truss final report

TRUSS DESIGNGEEN [2851] - SOLTYS[Group 7]

Slator Aplin

Thomas CeriseKristofer Holmquist

AJ Guerriro

TABLE OF CONTENTS

Contents

Introduction_________________________________________________________________________________________1(Kris)

Experimentation and Modelling_________________________________________________________________2(Slator)

Support Reactions______________________________________________________________________________3(Thomas)

Truss Force Calculations_______________________________________________________________________4(Thomas)

Calculation Results_______________________________________________________________________________5(Slator)

CALCULATION RESULTS (11, 12, 13)CALCULATION RESULTS (11, 12, 13)

Introduction

SCOPE OF PROJECTThe goal of this project is to design and build a truss capable of withstanding a predetermined load. All materials, designs and reactions are tested before any construction. Each team had the choice to select materials from a predetermined list and construct a design from the materials of a truss capable of withstanding a load of 1200 lbs.

Preliminary testing of the wood and glue selected gave clues to the construction of the truss on a larger scale. Results from four point bending of our selected Douglas Fir, and of shear testing Elmer’s wood glue indicated the maximum conditions our truss could withhold. With added data, technical drawings of our truss could be made in varying configurations in order to ensure its load capability.

Through extensive Excel calculations based on gathered material testing data, a truss geometry was designed to withstand the internal forces experienced by a real truss such as: internal force, normal stress, buckling, shear forces & stresses. By designing to withhold these stresses, safety factors of each members could be determined as well as the maximum load the truss could withstand, its strength-to-weight ratio, where and how it would fail.

Finally, a fully constructed truss based on materials testing, experimentation and simulation came to life. The truss was put to the test under varying load until it finally failed. Real versus theoretical limits of the truss design can now be compared with each other. Its eventual collapse denoted the success or failure of all calculations leading up to it.

Page 1

Figure 1. Bending and Shear test results


Experimentation and Modeling

σ=3PLbh2 =

3∗400∗1612

1.2512

∗0.7512

2 =3.93x 106( lbf t 2 )

τ= FA

= 525

2∗( 112 )∗( 0.5

12)=75600( lb

f t 2 )

Glue: Gorilla Glue

Wood: Douglas Fir

Figure 2. Beam Test

Page 2


Figure 2. Shear Test

Page 3


Page 4


Support Reactions

Within our truss there were a total of three external reactions: the two support reactions at joints A and E along with the load that was distributed on joint C. Our initial goal was to have the load distributed on C be

1200 lb or 1.2 kips. Thus with this we used the following two equations: ∑ F y=O∧∑ M=0 . We only

need to use F y because the support acts as a roller which means it has a force that acts normal/orthogonal

to both joint A and E. Thus with this the following calculations were made:

∑ F y=F Ay+FEy−1.2 Kips=0

F Ay+F Ey=1.2Kips

∑ M A=(−1.2 Kips )∗(11) + (FEy )*(22)=0

FEy∗¿

FEy=1.2 Kips∗(11/22 )=0.6 Kips

F Ay=(1.2 Kips )− (0.6 Kips )=0.6 Kips

Both joints A and E have the same reaction force of 0.6 kips, which also indicates that the forces will be symmetrical along the truss.

Page 5


Truss Force Calculations (7, 8, 9, 10)

Calculating Internal Forces, Normal Stress, and Safety Factor

After calculating the external forces the internal forces of each member can be calculated by using the method of joints or sections. Also since the external reactions on the truss are symmetrical about joint C we needed to only calculate the internal forces for half of the members and then apply those values to the corresponding members on the other side of joint C. Then to determine the forces in the x and y directions we needed to use simple trigonometry in which a ratio between the rise/run with the member length would be taken and then multiplied by the internal force of the member. The following two equations represent how to determine the x and y forces of each member:

F x=(Runof memberMember length )∗(internal forceof member )

F y=( Rise of memberMember length )∗(internal force of member )

Member Length (in) Rise (in) Run (in) Internal Force (kips)

F y (kips) F x (kips)

A-B 5.83 5 3 -0.7 -0.60 -0.31A-G 6.00 0 6 0.36 0.00 0.36B-G 5.83 5 3 0.55 0.47 0.28B-C 8.25 2 8 -0.66 -0.16 -0.64C-G 7.62 7 3 -0.58 -0.53 -0.23C-F 7.62 7 3 -0.58 -0.53 -0.23C-D 8.25 2 8 -0.66 -0.16 -0.64D-F 5.83 5 3 0.55 0.47 0.28D-E 5.83 5 3 -0.7 -0.60 -0.36E-F 6.00 0 6 0.36 0.00 0.36F-G 10.00 0 10 0.98 0.00 0.98

The next portion of each member that must be checked is the stress. Stress is determined by the force per area, which in this case is the cross sectional area of each member. Furthermore, each member has a unified width of 0.5” and all members have an actual height of 0.25”, except for G-F whose actual height is 0.375”. Then based on these values all members will have a cross sectional area of 0.13 ¿2 except for G-F

whose cross sectional area is 0.19¿2. Then based on these values the stress can be determined by dividing the internal force of the member by its cross sectional area. Then the final portion of this that must be determined is the safety factor of each truss which is determined by the following equation:

Safety Factor= Stressof themember|Internal force of themember|

The goal is to have the safety factor of each member be above 1 because this will indicate that the truss should not fail due to stress while supporting a load bearing weight of 1.2 kips. With these two equations here are the following stress and safety values for each member.

Page 6


Member Stress (ksi) Safety FactorA-B -5.60 2.9A-G 2.88 5.6B-G 4.40 3.6B-C -5.28 3.0C-G -4.64 3.4C-F -4.64 3.4C-D -5.28 3.0D-F 4.40 3.6D-E -5.60 2.9E-F 2.88 5.6F-G 5.23 3.1

Since every safety factor for each member is above 1 it is apparent that each member should not fail due to stress when given 1.2 kips of pressure on joint C.

Buckling Calculation and Safety Factor of All Compressive Members

The compressive members of our truss will be the ones who have a negative internal force. Those members are A-B, B-C, C-G, C-F, C-D, and D-E.

The calculation for the buckling, which is called Pcr, of each member was determined by the following equation:

Pcr (kips)=¿

The wood modulus for our wood is 1765 ksi, the truss width is 0.5”, and the h(actual) for each compressive member is 0.25” (2/8 “). Then from the Pcr equation we can then determine the safety factor of each compressive member by the following equation:

Saftey Factor=Pcr /(|Force of theMember|)

Member Buckling (kips) Safety FactorA-B 2.78 3.97B-C 1.96 2.98C-G 2.13 3.67C-F 2.13 3.67C-D 1.96 2.98D-E 2.78 3.97

Since all compressive members have a buckling safety factor above 1 none of those members should buckle/collapse when given 1.2 kips of pressure on joint C.

Page 7


Calculation of Shear Stress and Forces along With Safety Factor at Each Joint

The shear force is determined by the amount of force that is applied per cross sectional area. Furthermore, the shear force of each member is determined by the following equation:

τ=|Force of theMember|/(( Actualmember height )∗(overlap allowed ))

The overlap allowed is based upon the criteria that only 15% of a member on either side can be overlapped/covered by a gusset plate. That number is then transferred into eighths because it is too difficult to cut a piece of wood to sixteenths.

Here are the following overlaps allowed:

Member Calculated Overlap (inches) Overlap Allowed (inches)

A-B 14/16 7/8A-G 14/16 7/8B-G 14/16 7/8B-C 1 4/16 1 2/8C-G 1 2/16 1 1/8C-F 1 2/16 1 1/8C-D 1 4/16 1 2/8D-F 14/16 7/8D-E 14/16 7/8E-F 14/16 7/8F-G 1 8/16 1 4/8

Then with the actual overlap the shear stress can be calculated from the equation above and then from the shear stress the safety factor can be calculated by the following equation:

Safety Factor=(Glue Strength)/(Shear stress)

The glue strength is 1 Ksi.

With these two equations we are now able to determine the shear stress and safety factors of each member and they are as follows:

Member Shear Stress (ksi) Safety FactorA-B 1.60 0.62A-G 0.80 1.25B-G 1.26 0.80B-C 1.07 0.94C-G 1.01 0.99C-F 1.01 0.99C-D 1.07 0.94D-F 1.26 0.80D-E 1.60 0.62E-F 0.80 1.25F-G 0.87 1.15

Page 8


Since the majority of the members/joints have a safety factor that is below 1 it is apparent that the truss will fail due to shear stress at a joint before 1.2 kips of pressure will be distributed on joint C.

Prediction of Maximum Load, Strength to Weight Ratio, and Failure of Truss

Within our truss we determined that the maximum load was going to be 0.98 kips or 980 lb which occurs within member G-F. Then we were then able to determine our strength to weight ratio which determines how many pounds of force an object can with stand per pound of weight/mass. However in order to get the weight of the truss the weight of each individual member must first be calculated. The weight was calculated by the following equation:

W =(member length )∗(member width )∗(member height )∗(wood density )

The density of the wood is a constant which is 0.0185 lb /¿3 along with the width of the wood which was 0.5”. Then every member, except for G-F, had a height of 0.25” while member G-F had a height of 0.375”. With these values and the member lengths their weights are as follows:

Member Weight (lbs)A-B 0.0135A-G 0.0139B-G 0.0135B-C 0.0191C-G 0.0176C-F 0.0176C-D 0.0191D-F 0.0135D-E 0.0135E-F 0.0139F-G 0.0347

Based on these calculations the heaviest member is F-G at about 0.035 lb. Then by adding up the weight of all the members the total weight was determined to be 0.1897 lb which equals about 0.19 lb. However, this weight does not include the weight of the gusset plates nor the glue that will be attached to the both sides of each joint on the truss. Those factors cannot be determined until the truss is weighed as a whole. So with the weight being 0.1897 lb and our distributing load being 1200 lb the strength to weight ratio was calculated to be 6325.

Finally, with every truss it will eventually fail. Based upon all the safety factors that have been presented it has been determined that the truss will fail at either member A-B or D-E due to shear; both of these members had a safety factor of 0.62 for shear stress. With these safety factors of 0.62 the truss should fail at about 740 lb of pressure or 0.74 kips.

Page 9


Calculation Results (11, 12, 13)

Strength-to-Weight Ratio:(Load (lbs)/Weight(lbs))

Failure Type/Location:

Predicted: 3347Actual: 2450

Predicted: Shear failure at joint A/EActual: Shear failure at joint A and joint B

Within our design of the truss our group realized that drastic improvements/modifications needed to take place. Those were primarily with the height of the wood. Furthermore, initially the wood had a height of 0.25”, as stated in the above section. However, during the construction process we realized that we could not cut our wood that thin, as requested by the shop manager. So with a slight modification we increased our member height from 0.25” to 0.5”. This caused our weight of the truss to increase by nearly double, going from about 0.19 lb to about 0.4 lb, which then caused our predicted strength to weight ratio to go down by nearly half, from 6325 to 3347. However, with this modifications our truss still had several design flaws.

Our center joint truss was designed to withstand an applied load of 1,200 lbs and was short of this value by roughly 300 lbs. Our prediction of failure from shear forces at joint A was correct. During testing there was also another shear failure at joint B. The difference between our predicted and actual truss design is from a collection of construction and design errors. When building a wooden truss by hand, it is difficult to avoid tolerance build-up and other craftsmanship flaws. Our predicted truss was also designed as a ‘line-member and dot-joint’ truss. We did not account for the width of the members or the jointing of members at each gusset plate. This design flaw along with an accumulation of construction errors contributed to our predicted load becoming an overestimate.

Improvements for our truss design for a theoretical retest would be to increase the thickness of all of our members and increase the base length of our truss from 22 to 23 inches. We had trouble positioning our truss on the reaction bases due to its short length and an increase in thickness would increase the safety factor of the shear forces at each joint. Spending more time in construction would also lead to improvements in our design. More specifically we would account for the width of each member in the design process and determine angled cuts for the members going into each gusset plate. This would increase the accuracy of our actual model and lead to an overall improved truss design.

Page 10

truss final report

Documents