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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

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 analysis1

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

engr 1100 introduction to engineering analysis2

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

slide5

Lecture Outline

  • Rectangular components of a force
  • Resultant force by rectangular components
slide6

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.

slide7

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|

slide8

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.
solution

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

slide10

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

slide11

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

slide12

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

slide13

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

slide14

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=

example

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.
slide16

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

slide17

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

slide18

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

slide19

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

slide20

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);

example1

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

slide22

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)

slide23

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

slide24

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)

slide25

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

example2

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

slide27

Orthogonal (perpendicular) vectors

(w1 ,w2 ,w3)

z

w

(v1 ,v2 ,v3)

v

y

x

wand v are orthogonal if and only ifw·v=0

slide28

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

slide29

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.
solution1
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

slide31

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

slide32

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

slide33

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

slide34

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)

slide35

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

slide36

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

slide37

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

slide38

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