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Lecture 13

ENGR-1100 Introduction to Engineering Analysis. Lecture 13. Today Lecture Outline. - Equilibrium equations in 2-D - Solving problems of equilibrium of a rigid body in 2-D Free body diagram Application of physical laws - Statically indeterminate reaction and partial constraints. y. x.

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Lecture 13

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 13

  2. Today Lecture Outline - Equilibrium equations in 2-D - Solving problems of equilibrium of a rigid body in 2-D Free body diagram Application of physical laws - Statically indeterminate reaction and partial constraints

  3. y x A Equilibrium Equations in 2-D The necessary and sufficient conditions for a body to be in equilibrium in 2-D (3 independent equations) are: ‘A’ is ANY point on or off the rigid body. The moment equilibrium equation means that the resultant must pass through A Then apply the force equilibrium conditions along x and y to ensure that the resultant has zero magnitude (i.e. the body is in equilibrium)

  4. MB=0 Ry=0 Equilibrium Equations in 2-D: Alternate Forms • ‘A’ and ‘B’ are points on a line NOT PERPENDICULAR to the x-axis. • The condition MA=0 means that the resultant • must pass through A. • The condition Rx=0 means that the resultant (if • other than zero) is perpendicular to the x-axis. • Finally to MB=0 can only be satisfied if Ry=0. • In other words: x y B A Lid

  5. Equilibrium equations in 2D: Alternate Forms • ‘A’ ,‘B’ and ‘C’ three non-collinear points • The conditionMA=0means that the resultant • must pass through A. • The conditionMB=0means that the resultant • must pass through B. and thus must be along the lineAB. • Finally toMC=0can only be satisfied ifR=0 sinceCdoes not lie on AB. B C A Lid

  6. Don’t let the shape deceive you F F F F Special Force Systems Special case 1. A “two force member” is a body which is acted on by only two forces. The forces are equal, opposite and collinear (Newton’s Third Law!) F A strut

  7. F3 F2 F1 Special Force Systems Special case 2. A “three force member” is a body which is acted on by only three forces. The forces MUST BE CONCURRENT (otherwise there will be a resultant moment of the third force about the point of concurrency of the first two).

  8. y y x x A Ax B Lid Ay By Roller at B Can determine Ax, Ay and By from the 3 equilibrium equations Hinge at A Statically Determinate vs. Indeterminate Problems If the equilibrium equations are sufficient to determine all the support reactions, then the body is said to bestatically determinate with adequate constraints

  9. Statically Determinate vs. Indeterminate Problems If a body has more supports than are necessary for equilibrium then the equilibrium equations alone are not sufficient to determine all the support reactions, then the body is said to bestatically indeterminate Statically determinate Statically indeterminate y y x C x A A B Lid B Lid Roller at B Roller at B Hinge at A Hinge at A

  10. Problems with partial constraints Three support reactions in a 2D problem do not necessarily mean that the body is adequately constrained. Sometimes the body may bepartially constrainedand the equilibrium equations will not be sufficient to compute the support reactions. Statically indeterminate with inadequate constraints Statically determinate y y x x A B B Roller at B A Cannot determine Ax, Ay and Bx from the 3 equilibrium equations Hinge at A

  11. Problems with partial constraints A body with adequate number of reaction is improperly constrained when the constraints are arranged in such a way that the support forces are either concurrent or parallel.

  12. Example P6-40 An angle bracket is loaded and supported as shown in Fig. P6-40. Determine the reactions at supports A and B.

  13. First equation Ay unknown forces Bx=635 N Known forces Bx Second equation y 500 N x Ax=135 N Third equation Ax 350 N Ay=350 N Solution -350*0.22-500*0.1+Bx*0.2=0 -Bx+500+ Ax=0 Ay-350=0 A= 135 i + 350 j N

  14. Ay T T Ax Class Assignment: Exercise set 6-38 please submit to TA at the end of the lecture A beam is loaded and supported as shown in Fig. P6-38. The beam has a uniform cross section and a mass of 20 kg. Determine the reaction at support A and the tension T in the cable. Answer: T= 512 N A= -293 i + 256 j N.

  15. Example P6-63 GThe wrecker truck of Fig. P6-63 has a weight of 15,000 lb and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheel consists of a normal force Ay only. Determine the maximum pull P the wrecker can exert when 0 = 30 if Bx, cannot exceed 0.8 By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground).

  16. First equation To prevent tipping: Ay>0 Ay By W*8 – 10*P*sin(300) – 5*P*cos(300) =0 P=12.86 kip P y W=15 Kip Second equation By=26.1 kip x Third equation By-W- P*cos(300) =0 Bx Bx+ P*sin(300) =0 Bx=-6.43 kip Solution

  17. Bx=6.43 kip < 22.09 kip=0.8* By Which satisfy the requirement for Bx <0.8 By

  18. Class Assignment: Exercise set 6-64 please submit to TA at the end of the lecture Bar AB of Fig. P6-64 has a uniform cross section, a mass of 25 kg, and a length of 1m. Determine the angle q for equilibrium. Answer: q= 20.10

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