Info
Time/Location:
Exam Wed Oct 5 at 6:30pm
114 Richardson
Multiple Choice and workout
problems
Alternative Exam
(CVLB 421) 3:00pm and
(CVLB 418) 4:30pm (Same
Day) You arrange this with
me by Monday
Testable Topics (Problem
Solving:
Definition of Truss
Stable Truss
Force Components
Cartesian Coord translation
Moment (Torque)
Cables/pulleys
FBD
Reaction/Resultant Forces
Member Forces/Method of
Joints
Zero Force Members by
Inspection
Factor of Safety
ENGR 111 Track A
1
The Engineering Process:
Using quantification to make decisions!
Decision making from quantitative
data: Will the truss withstand the
necessary load?
Calculation of the Safe Loads in a Truss
TR-8
ENGR 111 Track A
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Learning Objectives
At the end of this module, you will be able to:
Identify the two common modes of structural failure.
Locate the links that are most likely to fail in a truss.
Apply a standard procedure using the measured
separation load on the links to calculate the allowable
safe operating load on a truss.
Apply Factor of Safety when necessary.
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Safe Load on a Truss
What is meant by the allowable design load on a truss?
Allowable loads (working loads) are the forces to which
the links can be safely subjected.
This is the calculated load that can be applied to the
truss which will just cause one of the links to fail.
For the trusses (bridges) that you build, it is the load at
which the link with the max tensile load separates from
the magnet.
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Quantification of Truss Failures
How do we quantify collapse and deformation?
The two principal modes of failure are
(1) Failure by separation (tensile failure)
(2) Failure by deflection (buckling)
Failure by separation is quantified by the minimum load
required to separate any portion of the truss from the
rest. Failure by separation is caused by tensile loads.
For the truss that we build, consider only tensile failure.
Failure by deflection is quantified by the maximum
amount of deflection that is allowed in any links of a
truss even though it is not yet separated from the other
links. Failure by deflection is caused by compressive
loads.
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Procedure for Identifying the Links at Risk
How do we find the links in a truss that are at risk of
failure?
1. Apply an external load of 1 lb or 1 N (or P) to the truss.
2. Find the internal forces in the links (numerical values or
in terms of P).
3. Identify links that support largest loads.
For our project, find the link(s) with the maximum
tension (positive) loads. These are the links that are
likely to fail in tension.
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Calculating Safe Load on a Truss
Max tensile load is the maximum load (force) that a
tension member can support. Also, called tensile
STRENGTH.
Max allowable external load is the safe load on a truss.
External Load
La rgest Tensile Force
=
Max Allowable External Load
Tensile STRENGTH
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Example
For the truss ABC
1. Find external reaction forces
and axial forces in all truss
members.
2. How much load can the truss
carry if the max tensile load
(strength) is 200 lb? Consider
only tensile loads.
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3
8
Example cont.
2. How much load can the truss carry if the max tensile
load (strength) is 200 lb?
Link
Axial Force
AB
1 lb (T)
BC
1.33 lb (T)
AC
1.67 lb (C)
LinkBCsupportsthelargesttensileforce.
LinkBCismorelikelytofailduetotension
thanlinkAB.
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Example cont.
2. How much load can the truss carry if
the max tensile load (tensile strength)
is 200 lb?
P = 1 lb
Internal force in BC = 1.33 lb
External Load
La rgest Tensile Force
=
Max Allowable External Load
Tensile STRENGTH
Pmax
200 lb
1 lb 1.33 lb
=
Pmax 200 lb
Pmax = 150.4 lb
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Example cont.
If external load is given as P, then we calculate all internal
forces in terms of P.
Link
P ( T)
7P
6
1.
P (T)
BC
1.33P (T)
AC
)
(C
Axial Force
AB
1.33P (T)
1.67P (C)
LinkBCsupportsthelargesttensileforce.
LinkBCismorelikelytofailduetotension
thanlinkAB.
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1.33P (T)
2. How much load can the truss carry
if the max tensile load (tensile
strength) is 200 lb?
Bar BC supports the largest
tensile force 1.33P.
If we apply maximum allowable
external load Pmax, bar BC will
have to support 1.33Pmax.
The max tensile load (tensile
strength) that bar BC can support
is 200 lb. If bar BC has to support
more than 200 lb it will fail.
ENGR 111 Track A
P ( T)
Example cont.
7P
6
1.
C)
(
1.33Pmax = 200 lb
Pmax = 150.4 lb
12
Load Scaling
Link Force
Axial Look at the internal forces
calculated in terms of P:
AB
P (T)
BC
1.33P (T)
AC
1.67P (C)
Load in the bars (axial force) is scaled with the applied
external Load!
Why?
All the equations of equilibrium were linear
equations for the forces!
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Factor of Safety
In project 1, we need to design a bridge that should be
able to carry a test load of at least 2 lb applied at (or
near) the center of top of the span. We probably would
like to make our bridges a little bit stronger than that.
Engineers account for various uncertainties by making
structures stronger than they really need to be. How?
They use a Factor of Safety in all analysis and design
calculations.
Failure Level
Factor of Safety =
Actual Level
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Factor of Safety for a Truss
A truss is only as strong as its weakest link.
For a truss, we can define the factor of safety as
Failure Level
Strength
Factor of Safety =
=
Actual Level
Internal Member Force
P (T)
1.33P (T)
P
67
1.
)
(C
Example: External load P = 100 lb. Tensile strength
of the bars is 200 lb. Compressive strength of the
bars is 180 kN.
FSAB = 200 lb / 100 lb = 2
FSBC = 200 lb / 133 lb = 1.5
FSAC = 180 lb / 167 lb = 1.08
Factor of Safety for the truss is 1.08
(the most conservative value)
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Example: for the truss below, internal forces in all truss
members have been calculated. Consider both modes
of failure and Determine factor of safety for the truss.
11.9kN
15kN
12kN
11.9kN
0.86kN
0.71kN
12kN
0.71kN
12kN
ENGR 111 Track A
15kN
12kN
16
AB, FH
15 kN (C)
Given: The tensile strength of the bars is 50
kN. The compressive strength of the bars is
45 kN.
BD, DF
11.9 kN (C)
Determine factor of safety for the truss.
BE, EF
0.71 kN (C)
Factor of Safety =
AC, CE,
EG, GH
12 kN (T)
DE
0.86 kN (T)
Member
BC, FG
Force
zero
Strength
Internal Member Force
The largest tensile force in the links is 12 kN.
FS = 50 kN / 12 kN = 4.17
The largest compressive force in the links is
15 kN.
FS = 45 kN / 15 kN = 3
Links AB and FH have the smallest factor of safety (compressive and
tensile loads). If AB and/or FH fail, the whole structure collapses.
Hence, the Factor of Safety for the truss is 3 we pick the
most conservative value.
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Exercise 1 - Individual
The max allowable tensile load for the links in the truss
shown is 75 N and the max allowable compressive load for
the links is 90 N. Find the max allowable value of P.
Determine the factor of safety for the structure if P = 100
N.
B
D
60
A
1m
60
C
2m
P/2
E
1m
H
P/2
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What if your truss is subjected to several
external forces? Example 1.
For the truss shown, lets assume we found that the
largest tensile force in the links is 3 kN.
The tensile strength is 4 kN.
How do we calculate the
max allowable load for the truss?
Total external load applied is 5 kN.
3 kN
2 kN
Make a proportion.
External load / Pmax = Largest Internal Force / STRENGTH
5 kN / Pmax = 3 kN / 4 kN
Solve for Pmax.
ENGR 111 Track A
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What if your truss is subjected to several
external forces? Example 2.
For the truss shown, we found that all blue links are in tension and
all grey links are in compression.
Internal force in each blue link is different:
P
T1 =0.55P, T2 = 1.2P, T3 = 0.8P
The largest tensile force is T2 = 1.2P
1.5P
How do we calculate the
max allowable load for the truss
T1
T2
T3
if we have to consider only tensile loads?
Total external load applied is P + 1.5P = 2.5P = Ptotal
Make a proportion:
Ptotal / Ptotalmax = 1.2P / STRENGTH
2.5P / Ptotalmax = 1.2P / STRENGTH
If tensile STRENGTH is given, you can calculate P totalmax.
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What if your truss is subjected to several
external forces? Example 3.
How do we calculate the max allowable load for the truss
if we have to consider all loads (tensile and compressive)?
Lets assume that we found that maximum tensile force in a truss member is
T = 1.2P and maximum compressive force in a truss member is C = 1.5P.
P
Given: max tensile STRENGTH is 500 lb
max compressive STRENGTH is 300 lb
1.5P
Total external load applied is P + 1.5P = 2.5P = Ptotal
Make a proportion:
2.5P / Ptotalmax = Largest Internal Force / STRENGTH
For the tensile member:
2.5*P / Ptotalmax = 1.2*P / 500 lb
Ptotalmax = 833.4 lb
For the compressive member:
2.5*P / Ptotalmax = 1.5*P / 300 lb
Ptotalmax = 500 lb
Pick the most conservative value for the max allowable load on the truss:
Ptotalmax = 500 lb.
ENGR 111 Track A
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