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Week ENGR 5 Wednesday Class 10 T. Pollock Fall, 2009
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Homework4
DuenextMONDAY!
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Unit Loads
Set the largest load to 1 unit and scale other loads accordingly. In this example, the magnitude of the external force applied at C is known to be 5/7ths of the magnitude of the external force applied at B.
First: Sketch the structure FBD. Second: Calculate the reactions at the supports.
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Now use the Method of Joints to solve for the internal forces in each element.
Start with a joint which has no more than two unknown magnitudes. Draw the joint FBD. Calculate angles. Choose the best order of solution. Solve for the unknown magnitudes.
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1 .0 0 0 lb f B 1 .3 4 7 lb f (C ) 1 .3 4 7 lb f (C ) 1 .0 0 0 lb f (C ) 0 .8 0 9 lb f A 0 .3 5 7 lb f 1 .3 4 7 lb f (C ) 1 .4 3 4 lb f (T ) C 1 .4 3 4 lb f (T ) 1 .0 7 7 lb f (T ) 1 .0 0 0 lb f (C )
0 .7 1 4 lb f
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Thecompletesolution:
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FactorofSafety
Afactorofsafetyisanumberthat: reducesmaximumallowableloadstovaluesthatguaranteealowfailurerate isgenerallyagreedtoby industry government professionalsocieties Inotherwords,weusuallydontwantthestructuretobeloadeduptothevery limitoffailure. Factorsofsafetyvarydependingonmanyfactors.Forexample: buckleonaparachuteharness: 15 1.5 wingofacargoaircraft: 1.2 wingofafighteraircraft: 1.1 airtoairmissile:
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The complete solution:
Now suppose that the maximum tensile force that each member can withstand without failing is 450 lbf. What applied load set would produce this maximum with a factor of safety of 1.5? (Remember that the external load at C is 5/7ths of the external load at B.) Factorofsafety
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MAXIMUM TENSILE LOADING CONDITION
209 lbf
B
282 lbf (C)
A
282 lbf (C) 209 lbf (C)
74.7 lbf
300 lbf (T) C 169 lbf
225 lbf (T) 149 lbf
D
169 lbf
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Beginning again:
Now suppose that the maximum compressive force that each member can withstand without failing is 330 lbf. What applied load set would produce this maximum with a factor of safety of 1.5? (Remember that the external load at C is 5/7ths of the external load at B.)
CORRECTED
compressive
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MAXIMUM COMPRESSIVE LOADING CONDITION
163 lbf
B
220 lbf (C)
A
220 lbf (C) 163 lbf (C)
58.3 lbf
234 lbf (T) C 132 lbf
176 lbf (T) 116 lbf
D
132 lbf
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WHICH CASE IS THE MAXIMUM ALLOWABLE LOAD? 282 lbf (C)
A
209 lbf
B
282 lbf (C) TENSION 209 lbf (C)
74.7 lbf
300 lbf (T) C 169 lbf
225 lbf (T) 149 lbf
D
169 lbf 163 lbf
B
THE LESSER ONE!
COMPRESSION 58.3 lbf
220 lbf (C)
A
220 lbf (C) 163 lbf (C)
234 lbf (T) C 132 lbf
176 lbf (T) 116 lbf
D
132 lbf
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TensileFailureLoad~Magnestix
Team 1 2 3 4 7 8 9 13 14 15 16 17 18 20 21 column av. test art. av. test art. SD global av. global SD L(a) 5.66 6.87 7.62 6.37 5.02 5.62 6.37 6.24 4.50 6.15 6.02 4.83 6.33 7.39 5.63 6.04 L(b) 6.01 6.74 8.04 6.37 5.18 5.84 6.20 6.31 4.10 6.26 5.76 2.24 6.39 7.69 5.81 5.93 5.99 1.13 SL(a) 5.20 6.07 5.55 5.64 5.19 5.23 5.47 5.62 4.45 6.61 5.10 5.35 5.85 8.12 4.96 5.63 SL(b) 5.87 4.34 5.27 5.30 4.11 5.40 5.03 5.00 4.30 6.62 5.25 5.66 5.64 7.39 4.96 5.34 5.49 0.85 SS(a) 4.89 4.99 4.85 4.92 4.03 5.15 4.70 4.90 3.04 6.24 5.92 4.51 5.03 7.52 4.81 5.03 SS(b) 4.65 5.60 5.20 5.34 4.05 4.69 4.77 4.94 3.20 6.09 5.52 4.36 5.15 7.14 4.60 5.02 5.03 0.94 Avg. 5.38 5.77 6.09 5.66 4.60 5.32 5.42 5.50 3.93 6.33 5.60 4.49 5.73 7.54 5.13 SD 0.55 0.99 1.37 0.60 0.59 0.40 0.72 0.65 0.65 0.23 0.37 1.21 0.57 0.34 0.48
SD:standarddeviation
largevalue=muchscatter smallvalue=littlerscatter
5.50 1.047749 FS 1.05
units=Newtons
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FactorofSafetyfortheMagnets
Thescariestminimum: =
Average Average SD 5.50 5.50 1.05 Average Average 2 SD 5.50 5.50 2 1.05
FS =
FS = 1.236
Whichwouldyou recommendfora pedestrianwalkway? Acomfortablemiddle: FS =
FS =
FS = 1.618
Conservative:
FS = FS =
Average Average 3 SD 5.50 5.50 3 1.05
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FS = 2.34
TestArticleForces ReducedbyaConservativeFS
5.99 N = 2.56 N 2.34 5.49 N = 2.346 N 2.34 5.03 N = 2.15 N 2.34
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Truss Performance Evaluation
Geometrical Specs Met? Predicted Failure Load Actual Load to failure Material Cost Construction Cost
The Evaluation scheme is as follows: 1. Stable truss meeting geometrical constraints 2. Truss capable of carrying 2 lbs 3. Failure load has been predicted 4. Actual load approximately matches failure load 5. Truss is not too strong (20 pts) (20 pts) (20 pts) (20 pts) (20 pts)
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1. Stable 2D truss meeting geometrical constraints
Recall that a 2D truss has vertical side walls, each of which carries half of the vertical load. Also, there are no horizontal force components. Stability criterion: L = 2J - 3 L: total number of links J: total number of joints
What is a link?
onelink onelink twolinks+onejoint
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1. Stable 3D truss meeting geometrical constraints
We have to support the truss sides. We seek a solution which has the minimum number of links required for structural stability. There are several kinds of links that we can add:
truss sides transverse links face diagonals body diagonals We do not need to add every possible stabilizing link. Stability criterion: L = 3J - 6 L: total number of links J: total number of joints
Because the loads are vertical, all of the transverse and diagonal links are zero force members. They fix the positions of the joints in space, but carry no internal forces.
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2. Truss capable of carrying 2 lbs.
Is your truss design too strong and therefore too costly?
3. Failure load has been predicted.
STEP 0: Sketch the truss STEP 1: Reactions draw truss FBD sum forces sum moments STEP 2: Internal Forces draw joint FBDs sum forces STEP 3: Identify links with the highest tensile and compressive loads STEP 4: Scale the loads so that the highest internal tensile force is equal to the failure load that was measured in the Physics Lab.
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4. Actual load matches failure load
STEP 0: Sketch the truss and apply a unit load STEP 1: Reactions draw the truss FBDs sum the forces sum the moments STEP 2: Internal Forces draw the joint FBDs sum the forces STEP 3: Identify links with the highest tensile and compressive loads STEP 4: Scale all of the loads so that the highest internal tensile force is equal to the failure load that was measured in class.
Including P Compare the calculated and experimental values of P.
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5. Truss is not too strong.
Holds more than 5 lbf (max load)/(weight) > some number (next lecture)
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Loading the truss
Actual 3D Model
Analytical 2D Model
We must now examine the specifications and interpret the word near because this word was not defined quantitatively. The problem is that joints near the abutments are not to be loaded, and the load was to be applied near the center. Standards of ethical engineering practice require the engineer to advise the client of the implications of this wording.
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Your Loading Conditions
P/10 P/10 P/10 P/10 P/10
P/4
P/4
If you have an arch or a peak, all of the load will be applied at the highest joint.
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TRUSS EVALUATION SHEET
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