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Unformatted text preview: RIGID BODY EQUILIBRIUM EQUILIBRIUM
Rigid Body Equilibrium 2D Rigid LEARNING OBJECTIVES LEARNING Be able to recognize twoforce Be members members Be able to apply equations of Be equilibrium to solve for unknowns equilibrium PREREQUISITE KNOWLEDGE PREREQUISITE Units of measurement Units Trigonometry concepts Trigonometry Vector concepts Vector Rectangular component concepts Rectangular RIGID BODY EQUILIBRIUM 2  D RIGID
In 2D, a body is in equilibrium if and only if the sum of In the forces in each direction is equal to zero and the sum of the moment at any point is equal to zero. of FX = ∑ FY = ∑ M = 0.0 ∑ STEPS FOR SOLVING 2D EQUILIBRIUM PROBLEMS
1. Establish a suitable xy coordinate system (if not given). 2. Draw a free body diagram (FBD) of the object under analysis. 3. Apply the three equations of equilibrium to solve for the unknowns. IMPORTANT NOTES
1. If we have more unknowns than the number of independent equations then the problem is a statically indeterminate. We cannot solve this problem using just static. 2. The order in which we apply equations may affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force, then solve for the horizontal force first using ∑ FX = 0. 3. If the answer for an unknown comes out as negative number then the direction of the unknown force is opposite to that drawn on the FBD. APPLICATIONS
For a given load on the platform, how can we determine the forces at joint A and the force in the link (cylinder) BC? A steel beam is used to support roof joists. How can we determine the support reactions at each end of the beam? TWOFORCE MEMBERS
The solution of some equilibrium problems can be simplified if we can recognize members that are subjected to forces at only two points (e.g., at points A and B). Applying the equations of equilibrium to such member one can determine the resultant forces at A and B. These resultants must be equal in magnitude and act in the opposite directions along the line joining points A and B. EXAMPLE
Twoforce Members In the above figures, since the directions of the resultant forces at A and B are known (along the line joining points A and B), all AB members can be considered twoforce members if their weights are neglected. This simplifies the equilibrium analysis of some rigid bodies. EQUATIONS OF EQUILIBRIUM
If A body is subjected to a system of forces lie in the xy plane, then the body is in equilibrium if and only if F 1 the net forces and net moments acting on the body are zero. This 2D condition can be represented by the three scalar equations
y F3 F4 O x ∑F X = ∑ FY = ∑ M O = 0.0
F2 where “O” is any arbitrary point. SUPPORT REACTIONS IN 2D SUPPORT REACTIONS IN 2D SUPPORT REACTIONS IN 2D In general, if a support prevents translation of a body in a given direction, a force acts on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body. Support Reactions in 2D
No support reactions against horizontal translation
100 lb F R1Y R2Y As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body. SUPPORT REACTIONS IN 2D
F .
R1Y R2Y RX As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body. SUPPORT REACTIONS IN 2D
No support reactions against rotation
F RX RY As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body. Support Reactions in 2D
M RX F RY As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body. MORE SUPPORT REACTIONS IN 2D MORE SUPPORT REACTIONS IN 2D Page 196, your book MORE SUPPORT REACTIONS IN 2D EXAMPLE 1
10 ft 5000 lb The jib crane is pinconnected at A and supported by a smooth collar at B. Sketch the reactions at Points A and B. SOLUTION 1
RBX 10 ft B 12 ft 10 ft RAX A RAY 5000 lb EXAMPLE 2
10 ft 5000 lb The jib crane is pinconnected at A and supported by a smooth collar at B. Determine all the support reactions. SOLUTION 2
10 ft RBX B 12 ft 5000 lb 10 ft 5000 lb RAX A RAY Step 1: Draw a FBD and Show all known external forces and the support reactions SOLUTION 2 Continued
10 ft RBX B 12 ft
5000 lb 5000 lb 10 ft RAX A RAY Step 2: Write the equilibrium equations Σ Fx = 0; RAX  RBX = 0 Σ Fy = 0; RAY – 5000 = 0 Σ MA = 0; 12 RBX – 10(5000) = 0 SOLUTION 2 Continued
10 ft RBX B 12 ft 5000 lb 10 ft 5000 lb RAX A RAY Step 3: Solve the system equations RAY = 5000 lb RBX = 4167 lb (Left Direction) RAX = 4167 lb FBD Each ball weighs 2943 N FBD – A FBD – B FBD EXAMPLE
Smooth wall Assume the ladder weight = W, Draw FBD of the system FBD SOLUTION
Smooth wall RBX Wladder Ffriction RAY FBD
Draw FBD FBD 10650 lb 12.48 ft RAX RAY RBY FBD 10 lb The idealized model A, B, and C are smooth contact points
FC FC 10 lb FC The free body diagram FBD
900 N 3.33 m 1000 N
RCY 2.5 m
RAX RAY FBD FBD
REX RAY REY FBD FBD
RAX RD RAY FBD FBD
MA RAX A RAY EXAMPLE 1 Determine all the support reactions at points A and E SOLUTION 1
Y A X RE X RAY REY Step 1: Draw FBD SOLUTION 1  continued
Y A X RA YR
A Y Y R E Y – REX = 0 R E RX E REX Step 2: write the equilibrium equations Σ FX = 0; Σ FY = 0; Σ MA = 0; RAY – 12 – 14 – 18 + REY = 0 – 3(12) – 6(14) – 9(18) + 12(REY) = 0 SOLUTION 1 Continued
Y A X RA YR
A Y Y R E Y R E RX E REX Step 3: Solve the three equilibrium equations REX = 0 kN REY = (3(12) + 6(14) + 9(18))/12 = 23.5 kN RAY = 12 + 14 + 18 – REY = 44 – 23.5 = 20.5 kN EXAMPLE 2 Determine all the support reactions at A and B SOLUTION 2
Step 1: Draw FBD RAY RB X RB SOLUTION 2 Continued
Step 2: Write the equilibrium equations ΣFX = 0; RBX – 0.5 = 0
RAY Σ FY = 0; RAY57102+RBY = 0 Σ MB = 0;
R B X 14(RAY)+14(5)+6(7)6(2)+6(0.5) = 0
RB Y SOLUTION 2 Continued
Step 3: Solve the system equations RBX = 0.5 kip RAY RB X RB Y RAY = (14(5) + 6(7) – 6(2) + 6(0.5))/14 = 7.36 kips RBY = 5 + 7 + 10 + 2 – RAY = 24 – 7.36 = 16.6 kips EXAMPLE 3 Determine all the support reactions at A and C SOLUTION 3
Step 1: Draw FBD after calculating F1, F2 and F3 Note that the distributed load at the top can be broken dawn to rectangular (where F2 can be calculated) and triangular (where F3 can be calculated) distributions as shown
F3
B F2 F1 RC Y RA X
A RA Y SOLUTION 3  continued
F3
B F2 F1 RC Y RA Y F1 = 5(200) = 1000 N; located at Y1 = 2.5 m (from A) F2 = 6(100) = 600 N; located at X2 = 3 m (from A or B) F3 = 0.5(6)(100) = 300 N; located at X3 = 2 m (from A or B) RA X
A SOLUTION 3  continued
F3
B F2 F1 RC Y RA Y Step 2: write the equilibrium equations Σ FX = 0; F1 – RAX = 1000 – RAX = 0 Σ FY = 0; RAY + RCY – F2 – F3 = RAY + RCY – 900 = 0 Σ MA = 0; F1X1 – F2X2 – F3X3 + 6(RCY) = 0 1000(2.5) – 600(3) – 300(2) + 6(RCY) = 0 RA X
A SOLUTION 3  continued
F3
B F2 F1 RC Y RA X
A RA Y Step 3: Solve the equilibrium equations RCY = (600(3) + 300(2) – 1000(2.5))/6 = – 16.7 N (downward) RAY = 600 + 300 – RCY = 916.7 N RAX = 1000 N EXAMPLE 4 600 lb Find the support reactions at points A and B if the weight of the boom is 125 lb, center of mass is at G and the load is 600 lb. SOLUTION 4
AY AX 1 ft 40° A 1 ft B FB 3 ft G 125 lb 600 lb 5 ft D FBD of the Boom 600 lb Note: CB is a twoforce member, the number of unknowns at B is reduced from two to one. Equations of equilibrium yield: ∑ MA = 125*4600∗ 9+(FBsin40°)(1)+(FBcos40°)(1)=0; ∑ FX = AX + 4188 cos40° = 0; ∑ FY = AY + 4188 sin40° – 125 – 600 = 0; F = 4188 lb A = –3210 lb A = –1970 lb EXAMPLE 5 The load on the bent rod is supported by a smooth inclined surface at point B and a collar at A. The collar is free to slide over the fixed inclined rod. Determine the support reactions at A and B. SOLUTION 5
MA 5 4 3 3 ft FBD of the Rod
100 lb 200 lbft 3 ft 5 13 12 NB 2 ft NA ∑ FX = (4/5) NA – (5/13) NB = 0 ∑ FY = (3/5) NA – (12/13) NB – 100 = 0 ∑ MA = MA – 100∗ 3 – 200 + (12/13) NB∗ 6 – (5/13) NB∗ 2 = 0 Solving the two equations yields N = 82.54 lb N = 39.68 lb M = 106 lbft EXAMPLE 6 EXAMPLE When holding the 5lb stone (center of mass at G) in equilibrium, the humorous H, assumed to be smooth, exerts normal forces FC and FA on the radius C and ulna A as shown. Determine these forces and the force FB that the biceps B exerts on the radius for equilibrium. SOLUTION 6 SOLUTION
Draw FBD
FB FC FA 5 lb 0.8 inch 2 inch 12 inch 75o B 5 lb ΣMB = 0; 5(12) + FA(2) = 0.0; F = 30 lb ΣFY = 0; FB sin75o – 5 – 30 = 0; F = 36.2 lb ΣFX = 0; FC – 36.2 cos75o=0; F = 9.38 lb QUIZ QUIZ When holding the 5lb stone (center of mass at G) in equilibrium, the humorous H, assumed to be smooth, exerts normal forces FC and FA on the radius C and ulna A as shown. Determine these forces and the force FB that the biceps B exerts on the radius for equilibrium. QUESTION QUESTION
What are the required conditions for a rigid body to achieve equilibrium? body A) Σ F = 0 A) B) Σ M = 0 B) C) Σ F = 0, and Σ M = 0 C) QUESTION QUESTION
A Free Body Diagram (FBD) of a rigid body represents _____________ represents A) all applied external forces and moments B) all internal forces and moments B) C) all internal and external forces and moments C) QUESTION A rigid body is subjected to forces as shown. This body can be considered as a ______ member. A) singleforce B) twoforce C) threeforce D) sixforce QUESTION
F F F F How many support reactions are there for the given beam and is the problem statically determinate? A) (2, Yes) B) (2, No) C) (3, Yes) D) (3, No) QUESTION For the given beam, find a) how many support reactions are there, b) is this problem statically determinate and c) is the structure stable? Fixed support F A) (4, Yes, No) B) (4, No, Yes) C) (5, Yes, No) D) (5, No, Yes)
Pin joints QUESTION
AX A AY FB B
θ 100 lb Which equation of equilibrium allows you to determine FB ? A) ∑ FX = 0 C) ∑ MA = 0 B) ∑ FY = 0 D) any one of the above QUESTION A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable for all types of loadings? A) (3, Yes) C) (4, Yes) B) (3, No) D) (4, No) ...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.
 Fall '10
 Baladi
 Equilibrium

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