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FBD: 3D Force Reaction 1 FBD : 3D Force Reaction 1 unknown 3 unknown 2 unknown No moment support? 2 3 unknowns 6 unknowns 6. smooth pin 5 unknown 3 7. bearing journal-bearing support 4 unknown thrust-bearing support 5 unknown For some problem, the couples in both case should be treated as zero to provide statical determinacy 4 If thrust barring If glue or friction exist 6 3D Equilibrium Supports #2 7 3D Equilibrium Supports #3 8 3D Equilibrium Supports #4 9 3D Equilibrium Supports #5 Comparison with 2D supports 10 Equilibrium in 3D R F 0 M M 0 or or Fx= 0 Fy= 0 Fz= 0 Mx= 0 My= 0 Mz= 0 any but only three independent (perpendicular) axis any but only one point Body (bodies) in Equilibrium OR any but only three independent axis The 3 axises need not interesect at one point y x z R 0 y x M position (where the axis pass) also matters. Direction of moment axis has a affect on solving problem. z O 0 12 Equilibrium in 3D or R F 0 M M 0 or Fx= 0 Fy= 0 Fz= 0 Mx= 0 My= 0 Mz= 0 at most 6 unknowns may be found. any but only three independent (perpendicular) axis any but only one point Body (bodies) in Equilibrium OR any but only three independent axis vector approach may be easier - Each of the equation may be applied independently; e.g., an accelerating car on a flat surface may be treated as in equilibrium in the vertical direction. Same for the moment equations. ax 0 z 0 - Not in this class, but be careful about the moment equations, things get very complicated if the body is not spinning in a single plane! 13 nˆ M F r r X position vector: from any point on line to any point on tline of action of the force. A d Y Z nˆ F5 M F , { ( r F ) nˆ }nˆ F3 F1 M ? { 5 {(r F ) nˆ } i }nˆ i=1 F4 F2 { (r F4 ) nˆ }nˆ Forces which interest or parallel with axis, do not cause the moment in that axis 15 Example 3/5 The uniform 7m steel shaft has a mass of 200kg and is supported by a ball-and-socket joint at A in the horizontal floor. The ball end B rests against the smooth vertical walls. Find the force exerted by the walls and the floor on the ends of the shaft. this is not FBD. 17 FBD: frequent mistake Correct FBD System Isolation Don’t forget axis Write force name Use (at least) 3 different colors No axis No system isolation (surrounding still exists) Caution: In some slides using in this class, FBD may be drawn wrongly according to the rule introduced, DO NOT IMITATE this style in your homework or examination. 18 W mg 200(9.81) 1962 N h 7 (6 2 ) 3 2 2 2 Fx 0 Fy 0 Fz 0 M M M x 0 y 0 z 0 any point but only one point A= (2,6,0) B=(0,0,3) G=(1,3,1.5) M A 0 iˆ ˆj 1 rAG rAB iˆ 3 ˆj 1.5kˆ 2 iˆ ˆj kˆ Vector Cross Product is useful in 3D Problem (3By 5890)iˆ (3Bx 1962) ˆj ( 2 By 6Bx )kˆ 0 2 6 3 1 3 1.5 0 Bx By 0 0 0 1962 F 0 Independent Eq. = 2 rAB 2iˆ 6 ˆj 3kˆ rAB ( Bx Bx ) rAG W 0 kˆ Use point A Bx 654 N By 1962 N Ans (654 Ax )iˆ (1962 Ay ) ˆj ( 1962 Az )kˆ 0 Ax 654 N A y 1962 N A y 1962 N | A | 654 1962 1962 2850 N 2 2 2 Ans 19 The homogeneous plate shown has a mass of 100 kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a balland-socket joint at B, and a cord at C, determine the components of reaction at the supports. 200 N-m 300 N z y x TC Hibbeler Ex 5-15 [ Fx 0] Bx 0 [ Fy 0] By 0 981 N By Az Bx Bz [ Fz 0] Az Bz TC 300 981 0 20 z TC 200 N-m 300 N y x [ Fx 0] Bx 0 [ Fy 0] By 0 3m [ Fz 0] 981 N M 0 By 2m Az Bx Bz [ M x 0] 2TC 2Bz (1)981 0 [ M y 0] Az Bz TC 300 981 0 Selection of moment axis OR (1.5)300 (1.5)981 OR [ M x 0] [ M y 0] +3TC (1.5)300 3Bx 3 Az 200 0 Az 790N Bz -217N 2 Az (2)300 (1)981 0 (1.5)981 200 0 TC 707N 21 Boom AB lies in vertical y-z plane, is supported by a balland-socket joint at B and by 2 cables at A. Calculate tension in each cable resulting from the 20-kN force acting in the horizontal plane at the midpoint M of the boom. Neglect the weight of the boom. 22 z 3m T1 A 20 M 4m P 4 iˆ 5 ˆj 10kˆ T2 T2 42 52 102 T1 ( 4 iˆ 5 ˆj 10kˆ ) 11.87 T2 (4 iˆ 5 ˆj 10kˆ ) 11.87 problem Bz By B D 4 iˆ 5 ˆj 10kˆ T1 T1 42 52 102 T2 10m F P 20(sin 20 iˆ cos20 ˆj ) 2m y There are 5 unknowns: Bx E 4m C use moment at B: x T1 T2 Bx By Bz MB 0 (need only 2 components (eq.) from 3) rBA 3 ˆj 10kˆ rBM 1 ˆ (3 j 10kˆ) 2 M B rBA T1 rBA T2 rBM P 0 component x , component z 23 Technique: Finding unknown in 1 equation The bent rod is supported at A by a journal bearing, at D by a ball-and-socket joint, at B by means of cable BC. Use only one equilibrium equation to obtain a direct solution for the tension in cable BC. Assume that the bearing at A is capable of exerting force components only in the z and y directions (since it is properly aligned on the shaft). Hibbeler Ex 5-19 30 Find T only in 1 scalar Equation. TB TB rB 1 ˆj Az 1 (0.2iˆ 0.3 ˆj 0.6kˆ) 0.7 Ay TB uˆ DA rB uˆ DA rE W 100(9.81)kˆ rE 0.5 ˆj rDA rDA 1 ˆ 1 ˆ i j 2 2 Dy W Dx Dz Scalar Equation M DA { uˆDA ( ri Fi ) } 0 TB W Dx Dy Dz Ay Az i MDA 0 uˆDA (rB TB rE W ) 0 31 uˆDA rB TB rE W 0 MDA 0 TB TB 1 (0.2iˆ 0.3 ˆj 0.6kˆ) 0.7 uˆDA (rB TB ) uˆDA (rE W ) 0 ˆj 1 ˆ 1 ˆ ˆ TB ˆ ˆ ˆ ( i j ) ( j ) (0.2i 0.3 j 0.6k ) ( ) ( W kˆ) 0 0.7 2 2 2 1 ˆ 1 ˆ TB 1 ˆ ˆ ( i j ) (2k 6i ) ( ) (W iˆ) 0 2 2 2 7 6TB W 0 7 2 You can use other rE and rB (Solve as 2D problem) TB 572.25 N Az TB (TB ) z 6 TB 7 1 2 1 1 (TB ) z W 0 2 2 2 (TB ) z W 1 0 2 6 1 (TB ) W 0 7 2 Ay uˆ DA 1 2 2 rB W rE Dx Dy Dz 32 3/90 Sign has a mass of 100 kg, center of mass at the center of the sign. - Support at C can be treated as a ball-andsocket joint. - At D, support is provided in y-direction only - Find the tensions, T1 and T2 - Find total force supported at C and the lateral force R supported at D. ( 4iˆ 1.5 ˆj 2.5kˆ) T1 T1 L1 L1 4 2 1.52 2.52 4.95 ( 2.5iˆ 1.5 ˆj 2.5kˆ) T2 T2 L2 L2 2.52 1.52 2.52 3.84 34 this is not a FBD, why? L1 42 1.52 2.52 4.95 T1 T1 ( 4iˆ 1.5 ˆj 2.5kˆ) L1 L2 2.52 1.52 2.52 3.84 T2 T2 (2.5iˆ 1.5 ˆj 2.5kˆ) L2 M R E Cx Cy M x F y 0 F z 0 2.5 2.5 T1 T2 Cz 0 L1 L2 Cy (1) 0 0 Cx (3.5) 100(9.81)(2) 0 z 0 x 0 AB M Cz 1.5 1.5 T1 T2 R 0 L1 L2 0 F Get R and C (1.5T1 ) (1.5T2 ) (4) (2.5) 0 L1 L2 4 2.5 T1 T2 Cx 0 L1 L2 Get T1 and T2 35 Properly Alignment and Moment Support A,B,C: journal bearings “Properly Alignment” If the line of pipe is initially placed without pre-torsion, the 6 couple reactions are not necessary to maintain equilibrium, and then can be considered as not existing. The 6 force reactions developed by the bearings are sufficient for maintaining the equilibrium since they prevent the shaft from rotating about any of the coordinate axes. 6 unknown reactions with 6 independent equations. 37 Properly Alignment and Moment Support “initially properly aligned” Even though the line of pipe is initially placed without pretorsion or properly aligned. The 4 force reactions developed at bearing A and tension T are not sufficient to maintain the equilibrium of the body. The couple reactions are developed to make the object in equilibrium 38 Properly Alignment and Moment Support “properly aligned” “properly aligned” The 6 force reactions developed by the bearing, hinge and ball are sufficient for maintaining the equilibrium, so the couple reactions will not be developed at the bearing and hinge. 39 thrust-bearing “initially properly aligned” 40 Sample 3/6 A 200-N force is applied to the handle of the hoist in the direction shown. The bearing A supports the thrust (force in the direction of the shaft axis), while bearing B supports only radial load (load normal to the shaft axis). Determine the mass m which can be supported and the total radial force exerted on the shaft by each bearing. Each bearing is initially properly align. 42 P 200(cos60sin 45iˆ sin60 ˆj cos60cos45kˆ) By (70.7iˆ 173.2 ˆj 70.7kˆ) Bx Ay P Ax F 0 F 0 F 0 x M x y 0 M at same point z y 0 M z 0 No need to be the same point Az m(9.81)100 – 173.2(250) = 0 -Bx(150)-70.7(175)+70.7(250)= 0 2D FBD Ay(150)+m(9.81)200-173.2(325)=0 Ax+Bx-70.7=0 Az-70.7=0 By+Ay-m(9.81)100-173.2=0 2D View convenient when almost forces are othorgonal.43 3/83 The shaft, level and handle are welded together and constitute a single rigid body. Their combined mass is 28 kg with mass center at G. The assembly is mounted in A and B and rotation is prevented by link CD. Determine the forces exerted on the shaft by bearing A and B, while the 30 N-m couple is applied to the handle as shown. BearingA,B : Journal bearing 45 ignore 2 force member T // plane x-y 2 force member Most forces are parallel with rectangular axis z Ay By T cos y T sin T z Ax Bearing A,B : Journal bearing mg Bx y M 30 mg Ax Bx 46 | A | Ax2 Ay2 | B | Bx2 By2 T=251N Ax=11.74N 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0 Bx=112.2N Ax+Bx+Tsin36.9-28(9.81)=0 300 300 600 Ay T cos 200 z By Ay(0.6)-Tcos36.9(0.5)=0 Ay+By-Tcos36.9=0 Ay=167.5N By=33.5N 200 y T sin T 100 200 Ax mg 300 z Bx y 380 M 30 220 mg Ax Bx 47 Bearing A,B : Journal bearing 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0 Ax+Bx+Tsin36.9-28(9.81)=0 Ay(0.6)-Tcos36.9(0.5)=0 Ay+By-Tcos36.9=0 Bearing A,B : thrust bearing Az+Bz=0 T=251N Ax=11.74N Bx=112.2N Ay=167.5N By=33.5N Cannot be solved to get exact value. 7 unknowns, 6 equations You will get exact value for some unknowns but not for all of them. 48 Statically determinate (SD) problems the problem that can be analyzed by using the equilibrium condition alone. number of unknowns F5 F4 For one obj, usually F 0 3 scalar eqs (2D) M 0 or 6 scalar eqs (3D) number of independent equilibrium equations maybe > (3 or 6) (many FBDs) y x F6 z F3 F1 F2 maybe < ( 3 or 6 ) FE (some specific problems i.e. point equilibrium) 54 Categories of Equilibrium (3D) F2 1) Concurrent at a point y F3 O F1 x 0 F x 0 M F dependant M y 0 F y 0 M 0 z z 0 M r A F1 r F2 r F3 r ( F1 F2 F3 ) 0 y x 2) Concurrent with a line z Independent Equations: F3 F1 Independent Equations: z F4 F2 x x 0 F x 0 M F M y 0 F y 0 M z z 0 0 All couples occuring when moving force |_ this direction M | _ r ( M r F) 55 Categories of Equilibrium (3D) F4 y x 3) Parallel Independent Equations: F2 F3 F1 F1 F4 y x F3 x 0 F x 0 M F M All couples occuring when moving force // the plane M F2 z z y 0 F y 0 M z z 0 0 M | _ F ( M r F) 4) General Independent Equations: x 0 F x 0 M F M y 0 F y 0 M z z Constraints and statical determinacy: see the book 0 0 56 SD/SI 1) SD Type of Statics Problem (statically determinate) the problem that can be solved for all unknown with only static equilibrium equations. 2) SI-redundant mass m bar Unknown 3 , Independent eq 3 (redundant statically indeterminate) the problem that can not be solved for all unknown with only static equilibrium equations. Unknown 4 , Independent eq 3 3) SI-Improper (improper statically indeterminate) the problem that can not be solved due to improper support. unknown 3 , independent 2 SD/SI Statically Indeterminacy Statically indeterminate (SI) problems the problem that can not be analyzed by using the equilibrium condition alone. SI-redundant SI-improper Problem with redundant supports Problem with imporper supports - Not able to maintain equilibrium No solution: (cant maintain moment) number of unknowns number of independent equilibrium equation unknown 3 , independent 2 59 3/91 The window is temporarily held open in the position shown. If a=0.8 m and b = 1.2 m and the mass of window is 50 kg with mass center at is geometric center, determine the compressive force F(CD) in the prop and all the component of the forces exerted by the hinges B on the windows. Assume that A is a thrust-bearing hinge but the hinge B is not. 61 D : (0.8sin50 , 0.8cos50 ,0) (0.613, 0.514,0) C : (0,0.8,0.3) By y 1.2 Ay Ax Az z F FnˆCD F (0.828iˆ 0.286ˆj 0.406kˆ) x ( M A ) z 0 : Bx mg 0.8 F solved 50(9.81)(0.4 sin 50) (0.828F )(0.8) 0 ( M A ) y 0 : Bx solved Bx (1.2) (0.828F )(0.3) 0 50 C ( M A ) x 0 : D By solved By (1.2) 50(9.81)(0.6) Ans (0.406F )(0.8) (0.386F )(0.3) 0 Fx 0 : F 0: F 0: y z Ax Bx 0.828F 0 Ax solved Ay By 0.386F 50(9.81) 0 Ay solved Az 0.406F 0 Az solved 62 Basic concept: Since all forces are given in orthogonal system, it is easier to solve the problem with orthogonal projections as sample problem 3/6. 63 Cx is the key for determining Ax and Bx On x-z plane: take moment at line AB to reduce unknowns M AB W 0.150 Cx 3 0 W 0.150 Cx 3 Cx W 0.050 176.6 N On x-y plane: take moment at point B to reduce unknowns y x A x z C x Az yW Bx M Az +Bz z Cx FBD x W Ax 3 Cx 2.4 0 Ax 3 Cx 2.4 Ax +Bx B B z Ax Cx 0.8 141.3N Ans Then sum forces on x-axis F x Ax Cx Bx 0 Ax Cx Bx Ans 64 Bx Ax Cx 35.3N Recommended Problem 3/73 3/82 3/92 3/93 65 Equilibrium Here ends the most important chapter of the subject. 66 Review Quiz #4 Review • Classify problems in equilibrium into SD and SI categories – How can I recognize the SD and SI problems? – Why do we need to differentiate between SD and SI problems? – What does it mean when an object has redundant supports? How about the improper supports? – What is the degree of redundancy? 67 Concepts Review • When a body is in equilibrium, the resultant force and couple about any point O are both zero. Problems can be analysed using free body diagrams (FBDs). • Statically determinate (SD) problems can be solved using the equilibrium conditions alone, while the statically indeterminate (SI) problems cannot due to too few or too many supports or constraints. 68 Chapter Objectives Descriptions #1 • Analyse objects (particles and rigid bodies) in equilibriums – Specify the condition of equilibrium – Describe objects in equilibrium in physical and idealized worlds – Choose appropriate bodies/parts of bodies for the analyses – Specify support reactions from physical worlds and vice versa – Specify loads from physical worlds and vice versa – Draw free-body diagrams – Calculate the support reactions from condition of equilibrium – Identify 2-force and 3-force members 69 Chapter Objectives Descriptions #2 • Classify problems in equilibrium into SD and SI categories – Specify types and conditions of SD and SI problems – Describe physical meanings of SI problems – Use FBDs differentiate between SD and SI problems – Obtain the degree of redundancy (when applicable) 70 Review Quiz #1 Review • Analyse objects (particles and rigid bodies) in equilibriums – What are the conditions of equilibrium for each type of bodies? – What are the physical meanings of 2D contact with cable, spring, smooth surface, contact with rough surface, roller, pin, fixed and slider supports? – What are the physical meanings of 3D contact with cable, smooth surface support, roller, ball & socket, journal bearing, thrust bearing, smooth pin, hinge and fixed support? – What are the differences between 2D and 3D FBDs? 71 Review Quiz #2 Review • Analyse objects (particles and rigid bodies) in equilibriums – Can we analyze objects in equilibrium without FBDs? Why or why not? – What are the steps in the sketching of an FBD? – In drawing the FBDs, how did we choose the directions of support reactions? – Why do we have to delete the physical supports, or isolate objects, before adding the support reactions? – How many independent equilibrium equations can you obtain from 2D and 3D diagrams? What are they? 72 Review Quiz #3 Review • Analyse objects (particles and rigid bodies) in equilibriums – Although we can choose the point about which the moments are evaluated arbitrarily, do you have some guidelines of choosing for simpler analyses? – Why it is worthwhile to recognize that an object is a two-force member? What about the three-force member? How can I identify 2-force and 3-force members in 2D and 3D problems? 73