Truss

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Analysis of Statically Determinate
Trusses
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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

• Roof Trusses

purlins
top cord
knee brace
bottom
cord gusset plate

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
top cord

sway
bracing
top lateral
bracing

portal
bracing
stringers
portal
end post

deck
panel

bottom cord

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

Ay = 500 N

500 N
45o

FBA

Cy = 500 N

+ ΣF = 0:
x

Joint B
y
B

2m

C

FBC

x

500 - FBCsin45o = 0
FBC = 707 N (C)
+

ΣFy = 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
+ ΣF = 0:
x

500 N

500 N

FAC

500 - FAC = 0
FAC = 500 N (T)

500 N

16

Zero-Force Members
0

B

P

C
0

0
0

A

E

Ey
C

FCB

D

Dx

Dy

+ ΣF = 0: F = 0
x
CB
ΣFy = 0: FCD = 0

+
FCD
FAB
A

θ

FAE

ΣFy = 0: FABsinθ = 0,

FAB = 0

+ ΣF = 0: F + 0 = 0,
x
AE

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

FDC
FDE

Joint D

y

+

θ

x

D

ΣFy = 0:

+ ΣF = 0:
x

E

0

0
D
0

P
FDCsinθ = 0,

FDC = 0

FDE + 0 = 0,

FDE = 0

FEC
E

FEF

0

Joint E
+ ΣF = 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

+ ΣF = 0:
y

H

FHB = 0

FHF
x

FGA
Gx

G

Joint G
FGF

+

ΣFy = 0:

FGA = 0

20

Example 3-5
• Determine all the member forces
• Identify zero-force members
5@3m = 15 m
1 kN
K

J

I

5m
A

B
2 kN

H

C
2 kN

D

G
E

F

2 kN

21

5@3m = 15 m

SOLUTION
Use method of joints
Kx

1 kN
K

J

I

H

0
Ky

5m

Ax A
Ay

C

B

2 kN

2 kN

4 18 2(11)
r + b = 2j,

+ ΣMA = 0:

D

G
E

F

2 kN

• Determinate
• Stable

K x (5) − 2(3) − 2(6) − 2(9) − 1(12) = 0

Kx = 7.6 kN,
+ ΣF = 0:
x
+

ΣFy = 0:

− 7.6 + Ax = 0,

Ax = 7.6 kN,

Ay − 2 − 2 − 2 − 1 = 0,

Ay = 7 kN,

22

5@3m = 15 m
1 kN
7.6 kN K

J

I

H

0
5m
A
7.6 kN
Use method of joint

B
2 kN

C
2 kN

D

G
E

0
0

F

2 kN

7 kN

• Joint F
+


FFG

F
θ

FFE

ΣF y´ = 0:

FFE sin θ = 0

FFE = 0


+ ΣF = 0:


FFG = 0

23

5@3m = 15 m
1 kN
7.6 kN K

J

I

H

G

0
5m
A
7.6 kN

B
2 kN

D

C
2 kN

0

E

0

F

2 kN

7 kN
• Joint E
+


FEG cosθ = 0

FEG = 0

FEG
θ

FED

ΣF y´ = 0:

0



+ ΣF = 0:


-FED = 0

E

24

5@3m = 15 m
1 kN
7.6 kN K

J

I

H

0

θ

0
5m
A
7.6 kN

G

B
2 kN

C
2 kN

D
2 kN

E

0

F
2
3

θ = tan −1 ( ) = 33.69 o

7 kN
• Joint G
+



G

0



+ ΣF = 0:


− FHG + 1.803 cos 33.69 = 0

FHG = 1.5 kN (T)
0

FDG

o

FDG sin 33.69 o − 1 = 0

FDG = 1.803 kN (C)

1 kN
FHG 33.69

ΣF y´ = 0:

25

5@3m = 15 m
1 kN
7.6 kN K

J

I

H
0

0
5m
A
7.6 kN

G

B
2 kN

C
2 kN

D

0

E

0
0

F

2 kN

7 kN
• Joint H
+

FHI

H

1.5 kN



ΣF y´ = 0:

+ ΣF = 0:


FHD = 0

− FHI + 1.5 = 0

FHI = 1.5 kN (T)
FHD

26

Use method of sections

3m

3m

1 kN

33.69o

FHI
I
FDI

FDC

3m

H

G
E

D

F
18.44o

2 kN
+ ΣMD = 0:

FHI (2) − 1(3) = 0

FHI = 1.5 kN (T)
+ ΣMF = 0:

− FDI sin 33.69(9) + 1(3) + 2(6) = 0

FDI = 3 kN (T)
+ ΣMI = 0:

Check : +

ΣF y = 0:

− FDC sin 18.44(9) − 1(6) − 2(3) = 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
a

B

Dy

C

Dx

D

2m
A
100 N

G

a

2m

B

2m

FBC

A
100 N

G
2m

Ex

2m
+ ΣMG = 0:
100(2) - FBC(2) = 0
FBC = 100 N (T)

C

FGC
45o

E

F

FGF

+

ΣFy = 0:
-100 + FGCsin45o = 0
FGC = 141.42 N (T)
+ ΣMC = 0:
100(4) - FGF(2) = 0
FGF = 200 N (C)

28

Example 3-6
• Determine member force CD, ID, and IH

5@3m = 15 m
1 kN
K

J

I

5m
A

B
2 kN

H

C
2 kN

D

G
E

F

2 kN

29

SOLUTION

3m

3m

Use method of sections

1 kN

33.69o

FHI
I
FDI

FDC

3m

H

+1.50E+00

G
E

D

F
18.44o

2 kN
+ ΣMD = 0:

FHI (2) − 1(3) = 0

-4.22E+00
3.00E+00

FHI = 1.5 kN (T)
+ ΣMF = 0:

− FDI sin 33.69(9) + 1(3) + 2(6) = 0

FDI = 3 kN (T)
+ ΣMI = 0:

Check : +

ΣF y = 0:

− FDC sin 18.44(9) − 1(6) − 2(3) = 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

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

31

G

SOLUTION

a

H

Ax = 0

F

A

E
B

Ay = 9 kN 6 kN
3m

C

4.5 m

a D

8 kN
3m

3m

2 kN
3m

Ey = 7 kN

3m

Section a-a
FFG

26.6o

FDG
FDC

+ ΣMD = 0:

F

56.3o

26.6o

D

E

2 kN
3m

O

Ey = 7 kN
3m

FFG sin26.6o(3.6) + 7(3) = 0,
FFG = -17.83 kN (C)
+ ΣMO = 0:
- 7(3) + 2(6) + FDG sin56.3o(6) = 0,
FDG = 1.80 kN (C)

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

0

A

13 kN

M

K
N

B

D

6 kN
5m

I

5m

H
P

O

C

6 kN
5m

J

E

G

F

8 kN
5m

3m
3m

7 kN
5m

5m

33

L

SOLUTION
A

0

a

M
a

6 kN
5m

J

N

B

13 kN

K

D

C

5m

H
P

O

6 kN
5m

I

E

G

F

8 kN
5m

3m
3m

7 kN
5m

5m

Section a-a
L
6m

FLM
FBM

A
B

13 kN

FLK
FBC

+ ΣML = 0:
FBC(6) - 13(5) = 0,
FBC = 10 kN (T)

6 kN
5m

34

L b

0

M

A

K
N

B

5m
FKL
FKM
FCM
10 kN

b
5m

D

6 kN
5m
+ ΣMK = 0:

G

F

7 kN

5m
I

5m

5m

H
P

O
E

G

F

8 kN
5m

3m
3m

P

8 kN

J

C

H

E

5m

N
31o

D

6 kN

K

I
O

C

6 kN

13 kN

J

3m
3m

7 kN
5m

5m

-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

G
J

I
Ay = 0

2m
F

K

A
B

C

4 kN
2 kN
Ay = 5 kN
2m
2m
2m

2m
E

D
4 kN

Ey = 5 kN
2m

37

SOLUTION

a

H

G

J

I
Ay = 0

4m
2m
F

K

2m

A
B

a

C

4 kN
2 kN
Ay = 5 kN
2m
2m
2m

E

D
4 kN

Ey = 5 kN
2m

FHG

H

+ ΣMC = 0:

J

I

4 sin60o m
FJC

A
B

FBC

4 kN
Ay = 5 kN
2m
2m

C

-5(4) + 4(2) + FHG(4sin60o) = 0
FHG = 3.46 kN (C)

38

4m
H

G
J

I
Ay = 0

2m
F

K

A
C

B

H

E

D

4 kN
2 kN
Ay = 5 kN
2m
2m
2m

2m

4 kN

Ey = 5 kN
2m

3.46 kN
+ ΣMA = 0:

4 sin60o m

J

I

FCK

60o

A
B

C

4 kN
2 kN
Ay = 5 kN
2m
2m

FCD

3.46(4sin60o) + FCKsin60o(4) - 4(2) - 2(4) = 0
FCK = 1.16 kN (T)
+ ΣF = 0:
x
-3.46 + 1.16cos60o + FCD = 0
FCK = 2.88 kN (T)

39

4m
H

G
J

I
Ay = 0

F

K

A
C

B

H

3.46 kN

60o

B

Ey = 5 kN
2m

Joint A : Determine FAB and FAI
FCK = 1.16

A

4 kN

Using the method of joints.

J

I

2m
E

D

4 kN
2 kN
Ay = 5 kN
2m
2m
2m

4 sin60o m

2m

C

4 kN
2 kN
Ay = 5 kN
2m
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

40

Example 3-10
Indicate how to analyze the compound truss shown in the fugure below. The
reactions at the supports have been calculated.

H

C

D

2m

Ax = 0 kN

45o

A

Ay = 15 kN
2m

4m

G
45o

B
15 kN
2m

F

45o

E
2m

2m

15 kN

Fy = 15 kN

2m

41

C

SOLUTION

a

H

D

2m

a

45o

Ax = 0 kN

A

Ay = 15 kN
2m

4m

G
45o

B
15 kN
2m

F

45o

E
2m

15 kN

2m

Fy = 15 kN

2m

+ ΣMB = 0:
C
45o

4m

D

2m

45o

A
Ay = 15 kN
2m

45o

B
15 kN
2m

FCE
FBH
FDG
2 sin 45o m

-15(2) - FDG(2 sin45o) - FCEcos45o(4)
- FCEsin45o(2) = 0 -----(1)
+ ΣFy = 0:
15 - 15 + FBHsin45o - FCEsin45o = 0
FBH = FCE-----(2)
+ ΣF = 0:
x
FBHcos45o + FDG + FCEcos45o = 0

From eq.(1)-(3): FBH = FCE = -13.38 kN (C)
FDG = 18.92 kN (T)

-----(3)

42

C

A

Ay = 15 kN
2m

a

45o

B
15 kN
2m

C
45o

FCE
FBH
FDG

D
45o

A
Ay = 15 kN
2m

45o

B
15 kN
2m

4m

G

45o

Ax = 0 kN

2m

H

D

2m

4m

a

F

45o

E
2m

2m

15 kN

Fy = 15 kN

2m

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

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

3 kN

F

5o

Ax = 0 kN

A

G

45o

D

5o

H

5o

45o

B

Ay = 4.62 kN

Fy = 4.62 kN

5 kN
6m

C

6m

44

E
3 kN

5o

3 kN

F

5o

Ax = 0 kN

G

45o

A

C
Fy = 4.62 kN

5 kN

FAE

6m
FEC
E
1.5 kN

3 kN

1.5 kN

F
G

D
H
C

A
FAE

45o

B
6m

3 kN

H

5o

Ay = 4.62 kN

E

D

5o

1.5 kN

1.5 kN

FEC

45

E
3 kN

5o

3 kN

F

5o

Ax = 0 kN

G

45o

A

D

5o

H

5o

45o

B

Ay = 4.62 kN

E

45o

Fy = 4.62 kN

5 kN
6m

1.5 kN

C

6m
FAE

1.5 kN

1.5 kN
45o

45o

45o

A
1.5 kN
A
4.62 kN

1.5 kN
45o

45o

B
5 kN

C
4.62 kN

+

ΣFy = 0:

FAB

4.62 kN

4.62 - 1.5sin45o - FAEsin45o = 0
FAE = 5.03 kN (C)

+ ΣF = 0:
x

1.5cos45o - 5.03cos45o + FAB = 0
FAB = 2.50 kN (T)
46

E

Complex Trusses
F

P
3 9 2(6)
r + b = 2j,

D
FAD

A

• Determinate
• Stable

C
B

F

P

+
C
B

P

F

D

A

E

f´EC

=

F´EC

E

1

Xx

F´EC + x f´EC = 0
F´EC
= FAD
x=
f´EC
Fi = F´i + x f´i

D

A

1

C
B

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

20 kN

1m
0.25 m

B

45o

F

45o

D

1m
A

E
2.5 m

48

SOLUTION

C

20 kN

1m
0.25 m

45o

B

F

45o

D

1m
A

E

Fi = F'i + x f´i

2.5 m

B
A

45o

20 kN
F

45o

C

=

C

F'BD+ x f´BD = 0
F'BD
x=
f´BD

D
E

B

+

x

A

45o

F

1 kN

45o

D
E

49

First determine reactions and next use the method of joint, start at join C, F, E, D, and B.
C
4
1
.
14

1m
0.25 m
B
1 m +10
20 kN A

20 kN
-1
4.1
4

F

45o

45o

-10

0

-

D
-18

0
0

4
21.5

C

E

+ x
0

2.5 m
18 kN

7
0
7
0.
45o

B
-0.3

0

A

7
.7 0

(20x2.25)/2.5 = 18 kN
C
7
0
.
7

0.25 m
1m

B
7
A

20 kN
18 kN

45o

7
0
.
7

16.15

-5.39

2.5 m

7.
07

-0.539

FBD + xf BD = 0

0.
70
7

D
-0.3
E
0

− 10 + x(1) = 0
x = 10

20 kN
-2
1.2
10
1
F
o
45

45o

1

9
-0.53

0

1m

F

-0
.70
7
1 kN

D
-21
E
18 kN

50

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

51

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

ball-and -socket

y

Fx
x

Fz
z

z

x

Fz

Fy

Fx
x

Fz

y

52

• x, y, z, Force Components.
z

l = x2 + y 2 + z 2

Fz
F

B

l

Fy

Fx z

A

x
Fx = F ( )
l

y

x
y

y
Fy = F ( )
l
2

z
Fz = F ( )
l

2

F = Fx + Fy + Fz

x

2

• Zero-Force Members
z

Case 1

Case 2

z

FD

B

D

x

FC

FD

C

A
B

y
B

x

A
FA = 0

FB

ΣFz = 0 ,

y
FA

D

FD = 0

FC = 0

FB

ΣFz = 0 ,

FB = 0

ΣFy = 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.
C

z
B

D
2.67 kN
2.44 m

1.22 m

E

A

1.22 m
2.44 m

y

x

54

SOLUTION

Cy

z

C

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)
ΣMy = 0:

-2.67(1.22) + Bx(2.44) = 0

ΣMz = 0:

Cy = 0 kN

ΣMx = 0:

By(2.44) - 2.67(2.44) = 0

By = 2.67 kN

ΣFx = 0:

-Ax + 1.34 = 0

Ax = 1.34 kN

ΣFy = 0:

Ay - 2.67 = 0

Ay = 2.67 kN

ΣFz = 0:

Az - 2.67 = 0

Az = 2.67 kN

Bx = 1.34 kN

55

0

0

z

C
z

0
By

B

0
3.66

Bx

m

0

2.44 m Ay

FDC

D
0

x

E
y

z
0

Joint C.

C
y

0
x

0

0

FCE

y
FDE

1.22 m

Az

2.44 m

x

D

2.67 kN
0

2.73 m

Ax

0

0

Joint D.
ΣFZ= 0:

FDC = 0

ΣFY = 0:

FDE = 0

ΣFx = 0:

FDA = 0

ΣFy = 0:

FCE = 0

ΣFz = 0:

FCA = 0

ΣFx = 0:

FCB = 0

56

0

0

z

C
z

0
By

B

3.66

Bx

x

2.44 m

m

0

2.44 m Ay
Ax

0

2.73 m
Az

0

0

B

2.67 kN

D
0

2.67 kN

1.34 kN

FBA

FBC
FBE

y

x

E
1.22 m
y

Joint B.
ΣFy = 0:

- 2.67 + FBE(2.44/3.66) = 0

FBE = 4 kN (T)

ΣFx = 0:

1.34 - FBC -4(1.22/3.66) = 0

FBC = 0

ΣFz = 0:

FBA - 4(2.44/3.66) = 0

FBA = 2.67 kN (C)

57

0

0

z

C

0
By

B

3.66

Bx

x

2.44 m

m

2.73 m
Az

FAC

0

0

0

2.44 m Ay
Ax

0

z

D
0

2.67 kN

2.67 kN
2.67 kN
A
1.34 kN

E
1.22 m

x

FAD
45o
2

1

FAE
y

2.67 kN

y

Joint A.
ΣFz = 0:
ΣFy = 0:
ΣFz = 0:

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

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