ENGR 102engr102.com/docs/Lecture20.pdf · ENGR 102 PROBLEM SOLVING FOR ENGINEERS Lecture # 20 April...

ENGR 102PROBLEM SOLVING FOR ENGINEERS

Lec

ture

# 2

0

April 14, 2020

I N T O / C S U P A R T N E R S H I P

PROJECT

Resume

7-Step Problem

Cover LetterDear company:

I am interested in the job that you have posted

and think that I would be good candidate for this

position for the following reasons:

. Blah blah blah

. Blah blah blah

.. Blah blah blah

I have attached my resume, which shows that I

have done . . . .

Let me know if you’d like further information or

an interview.

Respectfully,

Egor Student

Ch 14. Balance

Ch 13. Solids

Ch 12. Statics

Exam 4

Exam 3

Project Part 2

STATICS WHAT’s NEXT

22

MATLAB CORONAVIRUS PLOTS

Less coronavirus = More normal life

You can plot: New cases (or deaths) Cumulative cases (or deaths)

You can plot and compare: By country By state By county

Helps understand what next

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MATLAB CORONAVIRUS PLOTS

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Project Part 2 (Assignment 19)

55% of your project grade, which is 25% of class grade Due noon, Thursday, April 23 Largest single remaining part of grade in this class

Feedback on Assignment 19: Will give comments via Canvas

Conferences and help: Last conferences next Monday, April 20 Deadline for comments on draft, Tuesday, April 21, 3pm

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Exam 3 – Thursday, April 30(Can take day early if you have another exam on April 30)

Ch 14. Balance

Ch 13. Solids

Ch 12. Statics

• 7-step problem solving process Use for statics problem (Chapter 12)

• Mechanics of solids: Basic calculation problem (Chapter 13)

• Material balance: Use material balance 7-step (Chapter 14)

• Vocabulary: Similar exercise as Exam 2

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Recommended strategy for learning statics

• Today’s lecture Overview of all required concepts Will also post this with the assignment in Canvas

• Textbook Everything you need is explained Sometimes difficult to follow: dense with few examples

• Video tutorials I will try to create some

• Hein’s recitation tomorrow Try to do Assignment 20 before so you have questions

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Chapter 12 (Statics) Vocabulary

a. Scalar

b. Vector

c. Unit vector

d. Mechanics

e. Force

f. Statics

g. Dynamics

h. Collinear force

i. Concurrent force

j. Coplanar force

k. Principle of transmissibility

l. Line of action

m. Resultant force

n. Equilibrium

o. Compression

p. Tension

q. Moment

r. Couple

s. Moment arm

t. Free body diagram

u. Slender

v. Buckling

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Chapter 12 (Statics) Vocabulary

a. Scalar Magnitude only (temperature, mass, etc.)

b. Vector Direction and magnitude (weight, velocity, acceleration)

c. Unit vector Direction only / magnitude=1

d. Mechanics

e. Force

f. Statics

g. Dynamics

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Chapter 12 Vocabulary

d. Mechanics Force effects on objects

e. Force Direction + magnitude acting on object

f. Statics

Mechanics of not accelerating things (ΣF = 0)

g. Dynamics

Mechanics of accelerating things (ΣF = ma)

h. Collinear force

i. Concurrent force

j. Coplanar force

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h. Collinear force Action in same line

i. Concurrent force Action on same point

j. Coplanar force Action in same 2D (2 dimensions)


l. Line of action

m. Resultant force

n. Equilibrium

1111


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k. Principle of transmissibility Move force action line (can move force in same line of action)

l. Line of action Direction of vector

m. Resultant force Sum of actions

n. Equilibrium Resultant force = 0 Resultant moment = 0

o. Compression

p. Tension

1313



Collinear forces can be added, subtracted and moved along the line of action (vector direction) without changing the statics calculation.

Figure (a) and Figure (b) below are the same for statics.

1414


X component of force F

Resultant force F

Y c

om

po

ne

nt

of

forc

e F

Using trigonometry, you can calculate a resultant force from component forces and vice versa.

1515


o. Compression Pushing

p. Tension Pulling

q. Moment

r. Couple

s. Moment arm

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q. Moment Perpendicular distance from force to point

r. Moment arm Perpendicular distance from force to point

s. Couple Equal opposite parallel forces separated by moment arm


u. Slender

v. Buckling

1717

Moment Perpendicular distance from force to point

Moment arm Perpendicular distance from force to point

1818


In statics, moments must be equal

(45)(x) = (135)(2)x = (135)(2)/(45)

x = 6 ft

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Couple Equal opposite forces separate by a moment arm

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u. Slender

v. Buckling

Source: ah-engr.com

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t. Free body diagram Sketch forces on object

1kg

3 m1 m

3 m1 m

FAY FBY

9.8 N

A B

2222

Use 7-step problem solving process from chapter 4:

Step 1 Problem statementStep 2 Make a free body diagram

Step 3 Theory and equations:Sum of forces = 0Sum of moments = 0Understand the trigonometry

Step 4 Assumptions: things you did to simplify so you could calculate

Step 5 Solution steps: Carefully solve the equationsShow your work steps so you can find possible errors

Step 6 Verify accuracySignificant digitsCan you check your work in another way?

Step 7 Discussion: what does your answer and other work show?

HOW TO SOLVE STATICS PROBLEMS(of the type we do in this class)

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SAMPLE STATICS PROBLEM #1

1. Problem statement: What are the forces on the two table legs?

1kg

3 m1 m

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1. Problem statement: What are the forces on the two table legs?

1kg

3 m1 m

3 m1 m

FAY FBY

9.8 N2. Free body diagram:

A B


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3 m1 m

FAY FBY

9.8 N2. Free body diagram:

A B

3. Theory and equations:

Sum of forces = 0:

In Y direction: FAY + FBY = 9.8 N

Sum of moments = 0

About point A: (1m)(9.8N) = (4m)(FBY)


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4. Assumptions:

3 m1 m

FAY FBY

9.8 N

A B

5. Solution steps:

The 1 kg mass is a point load

The table legs are infinitely thin

The table top has no mass

(1m)(9.8N) = (4m)(FBY)

FBY = (1)(9.8)/(4) = 2.45

FAY = 9.8 N - FBY = 7.35

6. Results and accuracy:

FBY = 2.5 N (2 significant digits)

FAY = 7.4 N (2 significant digits)

7. Discussion:

When we move a load close to one leg of a table, the force on that leg increases as we get closer and closer to that leg of the table. The force on the other leg gets closer and closer to 0.


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BASIC STATICS THEORY EXAMPLE

FORCE BALANCE EQUATIONS:

MOMENT BALANCE EQUATIONS:

2 m 1 m

CY

2 N

AY

AX

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FORCE BALANCE EQUATIONS:

AX = 0

AY + 2N = CY

MOMENT BALANCE EQUATIONS:

Around point A: (2m)BY = (3m)(2N)

Therefore, the solution is:

CY = 3N

AY = 1N

AX = 0

2 m 1 m

CY

2 N

AY

AX

BASIC STATICS THEORY EXAMPLE

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1. Problem statement:What is the tension in the cable for this shelf?

10N

12 cm

Wall

8 cm

45OCable

Shelf


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1. Problem statement:What is the tension in the cable for this shelf?

10N

12 cm

2. Free body diagram:

Wall

8 cm

45OCable

Shelf

12 cm 8 cm

10 N

45O

A B

C

T

FBY

FAY

FAX FBX


3131

1. Problem statement:Find tension in cable


12 cm 8 cm

10 N

45O

A B

C

T

FBY

FAY

FAX FBX


Sum of forces = 0:In Y direction: FAY + FBY = 10NIn X direction: FAX = FBY

Sum of moments = 0About point A: (12cm)(10N) = (20cm)(FBY)

Trigonometry:Cable angle: T = FBX = (T)sin(45O)Cable angle: T = FBY = (T)sin(45O)


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1. Problem statement:Find tension in cable


12 cm 8 cm

10 N

45O

A B

C

T

FBY

FAY

FAX FBX




Trigonometry:Cable angle: FBX = (T)sin(45O)Cable angle: FBY = (T)sin(45O)

4. Assumptions:

The shelf and cable have no massThe 10N weight is loaded at a pointThe shelf can swivel at point A


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12 cm 8 cm

10 N

45O

A B

C

T

FBY

FAY

FAX FBX

5. Solution steps:



Trigonometry:Cable angle: FBX = (T)sin(45O)Cable angle: FBY = (T)sin(45O)

(12cm)(10N) = (20cm)(FBY)FBY = (12)(10)/(20) = 6NT = FBY /sin(45O)T = (6N)(1.414) = 8.485

6. Results and accuracy:

T = 8.5 N (2 significant digits)

7. Discussion:Most of the 10N load is carried by the cable because the load is closer to point B than point A.

The shelf is in compression and the cable is in tension.


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ASSIGNMENT 20(Due before noon on Thursday, April 16)

3535

MORE SAMPLE STATICS PROBLEMS

Problems 12.1 to 12.36 in textbook (answers for some on page 472):

12.1 to 12.4: X and Y components of a force (trigonometry)

12.5 to 12.10: Resultant force of two forces (trigonometry)

12.11 to 12.14: Complex resultant calculations (algebra + trigonometry)

12.15 to 12.20: Calculate moments about a point (trigonometry)

12.21 to 12.23: Resultant reactions of combination of forces and moments

12.24 to 12.30: Sample problems you should learn to solve

12.31 to 12.32: You don’t need to know how to solve these problems

12.33 to 12.36: Sample problems you should learn to solve

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NEXT 2-3 WEEKS(All times are in Colorado. Qatar/Kuwait +9 hours. Korea +15 hours)

Sunday Monday Tuesday Wednesday Thursday Friday

April 12 April 13 April 14Video class

3:30 pm

April 15Recitation(Statics)

6pm

April 16Assignment

20 noonVideo class

3:30pm

April 17Work on project

April 19Work on project

April 201:1 video

conferences

April 21Assignment

21 noonVideo class

3:30 pm

April 22Recitation(Solids+

Balance)6pm

April 23Project 2

noonVideo class

3:30pm

April 24

April 26 April 27 April 28Assignment

22 noonVideo class

3:30 pm

April 29Recitation

(Exam prep)6pm

April 30Exam 3 3:30pm

May 1

Exam 4: Monday, March 11, 9:40 to 11:40 am (Thermodynamics & Gas Law & 7-step)

ENGR 102engr102.com/docs/Lecture20.pdf · ENGR 102 PROBLEM SOLVING FOR ENGINEERS Lecture # 20 April...

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