Download presentation

Presentation is loading. Please wait.

Published byOmarion Bow Modified over 2 years ago

1
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş Statics of Particles

2
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 2 Forces on a Particle 50 N 30 o Line of Action Point of Application A direction magnitude Since the vector has a well defined point of application, it is a fixed vector, therefore can not be moved without modifications

3
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 3 Properties of Vector Addition Q P P+Q=Q+P Commutative A P Q P+Q Q+P Q P A A Associative A P Q S P+Q P+Q+S A Q Q+S P+Q+S P+Q+S=(P+Q)+S=P+(Q+S) P S

4
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 4 Resultant of several concurrent forces Q P S Using polygon rule A A P Q S P+Q+S

5
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 5 Resolution of a force into components A F P Q A F P Q A F Q P

6
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 6 A P Q R Example (Beer & Johnston) 25 o 20 o Q=60 N P=40 N A The two forces P and Q act on a bolt A. Determine their resultant. 20 o 25 o =155 o R 2 =P 2 +Q 2 -2PQcosB B R 2 =(40N) 2 +(60N) 2 -2(40N)(60N)cos155 0 R=97.73 N Law of cosines Law of sines

7
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 7 Example (cntd.) Law of sines A=15.04 o =20 o +A=35.04 o The answer is R=97.7 kN =35.0 o

8
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 8 Example (Beer and Johnston) B A C o 45 o A barge is pulled by two tugboats. If the resultant of the forces exerted by tugboats is a 25 kN force directed along the axis of barge, determine the tension in each of the ropes. R=25 kN 45 o 30 o T2T2 T1T1 105 o Using law of sines

9
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 9 Rectangular Components F x y FxFx FyFy

10
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 10 F x y FxFx FyFy

11
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 11 Unit Vectors x y i j Magnitude=1 F Fx=FxiFx=Fxi Fy=FyjFy=Fyj

12
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 12 Example (Beer and Johnston) F=(3.5 kN)i+(7.5kN)j A Determine the magnitude of the force and angle. A x y F x =3.5 kN F y =7.5 kN F

13
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 13 Addition of Forces by Summing x and y components A x y F 1 =150 N 30 o F 2 =80 N 20 o F 3 =110 N F 4 =100 N 15 o Determine the resultant of the forces on the bolt F 2 cos 20 o -(F 2 sin 20 o ) -F 3 F 1 sin 30 o F 1 cos 30 o F 4 cos 15 o -(F 4 sin 15 o ) Forces Magnitude, Nx Component, Ny Component, N F1F F2F F3F F4F R x =199.1 R y =+14.3

14
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 14 Example (cntd) R=Rxi+RyjR=Rxi+Ryj R=(199.1N)i+(14.30N)j R R x =199.1 N R y =14.30 N

15
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 15 Equilibrium of a Particle When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium. x y A F 1 =300 N F 2 =173.2 N F 3 =200 N F 4 =400 N F 1 =300 N F 2 =173.2 N F 3 =200 N F 4 =400 N R= F=0 ( F x )i+ ( F y )j=0 F x =0 F y =0 F x =300 N -(200 N) sin30 o -(400 N) sin 30 o =300 N -100 N -200 N = 0 F y = N -(200 N) cos30 o + (400 N) cos 30 o = N N N = 0 30 o

16
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 16 Free-Body Diagram A sketch showing the physical conditions of the problem is known as space diagram. Choose a significant particle and draw a separate diagram showing that particle and forces on it. This is the Free-Body diagram. A B C 50 o 30 o Space Diagram T AC 30 o 50 o T AB 736 N 60 o 80 o 40 o 736 N Free-Body Diagram T AC T AB Using law of sines T AB =647 N T AC =480 N

17
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 17 Analytical Solution Two unknowns T AB and T AC Equilibrium Equations F x =0 and F y =0 The system of equations then Solving the above system (2 unknowns 2 equations) T AB =647 N T AC =480 N

18
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 18 Sumary of Solution techniques equilibrium under three forces may use force triangle rule equilibrium under more than three forces may use force polygon rule If analytical solution is desired may use equations of equilibrium

19
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 19 Example (Beer and Johnston) E B C A 60 o 20 o Flow As a part of the design of a new sailboat, it is desired to determine the drag force which may be expected at a given speed. To do so, a model of the proposed hull is placed in a test channel and three cables are used to keep its bow on the centerline of the channel. Dynamometer reading indicate that for a given speed, the tension is 200 N in cable AB and 300 N in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. Unknowns; F D, T AC Start with drawing Free-Body Diagram T AE =300 N T AB =200 N FDFD T AC 60 o 20 o Equilibrium condition Resultant of the all forces should be zero R=T AB +T AC +T AE +F D =0 Lets write all the forces in x and y components T AB =-(200 N) sin 60 o i+(200 N) cos 60 o j T AB = i+100 j T AC = (T AC ) sin 20 o i+(T AC ) cos 20 o j T AC = 0.342(T AC ) i (T AC ) j T AE =-(300 N) j FD=FD iFD=FD i

20
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 20 Example (cntd.) R=T AB +T AC +T AE +F D =0 R=( N)i N j (T AC ) i (T AC ) j + (-300 N) j + F D i =0 R= { (T AC ) + F D } i + { (T AC ) + (-300 N) } j = N (T AC ) + F D = 0 F x =0 F y = (T AC ) + (-300 N) = 0 T AC =213 N F D =100 N

21
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 21 Forces in Space x y z V VzkVzk k j iVxiVxi VyjVyj x y z V=Vxi+Vyj+VzkV=Vxi+Vyj+Vzk V x = V cos x V y = V cos y V z = V cos z

22
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 22 Direction cosines l = cos x m = cos y n = cos z V x = l V V y = mV V z = nV V 2 = V x 2 + V y 2 + V z 2 l 2 + m 2 + n 2 =1

23
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 23 x y z B( x B, y B, z B ) A( x A, y A, z A ) V V= ( x B -x A ) i + ( y B -y A ) j + ( z B -z A ) k n Unit vector along AB

24
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 24 Rectangular Coordinates in Space y B A C O F FyFy FhFh y x z B C O FxFx FyFy y x z FhFh FzFz F y =F cos y F h =F sin y F x = F h cos = F sin y cos F z = F h sin = F sin y sin E D F 2 = F y 2 + F h 2 F h 2 = F x 2 + F z 2

25
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 25 Problems

26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 26 Problem 1- (Meriam and Kraige) A 4 3 F=1800 N x y z The 1800 N force F is applied at the end of the I-beam. Express F as a vector using the unit vectors i and j. First lets find the x and y components of the force F. F x = N 3/5 = N F y = N 4/5 = N F = F x i+ F y j = i j N

27
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 27 Problem 2 -(Meriam and Kraige) The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of the airfoil is 50 N, compute the magnitude of the resultant force R and the angle which it makes with the horizontal. C D=5 N L= 50 N Air flow D L C tan = 50/5 = tan =84.3 o F = ( ) 0.50 = 50.2 N

28
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 28 Problem 3 -(Meriam and Kraige) The gusset plate is subjected to the two forces shown. Replace them by two equivalent force, F x in the x -direction and F a in the a direction. Determine the magnitudes of F x and F a. y x 900 N 800 N 10 o 45 o 25 o A a A 800 N 900 N R 10 o 25 o 10 o 65 o From the law of cosines R 2 = (800)(900) cos75 R=1040 N From the law of sines 66.7 o 45 o o A FxFx FaFa 21.7 o 1040 N

29
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 29 Problem 4- (Meriam and Kraige) B A C 50 m 40 m 20 m The guy cables AB and AC are attached to the top of the transmission tower. The tension in cable AC is 8 kN. Determine the required tension T in cable AB such that the net effect of the two cable tensions is a downward force at point A. Determine the magnitude R of this downward force R 8 kN This time lets use another approach. Since the resultant force is downward the sum of horizontal components of the two forces should add up to zero. The magnitude of R

30
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 30 Problem 5- (Beer and Johnston) Two cables are tied together at C and loaded as shown. Determine the tension in AC and BC kg 500 mm 1375 mm 1200 mm A C B Lets draw the Free Body Diagram (FBD) at C C W=1600*9.81=15700N y x T BC T AC Writing Equilibrium Equations F x =0 F y =0 T AC =12470 N T CB =6370 N

31
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 31 Problem 6- (Beer and Johnston) A precast-concrete wall section is temporarily held by the cables shown. Knowing that tension is 4200 N in cable AB and 6000 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. D A B C 5.5 m 8 m 4 m 13.5 m x y z Coordinates of points A, B and C A (8, -4, -5.5)B (0, 0, 0) C (0, 0, -13.5) AB=(0-8) i + (0-(-4)) j + (0-(-5.5)) k = -8 i + 4 j k AC=(0-8) i + (0-(-4)) j + (-13.5-(-5.5)) k = -8 i + 4 j - 8 k

32
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 32 Problem 6- (Beer and Johnston) A precast-concrete wall section is temporarily held by the cables shown. Knowing that tension is 4200 N in cable AB and 6000 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. D A B C 5.5 m 8 m 4 m 13.5 m x y z R = F AB + F AC R =( ) i +( ) j + ( ) k = i j k x y z R = 8250 N Direction cosines l = m = n = 0.218

33
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 33 Problem 7- (Beer and Johnston) A crate is supported by three cables as shown. Determine the weight W of the crate knowing that the tension in cable AB is 4620 N. y x z B A O C D 700 mm 600 mm 1125 mm 650 mm 450 mm Coordinates of points A, B, C and D A (0, -1125, 0) Lets draw FBD first W F AB F AC F AD Unknowns : F AD, F AC and W B (700, 0, 0)C (0, 0, -600)D (-650, 0,450) AB=((700-0)i+(0-(-1125))j+0k=700i+1125j+0k AB=( ) 0.5 =1325 n AB =(700i+1125j+0k)/1325=0.5283i j F AB =4620(0.5283i j) N

34
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 34 Problem 7- (Beer and Johnston) y x z B A O C D 700 mm 600 mm 1125 mm 650 mm 450 mm Coordinates of points A, B, C and D A (0, -1125, 0) B (700, 0, 0)C (0, 0, -600)D (-650, 0,450) AC=(0i+(0-(-1125))j+(-600-0)k=0i+1125j-600k AC=( ) 0.5 =1275 n AC =(0i+1125j-600k)/1275=0.8824j k F AC =F AC (0.8824j k) N AD=((-650-0)i+(0-(-1125))j+(450-0)k=-650i+1125j+450k AD=( ) 0.5 =1375 n AD =(-650i+1125j+450k)/1375= i j k F AD =F AD ( i j k) N W=-W j

35
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 35 Problem 7- (Beer and Johnston) y x z B A O C D 700 mm 600 mm 1125 mm 650 mm 450 mm F AB +F AC +F AD +W=0 4620(0.5283i j) +F AC (0.8824j k)+F AD ( i j k)-W j= F AD =0 F AD = 5160 N F AC *5160=0 F AC = 3590 N W = 0 W= N

36
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/ Dr. Engin Aktaş 36 Problem 8- (Beer and Johnston) A 16 kg triangular plate is supported by three wires as shown. Determine the tension in each wire. y x z 600 mm 200 mm 400 mm 200 mm D B C A

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google