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# Lecture 5 (46)

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Concept of Particles and Free Body Diagram

Why FBD diagrams are used during the analysis?

It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.

Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.

Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.

A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.

Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.

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Are you sure you want to Yes No ### Lecture 5 (46)

1. 1. Equilibrium of Rigid Bodies
2. 2. FORCES ARE VECTORS THEREFORE WE NEED TO USE THE TECHNIQUES OF VECTOR ALGEBRA
3. 3. Catagories 2 D FORCE SYSTEMS
4. 4. EQUILIBRIUMEQUATIONS CONDITIONSOFEQUILIBRIUM STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
5. 5. Only ONE unknownOnly ONE unknown (Force component)(Force component) can be foundcan be found STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
6. 6. Example:Example: P STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS Determine the value of the force P so as toDetermine the value of the force P so as to satisfy the equilibrium?satisfy the equilibrium? F 0 -350+250-80+P=0 P=180 kN x + → = →∑
7. 7. Example Consider the particle subjected to two forces  Assume unknown force F acts to the right for equilibrium ∑Fx = 0 ; + F + 10N = 0 F = -10N  Force F acts towards the left for equilibrium STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
8. 8. F1 Example: If the stepped bar is in equilibrium find the force F1. Resultant of Collinear Forces STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
9. 9. EQUILIBRIUMEQUATIONS CONDITIONSOFEQUILIBRIUM STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
10. 10. A particle when is subjected to coconcurrentncurrent forcesforces in the x-y plane its equilibrium condition equation can be written as   ΣFx i + ΣFy j = 0 Both of these vector equations above to be valid, implies that both the x and the y components should be equal to zero. Hence,  +→ ΣFx = 0 F1x + F2x + ….. = 0  +↑ ΣFy = 0 F1y + F2y + ….. = 0 Both algebraic sums equal to zero. ∑ = 0F  Only TWO unknowns can be foundOnly TWO unknowns can be found 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem ‘‘CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces’’
11. 11. ∑ = 0F  0=∑ xF 0=∑ yF 0=+ ∑∑ jFiF yx  andand TwoTwo Force componentForce component unknownunknownss cancan be foundbe found 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
12. 12. • Resolve the given forces into i and j components and apply the equilibrium +→ ∑F∑Fxx = 0= 0 +↑ ∑F∑Fyy = 0= 0 • Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero. Only TWO unknowns can be foundOnly TWO unknowns can be found!! 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
13. 13. Determine the magnitudes of F1 and F2 for equilibrium. Set θ=60°. Example: 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
14. 14. FF11=1.827 kN F=1.827 kN F22=9.596 kN=9.596 kN Only TWO unknowns can be foundOnly TWO unknowns can be found 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
15. 15. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces Example: (T) 53° 24°
16. 16. Example Determine the tension in cables AB and AD for equilibrium of the 250kg engine. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
17. 17. SINCE the mass of the engine is given i.e. unit is ‘kg’ (scalar) and not the weight (FORCE) the calculations should be corrected to a vector having a unit of Newton. (mass * gravity ) 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
18. 18. Procedure for Analysis 1. Free-Body Diagram - Establish the x, y axes in any suitable orientation - Label all the unknown and known forces magnitudes and directions - Sense of the unknown force can be assummed 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
19. 19. Procedure for Analysis 2. Equations of Equilibrium - Apply the equations of equilibrium +→ ∑Fx = 0 +↑ ∑Fy = 0 - Components are positive if they are directed along the positive axis and negative, if directed along the negative axis 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
20. 20. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
21. 21. Solution FBD at Point A - Initially, two forces acting, forces of cables AB and AD - Engine Weight [W=m.g] = (250kg)(9.81m/s2 ) = 2.452 kN supported by cable CA - Finally, three forces acting, forces TB and TD and engine weight on cable CA FBD of the ring AFBD of the ring A 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
22. 22. Solution +→ ∑Fx = 0; TB cos30° - TD = 0 +↑ ∑Fy = 0; TB sin30° - 2.452 = 0 Solving, TB = 4.904 kN TD = 4.247 kN *Note: Neglect the weights of the cables since they are small compared to the weight of the engine FBD of the ring AFBD of the ring A 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
23. 23. Example If the sack at A has a weight of 20 N , determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
24. 24. Solution TEC 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces TEC
25. 25. FBD of the ring EFBD of the ring E FBD of the ring CFBD of the ring C TEC 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
26. 26. Solution FBD at Point E. Three forces acting, forces of cables EG and EC and the weight of the sack on cable EA FBD of the ring EFBD of the ring E 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
27. 27. Solution Use equilibrium at the ring to determine tension in CD and weight of B with TEC known +→ ∑Fx = 0; TEG sin30° - TECcos45° = 0 +↑ ∑Fy = 0; TEG cos30° - TECsin45° - 20 = 0 Solving, TEC = 38.637 N TEG = 54.641 N 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
28. 28. FBD of the ring EFBD of the ring EFBD of the ring CFBD of the ring C 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
29. 29. Solution FBD at Point C Three forces acting, forces by cable CD and EC (known) and weight of sack B on cable CB. FBD of the ringFBD of the ring CC 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
30. 30. Solution +→ ∑Fx = 0; 38.637cos45° - (4/5)TCD = 0 +↑ ∑Fy = 0; (3/5)TCD + 38.637sin45° – WB = 0 Solving, TCD = 34.151 N WB = 47.811 N *Note: components of TCD are proportional to the slope of the cord by the 3-4-5 triangle 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
31. 31. Example: (T) The 50-kg homogenous smooth sphere rests on the 30° incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B? 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
32. 32. FBD of the sphereFBD of the sphere 30° A B 30° A B 30° RRAA RRBB CC WW 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces Example:
33. 33. y A A + x B B W 50x9.81 490.5 F 0 F cos30 -490.5= 0 F 566.381 N (assumed direction correct) F 0 566.381sin30 -F 0 F 283.191 N + = = ↑ = ° ⇒ = → = ° = ⇒ = ∑ ∑ (assumed direction correct) 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces Example: N
34. 34. STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS
35. 35. STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS EQUILIBRIUMEQUATIONS CONDITIONSOFEQUILIBRIUM 3 unknowns3 unknowns 5 unknowns5 unknowns 3 unknowns3 unknowns 6 unknowns6 unknowns
36. 36. A particle when is subjected to coconcurrentncurrent forcesforces in the x-y-z axes, its equilibrium condition equation can be written as   ΣFx i + ΣFy j + ΣFz k = 0 Both of these vector equations above to be valid, implies that the x, the y and the z components be equal to zero separately. Hence,  +→ ΣFx = 0 F1x + F2x + ….. = 0  +↑ ΣFy = 0 F1y + F2y + ….. = 0  + ΣFz = 0 F1z + F2z + ….. = 0 ∑ = 0F  Only TOnly THREEHREE unknowns can be foundunknowns can be found Three-D Force Systems Concurrent at a point
37. 37. When the system of external 3 dimensional forces acting on an object in equilibrium: Σ F = (ΣFx) i + (ΣFy) j + (ΣFz) k = 0 so each component of this equation must be determined separately: ΣΣFFxx =0,=0, ΣΣFFyy =0=0,, ΣΣFFzz =0.=0. Three-D Force Systems Concurrent at a point
38. 38. • Resolve the given forces into i, j and k components and apply the equilibrium +→ ∑F∑Fxx = 0= 0 +↑ ∑F∑Fyy = 0= 0 + ∑F∑Fzz = 0= 0 • Equations of equilibrium require that the algebraic sum of x, y and z components must be equal to zero. TTHREEHREE unknowns can be foundunknowns can be found!! Three-D Force Systems Concurrent at a point
39. 39. The 100-kg cylinder is suspended from the ceiling by cables attached at points B, C and D. What are the tensions in cables AB, AC & AD ? Note that: the gravity effect is in –the gravity effect is in –veve y direction.y direction. Example: Three-D Force Systems Concurrent at a point
40. 40. Solution Strategy: •Isolate the part of the cable system near point A, •Obtain a free-body diagram subjected to forces due to the tensions in the cables. •Because the sums of the external forces in the x, y, and z directions must IN BALANCE, obtain 3 INDEPENDENT equations for the three unknown cables that are in tension. •To do so, express the forces exerted by the tensions in terms of their components. Three-D Force Systems Concurrent at a point
41. 41. Drawing the Free-Body Diagram and Applying the Equations Three-D Force Systems Concurrent at a point
42. 42. • Isolating the part of the cable system near point A and show the forces exerted by the tensions in the cables. The sum of the forces must equal zero: Σ F = TAB + TAC + TAD − (981 N)j = 0 • Writing the Forces in Terms of their Components • Obtain a unit vector that has the same direction as the force TAB by dividing the position vector rAB from point A to point B by its magnitude. rAB = (xB − xA)i + (yB − yA)j + (zB − zA)k = 4i + 4j +2k (m) Three-D Force Systems Concurrent at a point
43. 43. Solution kji r r e 333.0667.0667.0 ++== AB AB ABλ AB Three-D Force Systems Concurrent at a point
44. 44. Expressing the force TAB in terms of its components by writing it as the product of the tension TAB in cable AB and the unit vector eAB... TAB = TABeAB == TAB (0.667 i + 0. 667 j + 0.333 k) Express the forces TAC and TAD in terms of their components using the same procedure. TAC = TAC (−0.408 i + 0.816 j − 0.408 k) TAD = TAD (−0.514 i + 0.686 j + 0.514k ) λAB = λAB Three-D Force Systems Concurrent at a point
45. 45. Substituting these expressions into the equilibrium equation TAB + TAC + TAD − (981 N)j = 0 Because the i, j, and k components must each equal to zero, this results in three equations of: i-component: 0.667TAB − 0.408TAC − 0.514TAD = 0 j-component: 0.667TAB + 0.816TAC + 0.686TAD = 981 k-component: 0.333TAB − 0.408TAC + 0.514TAD = 0 Solving these 3 equations successively, the tensions are: TAB = 519 N TAC = 636 N Three-D Force Systems Concurrent at a point
46. 46. Three-D Force Systems Concurrent at a point Example: