Lecture 14Exam 2 Review

March 5th, 2018Introduction to Engineering Analysis

• Wednesday 3/7/2018• 8:00 – 9:50 am

Not 9:00 as stated in your SIS scheduleStudents arriving late will have to turn in their tests at 9:50

• Covers lectures 6 - 13

Test 2

• Sec-01 DARRIN 308• Sec-02 Darrin 318 – This is not a typo• Sec-03 DARRIN 308• Sec-04 DARRIN 308• Sec-05 DARRIN 308

Test 2 Rooms

Test 2

Four problems:• Matrices and determinants 25%• 3D particle equilibrium 25%• Moments/Couples (2D/3D) 25%• 2-D Rigid Body Equilibrium 25%

Test 2• Closed book and closed notes. • For maximum credit, show all set-ups and all details

necessary (see syllabus).• You can use only a hand-held calculator for calculations

(no smartphone, laptop, etc.). • Cell phones, audio devices, headphones and laptops must

be off and stored away. Any student found deviating from this policy will be asked to leave the exam room, given a grade of zero, and no retest will be allowed.

ALAC Review Session

• Location : DCC 330.• Time : 8 PM until 10 PM. This is later than the first one.• Date : March 5th, 2018

Test 2Students entitled to extra time should: • get a note from the Dean of Students

and give (or email) a copy to their instructor by 3/02/2018.

• remain in the test room until escorted by a TA to a different room (DCC 232) to finish their test.

Grade-Challenging Sessions

• Monday 3/19/18 in JEC 4309• Tuesday 3/20/18 in JEC 4309• 6:00 – 8:00 pm• This is a change from the first exam!

Makeup Test 2

• Wednesday 3/21/2018• 5:00 – 6:50 pm• Room: JONSSN 5119• Only students with a valid reason and

a note from the office of the Dean of Students will be allowed to take it

• No makeup for the makeup test

Definition : sum of matrices• If A and B are any two matrices of the same size, then the

sum A+B is the matrix obtained by adding together the corresponding entries in the two matrices. Matrices of different sizes can’t be added.

2 2

-2 0

-1 -1

A=

1 2

-2 0

3 -1

B=

2 2 1

-2 0 2

-1 -1 2

C=

3 4

-4 0

2 -2

A+B=A+C and B+C are undefined

Definition 4: Product of two matrices

1 2 4

2 6 0A=

4 1 4 3

0 -1 3 1

2 7 5 2

B=

A x B =?

=1 2 42 6 0

4

0

2

1

-1

7

4

3

5

3

1

2

26

(2x4)+(6x3)+(0x5)=26

=1 2 42 6 0

4

0

2

1

-1

7

4

3

5

3

1

2

13

26

(1x3)+(2x1)+(4x2)=13

The size of a product matrix

A B = AB

m x r r x n m x n

inside

outside

Remember, Rows then Columns. (Thankfully, your calculator and Matlab work like this too).

Transpose of a matrix AT (Sometimes this symbol is a dagger)

a11

a12

a13

a14

AT=

a21

a22

a23

a24

a31

a32

a33

a34

a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34

A=This is on the exam.Period. Full Stop

Rules of matrix arithmetic

-1 0

2 3A=

1 2

3 0B=

Multiply AB:

-1 -2

11 4AB=

3 6

-3 0BA=

Then: AB=BA

Multiply BA

Example 1

5 4

3 2A=

Then we define

5 4

3 2det(A)= = det(A)= (5x2)-(3x4)=-2

5 4

3 2

Inverse of a 2x2 matrixIf A is a 2x2 matrix

a11 a12

a21 a22A=

Then we define

a22 -a12

-a21 a11Inv(A)= A-1= / det(A)

THE EQUATIONS OF 3-D EQUILIBRIUM

This vector equation will be satisfied only whenΣFx = 0ΣFy = 0ΣFz = 0

These equations are the three scalar equations of equilibrium. They are valid for any point in equilibrium and allow you to solve for up to three unknowns.

When a particle is in equilibrium, the vector sum of all the forces acting on it must be zero (Σ F = 0 ) .This equation can be written in terms of its x, y and z components. This form is written as follows.

(Σ Fx) i + (Σ Fy) j + (Σ Fz) k = 0

EXAMPLE II

1) Draw a free body diagram of Point A. Let the unknown force magnitudes be FB, FC, FD .

2) Represent each force in its Cartesian vector form.

3) Apply equilibrium equations to solve for the three unknowns.

Given: A 600 N load is supported by three cords with the geometry as shown.

Find: The tension in cords AB, AC and AD.

Plan:

FB = FB (sin 30° i + cos 30° j) N= {0.5 FB i + 0.866 FB j} N

FC = – FC i N

FD = FD (rAD /rAD) = FD { (1 i – 2 j + 2 k) / (12 + 22 + 22)½ } N= { 0.333 FD i – 0.667 FD j + 0.667 FD k } N

FBD at AFCFD

A

600 N

z

y30˚

FBx

1 m2 m

2 m

EXAMPLE II (continued)

Solving the three simultaneous equations yields

FC = 646 N (since it is positive, it is as assumed, e.g., in tension)

FD = 900 N

FB = 693 N

y

Now equate the respective i , j , kcomponents to zero.

∑ Fx = 0.5 FB – FC + 0.333 FD = 0

∑ Fy = 0.866 FB – 0.667 FD = 0

∑ Fz = 0.667 FD – 600 = 0

FBD at A

FCFD

A

600 N

z

30˚

FBx

1 m2 m

2 m

EXAMPLE II (continued)

CROSS PRODUCT (Section 4.2)

In general, the cross product of two vectors A and B results in another vector, C , i.e., C = A × B. The magnitude and direction of the resulting vector can be written as

C = A × B = A B sin θ uC

As shown, uC is the unit vector perpendicular to both A and B vectors (or to the plane containing the A and B vectors).

CROSS PRODUCT (continued)

The right-hand rule is a useful tool for determining the direction of the vector resulting from a cross product.

For example: i × j = k

Note that a vector crossed into itself is zero, e.g., i × i = 0

CROSS PRODUCT (continued)

Also, the cross product can be written as a determinant.

Each component can be determined using 2 × 2 determinants.

MOMENT OF A FORCE – VECTOR FORMULATION (continued)

So, using the cross product, a moment can be expressed as

By expanding the above equation using 2 × 2 determinants (see Section 4.2), we get (sample units are N - m or lb - ft)

MO = (ry FZ - rZ Fy) i − (rx Fz - rz Fx ) j + (rx Fy - ry Fx ) k

The physical meaning of the above equation becomes evident by considering the force components separately and using a 2-D formulation.

This is important when finding the moment of a force couple:

�⃗�𝐹𝐴𝐴

�⃗�𝐹𝐵𝐵

𝑟𝑟𝐴𝐴𝐵𝐵

𝑀𝑀𝑜𝑜 = 𝑟𝑟𝐴𝐴𝐵𝐵 × �⃗�𝐹𝐵𝐵

EXAMPLE

1) Find F = F1 + F2 and rOA.

2) Determine MO = rOA × F .

o

Given: F1={100 i - 120 j + 75 k}lbF2={-200 i +250 j + 100 k}lb

Find: Resultant moment by the forces about point O.Plan:

EXAMPLE (continued)

Solution:First, find the resultant force vector F

F = F1 + F2

= { (100 - 200) i + (-120 + 250) j + (75 + 100) k} lb= {-100 i +130 j + 175 k} lb

Find the position vector rOA

rOA = {4 i + 5 j + 3 k} ft

i j k4 5 3

-100 130 175

MO = = [{5(175) – 3(130)} i – {4(175) –3(-100)} j + {4(130) – 5(-100)} k] ft·lb

= {485 i – 1000 j + 1020 k} ft·lb

Then find the moment by using the vector cross product.

Example PROBLEM SOLVING

1) Find F and rAC.

2) Determine MA = rAC × F

Given: The force and geometry shown.

Find: Moment of F aboutpoint A

Plan:

Solution:F ={ (80 cos30) sin 40 i

+ (80 cos30) cos 40 j − 80 sin30 k} N={44.53 i + 53.07 j − 40 k } N

rAC ={0.55 i + 0.4 j − 0.2 k } m

MA =

= { -5.39 i + 13.1 j +11.4 k } N·m

i j k0.55 0.4 − 0.2

44.53 53.07 − 40

Example PROBLEM SOLVING (continued)

Find the moment by using the cross product.

FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM(Section 4.8)

= =

In three special cases, concurrent, coplanar, and parallel systems of forces, the system can always be reduced to a single force.

If FR and MRO are orthogonal to each other, then the system can be further reduced to a single force, FR , by simply moving FR from O to P. (two vectors are orthogonal if and only if their dot products equal zero)

FREE-BODY DIAGRAMS (Section 5.2)

2. Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body.

Idealized model Free-body diagram (FBD)

1. Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape.

FREE-BODY DIAGRAMS (continued)

3. Label loads and dimensions on the FBD: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA. Indicate any necessary dimensions.

Idealized model Free-body diagram

I cannot emphasize this enough. Label all dimensions and give yourself an axis!!!

SUPPORT REACTIONS IN 2-D

As a general rule, if a support prevents translation of a body in a given direction, thena force is developed on the body in the opposite direction.

Similarly, if rotation is prevented, a couple moment is exerted on the body in the opposite direction.

A few example sets of diagrams s are shown above. Other support reactions are given in your textbook (Table 5-1).

EXAMPLE

1. Put the x and y axes in the horizontal and vertical directions, respectively.

2. Determine if there are any two-force members.

3. Draw a complete FBD of the boom.

4. Apply the E-of-E to solve for the unknowns.

Given: The 4kN load at B of the beam is supported by pins at A and C .

Find: The support reactions at A and C.

Plan:

EXAMPLE (continued)

Note: Upon recognizing CD as a two-force member, the number of unknowns at C are reduced from two to one. Now, using E-o-f E, we get,

→ + ∑FX = AX + 11.31 cos 45° = 0; AX = – 8.00 kN

↑ + ∑FY = AY + 11.31 sin 45° – 4 = 0; AY = – 4.00 kN

+ ∑MA = FC sin 45° × 1.5 – 4 × 3 = 0Fc = 11.31 kN or 11.3 kN

FBD of the beam:

AX

AY

A

1.5 m

C B

4 kN

FC

45°

1.5 m

Note that the negative signs means that the reactions have the opposite directions to that shown on FBD.