Presentation on theme: "Lecture 13 ENGR-1100 Introduction to Engineering Analysis."— Presentation transcript:
Lecture 13 ENGR-1100 Introduction to Engineering Analysis
- 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 Today Lecture Outline
Equilibrium Equations in 2-D The necessary and sufficient conditions for a body to be in equilibrium in 2-D (3 independent equations) are: x y A 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)
A and B are points on a line NOT PERPENDICULAR to the x-axis. -The condition M A =0 means that the resultant must pass through A. -The condition R x =0 means that the resultant (if other than zero) is perpendicular to the x-axis. -Finally to M B =0 can only be satisfied if R y =0. -In other words: Equilibrium Equations in 2-D: Alternate Forms Lid x y A B M B =0R y =0
Equilibrium equations in 2D: Alternate Forms Lid A B C A,B and C three non-collinear points -The condition M A =0 means that the resultant must pass through A. -The condition M B =0 means that the resultant must pass through B. and thus must be along the line AB. -Finally to M C =0 can only be satisfied if R=0 since C does not lie on AB.
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 (Newtons Third Law!) F F A strut F F Dont let the shape deceive you F
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). F3F3 F2F2 F1F1 Special Force Systems
Statically Determinate vs. Indeterminate Problems If the equilibrium equations are sufficient to determine all the support reactions, then the body is said to be statically determinate with adequate constraints x y AxAx AyAy ByBy Can determine A x, A y and B y from the 3 equilibrium equations Lid x y A B Hinge at A Roller at B
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 be statically indeterminate Lid x y A B Hinge at A Roller at B Statically determinate Lid x y A B Hinge at A Roller at B Statically indeterminate C Statically Determinate vs. Indeterminate Problems
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 be partially constrained and the equilibrium equations will not be sufficient to compute the support reactions. x y A B Hinge at A Roller at B Statically determinate x y A B Statically indeterminate with inadequate constraints Cannot determine A x, A y and B x from the 3 equilibrium equations
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. Problems with partial constraints
Example P6-40 An angle bracket is loaded and supported as shown in Fig. P6-40. Determine the reactions at supports A and B.
x Known forces unknown forces y AyAy AxAx 500 N 350 N BxBx Solution -350*0.22-500*0.1+B x *0.2=0 First equation Second equation -B x +500+ A x =0 Third equation A y -350=0 A= 135 i + 350 j N B x =635 N A x =135 N A y =350 N
Class Assignment: Exercise set 6-38 please submit to TA at the end of the lecture Answer: T= 512 N A= -293 i + 256 j N. 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. AyAy AxAx T T
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 B y and a tangential component B x, while the force exerted on the front wheel consists of a normal force A y only. Determine the maximum pull P the wrecker can exert when 0 = 30 if B x, cannot exceed 0.8 B y (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground).
Solution x y P AyAy BxBx ByBy W=15 Kip First equation W*8 – 10*P*sin(30 0 ) – 5*P*cos(30 0 ) =0 P=12.86 kip Second equation B y -W- P*cos(30 0 ) =0 B y =26.1 kip Third equation B x + P*sin(30 0 ) =0 B x =-6.43 kip To prevent tipping: A y >0
B x =6.43 kip < 22.09 kip=0.8* B y Which satisfy the requirement for B x <0.8 B y
Class Assignment: Exercise set 6-64 please submit to TA at the end of the lecture Answer: = 20.1 0 Bar AB of Fig. P6-64 has a uniform cross section, a mass of 25 kg, and a length of 1m. Determine the angle for equilibrium.