Points to Ponder for MECH 400 Prelims - PowerPoint PPT Presentation

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supplementary notes to statics of rigid bodies mech 400 n.
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  1. Supplementary Notes to Statics of Rigid BodiesMECH 400 ACCA

  2. REACTIONS • Always start by taking the sum of the moments at the SUPPORTS • It is usually advisable to convert the force(s) into its x- and y- components so that it will be easier to obtain the perpendicular distance(s) • Remember that the directions of the reactions are only ASSUMED at the beginning

  3. 6” W D C 8” B 4” F G 4” E A 9” 9” Reactions D and B are both pin-connected or pin-supported. The reactions at these two (2) joints are only internal. Therefore, when you take the moment at any point while considering the whole system, the only reactions will be Av and Ev plus the external load W. Av Ev Figure 1

  4. Reactions Only when isolating the member does the internal reactions of a pin-supported joint become exposed. Notice that the reaction at C is only vertical. This is due to the normal contact between the two members meeting at point C. C Bh B F FG A ISOLATED MEMBER ABC Av Cv Bv

  5. Normal Reactions D B A A C B C Isolating Member AB A B Isolating Member AC Av RB A B Isolating Member AC A C Note that RB is perpendicular to AC Av C

  6. NORMAL REACTIONS are normal to the surface of THE MEMBER whose side or surface is acted upon by the tip or end of the other member in contact

  7. Distinguishing Frames from Trusses • Problems about FRAMES are usually those involving members with THREE FORCES • Remember that three-force members (as in the case of frames) can only be solved using the Method of Members • In the Method of Members, we always solve the reactions by ISOLATING each members and drawing the FREE BODY DIAGRAMS (FBDs) of each isolated member • Know when to distinguish external from internal reactions (stated in previous slides) • ONLY EXTERNAL REACTIONS are exposed when considering the WHOLE SYSTEM • Internal forces (with respect to certain members and joints) only come out when the member being considered has been ISOLATED

  8. Trusses • Always start the solution by taking moments at the SUPPORTS • The forces on the trusses are found ONLY AT THE JOINTS • NEVER FORGET to include all the members connected to the JOINT considered when using the Method of Joints

  9. Trusses When using the Method of Sections, always include ALL the external forces and reactions AND the CUT MEMBER forces of the considered side of the cutting line Isolating Joint C 15KN C FBC FCD RE FCF FCG

  10. When To Eliminate Reactions • TRUSS SYSTEMS: When ALL LOADS are vertical AND the truss is SIMPLY SUPPORTED, the horizontal reaction can be eliminated. • A SIMPLY SUPPORTED system means the system has ONE ROLLER AND ONE HINGE/PINfor the supports • IF BOTH ROLLER AND HINGE/PIN are lying on the HORIZONTAL SURFACE, there are TWO VERTICAL REACTIONS and ONE HORIZONTAL REACTION. Therefore, the horizontal reaction can be equated to ZERO.

  11. Simple Reactions 10 k 5 k 10 k This truss system that is SIMPLY SUPPORTED has a hinge at A and a roller at G and both supports are lying on the horizontal surface. Because all loads are vertical, the horizontal reaction at A is equal to ZERO.

  12. F A 80 mm B 320 mm Simple Reactions For this example, even if the system is simply supported, the HORIZONTAL REACTIONS cannot just be cancelled or equated to zero! Note that the supports are stuck along the vertical. In this case, there are TWO HORIZONTAL REACTIONS. Hence, you cannot just cancel the horizontal reactions or equate them to ZERO even if there is NO HORIZONTAL LOAD on the system.

  13. Other Cases Example: The horizontal member AB has been isolated. Assuming both A and B are pin connected, how do you solve for the reactions? The member is carrying a vertical load W. W Horizontal Member A B Ah and Bh can both be cancelled BECAUSE the horizontal reactions pass through a horizontal member!

  14. Other Cases What happens if the isolated member AB is diagonal? B Diagonal Member W A Ah and Bhare NOT EQUAL TO ZERO because they don’t pass through a horizontal member.

  15. Other Cases What happens if the isolated diagonal member AB is carrying a horizontal loading? B W A Don’t forget to include the load W when solving for the reactions by taking the moment at any of the supports (or joints for this case)! Av and Bvare NOT EQUAL TO ZERO even if there is no vertical load and even if the weight has been neglected because they don’t pass through a vertical member. Note that if you take the moment at either A or B, the vertical reactions will have perpendicular distances respectively!

  16. Other Cases A and D are both supported by hinges. Similarly for trusses that have horizontal loads only, and there are vertical reactions at the supports, the vertical reactions CANNOT JUST BE CANCELLED! C B W D E A F Take note that there are TWO Vertical Reactions here and they don’t pass through just one vertical member. Taking moments at any of the supports, Av and Dv would each have its perpendicular distance.

  17. Other Cases A and D are both supported by hinges. For this case, there are no horizontal loads BUT the horizontal reactions of A and D are also NOT EQUAL TO ZERO. C W B D E A F Take note that there are TWO Horizontal Reactions here and they don’t support and pass through just one horizontal member. Taking moments at any of the supports, Ah and Dh would each have its perpendicular distance.

  18. UNDERSTAND EACH CONCEPT FIRST BEFORE JUMPING INTO MEMORIZING THE SOLUTION OF A CERTAIN PROBLEM. THAT WAY, IT WILL BE EASIER FOR YOU TO SOLVE JUST ABOUT ANY PROBLEMS THAT WILL COME OUT IN THE EXAMS. 

  19. God bless and best of luck on your preliminary examinations!