. 433 m r MO = F d =(F) (0. 433 m) = 75. 0 Nm O MO 60 o F = 75. 0 Nm / 0. 433 m = 173. 2 N or r = (0. 5 m cos 60 o) i + (0. 5 m sin 60 o) j = (0. 25 i + 0. 433 j) m F= Fi. N 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 25. 10. 2020 Dr. Engin Aktaş 18

IZMIR INSTITUTE OF TECHNOLOGY m A F Department of Architecture AR 231 Fall 12/13 c) Smallest Force 50 0 m Since Mo = F d, when d is max F is the smallest. d = (0. 5 m) MO = F d 60 o O 75. 0 Nm = F (0. 5 m) F = 75. 0 Nm / 0. 5 m = 150. 0 N MO F d) 750 N vertical force 0 m m A 30 o 50 B O 75 Nm = (750 N) d 750 N 60 o d d = 0. 1 m OB cos 60 o = d OB = 0. 1 m / cos 60 o = 200 mm MO 25. 10. 2020 Dr. Engin Aktaş 19

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture 800 N 60 o 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 AR 231 Fall 12/13 MB = r. A/B x F 800 N r. A/B = (- 0. 2 i +0. 16 j ) m Fx = (400 N) i r. A/B B F = (800 N) cos 60 o i + (800 N) sin 60 o j = 400 i + 693 j MB = (-0. 2 i + 0. 16 j) x 400 i + 693 j = (- 138. 6 Nm) k - 64. 0 Nm) k = (- 202. 6 Nm) k 25. 10. 2020 Dr. Engin Aktaş MB = 203 Nm 20

IZMIR INSTITUTE OF TECHNOLOGY 60 mm y 2 z 50 mm C B m 5 m 30 o Department of Architecture AR 231 Fall 12/13 200 N 60 o A 200 -N force is applied as shown to the bracket ABC. Determine the moment of the force about A. (Beer and x A Johnston) A(0, -50, 0) C(60, 25, 0) r. C/A = (0. 06 i +0. 075 j ) m MA = r. C/A x F F = -(200 N) cos 30 o j + (200 N) cos 60 o 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 25. 10. 2020 MAy MA Dr. Engin Aktaş 21

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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 25. 10. 2020 P • Q=Q • P P • (Q 1 + Q 2) = P • Q 1 + P • Q 2 ? 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 AR 231 Fall 12/13 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 25. 10. 2020 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 Sample Problem A cube of side a is acted upon by a force P as C D A a B P G E E O k a)Moment about A C D j shown. Determine the moment of P a) About A b) About the edge AB c) About the diagonal AG of the cube F y A AR 231 Fall 12/13 r. F/A = ai – aj B i G x F z 25. 10. 2020 Dr. Engin Aktaş 24

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 b) Moment about AB D C l B P A E 25. 10. 2020 c) Moment about diagonal AG. G F Dr. Engin Aktaş 25

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 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. B y -F r q M r. B A r. A O z r. A x F + r. B x (- F) = (r. A – r. B ) F r. A – r. B = r d F M=rx. F 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. 25. 10. 2020 Dr. Engin Aktaş 26

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture Equivalent Couples M y 100 mm y O 200 N z 100 mm y 150 mm O z Dr. Engin Aktaş M 300 N x 100 mm M z 200 N Two couples having the same moment M are equivalent, whether they are contained in the same plane or in parallel planes 25. 10. 2020 O x AR 231 Fall 12/13 300 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 direction they x have. The only thing which 100 mm counts is the moment of the couple. Couples with the same moment will have the same effect on the rigid body. 300 N 27

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Addition of Couples -R - F 1 Consider two intersecting planes P 1 and P 2. - F 2 M = r x R = r x (F 1 + F 2) P 2 B r P 1 A F 2 F 1 R M 1 M = r x F 1 + r x F 2 M M = M 1 + M 2 P 1 25. 10. 2020 M 2 Dr. Engin Aktaş 28

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Couples May Be Represented by Vectors M y M=Fd y y M -F O z d F x = O x z y My O = Mz Mx x z 25. 10. 2020 Dr. Engin Aktaş 29

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Resolution of a Given Force into a Force at a Different Point and a Couple F F F A O F A r = O -F r MO A = O 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. 25. 10. 2020 Dr. Engin Aktaş 30

IZMIR INSTITUTE OF TECHNOLOGY 240 140 mm Department of Architecture mm y Sample Problem (Beer and Johnston) B C Determine the components of the single couple equivalent to x the two couples shown 150 N E 180 mm AR 231 Fall 12/13 100 N A 100 N Mx = - (150 N) (0. 360 m) = - 54. 0 N. m My = + (100 N) (0. 240 m) = 24. 0 N. m 180 mm z 25. 10. 2020 D 150 N 100 N Mz = + (100 N) (0. 180 m) = 18. 00 N. m M = - (54. 0 N. m) i +(24. 0 N. m) j + (18. 0 N. m) k Dr. Engin Aktaş 31

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Sample Problem (Beer and Johnston) 30 0 m m B 60 mm O 150 mm F = -(400 N) j 260 mm 200 N 60 o 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 -(24 N. m) k 150 mm = -(24 N. m) k O -(60 N. m) k -(400 N) j 25. 10. 2020 Dr. Engin Aktaş 32

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 C -(400 N) j -(84 N. m) k = 60 o O O -(400 N) j -(84 N. m) k = OC x F = ( OC cos 60 o i + OC sin 60 o j ) x ( -400 N) j -(84 N. m) k = OC = 25. 10. 2020 - OC cos 60 o (400 N) k 84 N. m cos 60 o (400 N) Dr. Engin Aktaş = 0. 42 m = 420 mm 33

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Reduction of a System of Forces to One Force and a Couple F 2 M 3 A 2 F 1 A 1 r r 1 O 2 r 3 MO R F 3 F 1 = O R = F 3 O A 3 M 1 R = SF 25. 10. 2020 F 2 M 2 Dr. Engin Aktaş MOR = SMO = S(r x F) 34

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Equivalent System of Forces Two systems of forces F 1, F 2, F 3 , 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’ 25. 10. 2020 and SFx = SF’x SFy = SF’y SMx = SM’x SMy = SM’y Dr. Engin Aktaş SMO = SM’O SFz = SF’z SMz = SM’z 35

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 Sample Problem (Beer and Johnston) 250 N 150 N A B 1. 6 m A 4. 8 m beam is subjected to the forces shown. Reduce the given system of forces to 100 N 600 N 1. 2 m 2 m 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. 6 i) x (-600 j) + (2. 8 i) x (100 j) + (4. 8 i) x (-250 j) = - (1880 N. m)k - (600 N)j - (1880 N)k B A 25. 10. 2020 Dr. Engin Aktaş 36

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 b) An equivalent force couple system at B MRB = MRA + BA x R = - (1880 N. m)k + (- 4. 8 m)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 B A 25. 10. 2020 Dr. Engin Aktaş (1000 Nm)k 37

IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR 231 Fall 12/13 c) A single force or resultant x - (600 N)j B A r x R = MR A (x)i x (-600 N)j = - (1880 N. m)k - x(600 N)k = - (1880 N. m)k x = 3. 13 m 25. 10. 2020 Dr. Engin Aktaş 38

IZMIR INSTITUTE OF TECHNOLOGY 60 k. N 200 k. N O 100 k. N 40 k. N A 2 m 3 m y -(700 k. Nm)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 B R = SF -(400 k. N)j O (600 k. Nm)i x z Dr. Engin Aktaş MRO = S(r x F) r, m F, k. N r x F, k. N • m 0 -200 j 0 5 i -60 j -300 k 5 i + 2. 5 k 2 i + 5 k 25. 10. 2020 AR 231 Fall 12/13 Sample Problem (Beer and Johnston) y z Department of Architecture -40 j -100 j 100 i – 200 k 500 i – 200 k R = -400 j MRO = 600 i – 700 k 39

IZMIR INSTITUTE OF TECHNOLOGY y O -(400 k. N)j x (xi + zk) x (-400 j) = 600 i – 700 k - 400 xk + 400 zi = 600 i – 700 k - 400 x = – 700 x = 1. 750 m 25. 10. 2020 AR 231 Fall 12/13 r x R = MOR xi zk z Department of Architecture Dr. Engin Aktaş 400 z = 600 z = 1. 500 m 40