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ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis. Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276 6858 email: keblip@rpi.edu. ENGR-1100 Introduction to Engineering Analysis.

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ENGR-1100 Introduction to Engineering Analysis

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  1. ENGR-1100 Introduction to Engineering Analysis Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276 6858 email: keblip@rpi.edu

  2. ENGR-1100 Introduction to Engineering Analysis TA: Igor Bolotnov, Mechanical, Aerospace and Nuclear Engineering JEC5204 Office hours: Tuesday 3-5 phone: (518) 276 812 email:boloti@rpi.edu

  3. ENGR-1100 Introduction to Engineering Analysis SI (Supplemental Instruction) (Begins Wed. Sept. 8) Sun. 8pm-10pm - DCC 330 Wed 8pm-10pm - Sage 5510 Drop-in Tutoring - DCC 345 (Begins Tue. Sept. 7th) Sundays - 3-5pm M-Th - 7-9pm

  4. ENGR-1100 Introduction to Engineering Analysis Lecture 3

  5. Lecture Outline • Rectangular components of a force • Resultant force by rectangular components

  6. y F Fy x Fx Rectangular Components of a Force A force F can be resolved into a rectangular component Fx along the x-axis and a rectangular component Fy along the y-axis . The forces Fx and Fy are the vector components of the force F.

  7. q y F Fy x Fx Fx = F cos q F= Fx2+ Fy2 Fy = F sin q q=tan-1(Fy/Fx) The force F and its two dimensional vector components Fx and Fy can be written in Cartesian vector form by using unit vectors i and j. F = Fx + Fy= Fx i + Fy j F=| F|

  8. y F=275 lb 570 x Figure 2-47 Example • Determine the x and y scalar component of the force shown in figure 2-47. • Express the force in Cartesian vector form.

  9. y F=275 lb Fy 570 Fx Solution x Fx=cos(570) *275=149.8 lb Fy=sin(570) *275=230.6lb F= Fxi+ Fyj=149.8 i + 230.6 j

  10. Class Assignment: Exercise set 2-49 please submit to TA at the end of the lecture • Determine the x and y scalar components of the force shown in figure 2-49. • Express the force in Cartesian vector form. y Solution a) Fx=-440lb Fy=-177.9lb b)F=-440i- 177.9jlb x 220 F=475 lb Figure 2-49

  11. Where eF= cos qx i + cos qy j + cos qz kis a unit vector along the line of action of the force. z Fz=Fzk F qz Fy=Fyj qx qy y Fx=Fxi x Three Dimension Rectangular Components of a Force F = = Fxi + Fyj + Fzk = F cos qx i + F cos qy j + F cos qz k = FeF

  12. Fx = F cos qx; Fy = F cos qy; Fz = F cos qz; qx=cos-1(Fx/F); qy=cos-1(Fy/F); qz=cos-1(Fz/F); F= Fx2 + Fy2 + Fz2 z Fz cos2qx+cos2qy+cos2qz=1 0< q <1800 F qz Fy qx qy y Fx x The Scalar Components of a Force

  13. z Fz z y q Fx f Fxy q Fz F x Fy y Fx= Fxy cos q=Fcos f cos q Fxy = F cos f; Fz = F sin f ; Fxy Fy= Fxy sin q=Fcos f sin q x Azimuth Angle q- azimuth angle f- elevation angle

  14. z xB-xA (xB-xA)2+ (yB-yA)2+ (zB-zA)2 F cosqx= B A zB zA y xA yA xB yB x Finding the direction of a force by two points along its line of action A(xA, yA, zA) ; B(xB, yB, zB) yB-yA (xB-xA)2+ (yB-yA)2+ (zB-zA)2 cosqy= zB-zA (xB-xA)2+ (yB-yA)2+ (zB-zA)2 cosqz=

  15. z F=745 N 300 370 y x Example • For the force shown in the figure • Determine the x, y, and z scalar components of the force. • Express the force in Cartesian form.

  16. z F=745 N Fz 300 Fxy f 370 y x Fz = F sin f= 745 sin(600)=411.4 N Fxy = F cos f= 745 cos(600)=237.5 N Solution

  17. z Fz Fxy=237.5 N Fx 370 Fy y q x F=-142.9i – 189.7j + 411.4k b) Fx= Fxy cos q=-237.5*sin(370)=-142.9 N Fy= Fxy sin q=-237.5*cos(370)=-189.7 N Or if we follow the obtained formula: Fx= Fxy cos q=237.5*cos(2330)=-142.9 N Fy= Fxy sin q=237.5*sin(2330)=-189.7 N

  18. Class Assignment: Exercise set 2-55 please submit to TA at the end of the lecture • For the force shown in Fig. P2-58 • Determine the x,y, and z scalar components of the force. • Express the force in Cartesian form. z Solution a) Fx=-583 lb Fy=694lb Fz=423 lb b) F=-583 i +694 j + 423 k lb F=1000 lb 250 y 1300 x

  19. z A A = Ax i + Ay j + Azk y B = Bx i + Byj + Bzk B x The sum of the two forces are: R=A+B=(Ax i + Ayj + Azk) + (Bx i + Byj + Bzk) = (Ax +Bx )i + (Ay +By)j + (Az +Bz)k Rx= (Ax +Bx ); Ry= (Ay +By); Rz= (Az +Bz) Resultant Force by Rectangular Components

  20. R= Rx2 + Ry2 + Rz2 Rx= (Ax +Bx ); Ry= (Ay +By); Rz= (Az +Bz) The magnitude: The direction: qx=cos-1(Rx/R); qy=cos-1(Ry/R); qz=cos-1(Rz/R);

  21. F1=350 i+ 0 j + 0 k F2=500*cos(2100) i +500*sin(2100)j+0 k F3= 600*cos(1200) i + 600*sin(1200)j+0 k F1=350 i F2= -433 i -250 j F3= -300 i + 519.6 j Example Determine the magnitude and direction of the resultant force of the following three forces. Solution: R=F1+F2+F3=-383 i + 269.6 j

  22. qx R= Fx2 + Fy2 + Fz2 = 3832 + 269.62+02 The direction: qx=cos-1(Rx/R)= cos-1(-383/468.4)=144.80 qx=144.80 R=F1+F2+F3=-383 i + 269.6 j The force magnitude: R R= 468.4 N qx=cos-1(Rx/R)

  23. Class Assignment: Exercise set 2-71 please submit to TA at the end of the lecture Determine the magnitude R of the resultant of the forces and the angle qx between the line of action of the resultant and the x-axis, using the rectangular component method y F1=600 lb Solution: R=1696 lb 10.220 F2=700 lb 450 150 x 300 F3=800 lb

  24. A q B Finding the rectangular scalar component of vector A along the x-axis 0< q <1800 Ax=A•i=A cos(qx) et y A Along any direction n An=A•en=A cos(qn) q en At=A-An The scalar (or dot) product The scalar product of two intersecting vectors is defined as the product of the magnitudes of the vectors and the cosine of the angle between them A•B=B•A=AB cos(q)

  25. Since i, j, k are orthogonal: i•j= j•k= k•j=(1)*(1)*cos(900)=0 i•i= j•j= k•k=(1)*(1)*cos(00)=1 • The scalar product of the two vectors written in Cartesian form are: • A•B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k) = • Ax Bx (i•i) + Ax By (i•j) + Ax Bz (i•k)+ Ay Bx (j•i) + Ay By (j•j) + Ay Bz (j•k)+ • Az Bx (k•i) + Az By (k•j) + Az Bz (k•k) Therefore: A•B = Ax Bx + Ay By + Az Bz

  26. A•B=AB cos(q) cos(q)= A•B/ AB A = 32+42 = 5 AB= 28.7 B = 22 +22 +52 = 5.74 cos(q)= 26/28.7 q= 25.10 Example Determine the angle q between the following vectors: A=3i +0j +4k and B=2i -2j +5k A•B=3*2+0*(-2)+4*5=26

  27. Orthogonal (perpendicular) vectors (w1 ,w2 ,w3) z w (v1 ,v2 ,v3) v y x wand v are orthogonal if and only ifw·v=0

  28. If u , v, and w are vectors in 2- or 3- space and k is a scalar, then: • u·v= v·u • u·(v+w)= u ·v + u·w • k(u·v)= (ku) ·v = u·(kv) • v·v> 0 if v=0, and v·v= 0 if v=0 Properties of the dot product

  29. z F2 =240 lb F1 =300 lb 4.5 ft 600 y 1.5 ft 2 ft 6 ft x Using dot product for a force system • Determine: • The magnitude and direction (qx, qy, qz) • of the resultant force. • b) The magnitude of the rectangular • component of the forceF1along • the line of action of forceF2. • c) The angle a between forceF1andF2.

  30. F1= F1e1 z F2 =240 lb F1 =300 lb 4.5 ft 600 y 1.5 ft 2 ft 6 ft x F1 = 58.8i + 235.3j+ 176.5k lb Solution e1 = 1.5/(1.52+62+4.52)1/2i + 6/(1.52+62+4.52)1/2j+ 4.5/(1.52+62+4.52)1/2k e1 = 0.196i + 0.784j+ 0.588k

  31. z F2 =240 lb F1 =300 lb L2 4.5 ft 600 y L1 1.5 ft 2 ft 6 ft x F2= 72i - 96j+ 207.8k lb L1=(22+1.52)1/2=2.5 ft L2=2.5 tan(600)=4.33 ft • F2= F2e2 e2 = 1.5/(1.52+(-2)2+4.332)1/2i -2/(1.52+(-2)2+4.332)1/2j+ 4.33 /(1.52+(-2)2+4.332)1/2k e2 = 0.3i - 0.4j+ 0.866k

  32. R= Rx2 + Ry2 + Rz2 = 130.82 + 139.352+384.32 = 429 lb R= F1 +F2= 130.8i + 139.35j+ 384.3k lb R= 429 lb qx=cos-1(Rx/R); qy=cos-1(Ry/R); qz=cos-1(Rz/R); qx=cos-1(130.8/429) qx=72.20 qy=cos-1(139.35/429) qy=71.10 qz=cos-1(384.3/429) qz=26.40

  33. b) The magnitude of the rectangular component of the forceF1 along the line of action of forceF2. F1•e2=(58.8i + 235.3j+ 176.5k)•(0.3i - 0.4j+ 0.866k)= 58.8*0.3+235.3*(-0.4)+176.5*0.866=76 lb

  34. cos(a)= F1•F2/ F1 F2 F1•F2= 58.8i + 235.3j+ 176.5k)•(72i - 96j+ 207.8k )= 58.8*72+235.3*(-96)+176.5*207.8=18321 lb F1 F2=72000 cos(a)=18321/72000 a = 75.250 c) The angle a between forceF1andF2. F1•F2= F1 F2 cos(a)

  35. z 3 ft 3 ft 4 ft F1=900 lb 3 ft F2=700 lb 6 ft y x Class assignment: Exercise set 2-63 • Two forces are applied to an eye bolt as shown in Fig. P2-63. • Determine the x,y, and z scalar components of vector F1. • Express vector F1 in Cartesian vector form. • Determine the angle a between vectors F1 and F2. Fig. P2-63

  36. d1 = x12+ y12+ z12 d1 = (-6)2 +32 +72 z 3 ft 3 ft 4 ft F1=900 lb 3 ft F2=700 lb 6 ft y x Solution =9.7 ft F1x= F1 cos(qx) =900 *{(–6)/9.7}=-557 lb F1y= F1 cos(qy) =900 *{3/9.7}=278.5 lb F1z= F1 cos(qz) =900 *{7/9.7}=649.8 lb b) Express vector F1 in Cartesian vector form. F1 = -557i + 278.5j+ 649.8k lb

  37. d2 = x22+ y22+ z22 z 3 ft d2 = (-6)2 +62 +32 =9 ft 3 ft 4 ft F1=900 lb 3 ft F2=700 lb F1•e2= (-557)*(-2/3)+278.5*2/3+649.8*0.33=771.4lb 6 ft y x cos(a)=771.4/900 a=310 c) Determine the angle a between vectors F1 and F2. cos(a)= F1•e2/ F1 e2 F1 = -557i + 278.5j+ 649.8k lb e2 = -6/9i + 6/9j + 3/9k = -0.67 i + 0.67 j +0.33 k

  38. Class Assignment: Exercise set 2-78 please submit to TA at the end of the lecture Determine the magnitude R of the resultant of the forces and the angles qx, qy, qz between the line of action of the resultant and the x-, y-, and z-coordinate axes, using the rectangular component method. Solution: R=28.6 kN qx= 82.20 qy= 69.60 qz= 22.00

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