Lecture 6 Introduction to Engineering Approximate Running Time - 19 minutes

Slide 1 © 2006 Baylor University

EGR 1301

Lecture 6Introduction to Engineering

Approximate Running Time - 19 minutesDistance Learning / Online Instructional Presentation

Presented byDepartment of Mechanical Engineering

Baylor University

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EGR 1301

Introduction to Static Analysis – Part 3

EGR 1301 – Lecture 6

Prof. Dick Campbell Speaking


EGR 1301

Learning Objectives

• Understand why structures are built with triangles

• Understand truss bridges as an example of a structure built with triangles

• Apply static analysis to truss bridges• Understand the simplifying assumptions for the

truss analysis• Determine load in each member of truss bridge

that is supporting a load using static analysis


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Why structures are built with triangles

• Pinned triangles are naturally rigid• Joint strength becomes less critical• High stiffness can be achieved for minimal

amount of material used• Ease of construction

FF


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The truss is a simple skeletal structure. In design theory, the individual members of a simple truss are only subject to tension and compression forces and not bending forces.

There are both simple and continuous trusses. The small size of individual parts of a truss make it the ideal bridge for places where large parts or sections cannot be shipped. Because the truss is a hollow skeletal structure, the roadway may pass over or even through the structure allowing for clearance below the bridge often not possible with other bridge types.


EGR 1301

Truss Bridges

Trusses are constructed using triangles and are also classified by the basic design used.

The Warren truss is perhaps the most common truss for both simple and continuous trusses. For smaller spans, no vertical members are used lending the structure a simple look. For longer spans vertical members are added providing extra strength .

Warren trusses are typically used in spans of between 50-100m.


EGR 1301

Truss Bridges

• Pratt - The Pratt truss is identified by its diagonal members which, except for the very end ones, all slant down and in toward the center of the span. Except for those diagonal members near the center, all the diagonal members are subject to tension forces only while the shorter vertical members handle the compressive forces. This allows for thinner diagonal members resulting in a more economic design.

• Howe - The Howe truss is the opposite of the Pratt truss. The diagonal members face in the opposite direction and handle compressive forces (requiring thicker elements. This makes it very uneconomic design for steel bridges and its use is rarely seen.


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This bridge was designated as an international historical monument by the Canadian Society for Civil Engineering and the American Society of Civil Engineers. The Pont de Québec is formed by a 549 m (1702’) suspended span located between two main pillars, which makes this bridge the longest cantilever bridge in the world. While the bridge was under construction, the suspended span collapsed on two occasions (in 1907 and 1916), killing many workers. Trains began using the bridge in 1917 while automobiles were only allowed on it in 1929.

World's Longest Truss BridgeThe Pont de Quebec


EGR 1301

Concepts & Assumptions forStatic Analysis of Truss Bridges

• Sum of forces at each point (joint, or node) must equal zero• Each element is a “two-force” member, in that the direction of the

force is along the axis• If an element is in tension, it will pull on both points• If an element is in compression, it will push on both points• Joints are pinned and frictionless (pins will not support a Moment)• No deformation occurs to change angles or lengths• The external reactions are statically determinant, and the supports

are frictionless:

R1

R3

R2


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Truss Analysis

A

BC

D

E

1R

3R

2R

lbs 500P

Determine External Reactions using Newton’s 1st Law:

x

y

30 RΣFx

5000 21 RRΣFy

ft 10ft 10

ft 10

ft 10 ft 66.8h5*50020*0 1 RΣM D

The statically determinant truss has three external reactions, and we can write three equations of static equilibrium.

+

lbs 1251 R

lbs 3752 R

lbs 03 R

o60


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Truss Analysis

A

BC

E

125

03 R

lbs 500P

ft 10ft 10

ft 10

ft 10 ft 66.8h

x

y +

375

Determine the forces at node D, by drawing a “free body” diagram:

D

We assume that all unknown forces are in tension by drawing the arrow pulling on the node. Our sign convention will indicate if an element is in compression by a negative sign.

D

375

DCF

DEF


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Truss Analysis

D

Just as we used the equations of static equilibrium to determine the external reactions, we can apply the same method to each node:

375

DCF

DEF

cos0 DCDEx FFΣF

x

y +

375sin0 DCy FΣF

lbs 43360sin

375

DCF

lbs 21760cos)433(cos DCDE FF

n.compressioin is Element

thatmeanssign minus The

DCF

.in tension is Element DEF


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Truss Analysis Completed

A

BC

D

E

lbs 500P

The analysis is completed by moving to each node in turn, and applying the known forces to the free-body diagram of that node, until all elements have been solved. You should finish this analysis as an exercise.

x

y

ft 10ft 10

ft 10

ft 10 ft 66.8h

+

o60

lbs 433DCF

lbs 217DEF

03 R

3752 R1251 R


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Statics Homework Assignment #2

lbs 50P

ft 10

ft 10

ft 5h

x

y +

ft 10

lbs 50P

ft 10

ft 10

ft 10h

ft 10

1

2

3 4

5

6

7

1

2

3 4

5

6

7

Calculate the forces in the seven elements of both trusses, and indicate tension or compression.

Comparing the two results, how does changing the height to width ratio affect the force in each element?


EGR 1301

This Concludes Lecture 6

Documents

Lecture 6 Introduction to Engineering Approximate Running Time - 19 minutes