MEM202 Engineering Mechanics Statics

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Topics include problem formulation and solution methods; two- and ... Lectures are based on the textbook “Engineering Mechanics – STATICS,” 2nd.
MEM202 Engineering Mechanics - Statics

MEM

MEM202 Engineering Mechanics

Statics Course Web site: www.pages.drexel.edu/~cac542 COURSE DESCRIPTION This course covers intermediate static mechanics, an extension of the fundamental concepts and methods of static mechanics introduced in the freshmen course tDEC 111, and tDEC 113. Topics include problem formulation and solution methods; two- and three-dimensional vector representations of forces, moments, and couples; static equilibrium of particles, rigid bodies, and engineering structures; analysis of external and internal forces in structures via the methods of free-body diagrams; and properties of cross-sectional areas. PREREQUISITE(S) Sophomore standing; tDEC 111, tDEC 113. Lectures are based on the textbook “Engineering Mechanics – STATICS,” 2nd ed., by William F. Riley and Leroy D. Sturges, John Wiley & Sons, Inc. 1996. 1

MEM202 Engineering Mechanics - Statics

MEM

COURSE OBJECTIVES This course is aimed at providing the starting engineering students in their sophomore year a smooth transition from science-based mechanics problems to engineering-based mechanics problems, i.e. from equilibrium of relatively simple force systems to force systems in structures with connected members and with complex geometry. Conduct of the course emphasizes the correct and efficient free-body representation of the members in the structural system, along with formatted but logical solution techniques for the problems. The specific course objectives are: • Efficient use of trigonometric functions to decompose forces in selected coordinate directions; • Proficiency in computing moments by forces about a selected point; • Confidence in representing correctly a free-body diagram for a member in a loaded structure; • Confidence in setting up and the solution to the free-body diagram on hand; • Understand the physical nature of the internal force and moments in a structural member; • Confidence in handling the properties of a given cross-sectional area of any shape; • Familiar with the unique characteristics of tension, compression, shearing, bending, and torsion in structural members. 2

MEM202 Engineering Mechanics - Statics

MEM

COURSE REQUIREMENTS Class attendance (CA): Class attendance is mandatory, and is counted as 10% of your final term grade. A sign-up sheet will be circulated during each lecture and recitation to record the attendance. Please inform instructor/TA prior to the lecture/recitation that you can not attend. Homework assignments (HW): HW assignments and their due dates are listed in Course Outline. HW will be collected, recorded, but not checked. A grade will be given to each HW based on number of problems that have been completed and submitted. HW grade for the entire term will be counted as 10% of your final term grade. No late submissions of HW will be accepted as the solutions will be posted on the day they are collected. Mid-term Examinations (MT): There will be two 50-minute mid-term examinations; the date and material covered in each mid-term are listed in Course Outline. Each mid-term is counted as 25% of your final term grade. Final Examination (FL): Final examination will be a comprehensive examination, covering the material taught in the entire term. It will be counted as 30% of your final term grade. Term Grade: Term grade will be calculated based on 10% for CA, 10% for HW, 25% for each MT, and 30% for FL. Final letter grade will be assigned based on 100% ≥ A ≥ 90% > B ≥ 80% > C ≥ 70% > D ≥ 60% > F. There will be no grade curving. However, a student who earns a grade 90% or higher in his/her final examination will automatically receive an A grade for the course. 3

MEM202 Engineering Mechanics - Statics

MEM

Chapter 1 General Principles Book By hand Scalar : a a r Vector : a a SI Mass

U.S.

M kilogram kg slug

Length L Force F

meter Newton

m N

feet pound

Time

second

s

second sec

T

ft lb

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MEM202 Engineering Mechanics - Statics

MEM

Method of Problem Solving 1. 2. 3. 4.

Read the problem carefully. Identify the result requested. Identify the principles to be used to obtain the result. Prepare a scaled sketch (e.g., a free-body diagram) and tabulate the information provided. 5. Apply the appropriate principles and equations. 6. Report the answer with the appropriate number of significant figures and the appropriate units. 7. Study the answer and determine if it is reasonable.

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MEM202 Engineering Mechanics - Statics

Good

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Bad

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MEM202 Engineering Mechanics - Statics

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Chapter 2 Concurrent Force Systems

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MEM202 Engineering Mechanics - Statics

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2.1 Introduction

A physical body subjected to a pair of loads

The body is simplified to a particle – particle mechanics tDEC 111 Physics

The body deforms, may fail eventually

The Body does not deform – rigid body mechanics

MEM230 Mechanics of Materials

MEM202 Statics 8

MEM202 Engineering Mechanics - Statics

MEM

2.2 Forces and Their Characteristics A force is a vector; it has (1) magnitude, (2) direction, and (3) a point of application A free vector

A sliding vector

A bound vector

In Statics forces are treated as sliding vectors

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MEM202 Engineering Mechanics - Statics

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2.2 Forces and Their Characteristics Principle of Transmissibility The external effect of a force on a rigid body is the same for all points of application of the force along its line of action.

Principle of Transmissibility is applicable to rigid-body mechanics. Therefore, in Statics forces are treated as sliding vectors. 10

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MEM202 Engineering Mechanics - Statics

2.2 Forces and Their Characteristics Classification of Forces 1. Concurrent Forces

5. General Forces z

L

2. Coplaner Forces

O y r F1

x

3. Parallel Forces

4. Collinear Forces

r F2

r Fn

r ⎧0 ⇒ Statics ∑ Fi = ⎨ma ⇒ Dynamics ⎩ 11

MEM202 Engineering Mechanics - Statics

MEM

2.3 Resultant of Two Concurrent Forces (Parallelograms and Laws of Sines and Cosines) r F2 r F2 Principle of Transmissibility φ γ α ⇒ β φ r r F1 F1 r r φ : Angle between F1 and F2 r r r r r R = F1 + F2 = F2 + F1 r r β : Angle between R and F1

γ = π − φ;

r R

α =φ − β =π − β −γ 12

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MEM202 Engineering Mechanics - Statics

2.3 Resultant of Two Concurrent Forces r F2

(Parallelograms and Laws of Sines and Cosines)

γ α φ β φ r F1 γ = π −φ α =φ − β =π − β −γ

r R

c

β

γ α

b

a

Law of Sines :

a b c = = sin α sin β sin γ

Law of Cosines : c 2 = a 2 + b2 − 2ab cos γ

r Use Law of Cosines to determine the magnitude of R

R 2 = F12 + F22 − 2 F1F2 cos γ = F12 + F22 − 2 F1F2 cos(π − φ ) ⇒ R 2 = F12 + F22 + 2 F1F2 cos φ

r Use Law of Sines to determine the direction of R

R F F sin γ F2 sin (π − φ ) F sin φ = 2 ⇒ sin β = 2 = ⇒ sin β = 2 sin γ sin β R R R 13

MEM202 Engineering Mechanics - Statics

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2.3 Resultant of Two Concurrent Forces (Parallelograms and Laws of Sines and Cosines - Examples) r F2 = 800 lb

y

r R

φ = 180o − 56o = 124o R 2 = F12 + F22 + 2 F1F2 cos φ

= 5002 + 8002 + 2(500)(800) cos(124o ) = 442.6 × 103

φ β

56º

y

R = 665.3 lb

x

r F1 = 500 lb

φ = 40o

r R r F2 = 600 N 40º 35º

o ⎛ F2 sin φ ⎞ −1 ⎛ 800 × sin 124 ⎞ ⎟⎟ = 85.5o β = sin ⎜ ⎟ = sin ⎜⎜ 665.3 ⎝ R ⎠ ⎝ ⎠ −1

R 2 = F12 + F22 + 2 F1F2 cos φ

= 9002 + 6002 + 2(900 )(600) cos(40o ) = 1.997 × 103

r F1 = 900 N x

R = 1,413 N

o ⎛ F2 sin φ ⎞ −1 ⎛ 600 × sin 40 ⎞ ⎟⎟ = 15.8o β = sin ⎜ ⎟ = sin ⎜⎜ 1413 ⎝ R ⎠ ⎠ ⎝ −1

Homework: Problems 2-3, 2-11, 2-16 14

MEM202 Engineering Mechanics - Statics

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2.4 Resultant of Three or More Concurrent Forces (Parallelograms and Laws of Sines and Cosines) r F2

r r r R12 = F1 + F2 r F1

r r r R23 = F2 + F3 r F3

r R123 r r r R13 = F1 + F3

r r r R123 = R12 + F3 r r = R23 + F1 r r = R13 + F2 r r r = F1 + F2 + F3

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MEM202 Engineering Mechanics - Statics

MEM

2.4 Resultant of Three or More Concurrent Forces (Parallelograms and Laws of Sines and Cosines - Example) r r r Step 2 : R = R12 + F3

Determine the resultant of F1, F2, and F3 forces

r r r Step 1 : R12 = F1 + F2 φ1 = 30o + 30o = 60o

φ2 = 27o + 40o = 67o

R12 = F12 + F22 + 2 F1F2 cos φ1

R = R122 + F32 + 2 R12 F3 cos φ2 = 1,386 lb

= 954 lb ⎛ F1 sin φ1 ⎞ ⎟⎟ = 27.0o ⎝ R12 ⎠

β1 = sin −1 ⎜⎜

⎛ R12 sin φ2 ⎞ o ⎟ = 39.3 R ⎠ ⎝

β 2 = sin −1 ⎜

θ = 39.3o + 20o = 59.3o

Homework: Problems 2-19, 2-24, 2-28 16