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AC 2008-1952: BASSWOOD BRIDGES

Harvey Abramowitz, Purdue University CalumetHARVEY ABRAMOWITZ

Harvey Abramowitz received a BS in Materials Science, and MS and EngScD degrees inExtractive Metallurgy/Mineral Engineering, all from Columbia University. After graduating, hewas a Research Engineer for Inland Steel, where he worked on metal recovery from wastestreams. He is currently Professor of Mechanical Engineering at Purdue University Calumet. Prof.Abramowitz teaches courses in materials science and engineering, solid waste management,introduction to engineering design, and the freshman experience.

© American Society for Engineering Education, 2008

3.248.1

Basswood Bridges

Abstract

The “Elementary Engineering Design” course for freshmen students at Purdue University

Calumet consists of two components: one ME and one EE. Due to the two part structure and in

order to expose the students to the faculty, it is also team taught. The course counts as two

credits, with the format one hour lecture and three hours lab. The basswood bridge is the major

project of the ME half and counts for one quarter of the total course grade. The object, as is

usual with bridge projects, is to design, build and test a truss bridge having a high strength to

weight ratio. The design process includes statics analysis in combination with the tensile and

compressive properties of the basswood. The details of the project from initial design to final

testing are provided.

Background

At Purdue University Calumet (PUC), freshmen engineering students have been required to take

the course “Elementary Engineering Design” (ENGR190) for over three decades. The goals of

the course are:

1. To acquaint students with the design process and the creative

challenge inherent in design engineering through the medium of

individual design and construction projects.

2. To provide insight into what design engineers do.

The course is a two credit course that consists of a one hour lecture and a three hour laboratory.

Every semester the course is given. The Fall semester, which is the first semester for a typical

freshman entering college directly from high school, will have two to three sections. Each

section can handle 25 students, so for the Fall a maximum of 75 students can take the course.

For the Spring semester, the course is scheduled for late afternoon or evening to accommodate

students who work full time. One to two sections are usually on the schedule, so up to 50

students can fulfill the requirement in the Spring.

For many years, the laboratory projects were strictly mechanical in nature: a basswood bridge

and a mousetrap spring driven car. Since the projects were in a single discipline, the course was

taught by a single instructor for both the lectures and laboratories, with additional instructors

added to laboratory sections as needed. Around ten years ago, it was decided to split the course

in two, with half being oriented to mechanical engineering and the other half to electrical

engineering. This made sense since the Department of Engineering offered majors in

mechanical, electrical and computer engineering, and student surveys indicated a desire for an

electrical component in the course. In recent years, the single Department has been divided into

a Department of Mechanical Engineering and a Department of Electrical and Computer

Engineering. Therefore, it was decided to team teach the course using instructors from the

different disciplines. The first time this was tried, five instructors were used with each teaching

for 3 weeks. The three from ME had expertise in structures, heat transfer and fluid flow, and

Page 13.248.2

materials. Those from EE/CompE specialized in circuit design and electronic digital systems.

Each professor gave lectures and designed accompanying laboratory exercises. The use of five

instructors meant frequent changes in personnel and teaching methods. These changes were too

many for the students and the feedback asked for fewer instructors. Based on student surveys, a

professor from ME and one from EE were chosen to further develop the course.

Learning Objectives

Once the team teaching concept was established, specific learning objectives were set so that the

goals could be met. By the end of the course, each student should be able to:

1. Solve simple statics problems.

2. Analyze forces on trusses.

3. Design a basswood bridge, using statics analysis and material properties, that will have a

high strength to weight ratio.

4. Show that the design process is iterative in nature.

5. Write a technical laboratory report.

6. Determine simple types of equations that can represent a set of data, using x-y, semilog

and log-log plots.

7. Use EXCEL for analyzing data. Make x-y, semilog and log-log plots.

8. Solve simple DC circuits for voltage, current and power. This will include the use of

Ohm’s Law, series and parallel reduction, passive sign convention and Kirchoff’s

voltage and current laws.

9. Simulate simple DC circuits in PSpice.

10. Design timer circuits based on the LM555 timer.

11. Construct Truth Tables necessary for small scale design problems.

12. Implement logic functions with AND OR and NOT gates.

13. Perform longhand binary arithmetic operations including addition, subtraction using

complements, and multiplication.

Topics Covered

In order for the learning objectives to be met, the following are the lecture and laboratory topics

covered:

Week Lecture Topics Laboratory Topic

1. Introduction/Statics Statics Problem Set

2. Truss Calculations Bridge Problem Set

3. Bridge Design I Sample Truss Design

4. Bridge Design II Bridge Design Check-in

5. Determination of Tin M.P. Tin M.P. or Viscosity of Glycerin1

Experiment

Truss Completion Check-in

6. Data Analysis I Bridge Completion Check-in/

Bridge Testing

7. Data Analysis II Data Analysis/Bridge Critique Due

8. Electrical Introduction Mechanical Exam

9. DC Circuits 1 DC circuit lab, Pspice simulation

Page 13.248.3

10. RC circuits, 555 Timer, BCD counter 555 clock and 7447 BCD counter

11. Digital Logic 1 Counter with 7 segment display

12. Digital Logic 2 Gate implementation of adder circuit

13. System level engineering design Case study of 2 engineering

design processes

14. Review for Electrical final No Lab

Grading Policy

The final grade is based on an average of the electrical and mechanical engineering halves.

Material for the mechanical component (50% of Total) consists of:

Bridge Design 50%

Tin Melting Point or Viscosity of Glycerin 15%

Data Analysis Using EXCEL 15%

Exam 20%

Total 50%

Material for the electrical component (50% of Total) consists of:

Electrical Labs 25%

PSpice Simulations 5%

Electrical Homework 5%

Test over Electrical Material 15%

Total 50%

Basswood Bridge Project

The focus of this paper is to describe the bridge project from inception to final testing and

evaluation.

Basswood Bridge Project Description

A basswood bridge is to be designed, constructed, tested and critiqued. The design parameters

for the bridge are:

Design Parameters

(a) Truss type bridge consisting of two vertical parallel truss structures for the sides of the bridge

with necessary cross members and bracing to hold the sides in place.

(b) Top and bottom chords of the truss structures are to be parallel.

(c) Bridge is to span a 22 inch space between supports, so the length of the bridge should

be 24 inches.

(d) No bridge floor is required.

(e) A 3 ¾ inch x 3 ¾ inch block must be able to pass through the bridge with clearance.

(f) The load will be applied at the middle of the bridge using one or two ¼ or ½ inch diameter

steel rods.

(g) Maximum design load for credit is 100 lb.

(h) Best bridge is one with the highest Performance Value (PV) defined as L/W where L

Page 13.248.4

is the test load (lb) and W is the weight of the bridge (gm).

Bridge Grade

The bridge project counts for ½ of the grade for the mechanical engineering half. The project

grade is divided as follows:

%

Design Check In 10

Truss Check In 10

Bridge Check In 10

Bridge Testing 65

Bridge Critique 5

Total 100

Prerequisite Knowledge

The level of the course has been set to allow both those taking calculus and those taking pre-

calculus to succeed. Therefore, all the problem sets and designs can be calculated using

elementary algebra and similar triangles. Knowledge of trigonometry, while useful, is not a

requirement. By doing this, an attempt is made to encourage any student to major in

engineering.

Class Schedule

In the Topics Covered section the full semester’s lectures and laboratories were given. While

there are only four lectures necessary to learn how to make the bridge, it takes a full seven weeks

to finish the bridge project. Additional time may be required, if the bridge needs to be retested.

In Week 3 an assignment to design a bridge is given. The check-in and review of the design

takes place in the laboratory session of Week 4. The next week (5) an experiment is done in lab.

While this is being done, a truss completion check-in is made. Only one truss is to be

constructed. The check-in prevents erroneous construction. In Week 6, the completed bridge is

checked-in and tested. The following week (7) the bridge critique is due. The critique will be

delayed if the bridge needs to be retested. Retesting can be done if the bridge fell apart due to

glue failure, or there was a bad piece of wood that broke long before it should have.

Lectures and Laboratories

Lecture and Problem Set 1

For the introductory lecture (Appendix A), the students do not need knowledge of vector algebra.

In recent years, many of the students have taken physics in high school and do have this

understanding. Concepts covered are: (1) Parallelogram law; (2) Equilibrant; (3) Components of

a vector or force; (4) Determination of tension or compression in a member; (5) Definition of a

truss; (6) Free body diagrams; and (7) Method of similar triangles. The use of similar triangles

alleviates the need for using trigonometry and allows students with a variety of mathematical

levels to participate in the project. A number of examples are provided showing how to

Page 13.248.5

determine the forces in members of simple trusses using free body diagrams. Details for Lecture

1 are found in Appendix A. Problem Set 1 (Appendix B) consists of 2 simple statics problems.

They are similar to the examples given at the end of Lecture 1.

Lecture and Problem Set 2

This lecture (Appendix C) is focused on understanding trusses and truss bridges. Three basic

types of bridges are introduced – Pratt, Howe and Warren bridges. Bridges to be analyzed have

parallel top and bottom chords. Arch bridges are not allowed for the project. Examples using

free body diagrams to find the forces in the side trusses of each type of bridge are given.

Problem Set 2 (Appendix D) has three truss problems; one for each type of truss.

Lecture 3 – Design a Basswood Bridge

Lecture 3 (Appendix E) gives the steps needed to design and build the side trusses of the bridge.

The procedure to follow is: (1) Assume a design load; (2) Choose a truss design; (3) Calculate

forces in the members; (4) Determine which size basswood sticks to use, based on the load on

the member and whether it is in tension or compression; (5) Calculate the weight of one truss; (6)

Determine cross member size; (7) Calculate weight of the bridge; (8) Increase load to maximize

PV; and (9) Calculate final design PV. If the final design PV is not 2 or above, the design or

design load needs to be changed.

Additional information about past successful designs, materials allowed, material properties,

construction, and supplies and suppliers are also provided.

Based on this lecture, the student is to design a bridge and have the design checked during the

laboratory of week 4.

Lecture 4 – Detailed Construction Information

The construction lecture (Appendix F) covers all remaining items on the bridge construction,

with recommendations based on what has previously worked. Content includes: (1) Plates for

stressed joints; (2) Plates where loading rods are placed; (3) Cutting plates, (4) Cross members;

(5) Glues and gluing; and (6) Assembly.

Testing of Bridges

Testing Apparatus

Old

A homemade testing apparatus (Figs.1-3) was used until a few years ago. It consisted of a

hydraulic bottle jack for one support, with the other being a wooden stand which was sat on a

bathroom scale. Steel rods were placed in the middle of the bridge bottom. The rods were

attached to a plate under it by hooks and chains or ropes. The jack was pumped up and that side

moved upwards, with a force pulling the bridge in the middle. The scale showed the force on

Page 13.248.6

one support. That number was doubled to find the total load on the bridge. The student

controlled the rate of loading. Since the student was in control, there were many hesitating,

anticipatory and exciting moments while testing. Unfortunately, the load was not always applied

evenly, and sometimes the results were less than expected.

Fig. 1 Old Test Apparatus – Overall View Fig. 2 Old Test Apparatus – Steel Rod

Attachment and Scale Used to Measure

Loading

Fig. 3 Old Test Apparatus – Hydraulic Bottle Jack Used to Load Bridge

New

A 1000lb Q-Test tensile/compression machine was adapted for use in the bridge testing. The

bottom grip is removed and a support structure put in its place. This structure has a span of 22

inches to accommodate the 24 inch long bridge. The upper grip is replaced with a welded steel

U- shape. On the bottom of the U are openings for insertion of the steel rod(s). The machine is

then put into compression mode at a constant speed. The software controlling the machine is

Testworks 3 from MTS. The loading is much smoother than before and the bridge top stays

parallel to the ground. Since using this new method, the students are happier with the testing

procedure. More ‘A’ grades have resulted than with the old tester. The new tester is shown in

Figs.4 and 5. Adaptation of this machine is also being used for basswood bridge testing at

Baylor University.2

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Fig. 4 New Test Apparatus Structure Fig. 5 New Test Apparatus - Support Structure

and U for Holding Steel Rods

Bridge Testing Grade

A PV value of 2 will receive a numerical grade of 100 or an ‘A’. For a PV of 1.75, the grade is a

‘B’ or 85. A ‘C’ or 75 is gotten for a 1.50 value. The numerical grades for PVs from 0.19 and

above are listed in Appendix G. This grade constitutes 65% of the bridge project grade.

Bridge Critique

Once the student has finished testing the bridge, a bridge critique is required. The critique

consists of writing about the good and bad points of design, construction and performance;

stating what would be done differently if another bridge was to be built; and including the design

drawing and calculations. The critique (Appendix H) counts for 5% of the bridge project grade.

One thing that the students need to learn is that an engineer has to make it “work.” The proof of

the design is in the performance. So attention to the details in every aspect of the project is

required to achieve a high PV. If the bridge is not built squarely, for example, the PV will be

low.

Assessment of Bridge Project

To specifically assess the bridge project, two methods are being used. One is to survey the

students about fulfillment of the course objectives. As an example, the results of the Spring 2006

survey are listed in Table I. The course/learning objectives (1-4) for the bridge project, were

rated on a scale of 1-5, with 5 the highest. The scores ranged from 4.10-4.53, with an overall

average of 4.32, based on a survey population of 49 students.

The other method used is an Outcome Assessment Table. Such tables are prepared for all

courses in order to satisfy ABET requirements.3 For the ME half of the Design course the

Assessment Table (Table II) shows that the level of student achievement for the bridge project

ranges from 82-100%. Learning objectives 1, 2, 3 and 4 have been met. That is in Page 13.248.8

contradistinction with objective 5, writing a technical laboratory report, which only registered

62% achievement. Thus the bridge project has been successful in helping to meet ABET progam

Table I Student Survey Spring 2006 - Meeting Course Objectives for ME Half

Instructor Survey

Sec. 1 Sec. 2 Total Total

Total number of Surveys 22 27 49 49

How well were the following objectives met?

Course Objectives:

By the end of the course, each student should be able to: AVE AVE AVE AVE 1-4

4.32

1) Solve simple statics problems. 4.14 4.07 4.10

2) Analyze forces on trusses. 4.64 4.44 4.53

3) Design a bass wood bridge, using statics analysis 4.55 4.33 4.43

4) Show that the design process is iterative in nature. 4.23 4.19 4.21

5) Write a technical laboratory report. 3.82 3.74 3.78

6) Determine simple types of equations that can represnet 4.05 3.74 3.88

7) Using EXCEL for analyzing data. Make x-y, semilog and log-log plots 4.18 3.89 4.02

Rating Explanation

1= Strongly Disagree

2= Disagree

3= Undecided

4= Agree

5= Strongly Agree

outcomes a (an ability to apply knowledge of mathematics, science and engineering), c (an

ability to design a system, component, or process to meet desired needs) and e (ability to

identify, formulate and solve engineering problems). It is perhaps the most successful, in terms

of student achievement, of all the ME half activities. It is also, from personal discussions with

students, the most fun and enjoyable of all the ME activities. Alumni, years later, will still talk

about their bridges.

Conclusions

The methodology for designing, constructing and testing a basswood bridge having a high

strength to weight ratio is provided. Students only need to know basic algebra to successfully

complete the project. An outcome assessment shows that the project contributes to fulfilling

ABET progam outcomes a, c and e. Student surveys also show that the learning objectives for

the project are being met.

Page 13.248.9

Bibliography

1. Abramowitz, H., “Determination of Viscosity Using a Falling Sphere Viscometer,” National

Educators’ Workshop New Update 2000 Standard Experiments in Engineering, Materials

Science, and Technology, Kettering, OH, Oct./Nov. 2000, pp.183-196.

2. Skurla, C., Thomas, B. and Bradley, W.L., “Teaching Freshman Engineering Using Design Projects and

Laboratory Exercises to Increase Retention,” Proceedings of the 2004 American Society for Engineering

Education Annual Conference & Exposition, Salt Lake City, Utah, June 2004.

3. Houshangi, N., “Curriculum Outcome Assessment and Implementation Challenges,” Proceedings of ASEE 2006

Illinois-Indiana and North Central Joint Section Conference, Fort Wayne, Indiana, March 2006.

Appendices

Appendix A - Bridge Design Lecture 1

Appendix B – Bridge Design Problem Set 1

Appendix C – Bridge Design Lecture 2

Appendix D – Bridge Design Problem Set 2

Appendix E – Bridge Design Lecture 3

Appendix F – Bridge Design Lecture 4

Appendix G – Bridge Testing Grades

Appendix H – Bridge Critique

Page 13.248.10

TABLE II OUTCOME ASSESSMENT TABLE

(See notes at end of table to explain abbreviations)

Course Number: ENGR190 Course Name: Elementary Engineering Design

Evaluator: Semester Assessed: Spring 2006

Outcome

(ABET)

Expected

Level of

Contribution

(1, 2 or 3)

Performance Criteria Course Learning Objective Assessment Tool

BE VERY SPECIFIC

(eg. Problem 1 & 3 of Test 1, Quiz 2, or

Experiment 5)

Student Level

of

Achievement

(in percent)

a 3 Use appropriate

mathematical tools to

solve equations

1. Solve simple statics

problems.

2. Analyze forces on

trusses.

6. Determine simple types

of equations that can

represent a set a data , using

x-y, semilog and log-log

plots.

1. Statics Problem Set.

2. Bridge Problem Set.

Sample Truss Design

Bridge Design Check-in

6. Data Analysis Problem Set

100

100

92

93

63

AVE:90

a 2 Use concepts from

science to solve

engineering problems.

3. Design a bass wood

bridge, using statics

analysis and material

properties, that will have a

high strength to weight

ratio.



Bridge Completion Check-in

Bridge Testing

93

85

97

82

AVE:89

b 3 Conduct an experiment

and compare experimental

with predicted or expected

results.

5. Write a technical

laboratory report.

Tin melting point experiment

Tin melting point experiment report

94

62

AVE:78

b 3 Prepare reports that

present the data from an

experiment, interpret the

data/results, draw

conclusions, and make

recommendations.


laboratory report.

Tin melting point experiment report 62

c 3 Design components that

meet specifications and

constraints.




Sample Truss Design



92

93

85

Page 13.248.11



ratio.

4. Show that the design

process is iterative in

nature.

Bridge Completion Check-in

Bridge Testing

97

82

AVE:90

d 3 Function as a team leader

and/or team member in

laboratory and problem-

solving activities.


laboratory report.

Tin melting point experiment


94

62

AVE:78

e 3 Create sketches, figures,

flow-charts, and free-body

diagrams.

1. Solve simple statics

problems.

2. Analyze forces on

trusses.






ratio.

1. Statics Problem Set.

2. Bridge Problem Set.

Sample Truss Design

3. Bridge Design Check-in

100

100

92

93

AVE:96

e 2 Show understanding of the

applicable theories and

principles by

demonstrating the use of

relevant formulae and

relationships.

6. Determine simple types

of equations that can

represent a set a data , using

x-y, semilog and log-log

plots.

Data Analysis Problem Set 63

g 3 Write documents that are

well organized, properly

formatted, and clear.


laboratory report.

Tin melting point experiment report 62

g 3 Convey technical

information through the

use of data plots, graphs,

calculations, drawings, and

equations.


laboratory report.


Bridge Critique

62

60

AVE:61

Page 13.248.12

Notes for Outcome Assessment Table

Outcome One of the ABET outcomes

Level of Contribution 1 = Slightly, 2 = Moderately, 3 = substantially

Performance Criteria Performance Criteria for an outcome as given by ABET

Student Level of Achievement Average of the student scores for the specific assessment tool

.248.13

Appendix A - Bridge Design Lecture 1

ENGR 190 BRIDGE DESIGN LECTURE 1

1. Parallelogram Law

Forces OA)))&

and OB)))&

can be replaced by OR)))&

, their resultant.

PARALLELOGRAM LAW

2. Equilibrant

To put force system in equilibrium, add force equal and opposite

to OR)))&

. OE)))&

is equilibrant.

O

A R

B

O

A R

B

E

Page 13.248.14


3. Component(s) of force OR)))&

In general, pick coordinate system – can have 2 or more components.

3a. Special Case

Cartesian Coordinate System

More useful in machine and structural design, since many members are

vertical or horizontal.

R

O

Vertical

Component

VOR)))))&

Horizontal Component

HOR)))))&

O

A R

B

Page 13.248.15


4. Tension and Compression in Members

Rope has tension

force (stress) in it

Weight

Forces acting on rope

Rope is in tension

Stool in a coffee shop

has compression

force (stress) in it

Forces acting

on member

Member is in

compression

Page 13.248.16


4a. Further explanation of Tension/Compression

Newton’s 3rd

Law –

For every action, there is an equal and opposite reaction

Since there are equal and opposite reactions occurring, the members will

Exert forces on the connecting pins (joints) that will be equal and opposite

M2

M1 1 1F =M g&

Force of M2 on M1

Member in Tension

Member in Compression

Comrpression

(C)

Tension

(T)

Page 13.248.17


Demonstrations: Tension – Use rubber bands

Stretch rubber band between thumbs

What do your thumbs (pins or joints) feel?

Compression – Use springs

Compress spring between thumb and forefinger

What do your fingers (pins or joints) feel?

5. Truss Structures

To solve for forces in a typical joint of a truss structure [(define truss) – show

bridges from past semester] draw a free body diagram(FBD).

For example, below is the free body diagram for joint O. Joint O is in

equilibrium, which means that it is not moving.

What are forces OB)))&

and OD)))&

?

Given: (1) OA)))&

is 10 lbf or 10#

(2) the hypotenuse of a 45-45-90 right triangleis

2 = 1.4141… ≅ 1.4

FBD for joint O

A

B

D

45o

O

45o

45o

1

1

2

Page 13.248.18


The FBD is solved as follows, so that forces OB)))&

and OD)))&

can be

determined:

Step 1

Draw vertical (y direction) and horizontal (x direction) components for the

slant member (OD) as shown.

Label the components drawn. In this case V and H.

V stands for the vertical component of OD)))&

, and

H stands for the horizontal component of OD)))&

.

Step 2

The sum of the forces in any direction must equal zero for equilibrium to

exist.

Solve for V by setting the sum of the forces in the y direction to zero.

yF 0=ƒ&

+10# + V&

= 0

OA)))&

+ V&

= 0

V&

= -10# (-) = negative y direction

If the direction of V&

is known , then:

+10# - V = 0 V known to act in negative

V = 10# y direction

FBD for joint O

A

B

D

45o

V

H

O

Page 13.248.19


Step 3

Solve for H&

by setting the sum of the forces in the x direction to zero.

xF 0=ƒ&

H + OB 0=)))&&

can a value for H&

be found?

No, there are two unknowns ( H&

& OB)))&

) and only 1 equation

Step 4

Use geometry and trigonometry to solve for H&

.

Each side of a 45-45-90 right triangle has the same magnitude.

(See diagram in point 5)

So by proportion, method of similar triangles, H&

can be found

H H1 1;

1 10 1V= =

& &

& .

H 10∴ =

H = 10# magnitude of H&

Since V=-10#&

, H&

is acting to the left, or negative x direction

Thus H&

= -10#

Step 5

OD)))&

can now be found, using the similar triangle method

OD OD1.4 1.4

;1 10 1H

= =

)))& )))&

&

OD 14#∴ =)))&

Direction is SW, as shown on diagram

Alternatively:

OD OD1.4 1.4

;1 10 1V

= =

)))& )))&

&

OD 14#∴ =)))&

since both sides of the triangle are equal

Step 6

Find OB)))&

We can now go back to Step 3

xF 0=ƒ&

H + OB 0=)))&&

Page 13.248.20


-10# + OB)))&

= 0

OB)))&

= 10# positive x direction, or to the right

6. Other Triangles

Other triangles that will be used,

since they will make solutions a little simpler

3

4

5

5

12

13

1

2

3 1.73≅

Page 13.248.21


7. Method of Similar Triangles

Most solutions can be found using the method of similar triangles.

Trigonometry can also be used.

8. Directions of Force Components

Vertical and horizontal components of a slant member do the same thing as

the slant member, so directions of arrow heads are consistent. For instance,

arrow heads are up and to the right for both the slant member and its

components.

Arrow heads for components are always head, or tip, to tail; leading

around the corner such as:

Arrow heads are never:

Note: Drawing components using dashed or dotted lines avoids much

confusion. A different thickness line or a different color also can be used.

tail

slant member

components

or

or

Page 13.248.22


9. Example 1

Solution

Step 1

Draw Free Body Diagram of joint O.

Remember that the triangle below is the applicable triangle.

A

B

O

45o

Weight

Maximum force

1400# (tension)

Solve for force in OB and determine the maximum weight that can be held.

45o

45o

1

1

2 1.4≅

Page 13.248.23


The arrows are drawn for member OA and the weight due to the problem

statement. OA is in tension, which means that it is pulling joint O.

Step 2

Add components for all slant members.

Here only OA is a slant member.

Since the arrowhead was drawn on OA)))&

, the arrowheads for VOA)))))&

and HOA)))))&

can be drawn.

1400#

A

B

weight (w)

O

1400#

A

B

weight (w)

O

VOA)))))&

= 1000#

HOA)))))&

= 1000#

Page 13.248.24


Step 3

Sum up forces in x or y direction or determine value for slant member or

Components using ration of sides of triangle. In this case, components are

each 1000#. Label component values – see previous diagram.

Step 4

Solve for weight

yF 0=ƒ&

VOA)))))&

+ w = 0

1000 + w = 0

w = -1000# (-) sign means negative y direction

w = 1000#

Step 5

Solve for OB)))&

xF 0=ƒ&

HOA)))))&

+ OB)))&

= 0

-1000 + OB)))&

= 0

OB)))&

= 1000# Direction: positive x, or right

The force in OB is 1000# to the right, which means it is pushing on O.

So it’s in compression.

Note: Not necessary to show T or C on FBD. Arrow heads indicate this.

If arrow away from joint, member is in T. If arrow toward joint, member

is in C.

1400#

A

B = 1000#

weight (w) = 1000#

O

VOA)))))&

= 1000#

HOA)))))&

= 1000#

Page 13.248.25


Step 6

Finished problem should have notation as shown (no arrowheads)

A

B

O

45o

Weight

1000#

1400# T

1000# C T

Page 13.248.26


10. Example 2

Find the magnitude of the forces in members OA and OB. Also determine

If the member is in tension or compression.

Step 1

Draw Free Body Diagram at Joint O. This time you may leave out the # signs.

By inspection, the arrow heads can be drawn, since the force from w is in

the negative y direction.

3

4

5 A

O

B

w = 60#

3

4

5 A

O

B

60

Page 13.248.27


Step 2

Draw components for slant member.

Step 3

By inspection, solve for forces.

or:

solve for VOA)))))&

; VOA 60= ↑)))))&

solve for HOA)))&

; H

4OA 60 80

3

♣ •= =♦ ÷

♥ ≠

)))& ∋))

solve for OA; 5 5

OA 60 80 1003 4

♣ • ♣ •= = =♦ ÷ ♦ ÷

♥ ≠ ♥ ≠

)))&

(NW direction)

summing forces horizontally,

xF 0=ƒ&

OB 80=)))& ))&

3

4

5 A

O

B

60

80

100

HOA)))&

80

VOA)))))&

60

Page 13.248.28


Step 4

Completed solution

A

O

B

w = 60

80T

100T

T

Page 13.248.29

Appendix B – Bridge Design Problem Set 1

ENGR 190 BRIDGE PROJECT PROBLEM SET #1

NAME______________________________ DATE____________

1. Find tension force in AB and compression in AC

2. If force in AB is 1560 lbs T, find W. Also forces in BC, BE, and CD. Note

whether tension or compression.

Page 13.248.30

3. Determine forces in members AB, AC, BC, BD, CD, CE, DE, DF, EF and EG.

Note whether tension or compression.

Page 13.248.31

Appendix C – Bridge Design Lecture 2


1. Trusses

A truss structure is one in which any loads are applied at joints only and are

Comprised of one or more triangles such as:

2. Truss Bridges

Bridge will be composed of the two identical vertical sides plus necessary

crosspieces, etc., to hold it all together.

Force

or

Force

Page 13.248.32


3. Side Trusses

Typical side trusses:

Not authorized design:

Each joint on arch requires 2 simultaneous equations to solve for forces

In members.

Page 13.248.33


4. Pratt Bridge – Example

Find forces in the members and indicate whether the member is in

tension or compression. There is a force of 1600 lb acting on joint G.

Solution

Step 1

Apply forces in upward (positive y) direction of 800# to joints A and Al

(mirror image joint). The supports each carry half the force that is being

applied downward at the middle joint.

3

4 5

A

B D F

C E G

1600#

(given)

800 800

Page 13.248.34


Step 2

Draw a Free Body Diagram for each joint and solve for the force in each

member.

B

C A

800

600

1000

600

Joint A

A C

B

Joint C

600 600

0

800

Joint E

A C E

800

600 B

800

1200

600

1000

0

Joint B

Joint D

E

G

B

800 800

D

1000

600

600 1200 C E

B D

F

E G

800

1200 1800

800

600

1000

D |D

G

0

F

Joint F

1800

1200 1200

Page 13.248.35


Step 3

Write solution on diagram of truss indicating tension or compression

3

4 5

A

B F

C

G

1600#

(given)

800 800

1200C 1800C

D

00

600T 600T 1200

T E

1000C

1000T 1000T

800C

Page 13.248.36


5. Howe Bridge – Example



Solution

Step 1




1000#

(given)

500# 500#

45o

C E G A

B D F

1

1

1.4

Page 13.248.37


Step 2


member.

1500

Joint A

500 B

500

500

C A

500 700

D

C A

500

500

500 B

500 700

Joint B

Joint C

A

B

C

500

500 500

D

500

1000 E

700

D

Joint D

B F

C E

500 1000

700 500

500

500

E E|

G

F

Joint G

1000

1000

1500 1500

D F

C G

500

500

500 700

1000 E

Joint E

Page 13.248.38


Step 3


700C

500T

1000#

(given)

500# 500#

45o

G A

B

1

1

1.4 500C D 1000C F

500T C 1000T E 1500T

700C 700C

500T 1000T

Page 13.248.39


6. Warren Bridge – Example



Solution

Step 1




A C E

B D

1380#

(given)

690 690

1

2 1.73

Page 13.248.40


Step 2


member.

Joint A

400

B

C A

690 800

D

A

400 800 B

Joint B

Joint C

A

Joint E

C E|

E

D

1380

1200

690

690 690

C

400 800

800

1200

1380

C

B D 400 400

400

1200

690 690

800 800

Page 13.248.41


Step 3


A C E

B D

1380#

(given)

690 690

1

2 1.73

800C

800C 800C 1380T 800T

400T 1200T

Page 13.248.42

Appendix D – Bridge Design Problem Set 2

ENGR 190 BRIDGE PROJECT PROBLEM SET #2

NAME______________________________ DATE____________

1. All angle member 45 degrees. Find forces in all members on left half of

truss and in FG. List solutions on truss members. Note if Tension or

Compression.

Page 13.248.43

2. All panels are 3,4,5 proportion. Find forces in all members on left half of

truss and in FG. List solutions on truss members and state whether Tension

or Compression.

Page 13.248.44

3. Length of each member is 16 feet. Determine the forces in all members

between joints that are labeled. Indicate whether Tension or Compression.

Page 13.248.45

Appendix E – Bridge Design Lecture 3


Design a Basswood Bridge

Design Parameters

(a) Truss type bridge consisting of two vertical parallel truss structures for

the sides of the bridge with necessary cross members and bracing to

hold the sides in place.

(b) Top and bottom chords of the truss structures are to be parallel.

(c) Bridge is to span a 22 inch space between supports, so the length of the

bridge should be 24 inches.

(d) No bridge floor is required.

(e) A 3 ¾ inch x 3 ¾ inch block must be able to pass through the bridge with

clearance.

(f) The load will be applied at the middle of the bridge using 1 or 2

¼ or ½ inch diameter steel rods.

(g) Maximum design load for credit is 100 lb.

(h) Best bridge is one with the highest Performance Value (PV) defined as

L/W where L is the test load (lb) and W is the weight of the bridge (gm)

1. Assume a Design Load

for example - 36#

2. Choose a Truss Design

see truss chosen below

Steps 1 and 2 can be reversed

4

35

B 12C D

A 6” C 6” E 6” 6”

12T 24T

15C 9T 15C 18T

9# 9# 18#

½ total load

Page 13.248.46


3. Calculate Forces in Members

Draw FBDs for joints A,B,C,D and E

Joint A Joint B

Joint C Joint D

Joint E

9

9

C

B 12

12

15

A A

B D

C

12 12

9 9

15

C E C|

D

18

24 24

D B

C E

24 12 A

12

9 9

15

B B|

E

12 12 D

9 9 15 15

12 12

Joint A Joint B

Joint C Joint D

Joint E

9

9

C

B 12

12

15

A A

B D

C

12 12

9 9

15

C E C|

D

18

24 24

D B

C E

24 12 A

12

9 9

15

B B|

D

E

Page 13.248.47


4. Determine which Size Sticks to Use

Indicate on the structure – use information given

Member Force Length (inch) Size Max Load (lb) Weight (g)

AB 15C 6/4 x 5 = 7 1/2 3/16 25 (.24)(7.5) = 1.80

BC 9T 6/4 x 3 =4 1/2 3/32 20 (.06)(4.5) = 0.27

AC 12T 6 3/32 20 (.06)(6) = 0.36

BD 12C 6 5/32 22 (.17)(6) = 1.02

CE 24T 6 1/8 35 (.11)(6) = 0.66

DE 18T 4 1/2 3/32 20 (.06)(4½) = 0.27

CD 15C 7 1/2 3/16 25 (.24)(7.5) = 1.80

Since only 1 member (DE) is loaded to near its capacity at the 36 lb design load, the load can be

increased. Try a load increase. To determine the increase, notice the ratios of maximum load

(lb) / Force.

AB = 25/15 = 1.67

BC = 20/9 = 2.22

AC = 20/12 = 1.67

BD = 22/12 = 1.83

CE = 35/24 = 1.46

DE = 20/18 = 1.11

CD = 25/15 = 1.67

Increasing the force/load by 1.11 (DE) will not maximize efficiency of loading. Go to 1.67 ratio

(AB, AC & CD). So load is now increased by the factor of 1.67.

18 x 1.67 = 30# On the diagram, forces on the right side of the structure are for a 60# load.

Members will now be more efficiently loaded.

B 12C 5/32-22 D 20C

A 12T 24T

9T

3/32-20

15C

3/16-25

9# 9# 18# 30#

½ total load

15C

3/16-25

18T

3/32-20

30T

3/32-20 C 1/8-35 E 40T 20T

25C 25C 15T

Page 13.248.48


5. Calculate the Weight of 1 Truss

For 36# total load, wt =

((1.80 + .27 + .36 + 1.02 + .66 + .27 + 1.80) x 2) - .27 = 12.09

For 60# total load

Member Force Length (inch) Size Max Load (lb) Weight (g)

AB 25C 6/4 x 5 = 7 1/2 3/16 25 (.24)(7.5) = 1.80

BC 15T 6/4 x 3 =4 1/2 3/32 20 (.06)(4.5) = 0.27

AC 20T 6 3/32 20 (.06)(6) = 0.36

BD 20C 6 5/32 22 (.17)(6) = 1.02

CE 40T 6 5/32 54 (.17)(6) = 1.02

DE 30T 4 1/2 1/8 35 (.11)(4½) = 0.495

CD 25C 7 1/2 3/16 25 (.24)(7.5) = 1.80

For 60# total load, wt =

((1.80 + .27 + .36 + 1.02 + 1.02 + .495 + 1.80) x 2) - .495 = 13.035

Page 13.248.49


6. Determination of Cross Member Size

The following table and figures gives the size of cross members to choose

based on the type of cross member and the total design load.

A

B

E

D

Plates 1/32” thick for 50-100# design load

Can be 1/64” for less than 50#

0 0

0

C

These 2 zero force members can be omitted

24”

About

5.19”

tall

600

600 600

If you are doing this design, discuss cross

members with the instructor. B type cross

members may be too long and an X type

of bracing may be needed. Page 13.248.50


Table E-I: Sizing of Cross and Zero Load Members

Design Load Range (lb)

0-45 46-69 70-100

Cross Member (A) 3/32 1/8 5/32

Diagonal Cross Member (B) 3/32 3/32 1/8

0 Force Member (C) 3/32 3/32 1/8

0 Force of Stressed (D) 1/8 1/8 5/32

Horizontal Member at Ends (E) 1/8 5/32 3/16

7. Calculate Weight of Bridge

For 36# total load

2 trusses = 12.09 x 2 = 24.18

From Table E-I

Type Member Size Length

(in)

Weight

(g)

# Members Total wt

(g)

⊥ cross member

A

3/32 4 (.06)(4)=0.24 1 0.240

Diagonal cross

member B

3/32 7.2 (.06)(7.2)=0.432 2 0.864

Stressed D 1/8 4.5 (.11)(4.5)=0.495 4 1.980

Horiz. Members

at ends E

1/8 4 (.11)(4)=0.44 4 1.760

Total 4.844

-original BC(x4)=(.27)(4) -1.080

Total Cross & End Member Weight=3.764

Total Bridge Weight = 24.18 + 3.764 + plates + glue

= 27.944 + plates + glue

plates + glue ≈ 15-20% of wood member wt

15% = 4.192

= 27.944 + 4.192

= 32.136 g

PV = 36

1.1232.136

= ← low

Page 13.248.51


8. Increase Load to Maximize PV

For 60# total load

2 trusses = 13.035 x 2 = 26.07

From Table I

Type Member Size Length

(in)

Weight

(g)

# Members Total wt

(g)

⊥ cross member

A

1/8 4 (.11)(4)=0.44 1 0.440

Diagonal cross

member B

3/32 7.2 (.06)(7.2)=0.432 2 0.864

Stressed D 1/8 4.5 (.11)(4.5)=0.495 4 1.980

Horiz. Members

at ends E

5/32 4 (.17)(4)=0.68 4 2.720

Total 6.004

-original BC(x4)=(.27)(4) -1.080

Total Cross & End Member Weight=4.924

Total Bridge Weight = 26.07 + 4.924 + plates + glue

= 30.994 + plates + glue

plates + glue ≈ 15-20% of wood member wt

15% = 4.649

= 30.994 + 4.649

= 35.643 g

PV = 60

1.6835.643

= ← better

9. Calculate Final Design PV

see above PV for 60# total load

Page 13.248.52


10. Successful Designs

Successful designs have typically been similar to problem set #2

or one of the following:

Reminder: This type of truss is not permitted

11. Other Design Information

24”

About

5.19”

tall

600

600 600

Page 13.248.53


12. Construction Information

For top and bottom chords, do not use short pieces end to end.

Example – use one 16” piece of 5/32 sq and add1/32 plywood strips

to two adjacent sides where force is 60C

13. Authorized Materials for Bridge Construction

Members

Material – basswood for members

Sizes (square cross section) – 3/32, 1/8, 5/32, 3/16, 1/4 inch on a side

Plates and Lamination Material

Plates are used to reinforce joints and to laminate sticks to obtain proper size

Material – model aircraft plywood

Sizes – 1/64 & 1/32 inch thick

Glue

Ordinary wood glue, such as Elmer’s Carpenter’s Glue and some epoxies

Do not use glue from a glue gun (heated)

What has worked well in the past: Jet Glue with kicker, Titebond

Coatings – none allowed

14. Material Properties

Densities

Basswood Plywood

linear density square density

3/32 .06 g/in 1/64 .25 g/in2

1/8 .11 1/32 .39

5/32 .17

3/16 .24

1/4 .43

5/32-43

30C

4”

3/16-73

60C

4” 4” 4”

5/32 sq 1/32 sq plywood strips

best applied to top &

inside

Page 13.248.54


Tensile Strength for Basswood Sticks

Permissible tension loads for basswood (any length) are:

Size Tensile Strength

(in) (lb)

3/32 20

1/8 35

5/32 54

3/16 80

1/4 140

Compression Strengths for Basswood Sticks

Compression failure loads for basswood are given in the figure below.

Compression Strengths for Basswood Square Sections

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

180

0 1 2 3 4 5 6 7 8 9

Length (in)

Co

mp

ressio

n L

oad

at

Failu

re (

lb)

3 32

1 8

5 32

3 16

1 4

Page 13.248.55


15. Suppliers and Supplies

Go to your local hobby shop to purchase the wood.

Glue can be obtained at the hobby shop or local hardware store.

Hobby miter boxes and saws can be gotten from the hobby shop.

Hand files, sand paper, Exacto knives, hobby wood clamps and a Dremel tool

are also helpful in construction.

16. Project Grade

Project score will be derived as follows:

%

Bridge design checked in on schedule 10

One side truss completed and checked in 10

Bridge completed and checked in 10

Performance value (PV)* of tested material 65

Bridge critique submitted after testing 5

----

100

* Maximum design load for credit is 100#

If bridge holds more than 100# when tested, you only get credit for 100#

17. Assignment – Prepare Bridge Design

(a) Assume some design load

(b) Choose truss design (1 and 2 can be reversed)

(c) Calculate forces in members

Indicate on the structure the forces for each member

(d) Determine which size sticks you need for each member and indicate

on the structure also.

(e) Calculate weight of 1 truss

(f) Calculate weight of bridge – 2 trusses plus cross pieces plus plates

plus glue (estimate as best as possible)

(g) Calculate PV

load(lb)

PV=weight(g)

(h) Try more than one design

Page 13.248.56

Appendix F – Bridge Design Lecture 4


Detailed Construction Information

This lecture covers all remaining items on the bridge construction, with

recommendations based on what has worked before.

1. Plates for Stressed Joints

Plates should extend ¾” over each stressed member at joints. Zero force

members should have a little plate of some sort to assist the glue. Plates are

used only on outside of vertical trusses. None on the inside or horizontally on

top or bottom. Exception is the center joint for loading rods.

Note: These are the ideal conditions for minimum weight. Sometimes the

glue does not hold as well as expected and plates have been added to the top

and/or bottom cross members. This adds weight to the bridge but does

provide more support.

Use on both sides of each truss for loads > 65#. Total 4 for whole bridge

½ ”

rounded

Plates for Stresses Joints

Small plate

0

force

1/32” plates for loads � 50

#

1/64” plates for loads < 50# ¾ ” ¾ ”

¾ ”

1 3/8”

1 3/8” ½ x 1” ½ x 1”

Start with Page 13.248.57


2. Plates for Center Members Where Rod is Placed

For bars

Loading Plates

Sand members and plates lightly with coarse sand paper to rough up surface.

Glue holds better.

Cutting Plates

Clearance

1”

Don’t

cut

radius

or

Cover all three

again

File or use hobby

knife to get

round corner

Break

off

Page 13.248.58


3. End View

4. Cross Members

Construction

To construct: typical

(a) Put top and bottom chords on flat surface.

(b) Cut verticals and glue in. (Put weights on)

(c) After dry, put diagonals to fit loosely. Use glue sparingly. Don’t put in

gobs to fill corners, etc. Just enough to hold sticks together for check-in.

If diagonals are forced in, truss may end up warped.

Glue Glue

A little clearance end to end * Cut to 4” length

END VIEW

Approximately 45o 1/8” or 3/32” square material

Top only

Both sides each joint

¾” max

Page 13.248.59


To construct: no verticals

Place top and bottom chords on surface and add weight.

Cut diagonals to fit and glue in.

5. Tips

(a) Put wax paper or plastic down on surface, or glue may stick to it.

(b) Put weights on all members being glued, so bottom side will be flat.

(c) Don’t put any plates on before checking in one truss.

(d) Glues – Elmers Wood Glue, Elmers Por Bond,DAP Wood Glue,

Crazy Glue (gel type or slow setting), Titelock, Jet with or without kicker.

No model airplane glue, no flexible glue, no glue gun.

Jet with kicker and Titelock seem to work best.

(e) Can use hobby wood clamps, or clothespins or metal paper clamps.

(f)

vs

.

not much

advantage, but

OK

½ “

Etc.

Page 13.248.60


6. Assembling Two Sides

(a) Bottom at ends

(b) Top members

(c) Let set

(d) Diagonals (again – should not have to press in)

7. Typical

Typical etc.

¾ ”

¾ ”

¾ ”

½ ”

Books

Page 13.248.61

Appendix G – Bridge Testing Grades

Bridge Test Grades

PV GRADE Points for PV GRADE Points for PV GRADE Points for

ME half ME half ME half

2.00 100.00 32.50 1.65 81.00 26.33 1.30 67.00 21.78

1.99 99.00 32.18 1.64 81.00 26.33 1.29 67.00 21.78

1.98 98.00 31.85 1.63 81.00 26.33 1.28 66.00 21.45

1.97 97.00 31.53 1.62 80.00 26.00 1.27 66.00 21.45

1.96 96.00 31.20 1.61 80.00 26.00 1.26 65.00 21.13

1.95 95.00 30.88 1.60 80.00 26.00 1.25 65.00 21.13

1.94 94.00 30.55 1.59 79.00 25.68 1.24 64.00 20.80

1.93 93.00 30.23 1.58 79.00 25.68 1.23 64.00 20.80

1.92 92.00 29.90 1.57 78.00 25.35 1.22 64.00 20.80

1.91 91.00 29.58 1.56 78.00 25.35 1.21 63.00 20.48

1.90 90.00 29.25 1.55 77.00 25.03 1.20 63.00 20.48

1.89 89.00 28.93 1.54 77.00 25.03 1.19 63.00 20.48

1.88 89.00 28.93 1.53 76.00 24.70 1.18 62.00 20.15

1.87 89.00 28.93 1.52 76.00 24.70 1.17 62.00 20.15

1.86 88.00 28.60 1.51 75.00 24.38 1.16 62.00 20.15

1.85 88.00 28.60 1.50 75.00 24.38 1.15 61.00 19.83

1.84 88.00 28.60 1.49 74.00 24.05 1.14 61.00 19.83

1.83 87.00 28.28 1.48 74.00 24.05 1.13 61.00 19.83

1.82 87.00 28.28 1.47 74.00 24.05 1.12 60.00 19.50

1.81 87.00 28.28 1.46 73.00 23.73 1.11 60.00 19.50

1.80 86.00 27.95 1.45 73.00 23.73 1.10 60.00 19.50

1.79 86.00 27.95 1.44 73.00 23.73 1.09 59.00 19.18

1.78 86.00 27.95 1.43 72.00 23.40 1.08 59.00 19.18

1.77 85.00 27.63 1.42 72.00 23.40 1.07 58.00 18.85

1.76 85.00 27.63 1.41 72.00 23.40 1.06 58.00 18.85

1.75 85.00 27.63 1.40 71.00 23.08 1.05 57.00 18.53

1.74 84.00 27.30 1.39 71.00 23.08 1.04 57.00 18.53

1.73 84.00 27.30 1.38 71.00 23.08 1.03 56.00 18.20

1.72 84.00 27.30 1.37 70.00 22.75 1.02 56.00 18.20

1.71 83.00 26.98 1.36 70.00 22.75 1.01 55.00 17.88

1.70 83.00 26.98 1.35 70.00 22.75 1.00 55.00 17.88

1.69 83.00 26.98 1.34 69.00 22.43 0.99 54.00 17.55

1.68 82.00 26.65 1.33 69.00 22.43 0.98 54.00 17.55

1.67 82.00 26.65 1.32 68.00 22.10 0.97 54.00 17.55

1.66 82.00 26.65 1.31 68.00 22.10 0.96 53.00 17.23

Page 13.248.62

Bridge Test Grades

PV GRADE Points for PV GRADE Points for

ME half ME half

0.95 53.00 17.23 0.60 40.00 13.00

0.94 53.00 17.23 0.59 39.00 12.68

0.93 52.00 16.90 0.58 39.00 12.68

0.92 52.00 16.90 0.57 38.00 12.35

0.91 52.00 16.90 0.56 38.00 12.35

0.90 51.00 16.58 0.55 37.00 12.03

0.89 51.00 16.58 0.54 37.00 12.03

0.88 51.00 16.58 0.53 36.00 11.70

0.87 50.00 16.25 0.52 36.00 11.70

0.86 50.00 16.25 0.51 35.00 11.38

0.85 50.00 16.25 0.50 35.00 11.38

0.84 49.00 15.93 0.49 34.00 11.05

0.83 49.00 15.93 0.48 34.00 11.05

0.82 48.00 15.60 0.47 34.00 11.05

0.81 48.00 15.60 0.46 33.00 10.73

0.80 47.00 15.28 0.45 33.00 10.73

0.79 47.00 15.28 0.44 33.00 10.73

0.78 46.00 14.95 0.43 32.00 10.40

0.77 46.00 14.95 0.42 32.00 10.40

0.76 45.00 14.63 0.41 32.00 10.40

0.75 45.00 14.63 0.40 31.00 10.08

0.74 44.00 14.30 0.39 31.00 10.08

0.73 44.00 14.30 0.38 31.00 10.08

0.72 44.00 14.30 0.37 30.00 9.75

0.71 43.00 13.98 0.36 30.00 9.75

0.70 43.00 13.98 0.35 30.00 9.75

0.69 43.00 13.98 0.34 29.00 9.43

0.68 42.00 13.65 0.33 29.00 9.43

0.67 42.00 13.65 0.32 28.00 9.10

0.66 42.00 13.65 0.31 28.00 9.10

0.65 41.00 13.33 0.30 27.00 8.78

0.64 41.00 13.33 0.29 27.00 8.78

0.63 41.00 13.33 0.28 26.00 8.45

0.62 40.00 13.00 0.27 26.00 8.45

0.61 40.00 13.00 0.26 25.00 8.13

Page 13.248.63

Appendix H – Bridge Critique

Engineering 190

BRIDGE CRITIQUE

Good Points (1.5 pts)

of design

of construction

of performance

Bad Points (1.5 pts)

of design

of construction

of performance

What would you do differently, if you built another bridge? (1 pt)

Include your design drawing and calculations (1 pt)

Page 13.248.64

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