isostatica IV

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Análise de Treliças pelo Método 
das Seções 
http://arch.umd.edu/Tech/Tech_II/Lectures 
Truss Analysis – Method of Sections 
4 @ 12’ = 48’ 
6’ 
6K 6K 6K 3K 3K 
A 
B 
C D E 
F G H I J 
12K 12K 
Truss Analysis – Method of Sections 
• Method of joints uses ∑FH = 0, and ∑Fv = 0 
• Method of sections adds ∑M = 0 
4 @ 12’ = 48’ 
6’ 
6K 6K 6K 3K 3K 
A 
B 
C D E 
F G H I J 
Step #1: Find Reactions (always 1st!) 
∑FV=2(3K)+3(6K)=24K 
By symmetry, RL=RR=24K/2=12K 
12K 12K 
Step #2: 
Cut section through truss where looking for forces 
Truss Analysis – Method of Sections 
Step #3: 
Replace internal forces at cut with external “loads” 
on the free body diagram 
12’ 
6K 6K 3K 
A 
C D E 
F G H I 
12K 12’ 
HI 
EH 
DE 
Step #4: 
Sum moments about various points to solve for forces 
— Need to intelligently chose points to minimize unknowns 
— Process will yield a direct solution for member forces 
— Can also use ∑FX=0 and ∑FY=0 
Truss Analysis – Method of Sections 
Step #4: 
Choose joint ‘E’ as moment center, since two members 
pass through this point, ∴drop out of equation 
∑ME=0: 
-12K(36’)+3K(36’)+6K(24’)+6K(12’)+HI(6’)=0 
HI=18K 
+ 
Next, chose point ‘H’ since HI & EH drop out 
∑MH=0: 
-12K(24’)+3K(24’)+6K(12’)+6K(0’)+DE(6’)=0 
DE=24K 
+ 
12K 12’ 
6K 6K 3K 
A 
C D E 
F G H 
I 
12’ 
HI 
EH 
DE 
12’ 
6’ 
Lastly, select joint “D” as moment center 
– Force in HI already computed, DE drops out 
– “Trick” is to intelligently relocate components of EH 
 so that one drops out 
EHV 
EHH 
EHV 
EHH 
Truss Analysis – Method of Sections 
∑MD=0: 
-12K(24’)+3K(24’)+6K(12’)+6K(0’)+18K(6’)+ EHV(12’)=0 
 EHV =3K 
+ 
12’ 
6K 6K 3K 
A 
C D 
E 
F G H 
I 
12K 12’ 
HI=18K 
DE = 24K 
12’ 
6’ EHV 
– Relocating components to joint E makes EHH drop out, 
 solve directly for EHV 
Truss Analysis – Method of Sections 
Can also find (or verify) vertical force EHV by summing forces vertically! 
∑FV=0: 
12K - 3K - 6K - 6K + EHV = 0 
 EHV = 3K 
Horizontal component EHH can be found by slope of member, by 
summing forces horizontally, or by choosing a new moment center! 
2 
1 EHv = EHH 
1 2 
EHH = 2EHv 
EHH = 2(3K)=6K 
∑FH=0: 
18K - 24K - EHH = 0 
EHH = 6K 
12’ 
6K 6K 3K 
A 
C D E 
F G H 
I 
12K 12’ 
HI=18K 
DE=24K 
12’ 
6’ EHV=3
K
 
EHH=6K 
3o Trabalho Prático: 
Exercício 2: 
4K 
4K 4K 
2K 2K 
A B 
F 
E 
D C 
G 
16’ 16’ 
10’ 
48’ 
12’ 
10’ 
16’ 8k 8k 
Remember that you can cut a section 
anywhere in a truss to obtain forces in 
the members that are cut through 
Truss Analysis – Method of Sections 
4K 
4K 4K 
2K 2K 
A B 
F 
E 
D C 
G 
16’ 16’ 
10’ 
48’ 
12’ 
10’ 
16’ 8k 8k 
1) Cut section through members that you want to find forces of 
Truss Analysis – Method of Sections 
4K 
2K 
A 
E 
F 
C 
8k 
EF 
EC 
AC=7.2K (From original joint equilibrium 
calcs at joint “A”) 
ECV 
ECH 
EFH 
EFV 
2) Apply forces of those members as external “loads” to truss 
section 
3) Resolve those forces into their components 
4) Select moment center that will 
make all but one unknown go 
away. Remember that any 
component can be translated 
along its line of action and there 
will be no change in moment 
about that point 
Truss Analysis – Method of Sections 
EFH 
EFV 
Chose joint “C” to solve for EFV, 
Remembering that EFH will drop 
out of the calculation when the 
force pair is translated to joint “A” 
EFH 
EFV 
4K 
2K 
A 
E 
F 
C 
8k 
AC=7.2K 
ECV 
ECH 
16’ 
12’ 
Even though the force EF is not 
located at joint “A”, we can do this 
because the Varignon’s Theorem 
and the principle of transmissibility 
say that it can be relocated 
anywhere along its line of action 
with no net change 
Truss Analysis – Method of Sections 
Think in terms of lines of actions of 
forces and for now just forget 
about the truss itself! 
It’s only those rotations about 
the point “C” that matter!! 
8k 
EFV 
4K 
2K 
A 
E 
C AC=7.2K 
16’ 
12’ 4’ 
EFH 
∑MC=0: 
-8K(16’)+2K(16’)+4K(4’)+EFV(16’)=0 
 EFV =5K 
+ 
Truss Analysis – Method of Sections 
∑MC=0: 
-8K(16’)+2K(16’)+4K(4’)+EFV(16’)=0 
 EFV =5K 
+ 
∑FV=0: 
8K - 2K - 4K - EFV +ECV = 0 
ECV = 3K 
5K 
Now let’s look at it with the members 
back in place and solve for EC: 
EFH = 6 (5K) = 6K 
20’ 
24’ 6 
5 
EFH = EFV 
 6 5 
EF = 
5 
2 2 k5 6 7.81+ =
=5K 
ECV 
ECH 
EFH = 6K 
EFV 
8k 
EFV 
4K 
2K 
A 
E 
C AC=7.2K 
16’ 
12’ 4’ 
EFH 
∑FH=0: 
7.2K - EFH = 0, ECH = 1.2K 
6K 
∴EC = 
2 2 k3 1.2 3.2+ =
Truss Analysis – Method of Sections 
4K 
2K 
A 
E 
F 
D 
8k 
CD 
CFV 
CFH 
16’ 
12’ 
EFH=6K 
EFV=5K 
C 
16’ 
Now, on your own or with a 
partner, solve for the forces in 
members CF and CD by 
cutting a section through 
those members 
Remember that you can cut a section 
anywhere in a truss to obtain forces in 
the members that are cut through 
Answers: CFv=3K, CFH=1.2K 
 CD=4.8K 
	Análise de Treliças pelo Método das Seções
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