Transcript General Problem Solving Methodology

EGR 106 – Project Description (cont.)
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Analysis of truss structures
A simple example
Factor of Safety calculations
This week’s assignment
Truss Analysis
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A function, analyze_truss.m, has been developed
and is available for download from the course web
page:
[displacements,stress]=analyze_truss(joint_def,member_def)
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Input variables: joint_def (joint definitions) and
member_def (member definitions)
Output variables: disp (joint displacements) and
stress (member stresses)
Truss Analysis – A simple example
Cy = -50 N
Cx = 100 N
150 mm
300 mm
A simple example – joint definitions
Cy = -50 N
Cx = 100 N
150
mm
300
mm
joint_def=[0,
0,
50, 50,
150, 150,
250, 50,
300,
0,
200,
0,
100,
0,
1,
0,
0;
0,
0,
0;
0, 100, -50;
0,
0,
0;
3,
0,
0;
0,
0,
0;
0,
0,
0];
Column 1 – x coordinates of joints
Column 2 – y coordinates of joints
Column 3 – bcdof, where
bcdof=0 for unconstrained
bcdof=1 for X&Y–displacement=0
bcdof=2 for X-displacement=0
bcdof=3 for Y-displacement=0
Column 4 – force in X-direction
Column 5 – force in Y-direction
A simple example – member definitions
Cy = -50
Cx = 100
150
mm
300
mm
member_def=[1,
2,
3,
4,
5,
6,
7,
2,
7,
3,
6,
Column
Column
Column
Column
1
2
3
4
–
–
–
–
2,
3,
4,
5,
6,
7,
1,
7,
3,
6,
4,
10,
10,
10,
10,
10,
10,
10,
10,
10,
10,
10,
10;
10;
10;
10;
10;
10;
10;
10;
10;
10;
10];
first joint in member
second joint in member
width of member
thickness of member
Simple example – displacement results
Cy = -50 N
Cx = 100 N
Joint Displacments
displacments =
0.0000
X & Y displacements – joint 1 (A)
0.0000
0.2457
X & Y displacements – joint 2 (B)
-0.1750
0.6493
X & Y displacements – joint 3 (C)
-0.4371
0.2043
X & Y displacements – joint 4 (D)
-0.4578
0.4500
X & Y displacements – joint 5 (E)
-0.0000
0.3000
X & Y displacements – joint 6 (F)
-0.5536
0.1500
X & Y displacements – joint 7 (G)
-0.2707
}
}
150
mm
300
mm
}
}
}
}
}
Simple example – stress results
Member Stresses
Cy = -50
Cx = 100
150
mm
300
mm
stress =
0.3536
0.3536
-1.0607
-1.0607
0.7500
0.7500
0.7500
-0.0000
0.0000
0.0000
-0.0000
}
}
}
}
}
}
}
}
}
}
}
stress
stress
stress
stress
stress
stress
stress
stress
stress
stress
stress
in
in
in
in
in
in
in
in
in
in
in
member
member
member
member
member
member
member
member
member
member
member
AB
BC
CD
DE
EF
FG
AG
BG
CG
CF
DF
Strength Analysis
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The stress results give the force per unit area in
each member (N/mm2)
If the stress is positive, the member is being
stretched in tension
If the stress is negative, the member is being
compressed
These stresses need to be compared to material
strengths
To quantify the relation between stress and strength,
we introduce the Factor of Safety (FOS)
Factor of Safety – members in tension
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For members in tension (stress > 0), the factor of
safety is the ratio of the material’s tensile strength,
ST, to the stress in that member:
FOS = ST / stress
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For example, if stress = 100 N/mm2 and ST=200
N/mm2, then FOS =2 (i.e. the material is twice as
strong as needed)
Factor of Safety - members in compression
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For members in compression (stress < 0), two failure
modes must be considered:
–
–
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Failure may occur when the stress reaches the compressive
strength
Or, failure may occur due to buckling of long slender
members
Hence, we must compute two Factor of Safety values
The smaller of these values is the critical Factor of
Safety to be used in designing the truss
Factor of Safety –
members in compression (cont.)
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For members in compression (stress < 0), the first
factor of safety is the ratio of the material’s
compressive strength, Sc, to the stress in the
member (times negative one so that FOS > 0):
FOS1 = - SC / stress
Factor of Safety –
members in compression (cont.)
The second compression factor of safety is the ratio of
the material’s buckling strength to the stress in the
member. The buckling strength is inversely
proportional to the length of the member and is
defined as a buckling coefficient divided by the
member length, KB/L2. Hence, the second factor of
safety for compression members is given by:
FOS2 = - (KB/L2) / stress
Factor of Safety –
members in compression (cont.)
The critical failure mode is determined by
comparing FOS1 and FOS2. The smaller
value is the one to be used in designing the
truss
FOS = minimum (FOS1, FOS2)
Factor of Safety – zero stress members
If a member has a stress of zero, then its
Factor of Safety in infinite:
FOS = 
Overall Factor of Safety
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After determining the Factors of Safety for
each member, the overall Factor of Safety is
the smallest of the individual member
Factors of Safety.
If the Overall Factor of Safety is less than
one, than the truss is expected to fail when
the forces are applied.
The larger the Overall Factor of Safety, the
safer the design.
This Week’s Assignment*
For the truss shown, write a Matlab program that:
Determines the joint displacements
100
mm
Determines the member stresses
Computes a factor of safety for each member
Computes an overall factor of safety
Displays the joints and elements in a figure window
Displays the deformed truss in a 2nd figure window
* See assignment handout for more details
100
mm
100
mm
100
mm