No Slide Title

Transcript No Slide Title

Analysis of Statically Determinate
Trusses








Common Types of Trusses
Classification of Coplanar Trusses
The Method of Joints
Zero-Force Members
The Method of Sections
Compound Trusses
Complex Trusses
Space Trusses
1
Common Types of Trusses
gusset plate
roof
• Roof Trusses
purlins
top cord
knee brace
bottom
cord gusset plate
span
bay
2
Howe truss
Pratt truss
howe truss
Warren truss
saw-tooth truss
Fink truss
three-hinged arch
3
• Bridge Trusses
top cord
sway
bracing
top lateral
bracing
portal
bracing
stringers
portal
end post
deck
bottom cord
panel
floor beam
4
trough Pratt truss
Warren truss
deck Pratt truss
parker truss
(pratt truss with curved chord)
Howe truss
baltimore truss
K truss
5
Assumptions for Design
1. All members are connected at both ends by smooth frictionless pins.
2. All loads are applied at joints (member weight is negligible).
Notes: Centroids of all joint members coincide at the joint.
All members are straight.
All load conditions satisfy Hooke’s law.
6
Classification of Coplanar Trusses
• Simple Trusses
C
P
A
P
B
a
b
D
C
A
P
D
B
A
C
B
new members
d (new joint)
c
7
• Compound Trusses
simple truss
simple truss
simple truss
Type 1
simple truss
Type 2
secondary
simple truss
secondary
simple truss
secondary
simple truss
secondary
simple truss
main simple truss
Type 3
8
• Complex Trusses
• Determinacy
b + r = 2j
b + r > 2j
statically determinate
statically indeterminate
In particular, the degree of indeterminacy is specified by the difference in the
numbers (b + r) - 2j.
9
• Stability
b + r < 2j
b + r >2j
unstable
unstable if truss support reactions are concurrent or parallel
or if some of the components of the truss form a collapsible
mechanism
External Unstable
Unstable-parallel reactions
Unstable-concurrent reactions
10
Internal Unstable
F
C
O
8 + 3 = 11
< 2(6)
D
A
B
E
AD, BE, and CF are concurrent at point O
11
Example 3-1
Classify each of the trusses in the figure below as stable, unstable, statically
determinate, or statically indeterminate. The trusses are subjected to arbitrary
external loadings that are assumed to be known and can act anywhere on the
trusses.
12
SOLUTION
Externally stable, since the reactions are not concurrent or parallel. Since b = 19,
r = 3, j = 11, then b + r = 2j or 22 = 22. Therefore, the truss is statically determinate.
By inspection the truss is internally stable.
Externally stable. Since b = 15, r = 4, j = 9, then b + r > 2j or 19 > 18. The truss
is statically indeterminate to the first degree. By inspection the truss is internally
stable.
13
Externally stable. Since b = 9, r = 3, j = 6, then b + r = 2j or 12 = 12. The truss is
statically determinate. By inspection the truss is internally stable.
Externally stable. Since b = 12, r = 3, j = 8, then b + r < 2j or 15 < 16. The truss
is internally unstable.
14
The Method of Joints
B
500 N
2m
Ax = 500 N
45o
A
2m
Ay = 500 N
500 N
45o
FBA
Cy = 500 N
+ SF = 0:
x
Joint B
y
B
C
FBC
x
500 - FBCsin45o = 0
FBC = 707.11 N (C)
+
SFy = 0:
- FBA + FBCcos45o = 0
FBA = 500 N (T)
15
B
500 N
2m
Ax = 500 N
45o
A
Ay = 500 N
2m
C
Cy = 500 N
Joint A
+ SF = 0:
x
500 N
500 - FAC = 0
500 N
FAC
FAC = 500 N (T)
500 N
16
Zero-Force Members
P
B
C
D
E
A
Dx
Dy
Ey
C
FCB
+ SF = 0:
x
FCB = 0
SFy = 0:
FCD = 0
+
FCD
FAB
q
A
FAE
+
SFy = 0:
+ SF = 0:
x
FABsinq = 0,
FAB = 0
FAE + 0 = 0, FAE = 0
17
Example 3-4
Using the method of joints, indicate all the members of the truss shown in the
figure below that have zero force.
A
B
C
H
G
E
F
D
P
18
SOLUTION
A
Ax
B
Ax
C
H
Gx
F
G
E
0
0
D
0
P
FDC
FDE
Joint D
y
+
q
SFy = 0:
FDCsinq = 0,
FDC = 0
x
D
+ SF = 0:
x
FDE + 0 = 0, FDE = 0
FEC
E
FEF
0
Joint E
+ SF = 0:
x
FEF = 0
P
19
A
Ax
B
0
Ax
C
0
H
Gx
F
G
0
D
0
P
y
Joint H
FHB
FHA
E
0
+ SF = 0:
y
FHB = 0
H
FHF
x
FGA
Gx
G
Joint G
FGF
+
SFy = 0:
FGA = 0
20
The Method of Sections
a
B
Dy
C
D
Dx
2m
A
100 N
G
a
2m
2m
E
F
Ex
2m
+ SMG = 0:
FBC
B
FGC
45o
A
100 N
G
2m
100(2) - FBC(2) = 0
FBC = 100 N (T)
C
FGF
+
SFy = 0:
-100 + FGCsin45o = 0
FGC = 141.42 N (T)
+ SMC = 0:
100(4) - FGF(2) = 0
FGF = 200 N (C)
21
Example 3-6
Determine the force in members GF and GD of the truss shown in the figure
below. State whether the members are in tension or compression. The reactions at
the supports have been calculated.
G
H
Ax = 0
F
A
E
B
Ay = 9 kN 6 kN
3m
C
D
8 kN
2 kN
3m
3m
3m
4.5 m
Ey = 7 kN
3m
22
G
SOLUTION
a
H
F
A
Ax = 0
E
B
Ay = 9 kN 6 kN
3m
C
a D
8 kN
3m
3m
4.5 m
2 kN
3m
Ey = 7 kN
3m
Section a-a
FFG
26.6o
FDG
FDC
+ SMD = 0:
F
56.3o
26.6o
D
E
2 kN
3m
O
Ey = 7 kN
3m
FFGsin26.6o(3.6) + 7(3) = 0,
FFG = -17.83 kN (C)
+ SMO = 0:
- 7(3) + 2(6) + FDGsin56.3o(6) = 0,
FDG = 1.80 kN (C)
23
Example 3-7
Determine the force in members BC and MC of the K-truss shown in the figure
below. State whether the members are in tension or compression. The reactions at
the supports have been calculated.
L
0
A
12.9 kN
M
B
K
J
N
C
I
P
O
D
H
E
F
G
3m
3m
5.34 kN 6.67 kN 8 kN
4.6 m 4.6 m
7.11 kN
4.6 m 4.6 m 4.6 m 4.6 m
24
a
SOLUTION
0
L
M
A
B
12.9 kN
K
J
N
a C
I
P
O
D
H
E
F
G
3m
3m
5.34 kN 6.67 kN 8 kN
4.6 m 4.6 m
7.11 kN
4.6 m 4.6 m 4.6 m 4.6 m
Section a-a
L
FLK
6m
FLM
FBM
A
B
12.9 kN
FBC
+ SML = 0:
FBC(6) - 12.9(4.6) = 0,
FBC = 9.89 kN (T)
5.34 kN
4.6 m
25
b
L
0
M
A
K
J
N
I
D
E
C
b
5.34 kN 6.67 kN 8 kN
FKM
FCM
9.89 kN
K
J
N
33.1o
C
G
F
7.11 kN
4.6 m 4.6 m 4.6 m 4.6 m
4.6 m 4.6 m
FKL
3m
3m
P
O
B
12.9 kN
H
I
3m
P
O
D
H
E
F
G
3m
6.67 kN 8 kN
7.11 kN
4.6 m 4.6 m 4.6 m 4.6 m
+ SMK = 0:
-FCMcos33.1o(6) - 9.89(6) - 8(4.6) + 7.11(18.4) = 0
FCM = 6.90 kN (T)
26
Compound Trusses
Procedure for Analysis
Step 1. Identify the simple trusses
Step 2. Obtain external loading
Step 3. Solve for simple trusses separately
27
Example 3-8
Indicate how to analyze the compound truss shown in the figure below. The
reactions at the supports have been calculated.
4m
H
G
J
I
Ay = 0
2m
F
K
A
B
4 kN
Ay = 5 kN
2m
C
2m
E
D
2 kN
4 kN
2m
2m
Ey = 5 kN
2m
28
4m
SOLUTION
a
H
G
J
I
2m
F
K
A
Ay = 0
B
4 kN
Ay = 5 kN
2m
C
a
2m
E
D
2 kN
4 kN
2m
2m
Ey = 5 kN
2m
FHG
H
+ SMC = 0:
J
I
4 sin60o m
FJC
A
B
4 kN
Ay = 5 kN
2m
FBC
2m
-5(4) + 4(2) + FHG(4sin60o) = 0
FHG = 3.46 kN (C)
C
29
4m
H
2m
G
J
I
F
K
A
Ay = 0
B
4 kN
Ay = 5 kN
2m
H
C
2m
E
D
2 kN
4 kN
2m
2m
Ey = 5 kN
2m
3.46 kN
+ SMA = 0:
4 sin60o m
J
I
FCK
60o
A
C
B
4 kN
Ay = 5 kN
2m
2 kN
2m
FCD
3.46(4sin60o) + FCKsin60o(4) - 4(2) - 2(4) = 0
FCK = 1.16 kN (T)
+ SF = 0:
x
-3.46 + 1.16cos60o + FCD = 0
FCK = 2.88 kN (T)
30
4m
H
G
J
I
F
K
A
Ay = 0
B
4 kN
Ay = 5 kN
2m
H
4 sin60o m
2m
C
2m
60o
A
4 kN
Ay = 5 kN
2m
C
2 kN
2m
2m
Joint A : Determine FAB and FAI
FCK = 1.16
B
Ey = 5 kN
Using the method of joints.
J
I
4 kN
2m
3.46 kN
E
D
2 kN
2m
2.88 kN
Joint H : Determine FHI and FHJ
Joint I : Determine FIJ and FIB
Joint B : Determine FBC and FBJ
Joint J : Determine FJC
31
Example 3-9
Indicate how to analyze the compound truss shown in the fugure below. The
reactions at the supports have been calculated.
C
H
D
Ax = 0 kN
45o
A
Ay = 13.3 kN
1.8 m
3.7 m
G
45o
B
13.3 kN
1.8 m
F
45o
E
1.8 m
1.8 m
13.3 kN Fy = 13.3 kN
1.8 m
32
C
SOLUTION
a
H
D
Ax = 0 kN
a
45o
A
Ay = 13.3 kN
1.8 m
45o
B
13.3 kN
1.8 m
E
1.8 m
45o
A
Ay = 13.3 kN
1.8 m
45o
B
13.3 kN
1.8 m
1.8 m
13.3 kN Fy = 13.3 kN
1.8 m
+ SMB = 0:
45o
D
F
45o
C
3.7 m
3.7 m
G
FCE
FBH
FDG
1.8 sin 45o m
-13.3(1.8) - FDG(1.8sin45o) - FCEcos45o(3.7)
- FCEsin45o(1.8) = 0 -----(1)
+
SFy = 0:
13.3 - 13.3 + FBHsin45o - FCEsin45o = 0
FBH = FCE-----(2)
+ SF = 0:
x
FBHcos45o + FDG + FCEcos45o = 0
-----(3)
33
C
a
H
D
Ax = 0 kN
a
45o
A
Ay = 13.3 kN
1.8 m
45o
B
13.3 kN
1.8 m
C
45o
3.7 m
D
45o
A
Ay = 13.3 kN
1.8 m
45o
B
13.3 kN
1.8 m
3.7 m
G
F
45o
E
1.8 m
1.8 m
13.3 kN Fy = 13.3 kN
1.8 m
From eq.(1)-(3): FBH = FCE = -11.45 kN (C)
FDG = 16.19 kN (T)
FCE
FBH
FDG
1.8 sin 45o m
Analysis of each connected simple truss
can now be performed using the method of
joints.
Joint A : Determine FAB and FAD
Joint D : Determine FDC and FDB
Joint C : Determine FCB
34
Example 3-10
Indicate how to analyze the symmetrical compound truss shown in the figure
below. The reactions at the supports have been calculated.
E
3 kN
5o
3 kN
F
Ax = 0 kN
A
D
5o
5o
G
45o
H
5o
45o
C
B
Fy = 4.62 kN
5 kN
Ay = 4.62 kN
6m
6m
35
E
3 kN
5o
3 kN
F
Ax = 0 kN
G
45o
A
D
5o
5o
H
5o
45o
C
B
Fy = 4.62 kN
5 kN
Ay = 4.62 kN
6m
6m
FEC
E
3 kN
E
1.5 kN
3 kN
1.5 kN
F
G
D
H
C
A
FAE
FAE
1.5 kN
1.5 kN
FEC
36
E
3 kN
5o
3 kN
F
Ax = 0 kN
D
5o
5o
G
45o
A
H
5o
45o
B
6m
E
45o
Fy = 4.62 kN
5 kN
Ay = 4.62 kN
1.5 kN
C
6m
1.5 kN
1.5 kN
45o
45o
45o
A
1.5 kN
A
1.5 kN
45o
45o
B
4.62 kN
5 kN
FAE
C
4.62 kN
+
SFy = 0:
FAB
4.62 kN
4.62 - 1.5sin45o - FAEsin45o = 0
FAE = 5.03 kN (C)
+ SF = 0:
x
1.5cos45o - 5.03cos45o + FAB = 0
FAB = 2.50 kN (T)
37
E
Complex Trusses
P
F
3 9 2(6)
r + b = 2j,
D
FAD
A
• Determinate
• Stable
C
B
E
E
F
F
D
C
B
Xx
FÉC + x fÉC = 0
x=
Fi = Fí + x fí
D
1
+
A
P
fÉC
=
P
FÉC
FÉC = F
AD
fÉC
A
1
C
B
38
Example 3-11
Determine the force in each member of the complex truss shown in the figure
below. Assume joints B, F, and D are on the same horizontal line. State whether
the members are in tension or compression.
C
22.24 kN
1.22 m
B
45o
F
45o
D
0.91 m
A
E
2.44 m
39
SOLUTION
C
22.24 kN
1.22 m
45o
B
F
45o
F´BD+ x f´BD = 0
F´BD = F
x=
CF
f´BD
D
0.91 m
A
E
Fi = Fí + x fí
2.44 m
B
A
45o
F
22.24 kN
45o
D
E
C
=
C
+
B
x
A
45o
1 kN
F
45o
D
E
40
C
B
11.12
22.24 kN
F
D
0
A
0
0
23.72
-19.41
+ x
1 kN
1.170
F
B
-0.25
A
E
-0.715
-0.715
D
-0.25
E
=
-11.10
C
F´DB + x f´DB = 0
-11.10 + x(1.17) = 0
C
x = 9.487
9.487
F
B
8.75
A
22.24 kN
16.94
-6.78
D
-21.78
E
41
Space Trusses
• Determinacy and Stability
P
b + r < 3j
unstable truss
b + r = 3j
statically determinate-check stability
b + r > 3j statically determinate-check stability
42
z
z
y
Fy
short link
x
y
x
z
z
y
y
x
roller
x
z
z
slotted roller
y constrained
in a cylinder
x
y
Fx
x
Fz
z
z
y
x
Fz
ball-and -socket
Fy
Fx
x
Fz
y
43
• x, y, z, Force Components.
z
l  x2  y2  z 2
Fz
F
B
l
Fy
Fx z
A
x
Fx  F ( )
l
y
x
y
y
Fy  F ( )
l
z
Fz  F ( )
l
F  Fx  Fy  Fz
2
x
2
2
• Zero-Force Members
z
Case 1
Case 2
z
FC = 0
FD
B
D
x
FD
C
FC
A
B
FD = 0
y
B
x
A
FA = 0
FB
SFz = 0 ,
y
FA
D
FB
SFz = 0 ,
FB = 0
SFy = 0 ,
FD = 0
44
Example 3-12
Determine the force in each member of the space truss shown in the figure below.
The truss is supported by a ball-and-socket joint at A, a slotted roller joint at B,
and a cable at C.
C
z
B
D
2.67 kN
2.44 m
1.22 m
E
A
1.22 m
2.44 m
y
x
45
Cy
SOLUTION
C
z
B
By
D
Bx
2.67 kN
2.44 m Ay
Ax
E
Az
1.22 m
y
2.44 m
x
The truss is statically determinate since b + r = 3j or 9 + 6 = 3(5)
SMy = 0:
-2.67(1.22) + Bx(2.44) = 0
SMz = 0:
Cy = 0 kN
SMx = 0:
By(2.44) - 2.67(2.44) = 0
SFx = 0:
-Ax + 1.34 = 0
SFy = 0:
Ay - 2.67 = 0
Ay = 2.67 kN
SFz = 0:
Az - 2.67 = 0
Az = 2.67 kN
Bx = 1.34 kN
By = 2.67 kN
Ax = 1.34 kN
46
0
0
z
C
z
0
B
By
0
0
0
FDC
D
Bx
0
0
2.73 m
2.44 m Ay
Ax
E
Az
y
Joint C.
C
y
0
x
0
0
FCE
y
FDE
1.22 m
z
0
0
x
2.44 m
x
D
2.67 kN
Joint D.
SFZ= 0:
FDC = 0
SFY = 0:
FDE = 0
SFx = 0:
FDA = 0
SFy = 0:
FCE = 0
SFz = 0:
FCA = 0
SFx = 0:
FCB = 0
47
0
0
z
C
z
0
B
By
0
0
0
B
2.67 kN
D
Bx
0
2.73 m
2.44 m Ay
Ax
x
0
2.67 kN
1.34 kN
FBA
FBC
FBE
y
x
E
1.22 m
Az
y
2.44 m
Joint B.
SFy = 0:
- 2.67 + FBE(2.44/3.66) = 0
SFx = 0:
1.34 - FBC -4(1.22/3.66) = 0
SFz = 0:
FBA - 4(2.44/3.66) = 0
FBE = 4 kN (T)
FBC = 0
FBA = 2.67 kN (C)
48
0
0
z
C
0
B
By
0
0
z
FAC
0
2.67 kN
2.67 kN
A
D
Bx
0
2.73 m
2.44 m Ay
Ax
x
0
2.67 kN
1.34 kN
E
45o
2
1
FAE
y
2.67 kN
x
1.22 m
Az
FAD
y
2.44 m
Joint A.
SFz = 0:
SFy = 0:
SFz = 0:
2.67 - 2.67 - FACsin45o = 0
2
- FAE(
) + 2.67 = 0
5
- 1.34 + FAD + 2.99(
FAC = 0, OK
FAE = 2.99 kN (C)
1
)=0
5
FAD = 0, OK
49