truss
Content
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
• Roof Trusses top cord knee brace bottom cord gusset plate roof purlins
bay, 5-6 m typical span, 18 - 30 m, typical
gusset plate
2
Howe truss 18 - 30 m
Pratt truss 18 - 30 m
Howe truss flat roof
Warren truss flat roof
saw-tooth truss skylight
Fink truss > 30 m
three-hinged arch hangar, gymnasium
3
• Bridge Trusses sway bracing top lateral bracing deck panel floor beam bottom cord top cord
portal bracing stringers portal end post
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
P C D C P C
P
D A
A
B
A
B
B
a
new members d (new joint)
b
c
7
• Compound Trusses
simple truss
simple truss
simple truss
simple truss
Type 1
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 A 45o 2m C Cy = 500 N
Ax = 500 N
Ay = 500 N
Joint B y B 45o FBA 500 N FBC x
+ "F = 0: x 500 - FBCsin45o = 0 FBC = 707 N (C) + "Fy = 0:
- FBA + FBCcos45o = 0 FBA = 500 N (T)
15
B
500 N
2m A 45o 2m C Cy = 500 N
Ax = 500 N
Ay = 500 N
Joint A 500 N + "F = 0: x 500 - FAC = 0 FAC = 500 N (T)
500 N 500 N
FAC
16
Zero-Force Members
P 0 0 B 0 C 0 E D
A
Dx
Ey FCB C + "F = 0: F = 0 x CB + FCD FAB A # FAE + "Fy = 0: FCD = 0
Dy
"Fy = 0: FABsin# = 0,
FAB = 0 FAE = 0
+ "F = 0: F + 0 = 0, x AE
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 D
G
F P
E
18
SOLUTION
Ax Ax
A
B C H
0 D 0
Gx G y
#
F Joint D + "Fy = 0:
0 P
E
FDC FDE
FDCsin# = 0, FDE + 0 = 0,
FDC = 0 FDE = 0
D FEC
x
+ "F = 0: x
FEF P
E
0
Joint E + "F = 0: x
FEF = 0
19
Ax Ax Gx
A 0 0
B C H F 0 P Joint H + "F = 0: y FHB = 0 E
0 D 0
G y FHA H FHF FGA Gx FGF x Joint G + FHB
G
"Fy = 0:
FGA = 0
20
Example 3-5 • Determine all the member forces • Identify zero-force members 5@3m = 15 m 1 kN K 5m A B 2 kN J I H D 2 kN E G F
C 2 kN
21
SOLUTION Use method of joints Kx 5m K 0 Ky Ax A Ay B 2 kN J
5@3m = 15 m 1 kN I H D 2 kN E G F
C 2 kN
4 18 2(11) r + b = 2j,
• Determinate • Stable
+ "MA$= 0: + "F = 0: x + "Fy = 0:
K x (5) & 2(3) & 2(6) & 2(9) & 1(12) % 0
Kx = 7.6 kN,
& 7.6 ' Ax % 0,
Ax = 7.6 kN, Ay = 7 kN,
Ay & 2 & 2 & 2 & 1 % 0,
22
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN Use method of joint • Joint F y´ FFG
#
J
I
H D 2 kN E
G
0 0
F
B 2 kN
C 2 kN
7 kN
+ x´
"F y´ = 0:
FFE sin # % 0
F
FFE = 0 + "F = 0: x´ FFG = 0
FFE
23
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN 7 kN • Joint E + y´
#
J
I
H D 2 kN
0
G E
0
F
B 2 kN
C 2 kN
"F y´ = 0:
FEG cos# % 0
FEG 0 x´ + "F = 0: x´
FEG = 0 -FED = 0
FED
E
24
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN 7 kN • Joint G y´ 1 kN FHG 33.69 FDG
o
J
I
H
#
G
0
0
B 2 kN
C 2 kN
D 2 kN
E
F
# % tan &1 ( ) % 33.69 o
2 3
+
"F y´ = 0:
FDG sin 33.69 o & 1 % 0
FDG = 1.803 kN (C) + "F = 0: x´
& FHG ' 1.803 cos 33.69 % 0
G
0
x´
FHG = 1.5 kN (T) 0
25
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN 7 kN • Joint H + y´ FHI H 1.5 kN "F y´ = 0: FHD = 0 B 2 kN C 2 kN J I H
0
G
0
0 0
D 2 kN
E
F
x´
+ "F = 0: x´
& FHI ' 1.5 % 0
FHI = 1.5 kN (T) FHD
26
Use method of sections
3m 33.69o H
3m
3m G E F 18.44o
1 kN
FHI I FDI
FDC + "MD$= 0:
D 2 kN
FHI (2) & 1(3) % 0
FHI = 1.5 kN (T) + "MF$= 0:
& FDI sin 33.69(9) ' 1(3) ' 2(6) % 0
FDI = 3 kN (T) + "MI$= 0:
& FDC sin 18.44(9) & 1(6) & 2(3) % 0
Check : +
"F y = 0:
FDC = -4.25 kN (C) 3 -4.25 FDI sin 33.69 & FDC sin 18.44 & 2 & 1 % 0
O.K.
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The Method of Sections
B a C D
Dy Dx 2m
A 100 N 2m
G
a 2m
F 2m
E
Ex
B
FBC
+ "MG$%$() C + 100(2) - FBC(2) = 0 FBC = 100 N (T) "Fy = 0: -100 + FGCsin45o = 0 FGC = 141.42 N (T) + "MC$%$() 100(4) - FGF(2) = 0 FGF = 200 N (C)
FGC A 100 N 2m
45o
G
FGF
28
Example 3-6 • Determine member force CD, ID, and IH
5@3m = 15 m 1 kN K 5m A B 2 kN J I H D 2 kN E G F
C 2 kN
29
SOLUTION Use method of sections
3m FHI I FDI 33.69o H
3m
3m G E F 18.44o
1 kN
+1.50E+00 -4.22E+00 3.00E+00
FDC + "MD$= 0:
D 2 kN
FHI (2) & 1(3) % 0
FHI = 1.5 kN (T) + "MF$= 0:
& FDI sin 33.69(9) ' 1(3) ' 2(6) % 0
FDI = 3 kN (T) + "MI$= 0:
& FDC sin 18.44(9) & 1(6) & 2(3) % 0
Check : +
"F y = 0:
FDC = -4.25 kN (C) 3 -4.25 FDI sin 33.69 & FDC sin 18.44 & 2 & 1 % 0
O.K.
30
Example 3-7 Determine the force in members GF and GD of the truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated. G H A B C 8 kN 3m 3m D 2 kN 3m Ey = 7 kN F E 3m 4.5 m
Ax = 0
Ay = 9 kN 6 kN 3m
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SOLUTION H A B C
G
a F E a D 2 kN 3m 3m Ey = 7 kN 3m 4.5 m
Ax = 0
Ay = 9 kN 6 kN 3m Section a-a FFG FDG FDC
56.3o 26.6o
8 kN 3m
F
26.6o
+ "MD$%$() FFG sin26.6o(3.6) + 7(3) = 0, FFG = -17.83 kN (C) + "MO$%$() - 7(3) + 2(6) + FDG sin56.3o(6) = 0, FDG = 1.80 kN (C)
D 2 kN
E 3m 3m
O
Ey = 7 kN
32
Example 3-8 Determine the force in members BC and MC of the K-truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated. L A M B 6 kN 5m 5m N C 6 kN 5m D 8 kN 5m 5m 5m K J O E I P F G 7 kN H 3m 3m
0
13 kN
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SOLUTION A
L M B 6 kN
a N a 5m C
K
J O D 8 kN 5m E
I P F
H 3m 3m
0
G 7 kN
13 kN
6 kN 5m
5m
5m
5m
Section a-a L 6m A B 13 kN 6 kN 5m FLM FBM FLK FBC
+ "ML$%$() FBC(6) - 13(5) = 0, FBC = 10 kN (T)
34
L b A M B 6 kN 5m FKL FKM FCM 10 kN
31o
K N C
J O D 8 kN 5m I O P E F E
I P F
H 3m 3m
0
G 7 kN
b 5m
13 kN
6 kN 5m J
5m H
5m
K N C 6 kN 5m
D 8 kN
G 7 kN
3m 3m
5m
5m
5m
+ "MK$%$()
-FCMcos31o(6) - 10(6) - 8(5) + 7(20) = 0 FCM = 7.77 kN (T)
35
Compound Trusses
Procedure for Analysis Step 1. Identify the simple trusses Step 2. Obtain external loading Step 3. Solve for simple trusses separately
36
Example 3-9 Indicate how to analyze the compound truss shown in the figure below. The reactions at the supports have been calculated. 4m H I Ay = 0 A B C D 4 kN 2m 4 kN 2 kN Ay = 5 kN 2m 2m 2m J K G 2m F 2m E Ey = 5 kN
37
SOLUTION
H I J
a
4m G K 2m F 2m E Ey = 5 kN 2m
Ay = 0
A B a C D 4 kN 4 kN 2 kN Ay = 5 kN 2m 2m 2m
H I A B 4 kN Ay = 5 kN 2m 2m J
FHG + "MC$%$() 4 sin60o m FJC FBC C -5(4) + 4(2) + FHG(4sin60o) = 0 FHG = 3.46 kN (C)
38
4m H I Ay = 0 A B C D 4 kN 2m 4 kN 2 kN Ay = 5 kN 2m 2m 2m H 4 sin60o m A B I J K G 2m F 2m E Ey = 5 kN
3.46 kN + "MA$%$() J
60o
FCK FCD
3.46(4sin60o) + FCKsin60o(4) - 4(2) - 2(4) = 0 FCK = 1.16 kN (T) + "F = 0: x -3.46 + 1.16cos60o + FCD = 0 FCK = 2.88 kN (T)
C
4 kN 2 kN Ay = 5 kN 2m 2m
39
4m H I Ay = 0 A B C D 4 kN 2m 4 kN 2 kN Ay = 5 kN 2m 2m 2m H 4 sin60o m A B I 3.46 kN J
60o
G J K
2m F 2m E Ey = 5 kN
Using the method of joints. Joint A : Determine FAB and FAI FCK = 1.16 2.88 kN Joint H : Determine FHI and FHJ Joint I : Determine FIJ and FIB Joint B : Determine FBC and FBJ Joint J : Determine FJC
C
4 kN 2 kN Ay = 5 kN 2m 2m
40
Example 3-10 Indicate how to analyze the compound truss shown in the fugure below. The reactions at the supports have been calculated.
C
H
2m
D
45o 45o 45o
G F E 2m 2m 15 kN 2m B 15 kN 2m
4m
Ax = 0 kN
A
Ay = 15 kN 2m
Fy = 15 kN
41
SOLUTION
2m
C
a
H
D
45o 45o
G a
45o
4m F
Ax = 0 kN
A
Ay = 15 kN 2m
B 15 kN 2m
E 2m 2m
15 kN 2m
Fy = 15 kN
+ "MB$%$() C 4m FCE FBH FDG 2 sin 45o m
45o
-15(2) - FDG(2 sin45o) - FCEcos45o(4) - FCEsin45o(2) = 0 -----(1) + "Fy = 0: 15 - 15 + FBHsin45o - FCEsin45o = 0 FBH = FCE-----(2) + "F = 0: x FBHcos45o + FDG + FCEcos45o = 0 -----(3)
2m
D
45o 45o
A Ay = 15 kN 2m
B 15 kN 2m
From eq.(1)-(3): FBH = FCE = -13.38 kN (C) FDG = 18.92 kN (T)
42
C
a
H
2m
D
45o 45o
G a
45o
4m F
Ax = 0 kN
A
Ay = 15 kN 2m
B 15 kN 2m
E 2m 2m
15 kN 2m
Fy = 15 kN
C 4m FCE FBH FDG
45o 45o
From eq.(1)-(3): FBH = FCE = -13.38 kN (C) FDG = 18.92 kN (T) 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
2m
D
45o
A Ay = 15 kN 2m
B 15 kN 2m
43
Example 3-11 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 5o 5o
3 kN D H
45o
F
45o 5o
G
Ax = 0 kN
A
B 5 kN 6m 6m
C Fy = 4.62 kN
Ay = 4.62 kN
44
E 3 kN
5o 5o 5o
3 kN D H
45o
F
45o 5o
G
Ax = 0 kN
A
B 5 kN 6m 6m FEC
C Fy = 4.62 kN
Ay = 4.62 kN
3 kN F G A FAE 1.5 kN
E
FAE 1.5 kN
E 1.5 kN D H C 1.5 kN FEC 3 kN
45
E 3 kN
5o 5o 5o
3 kN D H
45o
F
45o 5o
G
Ax = 0 kN
A
B 5 kN 6m 6m
C Fy = 4.62 kN FAE
45o
Ay = 4.62 kN
1.5 kN
45o
E
1.5 kN
45o
1.5 kN
45o
A 1.5 kN A 4.62 kN
45o 45o
FAB
1.5 kN B 5 kN C 4.62 kN
+
"Fy = 0:
4.62 kN
+ "F = 0: x
4.62 - 1.5sin45o - FAEsin45o = 0 FAE = 5.03 kN (C)
1.5cos45o - 5.03cos45o + FAB = 0 FAB = 2.50 kN (T) 46
Complex Trusses
F
E
P D FAD 3 9 2(6) r + b = 2j,
A B F´EC F E
C
• Determinate • Stable
=
P D
f´EC F
E
P D 1
+
A B C F´EC + x f´EC = 0
Xx
A
1 B
C
F´EC = FAD x= f´EC Fi = F´i + x f´i
47
Example 3-12 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 1m 0.25 m 1m A 2.5 m B
45o
20 kN F
45o
D E
48
SOLUTION 1m 0.25 m 1m A B
45o
C F
20 kN
45o
D E
F'BD+ x f´BD = 0 F'BD x= f´BD Fi = F'i + x f´i C 1 kN
2.5 m 20 kN F
45o
=
C
B A
45o
D E
B
45o
F
45o
D E
+
x
A
49
First determine reactions and next use the method of joint, start at join C, F, E, D, and B. C 1m 0.25 m B 1 m +10 20 kN A
0
4 21.5
4 1 . 14
45o
F -10
0
20 kN -1 4.1 4
45o
C
7 0 7 0.
45o
0
D -18 E
F 1
-0 .70 7 1 kN
45o
+ x
0
B -0.3 A
0
7 .7 0
9 -0.53
-0.539
0. 70 7
D -0.3 E 0
2.5 m 18 kN (20x2.25)/2.5 = 18 kN C 1m 0.25 m 1m B 7 A
7 0 . 7
45o
0 20 kN -2 1.2 10 1 F o
45
FBD ' xf BD % 0 & 10 ' x(1) % 0 x % 10
7 0 . 7
20 kN 18 kN
16.15
-5.39
7. 07
D -21 E 18 kN
2.5 m
50
Space Trusses
• Determinacy and Stability
P
b + r < 3j b + r = 3j b + r ! 3j
unstable truss statically determinate-check stability statically determinate-check stability
51
z
z y
Fy x z
y
x z
short link
y x
y x Fz z
roller
z
x z
slotted roller y constrained in a cylinder
Fx x Fz z
y
y x
Fx x Fz
Fy
y
ball-and -socket
52
• x, y, z, Force Components.
z Fz
B
l % x2 ' y 2 ' z 2
F Fy x y
l A y x
Fx z
x Fx % F ( ) l
y Fy % F ( ) l
F % Fx ' Fy ' Fz
2 2 2
z Fz % F ( ) l
• Zero-Force Members Case 1
FD D FD y FA x D B A FA = 0 FB z
Case 2
z B FC = 0 y
x
FC
C B FB
A
"Fz = 0 ,
FD = 0
"Fz = 0 , "Fy = 0 ,
FB = 0 FD = 0
53
Example 3-13 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. z B D 2.67 kN 2.44 m E 1.22 m 2.44 m x y 1.22 m C
A
54
SOLUTION By Bx 2.44 m Ay Ax
z B
Cy
C
D 2.67 kN E Az 1.22 m y
2.44 m x The truss is statically determinate since b + r = 3j or 9 + 6 = 3(5) "My$%$() "Mz$%$() "Mx$%$() "Fx = 0: "Fy = 0: "Fz = 0: -2.67(1.22) + Bx(2.44) = 0 Cy = 0 kN By(2.44) - 2.67(2.44) = 0 -Ax + 1.34 = 0 Ay - 2.67 = 0 Az - 2.67 = 0 By = 2.67 kN Ax = 1.34 kN Ay = 2.67 kN Az = 2.67 kN Bx = 1.34 kN
55
z By Bx 2.44 m Ay Ax x 2.44 m z 0 0 x 0 0 FCE C Az B
0 0 0
3.66
0
C z 0 D 0 2.67 kN 0 E 1.22 m y Joint D. "FZ= 0: "FY = 0: Joint C. "Fx = 0: FCE = 0 FCA = 0 FCB = 0 FDC = 0 FDE = 0 FDA = 0 x FDE FDC D y
0
m
0 2.73 m
y
"Fy = 0: "Fz = 0: "Fx = 0:
56
z By Bx 2.44 m Ay Ax x 2.44 m Joint B. "Fy = 0: "Fx = 0: "Fz = 0: Az B
0 0 0
3.66
0
C z 0 D 0 2.67 kN E 1.22 m y x 2.67 kN 1.34 kN B FBA FBC FBE y
0
m
0 2.73 m
- 2.67 + FBE(2.44/3.66) = 0 1.34 - FBC -4(1.22/3.66) = 0 FBA - 4(2.44/3.66) = 0
FBE = 4 kN (T) FBC = 0 FBA = 2.67 kN (C)
57
z By Bx 2.44 m Ay Ax x 2.44 m Joint A. "Fz = 0: "Fy = 0: "Fz = 0: Az B
0 0 0
3.66
0
C 0 D 0 2.67 kN E 1.22 m y
z FAC 2.67 kN 2.67 kN A 1.34 kN x FAD
45o
2 1
0
m
0 2.73 m
FAE y
2.67 kN
2.67 - 2.67 - FACsin45o = 0 2 ) + 2.67 = 0 - FAE( 5 1 - 1.34 + FAD + 2.99( 5 ) = 0
FAC = 0, OK FAE = 2.99 kN (C) FAD = 0, OK
58
! ! ! ! ! ! ! !
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
• Roof Trusses top cord knee brace bottom cord gusset plate roof purlins
bay, 5-6 m typical span, 18 - 30 m, typical
gusset plate
2
Howe truss 18 - 30 m
Pratt truss 18 - 30 m
Howe truss flat roof
Warren truss flat roof
saw-tooth truss skylight
Fink truss > 30 m
three-hinged arch hangar, gymnasium
3
• Bridge Trusses sway bracing top lateral bracing deck panel floor beam bottom cord top cord
portal bracing stringers portal end post
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
P C D C P C
P
D A
A
B
A
B
B
a
new members d (new joint)
b
c
7
• Compound Trusses
simple truss
simple truss
simple truss
simple truss
Type 1
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 A 45o 2m C Cy = 500 N
Ax = 500 N
Ay = 500 N
Joint B y B 45o FBA 500 N FBC x
+ "F = 0: x 500 - FBCsin45o = 0 FBC = 707 N (C) + "Fy = 0:
- FBA + FBCcos45o = 0 FBA = 500 N (T)
15
B
500 N
2m A 45o 2m C Cy = 500 N
Ax = 500 N
Ay = 500 N
Joint A 500 N + "F = 0: x 500 - FAC = 0 FAC = 500 N (T)
500 N 500 N
FAC
16
Zero-Force Members
P 0 0 B 0 C 0 E D
A
Dx
Ey FCB C + "F = 0: F = 0 x CB + FCD FAB A # FAE + "Fy = 0: FCD = 0
Dy
"Fy = 0: FABsin# = 0,
FAB = 0 FAE = 0
+ "F = 0: F + 0 = 0, x AE
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 D
G
F P
E
18
SOLUTION
Ax Ax
A
B C H
0 D 0
Gx G y
#
F Joint D + "Fy = 0:
0 P
E
FDC FDE
FDCsin# = 0, FDE + 0 = 0,
FDC = 0 FDE = 0
D FEC
x
+ "F = 0: x
FEF P
E
0
Joint E + "F = 0: x
FEF = 0
19
Ax Ax Gx
A 0 0
B C H F 0 P Joint H + "F = 0: y FHB = 0 E
0 D 0
G y FHA H FHF FGA Gx FGF x Joint G + FHB
G
"Fy = 0:
FGA = 0
20
Example 3-5 • Determine all the member forces • Identify zero-force members 5@3m = 15 m 1 kN K 5m A B 2 kN J I H D 2 kN E G F
C 2 kN
21
SOLUTION Use method of joints Kx 5m K 0 Ky Ax A Ay B 2 kN J
5@3m = 15 m 1 kN I H D 2 kN E G F
C 2 kN
4 18 2(11) r + b = 2j,
• Determinate • Stable
+ "MA$= 0: + "F = 0: x + "Fy = 0:
K x (5) & 2(3) & 2(6) & 2(9) & 1(12) % 0
Kx = 7.6 kN,
& 7.6 ' Ax % 0,
Ax = 7.6 kN, Ay = 7 kN,
Ay & 2 & 2 & 2 & 1 % 0,
22
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN Use method of joint • Joint F y´ FFG
#
J
I
H D 2 kN E
G
0 0
F
B 2 kN
C 2 kN
7 kN
+ x´
"F y´ = 0:
FFE sin # % 0
F
FFE = 0 + "F = 0: x´ FFG = 0
FFE
23
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN 7 kN • Joint E + y´
#
J
I
H D 2 kN
0
G E
0
F
B 2 kN
C 2 kN
"F y´ = 0:
FEG cos# % 0
FEG 0 x´ + "F = 0: x´
FEG = 0 -FED = 0
FED
E
24
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN 7 kN • Joint G y´ 1 kN FHG 33.69 FDG
o
J
I
H
#
G
0
0
B 2 kN
C 2 kN
D 2 kN
E
F
# % tan &1 ( ) % 33.69 o
2 3
+
"F y´ = 0:
FDG sin 33.69 o & 1 % 0
FDG = 1.803 kN (C) + "F = 0: x´
& FHG ' 1.803 cos 33.69 % 0
G
0
x´
FHG = 1.5 kN (T) 0
25
5@3m = 15 m 1 kN 7.6 kN K 0 5m A 7.6 kN 7 kN • Joint H + y´ FHI H 1.5 kN "F y´ = 0: FHD = 0 B 2 kN C 2 kN J I H
0
G
0
0 0
D 2 kN
E
F
x´
+ "F = 0: x´
& FHI ' 1.5 % 0
FHI = 1.5 kN (T) FHD
26
Use method of sections
3m 33.69o H
3m
3m G E F 18.44o
1 kN
FHI I FDI
FDC + "MD$= 0:
D 2 kN
FHI (2) & 1(3) % 0
FHI = 1.5 kN (T) + "MF$= 0:
& FDI sin 33.69(9) ' 1(3) ' 2(6) % 0
FDI = 3 kN (T) + "MI$= 0:
& FDC sin 18.44(9) & 1(6) & 2(3) % 0
Check : +
"F y = 0:
FDC = -4.25 kN (C) 3 -4.25 FDI sin 33.69 & FDC sin 18.44 & 2 & 1 % 0
O.K.
27
The Method of Sections
B a C D
Dy Dx 2m
A 100 N 2m
G
a 2m
F 2m
E
Ex
B
FBC
+ "MG$%$() C + 100(2) - FBC(2) = 0 FBC = 100 N (T) "Fy = 0: -100 + FGCsin45o = 0 FGC = 141.42 N (T) + "MC$%$() 100(4) - FGF(2) = 0 FGF = 200 N (C)
FGC A 100 N 2m
45o
G
FGF
28
Example 3-6 • Determine member force CD, ID, and IH
5@3m = 15 m 1 kN K 5m A B 2 kN J I H D 2 kN E G F
C 2 kN
29
SOLUTION Use method of sections
3m FHI I FDI 33.69o H
3m
3m G E F 18.44o
1 kN
+1.50E+00 -4.22E+00 3.00E+00
FDC + "MD$= 0:
D 2 kN
FHI (2) & 1(3) % 0
FHI = 1.5 kN (T) + "MF$= 0:
& FDI sin 33.69(9) ' 1(3) ' 2(6) % 0
FDI = 3 kN (T) + "MI$= 0:
& FDC sin 18.44(9) & 1(6) & 2(3) % 0
Check : +
"F y = 0:
FDC = -4.25 kN (C) 3 -4.25 FDI sin 33.69 & FDC sin 18.44 & 2 & 1 % 0
O.K.
30
Example 3-7 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 A B C 8 kN 3m 3m D 2 kN 3m Ey = 7 kN F E 3m 4.5 m
Ax = 0
Ay = 9 kN 6 kN 3m
31
SOLUTION H A B C
G
a F E a D 2 kN 3m 3m Ey = 7 kN 3m 4.5 m
Ax = 0
Ay = 9 kN 6 kN 3m Section a-a FFG FDG FDC
56.3o 26.6o
8 kN 3m
F
26.6o
+ "MD$%$() FFG sin26.6o(3.6) + 7(3) = 0, FFG = -17.83 kN (C) + "MO$%$() - 7(3) + 2(6) + FDG sin56.3o(6) = 0, FDG = 1.80 kN (C)
D 2 kN
E 3m 3m
O
Ey = 7 kN
32
Example 3-8 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 A M B 6 kN 5m 5m N C 6 kN 5m D 8 kN 5m 5m 5m K J O E I P F G 7 kN H 3m 3m
0
13 kN
33
SOLUTION A
L M B 6 kN
a N a 5m C
K
J O D 8 kN 5m E
I P F
H 3m 3m
0
G 7 kN
13 kN
6 kN 5m
5m
5m
5m
Section a-a L 6m A B 13 kN 6 kN 5m FLM FBM FLK FBC
+ "ML$%$() FBC(6) - 13(5) = 0, FBC = 10 kN (T)
34
L b A M B 6 kN 5m FKL FKM FCM 10 kN
31o
K N C
J O D 8 kN 5m I O P E F E
I P F
H 3m 3m
0
G 7 kN
b 5m
13 kN
6 kN 5m J
5m H
5m
K N C 6 kN 5m
D 8 kN
G 7 kN
3m 3m
5m
5m
5m
+ "MK$%$()
-FCMcos31o(6) - 10(6) - 8(5) + 7(20) = 0 FCM = 7.77 kN (T)
35
Compound Trusses
Procedure for Analysis Step 1. Identify the simple trusses Step 2. Obtain external loading Step 3. Solve for simple trusses separately
36
Example 3-9 Indicate how to analyze the compound truss shown in the figure below. The reactions at the supports have been calculated. 4m H I Ay = 0 A B C D 4 kN 2m 4 kN 2 kN Ay = 5 kN 2m 2m 2m J K G 2m F 2m E Ey = 5 kN
37
SOLUTION
H I J
a
4m G K 2m F 2m E Ey = 5 kN 2m
Ay = 0
A B a C D 4 kN 4 kN 2 kN Ay = 5 kN 2m 2m 2m
H I A B 4 kN Ay = 5 kN 2m 2m J
FHG + "MC$%$() 4 sin60o m FJC FBC C -5(4) + 4(2) + FHG(4sin60o) = 0 FHG = 3.46 kN (C)
38
4m H I Ay = 0 A B C D 4 kN 2m 4 kN 2 kN Ay = 5 kN 2m 2m 2m H 4 sin60o m A B I J K G 2m F 2m E Ey = 5 kN
3.46 kN + "MA$%$() J
60o
FCK FCD
3.46(4sin60o) + FCKsin60o(4) - 4(2) - 2(4) = 0 FCK = 1.16 kN (T) + "F = 0: x -3.46 + 1.16cos60o + FCD = 0 FCK = 2.88 kN (T)
C
4 kN 2 kN Ay = 5 kN 2m 2m
39
4m H I Ay = 0 A B C D 4 kN 2m 4 kN 2 kN Ay = 5 kN 2m 2m 2m H 4 sin60o m A B I 3.46 kN J
60o
G J K
2m F 2m E Ey = 5 kN
Using the method of joints. Joint A : Determine FAB and FAI FCK = 1.16 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
C
4 kN 2 kN Ay = 5 kN 2m 2m
40
Example 3-10 Indicate how to analyze the compound truss shown in the fugure below. The reactions at the supports have been calculated.
C
H
2m
D
45o 45o 45o
G F E 2m 2m 15 kN 2m B 15 kN 2m
4m
Ax = 0 kN
A
Ay = 15 kN 2m
Fy = 15 kN
41
SOLUTION
2m
C
a
H
D
45o 45o
G a
45o
4m F
Ax = 0 kN
A
Ay = 15 kN 2m
B 15 kN 2m
E 2m 2m
15 kN 2m
Fy = 15 kN
+ "MB$%$() C 4m FCE FBH FDG 2 sin 45o m
45o
-15(2) - FDG(2 sin45o) - FCEcos45o(4) - FCEsin45o(2) = 0 -----(1) + "Fy = 0: 15 - 15 + FBHsin45o - FCEsin45o = 0 FBH = FCE-----(2) + "F = 0: x FBHcos45o + FDG + FCEcos45o = 0 -----(3)
2m
D
45o 45o
A Ay = 15 kN 2m
B 15 kN 2m
From eq.(1)-(3): FBH = FCE = -13.38 kN (C) FDG = 18.92 kN (T)
42
C
a
H
2m
D
45o 45o
G a
45o
4m F
Ax = 0 kN
A
Ay = 15 kN 2m
B 15 kN 2m
E 2m 2m
15 kN 2m
Fy = 15 kN
C 4m FCE FBH FDG
45o 45o
From eq.(1)-(3): FBH = FCE = -13.38 kN (C) FDG = 18.92 kN (T) 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
2m
D
45o
A Ay = 15 kN 2m
B 15 kN 2m
43
Example 3-11 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 5o 5o
3 kN D H
45o
F
45o 5o
G
Ax = 0 kN
A
B 5 kN 6m 6m
C Fy = 4.62 kN
Ay = 4.62 kN
44
E 3 kN
5o 5o 5o
3 kN D H
45o
F
45o 5o
G
Ax = 0 kN
A
B 5 kN 6m 6m FEC
C Fy = 4.62 kN
Ay = 4.62 kN
3 kN F G A FAE 1.5 kN
E
FAE 1.5 kN
E 1.5 kN D H C 1.5 kN FEC 3 kN
45
E 3 kN
5o 5o 5o
3 kN D H
45o
F
45o 5o
G
Ax = 0 kN
A
B 5 kN 6m 6m
C Fy = 4.62 kN FAE
45o
Ay = 4.62 kN
1.5 kN
45o
E
1.5 kN
45o
1.5 kN
45o
A 1.5 kN A 4.62 kN
45o 45o
FAB
1.5 kN B 5 kN C 4.62 kN
+
"Fy = 0:
4.62 kN
+ "F = 0: x
4.62 - 1.5sin45o - FAEsin45o = 0 FAE = 5.03 kN (C)
1.5cos45o - 5.03cos45o + FAB = 0 FAB = 2.50 kN (T) 46
Complex Trusses
F
E
P D FAD 3 9 2(6) r + b = 2j,
A B F´EC F E
C
• Determinate • Stable
=
P D
f´EC F
E
P D 1
+
A B C F´EC + x f´EC = 0
Xx
A
1 B
C
F´EC = FAD x= f´EC Fi = F´i + x f´i
47
Example 3-12 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 1m 0.25 m 1m A 2.5 m B
45o
20 kN F
45o
D E
48
SOLUTION 1m 0.25 m 1m A B
45o
C F
20 kN
45o
D E
F'BD+ x f´BD = 0 F'BD x= f´BD Fi = F'i + x f´i C 1 kN
2.5 m 20 kN F
45o
=
C
B A
45o
D E
B
45o
F
45o
D E
+
x
A
49
First determine reactions and next use the method of joint, start at join C, F, E, D, and B. C 1m 0.25 m B 1 m +10 20 kN A
0
4 21.5
4 1 . 14
45o
F -10
0
20 kN -1 4.1 4
45o
C
7 0 7 0.
45o
0
D -18 E
F 1
-0 .70 7 1 kN
45o
+ x
0
B -0.3 A
0
7 .7 0
9 -0.53
-0.539
0. 70 7
D -0.3 E 0
2.5 m 18 kN (20x2.25)/2.5 = 18 kN C 1m 0.25 m 1m B 7 A
7 0 . 7
45o
0 20 kN -2 1.2 10 1 F o
45
FBD ' xf BD % 0 & 10 ' x(1) % 0 x % 10
7 0 . 7
20 kN 18 kN
16.15
-5.39
7. 07
D -21 E 18 kN
2.5 m
50
Space Trusses
• Determinacy and Stability
P
b + r < 3j b + r = 3j b + r ! 3j
unstable truss statically determinate-check stability statically determinate-check stability
51
z
z y
Fy x z
y
x z
short link
y x
y x Fz z
roller
z
x z
slotted roller y constrained in a cylinder
Fx x Fz z
y
y x
Fx x Fz
Fy
y
ball-and -socket
52
• x, y, z, Force Components.
z Fz
B
l % x2 ' y 2 ' z 2
F Fy x y
l A y x
Fx z
x Fx % F ( ) l
y Fy % F ( ) l
F % Fx ' Fy ' Fz
2 2 2
z Fz % F ( ) l
• Zero-Force Members Case 1
FD D FD y FA x D B A FA = 0 FB z
Case 2
z B FC = 0 y
x
FC
C B FB
A
"Fz = 0 ,
FD = 0
"Fz = 0 , "Fy = 0 ,
FB = 0 FD = 0
53
Example 3-13 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. z B D 2.67 kN 2.44 m E 1.22 m 2.44 m x y 1.22 m C
A
54
SOLUTION By Bx 2.44 m Ay Ax
z B
Cy
C
D 2.67 kN E Az 1.22 m y
2.44 m x The truss is statically determinate since b + r = 3j or 9 + 6 = 3(5) "My$%$() "Mz$%$() "Mx$%$() "Fx = 0: "Fy = 0: "Fz = 0: -2.67(1.22) + Bx(2.44) = 0 Cy = 0 kN By(2.44) - 2.67(2.44) = 0 -Ax + 1.34 = 0 Ay - 2.67 = 0 Az - 2.67 = 0 By = 2.67 kN Ax = 1.34 kN Ay = 2.67 kN Az = 2.67 kN Bx = 1.34 kN
55
z By Bx 2.44 m Ay Ax x 2.44 m z 0 0 x 0 0 FCE C Az B
0 0 0
3.66
0
C z 0 D 0 2.67 kN 0 E 1.22 m y Joint D. "FZ= 0: "FY = 0: Joint C. "Fx = 0: FCE = 0 FCA = 0 FCB = 0 FDC = 0 FDE = 0 FDA = 0 x FDE FDC D y
0
m
0 2.73 m
y
"Fy = 0: "Fz = 0: "Fx = 0:
56
z By Bx 2.44 m Ay Ax x 2.44 m Joint B. "Fy = 0: "Fx = 0: "Fz = 0: Az B
0 0 0
3.66
0
C z 0 D 0 2.67 kN E 1.22 m y x 2.67 kN 1.34 kN B FBA FBC FBE y
0
m
0 2.73 m
- 2.67 + FBE(2.44/3.66) = 0 1.34 - FBC -4(1.22/3.66) = 0 FBA - 4(2.44/3.66) = 0
FBE = 4 kN (T) FBC = 0 FBA = 2.67 kN (C)
57
z By Bx 2.44 m Ay Ax x 2.44 m Joint A. "Fz = 0: "Fy = 0: "Fz = 0: Az B
0 0 0
3.66
0
C 0 D 0 2.67 kN E 1.22 m y
z FAC 2.67 kN 2.67 kN A 1.34 kN x FAD
45o
2 1
0
m
0 2.73 m
FAE y
2.67 kN
2.67 - 2.67 - FACsin45o = 0 2 ) + 2.67 = 0 - FAE( 5 1 - 1.34 + FAD + 2.99( 5 ) = 0
FAC = 0, OK FAE = 2.99 kN (C) FAD = 0, OK
58
