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ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis. Lecture 7. Lecture Outline. Equilibrium of a particle - three dimensional problems. Equilibrium of a particle – three dimensions. R = R x + R y + R z = = R x i + R y j + R z k = S F x i + S F y j + S F k k.

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ENGR-1100 Introduction to Engineering Analysis

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 7

  2. Lecture Outline • Equilibrium of a particle - three dimensional problems.

  3. Equilibrium of a particle – three dimensions R= Rx + Ry + Rz= = Rxi + Ryj+ Rzk =SFxi + SFyj + SFkk The equation can only be satisfied if: Rx = Rxi=SFxi=0 Ry = Ryj=SFyj=0 Rz = Rzj=SFzj=0

  4. Example – P3-22 Struts AB and AC of Fig. P3-22 can transmit only axial tensile compressive forces. Determine the forces in struts AB and AC and the tension in cable AD when force F=1250 N.

  5. z y x TAD FAC FAB F Solution 1) Free body diagram – body #1 TAD= (-6/(32+62)1/2j + 3/(32+62) 1/2k)TAD=(-0.89j + 0.477k )TAD FAC= (-3/(32 +62 +22) 1/2i -6/(32 +62 +22) 1/2j -2/(32+62 +22) 1/2k )FAC = (-0.43i - 0.86j - 0.29k )FAC FAB= (2/(22 +62 +22) 1/2i -6/(22 +62 +22) 1/2j -2/(22+62 +22) 1/2k )FAB =( 0.3i - 0.9j - 0.3k) )FAB F= (-1250 k) N

  6. SFx=0 -0.43FAC +0.3FAB =0 z SFy=0 y x -0.89TAD-0.86FAC -0.9FAB =0 SFz=0 0.477TAD-0.29FAC -0.3FAB =1250 lb TAD FAC FAB F

  7. Using Maple: TAD=1606.5 N; FAC=-665.1 N; FAB =-953.2 N; 0 -0.43 0. 3 0 -0.89 -0.86 -0.9 0 0.48 -0.29 -0.3 1250 A=

  8. Class Assignment: Exercise set 3-21 please submit to TA at the end of the lecture The block shown in fig. P3-21 weighs 500 lb. Determine the tensions in cable AB, AC, and AD. Solution: TAB=267.4 lb; TAC=142.5 lb; TAD =164 lb;

  9. TAD z TAC TAB y x W=500 lb Solution TAB= (6/(62+122) 1/2j + 12/(62+122)1/2k)TAB=(0.447j + 0.89k )TAB TAC= (4/(42 +32 +122) 1/2i -3/(42 +32 +122) 1/2j -12/(42 +32 +122) 1/2k )TAC = (0.31i – 0.23j + 0.92k )TAC TAD= (-4/(42 +82 +122)1/2i-8/(42 +82+122)1/2j+12/(42 +82 +122)1/2k)TAC = (-0.27i - 0.53j + 0.8k )TAC

  10. TAB= (0.447j + 0.89k )TAB TAC= (0.31i – 0.23j + 0.92k )TAC TAD= (-0.27i - 0.53j + 0.8k )TAD W= (-500 k) lb 0TAB +0.31TAC -0.27TAD=0 SFx=0: 0.447TAB-0.23TAC -0.53TAD =0 SFy=0: 0.89TAB+0.92TAC +0.8TAD =500 SFz=0:

  11. 0 -0.31 -0.27 0 0.45 -0.23 -0.53 0 0.89 0.92 0.8 500 A= Using Maple: TAB=267.4 lb; TAC=142.5 lb; TAD =164 lb;

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