ENGINEERING DESIGN LAB IIENGR-102 WEEK 2 LECTURE – BRIDGE MODULE

Pramod Abichandani, Ph.D.

Richard Primerano, Ph.D.

OVERVIEW OF WEEK 2 LAB ACTIVITIESTHE METHOD OF JOINTSFREE BODY DIAGRAMS

2

THE TRUSS

A structural member composed of triangular sections

The members of a truss are connected at nodes Each member is in either tension or compression A technique called the method of joints can be

used to determine the forces in all members under a given load.

3

EXAMPLE: FIND THE FORCES IN THIS STRUCTURE

We know the applied loads, the types of supports, and the geometry of the structure

From this, we can determine the tension and compression forces in the members AB, BC, and AC

100lb

A

B

C

pin joint – prevents motion in the X and Y directions.

roller joint – prevents motion in the Y direction.

1ft

1ft1ft

4

ANALYZING FORCES IN A TRUSSTHE METHOD OF JOINTS

1. Determine the support reactions.2. Draw a free body diagram for each joint.3. Write an equilibrium equation for each joint.4. Solve for the forces at each joint.

A good video tutorial on the method of joints http://www.youtube.com/watch?v=ZOtSRbzKCC8

5

STEP 1: DETERMINE SUPPORT REACTIONS

Employ Newton’s law to find the reaction forces keeping the structure stationary

100lb

A

B

C1ft

1ft1ft

FcyFay

Fcx

Σ forces in X direction = 0

Σ forces in Y direction = 0

Applying Σ of moments about A will tell us that:

6

STEPS 2&3: DRAW FREE BODY DIAGRAMS

Draw all the forces on node A Each force must be resolved into its X and Y

components

50

0 Fac

Fab

60º

Fab • sin 60º

Fab • cos 60º

Sum of forces in X direction

Sum of forces in Y direction

Fab = - 50 / 0.87 = -57.7 lb Fac = 57.7 * 0.5 = 28.9 lb

7

BRIDGE DESIGN MODULE WEEK 2SOFTWARE TOOLS FOR BRIDGE DESIGN

We introduce one of the three design tools (West Point Bridge Designer) that will be used in this module for bridge design and analysis. AutoCAD and Visual Analysis are the others

WPBD is an educational package developed and distributed by the United States Military Academy.

It allows you to quickly design and simulate bridges using a variety of common construction materials.

Using this, you can visualize the forces in a truss for given loads.

8

SOME BASIC PHYSICSSTATIC EQUILIBRIUM When an object is in static equilibrium, all of the particles

that make up the object are stationary and there is no net force acting on the object. Basically, the object is at rest and not accelerating All applied forces are balanced by reaction forces.

100lb

A

B

C1ft

1ft1ft

50lb50lb

0 lb

9

EXTERNAL FORCES ACTING ON A STRUCTURE

Applied forces (loads) – these are the forces that the structure is designed to carry: cars on a bridge, shingles on a roof, lift force on a wing Dead loads – loads permanently acting on a

structure: the weight of members, road decking, roofing material

Live loads – time varying loads acting on a structure: cars on a bridge, people on a floor, snow on a roof

The dead loads can usually be calculated with high accuracy.

The live loads are generally estimated based on experience of calculation. video

10

EXTERNAL FORCES ACTING ON A STRUCTURE

Reaction forces – exerted by support structures and responsible for keeping a structure in static equilibrium These forces “adjust themselves” as the applied

forces change.100lb

A

B

C1ft

1ft1ft

50lb50lb

0 lb

100lb

A

B

C1ft

1ft1ft

43lb43lb

100 lb

11

EXAMPLE: TUG OF WAR Consider a 3-way tug of war.

Force A is exerted along the -X axis Force B is exerted along the -Y axis Force C is exerted at a 45˚ angle w.r.t. the +X axis The knot is in static equilibrium

If for force exerted by player A is 150lb, find all remaining forces.

+Y

+X

FA = 150lb

FB

FC

45˚

12

SOME INITIAL OBSERVATIONS

FA is resisted (balanced) by a component of FC

FB plays no role in balancing FA

FB is also balanced by a component of FC

FA plays no role in balancing FB

FA = 150lb

FB

FC

45˚

+Y

+X

13

RESOLVING A VECTOR INTO ITS COMPONENTS

The force FC can be considered as the sum of two component forces, one acting in the X direction and the other acting in the Y direction.

FC

θ+Y

+X

FCY

FCX

FCY = FC•sin(θ)FCX = FC•cos(θ)

FC FCY

FCX

θ

As FC becomes closer to horizontal:• FCX → FC

• FCY → 0

14

RESOLVING A VECTOR INTO ITS COMPONENTS

Note that we usually draw both component vectors with their tails at the origin, not head to tail.

FC

θ

+Y

+X

FCY

FCX

FC

θ

FCY

FCX

same thing

15

BACK TO OUR PROBLEM

All forces are now decomposed into their X and Y components.

+Y

+X

FA = 150lb

FB

45˚

FCFCY

FCX

16

THE CONDITIONS NEEDED FOR STATIC EQUILIBRIUM

An object in static equilibrium is one that “isn’t moving” Neither translating nor rotating

Assuming the ropes can only move in the plane of the screen, this means All forces in the X direction must sum to zero All forces in the Y direction must sum to zero

We will come back to the rotation part

17

THE EQUATIONS OF STATIC EQUILIBRIUM

The net force acting on the object is zero If it were non-zero, the object would be

accelerating

+Y

+X

FA = 150lb

FB

45˚

FCFCY

FCX

FCY = FC•sin(45) = FC / √2 FCX = FC•cos(45) = FC / √2 ΣFX = 0

FCX – FA = 0FA = FCX = FC / √2FC = 150 √2 lb

ΣFY = 0FCY – FB = 0FB = FCY = FC / √2FB = 150 lb

18

WHAT IS MOMENT OF FORCE

Also called moment or torque.

5lb

1ft

5lb

0.707ft

45˚

5 ft-lbs of torque 3.54 ft-lbs of torque

5lb

0 ft-lbs of torque

1ft

19

CALCULATING THE REACTION FORCES ACTING ON A STRUCTURE

The reaction forces keep the object fixed both transnationally (X and Y directions) and rotationally (about the Z axis)

100lb

A

B

C

2ft

1ft

FCYFAY

FAX

20

THE RELEVANT EQUATIONS

The object is not translating: ΣFX = 0, ΣFY = 0 The object is not rotating:

Σ torques about any point = 0 Lets choose point A: ΣMA = 0

100lb

A

B

C

2ft

1ft

FCYFAY

FAX

ΣFX = 0FAX – 100 = 0FAX = 100

ΣFY = 0FAY + FCY = 0FAY = -FCY

ΣMA = 01ft*100lb + 2ft*FCY = 0FCY = -50 ft-lbFAY = 50 ft-lb

21

ANALYZING FORCES IN A TRUSSTHE METHOD OF JOINTS

1. Determine the support reactions.2. Draw a free body diagram for each joint.3. Write an equilibrium equation for each joint.4. Solve for the forces at each joint.

A good video tutorial on the method of joints http://www.youtube.com/watch?v=ZOtSRbzKCC8

22

FINDING THE SUPPORT REACTIONS Last time we had a symmetrical structure that was

symmetrically loaded. What observations can we make

Weight distributions between support reactions Sign of each support reaction

FCYFAY

FAX

100lb

50lb

5 ft 5 ft

A C

10 ft

23

FINDING THE SUPPORT REACTIONS

Last time we had a symmetrical structure that was symmetrically loaded.

FCYFAY

FAX

100lb

50lb

5 ft 5 ft

ΣFX = 0FAX = 0

ΣFY = 0FAY + FCY = 150lb

ΣMA = 020FCY – 500 – 250= 0FCY = FAY =

A C

10 ft

24

THANK YOU!