Slide 1 © 2006 Baylor University

EGR 1301

Lecture 5Introduction to Engineering

Approximate Running Time - 15 minutesDistance Learning / Online Instructional Presentation

Presented byDepartment of Mechanical Engineering

Baylor University

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Slide 2 © 2006 Baylor University

EGR 1301

Introduction to Static Analysis – Part 2

EGR 1301 – Lecture 5

Prof. Dick Campbell Speaking

Slide 3 © 2006 Baylor University

EGR 1301

Learning Objectives

• Understand the concept of Force as a vector.• Understand separating a vector into

components.• Apply this concept to analyzing sums of Forces.• Determine the load in structural elements.• Understand the concept of a “Factor of Safety”.

Slide 4 © 2006 Baylor University

EGR 1301

Force as a Vector

• Forces – Have magnitude and direction– in an x-y coordinate system, the force may be broken

down into “components” along the coordinate axes.

•The magnitude of a vector can be found by the magnitude of its components using Pythagorean theorem.

x

y

F

xF

yF

22yx FFF

Slide 5 © 2006 Baylor University

EGR 1301

Setting Up the Analysis

• Draw a sketch of the Forces• Write each force in terms of i and j components (components

perpendicular to each other can be treated separately) • Sum of i and j components = zero (Newton’s 1st Law)• Solve two equations, two unknowns• Find tension force in cable• Calculate safety factor

x

y

1F

xF1

yF1

jFiFF yxˆˆ

111

i

j

jFiFF yxˆˆ

222

2F

Slide 6 © 2006 Baylor University

EGR 1301

Force Vectors and Static Analysis

• Consider the foot bridge– Loaded by six persons (approx. 1000 lbs)– Loaded at the center of the bridge

• Simplifying assumption:– Neglect the weight of the bridge

lbsF 10003

?1 F ?2 F

i

j

ft 130

ft 5

Slide 7 © 2006 Baylor University

EGR 1301

• Resolve the three forces into i,j components.

• Our unknowns are F1 and F2.

• Since the i,j directions are independent, we can solve these two equations for the unknowns

Static Analysis of the Bridge Cable

lbsF 10003 ?1 F

?2 F

ft 130

ft 5)(tan 4.4 65

51 o

jiF ˆ1000ˆ03

jfifF ˆsinˆcos 222

jfifF ˆsinˆcos 111

so, 0321 FFF

0ˆ0ˆcosˆcosˆ21 iififiF

0ˆ1000ˆsinˆsinˆ21 jjfjfjF

Slide 8 © 2006 Baylor University

EGR 1301

Static Analysis of the Bridge Cable (cont.)

0ˆ0ˆcosˆcosˆ21 iififiF

0ˆ1000ˆsinˆsinˆ21 jjfjfjF

substituting:

2121 ),4.4cos()4.4cos( ffff

1000)4.4sin()4.4sin( 21 ff

lbs 651721 ff

Safety Factor (Cable Strength - 16,000 lbs):

46.2lbs 6517

lbs 16000.. FS

Slide 9 © 2006 Baylor University

EGR 1301

Static Analysis of the Bridge Cable (cont.)

Then: 1000)10sin(2 1 f lbs 287921 ff

56.5lbs 8792

lbs 16000.. FS

What happens to the Safety Factor if we increase the sag? Let o10

What happens to the Safety Factor if we include the weight of the bridge? Let lbs 600 and 10 Wo

Then: 1600)10sin(2 1 f lbs 460721 ff

47.3lbs 6074

lbs 16000.. FS

Slide 10 © 2006 Baylor University

EGR 1301

Homework Assignment #1Problem #1

Given a load of 500 N (Newtons) supported by two cables as shown, determine the force in each cable.

N 500

?1 F

?2 F

i

j

m 20

m 5

m 5

Slide 11 © 2006 Baylor University

EGR 1301

Homework Assignment #1Problem #2

Given

?1 F

?2 Fi

j

m 8

m 5

magnitude N 1000ˆ447ˆ8943 jiF

m 10

m 10

m 4

3F

Determine the required forces (magnitudes) in cables 1 & 2 so that the system will remain in static equilibrium.

Slide 12 © 2006 Baylor University

EGR 1301

This Concludes Lecture 5