Contents 4-2 Intoduction Fee-Body Diagam Reactions at Suppots and Connections fo a Two-Dimensional Stuctue Equilibium of a Rigid Body in Two Dimension

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1 Chapte 4: Equilibium of Rigid Bodies ( 강체의평형 ) 최해진 hjchoi@cau.ac.k

2 Contents 4-2 Intoduction Fee-Body Diagam Reactions at Suppots and Connections fo a Two-Dimensional Stuctue Equilibium of a Rigid Body in Two Dimensions Statically Indeteminate Reactions Sample Poblem 4.1 Sample Poblem 4.3 Sample Poblem 4.4 Equilibium of a Two-Foce Body Equilibium of a Thee-Foce Body Sample Poblem 4.6 Equilibium of a Rigid Body in Thee Dimensions Reactions at Suppots and Connections fo a Thee-Dimensional Stuctue Sample Poblem 4.8

3 4.1 Intoduction 4-3 Fo a igid body in static equilibium, the extenal foces and moments ae balanced and will impat no tanslational o otational motion to the body. The necessay and sufficient condition fo the static equilibium of a body ae that the esultant foce and couple fom all extenal foces fom a system equivalent to zeo, F 0 M F 0 O ( ) Resolving each foce and moment into its ectangula components leads to 6 scala equations which also expess the conditions fo static equilibium, F x M x 0 0 F y M y 0 0 F z M z 0 0

4 4.1 Intoduction 4-4 만약물체에작용하는힘계의합력이 0 이라면, 물체는평형 (equilibium) 이라고말한다. 평형은합력과합우력이모두 0 임을의미한다. 합력이 0 이라는것 힘계가초기에정지하여있는물체에작용할때, 물체가이동하려는경향이없음을의미한다. 동역학 정역학 평형이아닌힘계에대한물체의응답에관심 공학적구조물의평형상태를유지하기위한힘의상태에대한필요충분조건을제시

5 4.1 Intoduction 4-5 평형상태 : 어떤물체에여러개의힘이작용하여그물체가이동하거나회전하지않고정지되어있는상태 * Equilibium Æ ïì R í ïî M Fi 0 M i ö 0 ø F x M x 0 0 F y M y 0 0 F z M z 0 0 ; necessay & sufficient conditions fo equilibium 평형해석의첫단계 물체에작용하는모든힘들을찾아내는것, 이것은자유물체도에의해이루어진다.

6 4.2 Fee-Body Diagam 4-6 물체의자유물체도 (Fee-Body Diagam ; FBD) 는물체에작용하는모든힘을나타내는물체의개략도이다. 자유 (fee) 라는용어는모든지지부가제거되고, 물체에작용하는힘 ( 반력 ) 에의해대체되는것을의미. 물체에작용하는힘 1. 작용력 (applied foce) 2. 반작용력 (eactive foce) 작용력 지지부에의해주어지지않고물체에작용하는힘 반작용력단순히반력 (eactions) 이라고하고반력은물체에결합되어있는지지부에의해물체에작용하는힘

7 4.2 Fee-Body Diagam 4-7 Fist step in the static equilibium analysis of a igid body is identification of all foces acting on the body with a fee-body diagam. Select the extent of the fee-body and detach it fom the gound and all othe bodies. Indicate point of application, magnitude, and diection of extenal foces, including the igid body weight. Indicate point of application and assumed diection of unknown applied foces. These usually consist of eactions though which the gound and othe bodies oppose the possible motion of the igid body. Include the dimensions necessay to compute the moments of the foces.

8 4.2 Fee-Body Diagam 4-8 자유물체도를구성하는일반적인과정 1. 모든지지부 ( 접촉면, 지지용케이블등 ) 가제거된것으로가정한물체의개략도를그린다. 2. 개략도에모든작용력을그리고, 기호로표시한다. 물체의무게는무게중심에작용하는힘으로간주한다. 균질 (homogenous) 인물체의무게중심 (cente of gavity) 은체적의도심 (centoid) 와일치한다. 3. 개략도에각지지부에기인하는반력을그리고, 기호로표시한다. 만약반력의방향이미지이면가정되어야한다. 이는구해진해로부터정확한방향이결정될것이다. 양 (+) 의결과는가정된방향이올바르다는것을나타내고, 반면에 (-) 의결과는정확한방향이가정된방향과반대라는것을의미한다 개략도에모든관련된각도와차원을나타낸다.

9 4.3 Reactions at Suppots and Connections fo a Two-Dimensional Stuctue 4-9 Reactions equivalent to a foce with known line of action.

10 4.3 Reactions at Suppots and Connections fo a Two-Dimensional Stuctue 4-10 Reactions equivalent to a foce of unknown diection and magnitude. Reactions equivalent to a foce of unknown diection and magnitude and a couple.of unknown magnitude

4.4 Equilibium of a Rigid Body in Two Dimensions F x M x 0 0 F y M y 0 0 F z M z 0 0 4-11 Fo all foces and moments acting on a twodimensional stuctue, F 0 M M 0 M M z x Equations of

11 4.4 Equilibium of a Rigid Body in Two Dimensions F x M x 0 0 F y M y 0 0 F z M z Fo all foces and moments acting on a twodimensional stuctue, F 0 M M 0 M M z x Equations of equilibium become F F 0 M 0 x y 0 y A whee A is any point in the plane of the stuctue. The 3 equations can be solved fo no moe than 3 unknowns. The 3 equations can be eplaced F M 0 M 0 x 0 A B z O

12 4.5 Statically Indeteminate Reactions 4-12 평형방정식의수 < 미지수의수평형방정식의수 > 미지수의수평형방정식의수 미지수의수 Moe unknowns than equations Fewe unknowns than equations, patially constained Equal numbe unknowns and equations but impopely constained

13 4.5 Statically Indeteminate Reactions 4-13 구속과정정 ( Constaints and statically deteminate ) 1. 독립된평형방정식의수 미지수의수 정역학적으로풀수있고, 정정 (statically deteminate) 문제이다. 정정문제는평형해석자체만으로해결될수있다. 2. 독립된평형방정식의수 < 미지수의수 정역학적으로풀수없고, 부정정 (statically indeteminate) 문제이다. 평형을유지하는데필요한구속만을가지고또한미지의지지력 ( unknown eaction foces) 이독립된평형방정식에의해서완전히결정될수있는정정 (statically deteminate) 인물체에국한하여취급함

14 Sample Poblem SOLUTION: Ceate a fee-body diagam fo the cane. Detemine B by solving the equation fo the sum of the moments of all foces about A. Note thee will be no contibution fom the unknown eactions at A. A fixed cane has a mass of 1000 kg and is used to lift a 2400 kg cate. It is held in place by a pin at A and a ocke at B. The cente of gavity of the cane is located at G. Detemine the components of the eactions at A and B. Detemine the eactions at A by solving the equations fo the sum of all hoizontal foce components and all vetical foce components. Check the values obtained fo the eactions by veifying that the sum of the moments about B of all foces is zeo.

15 Sample Poblem Ceate the fee-body diagam. 3. Detemine the eactions at A by solving the equations fo the sum of all hoizontal foces and all vetical foces. F A x x 0 : A + B 0 x kN F y 0 : A kN kn A y y kn 0 2. Detemine B by solving the equation fo the sum of the moments of all foces about A. + B M A 0 : ( 1.5m) kN( 2m) kn( 6m) 0 B kN 4. Check the values obtained. 점 B 에대한모멘트가제로이면확인됨

Sample Poblem 4.3 4-16 SOLUTION: Ceate a fee-body diagam fo the ca with the coodinate system aligned with the tack.

16 Sample Poblem SOLUTION: Ceate a fee-body diagam fo the ca with the coodinate system aligned with the tack. Detemine the eactions at the wheels by solving equations fo the sum of moments about points above each axle. A loading ca is at est on an inclined tack. The goss weight of the ca and its load is 25 kn, and it is applied at at G. The cat is held in position by the cable. Detemine the tension in the cable and the eaction at each pai of wheels. Detemine the cable tension by solving the equation fo the sum of foce components paallel to the tack. Check the values obtained by veifying that the sum of foce components pependicula to the tack ae zeo.

Sample Poblem 4.3 4-17 2. Detemine the eactions at the wheels. M A 0 : ( 10 kn) 625 mm ( 22.65 kn) 150 mm + R ( 1250 mm) 0 R 2 8 kn 2 M B 0 : + ( 10.5 kn) 625 mm ( 22.

17 Sample Poblem Detemine the eactions at the wheels. M A 0 : ( 10 kn) 625 mm ( kn) 150 mm + R ( 1250 mm) 0 R 2 8 kn 2 M B 0 : + ( 10.5 kn) 625 mm ( kn) 150 mm R ( 1250 mm) 0 R kn 1 1. Ceate a fee-body diagam W W x y + ( 25 kn) kn - ( 25 kn) 10.5 kn cos25 sin 25 o o 3. Detemine the cable tension. T F x 0 : kn T kn 4. Check the values obtained. y- 방향의힘의합이제로이거나, 점 A 또는 B 이외의한점에대한모멘트가제로이면확인됨

Sample Poblem 4.4 4-18 SOLUTION: Ceate a fee-body diagam fo the fame and cable.

18 Sample Poblem SOLUTION: Ceate a fee-body diagam fo the fame and cable. Solve 3 equilibium equations fo the eaction foce components and couple at E. The fame suppots pat of the oof of a small building. The tension in the cable is 150 kn. Detemine the eaction at the fixed end E.

19 Sample Poblem Solve 3 equilibium equations fo the eaction foce components and couple. 4.5 F x 0 : Ex kn E x ( 150kN) 0 1. Ceate a fee-body diagam fo the fame and cable. 6 F y 0 : E y kn E y 0 : M E ( 20 kn) - ( 150kN) 0 M E + 20 kn( 7.2 m) + 20 kn( 5.4 m) + 20 kn( 3.6 m) + 20 kn( 1.8m) kn m ( 150 kn) 4.5m + 0 M E

Sample Poblem 4.5 4-20 F F F A W 1800 N kq x y M O R R x y 1800 -W q 0, q + kq 0 0 l 200 mm B O 75 mm Þ Þ Wl sinq - kq R R 0 N 200mm sinq - 45 x y q 80.

20 Sample Poblem F F F A W 1800 N kq x y M O R R x y W q 0, q + kq 0 0 l 200 mm B O 75 mm Þ Þ Wl sinq - kq R R 0 N 200mm sinq - 45 x y q 80.3 C k 45 N/mm -kq W N / mm (75 mm) when q 0, sping is unstetched. Equilibium position? SOLUTION: Ceate a fee-body diagam fo the fame and cable. A W W 1800 N 2 l sin q q 0 q R x O R y F ks 연습문제 #1, #3, #5 s Undefomed position

21 4.6 Equilibium of a Two-Foce Body 4-21 Conside a plate subjected to two foces F 1 and F 2 Fo static equilibium, the sum of moments about A must be zeo. The moment of F 2 must be zeo. It follows that the line of action of F 2 must pass though A. Similaly, the line of action of F 1 must pass though B fo the sum of moments about B to be zeo. Requiing that the sum of foces in any diection be zeo leads to the conclusion that F 1 and F 2 must have equal magnitude but opposite sense.

22 4.7 Equilibium of a Thee-Foce Body 4-22 Conside a igid body subjected to foces acting at only 3 points. Assuming that thei lines of action intesect, the moment of F 1 and F 2 about the point of intesection epesented by D is zeo. Since the igid body is in equilibium, the sum of the moments of F 1, F 2, and F 3 about any axis must be zeo. It follows that the moment of F 3 about D must be zeo as well and that the line of action of F 3 must pass though D. The lines of action of the thee foces must be concuent o paallel. 세힘이작용하는물체가평형을이루자면세힘의작용선은동일점에작용하거나, 평행하여야함

Sample Poblem 4.6 4-23 SOLUTION: Ceate a fee-body diagam of the joist. Note that the joist is a 3 foce body acted upon by the ope, its weight, and the eaction at A.

23 Sample Poblem SOLUTION: Ceate a fee-body diagam of the joist. Note that the joist is a 3 foce body acted upon by the ope, its weight, and the eaction at A. A man aises a 10 kg joist, of length 4 m, by pulling on a ope. Find the tension in the ope and the eaction at A. The thee foces must be concuent fo static equilibium. Theefoe, the eaction R must pass though the intesection of the lines of action of the weight and ope foces. Detemine the diection of the eaction foce R. Utilize a foce tiangle to detemine the magnitude of the eaction foce R.

Sample Poblem 4.6 4-24 Ceate a fee-body diagam of the joist. Detemine the diection of the eaction foce R.

24 Sample Poblem Ceate a fee-body diagam of the joist. Detemine the diection of the eaction foce R. AF CD BD CE tan a o a 58.6 ABcos45 AE 1 2 ( m) BF - CE AE BD AF ( 4m) ( ) m tan 20 cos m m 2.828m m

25 Sample Poblem B C Detemine the magnitude of the eaction foce R. T sin 31.4 o R sin110 o 98.1 N sin 38.6 o T 81.9 N A R N 연습문제 #67, (#73, #75 )

26 4.8 Equilibium of a Rigid Body in Thee Dimensions 4-26 Six scala equations ae equied to expess the conditions fo the equilibium of a igid body in the geneal thee dimensional case. F 0 x M 0 x F 0 y M 0 y F 0 z M 0 z 3 차원강체의평형조건 These equations can be solved fo no moe than 6 unknowns which geneally epesent eactions at suppots o connections. The scala equations ae conveniently obtained by applying the vecto foms of the conditions fo equilibium, 0 ( F) F 0, M O

27 4.9 Reactions at Suppots and Connections fo a Thee-Dimensional Stuctue 4-27

28 4.9 Reactions at Suppots and Connections fo a Thee-Dimensional Stuctue 4-28

Sample Poblem 4.8 4-29 SOLUTION: Ceate a fee-body diagam fo the sign. Apply the conditions fo static equilibium to develop equations fo the unknown eactions.

29 Sample Poblem SOLUTION: Ceate a fee-body diagam fo the sign. Apply the conditions fo static equilibium to develop equations fo the unknown eactions. A sign of unifom density weighs 1200 N and is suppoted by a balland-socket joint at A and by two cables. Detemine the tension in each cable and the eaction at A.

30 Sample Poblem 4.8 Ceate a fee-body diagam fo the sign. Since thee ae only 5 unknowns, the sign is patially constain. It is fee to otate about the x axis. It is, howeve, in equilibium fo the given loading. T T BD EC T T T T T T BD BD BD EC EC EC D - B D - B - 8i + 4 j - 8k 12 (- 2 i + 1 j - 2 k ) 3 C - E C - E - 6i + 3 j + 2k 7 (- 6 i + 3 j + 2 k ) 7 평형방정식의수 > 미지수의수 부분적인구속 4-30

31 Sample Poblem F i : j : k : A + T A x A A y z BD T T T + T BD BD BD - EC T T T ( 1200 N) EC EC EC 0 0 j N 0 Apply the conditions fo static equilibium to develop equations fo the unknown eactions. 연습문제 #96 M j : k : A B 1.6T 0.8T T BD BD BD + E T T N v T EC EC EC + ( 1.2 m) i (-1200 N) N m Solve the 5 equations fo the 5 unknowns, TBD A T EC 1401 N ( 1503 N) i + ( N) j - ( N)k 0 j 0

32 Sample Poblem m 1.8 m 3.6 m 1.8 m 3.6 m 2000 N 3.6 m W2000 N 1.8 m 3.6 m 1.8 m 3.6 m Detemine (a) when G should be located if the tension in the cable is to be minimum (b) the coesponding minimum value of the tension

33 4 장 EX

34 EX

35 EX

36 EX

37 EX

38 EX. 67 교정 : 원본에서도 1.05 m 임, 실제 105mm 4-38 세힘이작용하는물체가평형을이루자면세힘의작용선은동일점에작용하거나, 평행하여야함

39 EX

40 EX

4.1 힘의모멘트 스칼라공식 4.1 힘의모멘트 스칼라공식 모멘트크기 (resultant moment) 2

Engineering Mechanics 정역학 (Statics) 4장힘계의합력 1 GeoPave Lab. 4.1 힘의모멘트 스칼라공식 1 4.1 힘의모멘트 스칼라공식 4.1 힘의모멘트 스칼라공식 모멘트크기 (resultant moment) 2 4.1 힘의모멘트 The moment does not always cause r otation. The actual