Analysis of statically determinate truss structures

1. Chapter 2
Analysis of Statically
Determinate Structures

2. Idealized Structure
• In real sense exact analysis of a structure can
never be carried out.
Structural Analysis IDr. Mohammed Arafa

3. Support connection
• Pin support and pin connection

4. • Fixed support & fixed connection

5. • Roller support

6. Notice that the deck of this concrete bridge is
made so that one section can be considered
roller supported on the other section.

12. Support for coplanar structures

13. Support for coplanar structures

14. Idealized Structure
Structural Analysis IDr. Mohammed Arafa

17. Structural Analysis IDr. Mohammed Arafa

25. The flat roof of the steel-frame building shown in the photo is intended
to support a total load of 2 kN/m2 over its surface. Determine the roof
load within region ABCD that is transmitted to beams BC and DC.
Example 2.2
  
2 1
2 1
2
7 4
/ 7 / 4 1.75 2
int 2 / 2 4 /
L m and L m
L L Two way slab action
The peak ensity kN m kN m
 
   
 

26. Example 2.2
Loading on Beam BC from Area 1
Loading on Beam DC from Area 3
Loading on Beam BC from Area 1 & 2
8 kN/m

27. Example 2.3
The concrete girders shown in the photo of the passenger car parking
garage span 30 ft and are 15 ft on center. If the floor slab is 5 in.
thick and made of reinforced stone concrete, and the specified live
load is 50 lb/ft2 , determine the distributed load the floor system
transmits to each interior girder.

28. Example 2.3
L2=30ft and L1=15ft , so that L2/L1=2 We have a two-way slab.
From Table 1–2, for reinforced stone concrete, the specific weight of
the concrete is 150 lb/ft3.
Thus the design floor loading is
 3 2 25
150 lb/ft ft 50lb/ft 112.5lb/ft
12
p
 
   
 

29. Example 2.3
A trapezoidal distributed loading is transmitted to each interior girder AB
from each of its sides. The maximum intensity of each of these distributed
loadings is (112.5 lb/ft2)(7.5ft)=843.75 lb/ft, so that on the girder this
intensity becomes , 2(843.75 lb/ft)=1687.5 lb/ft
Note: For design, consideration should also be given to the weight of the girder.

30. Problem

31. Load Path

32. Load Path

33. Equation of Equilibrium
• In x-y plane
0
0
0






o
y
x
M
F
F
Internal loading

34. Determinacy & Stability
• Determinacy: when all the forces in structure can be
determined from equilibrium equation , the structure is
referred to as statically determinate. Structure having more
unknown forces than available equilibrium equations
called statically indeterminate
• If nis number of structure parts & ris number of
unknown forces:
r = 3n, statically determinate
r > 3n, statically indeterminate

35. Classify determinate & indeterminate structure
eDeterminatStatically)1(33
1
3



n
r

36. degree2
ateindeterminStatically)1(35
1
5
nd



n
r
edeterminatStatically)2(36
2
6



n
r

37. degree1
ateindeterminStatically)3(310
3
10
st



n
r
degree1ate,indeterminStatically)2(37
2
7
st



n
r

38. degree4
ateindeterminStatically)2(310
2
10
th



n
r

41. • Stability
Partial Constraints
caseloadingwith thisunstablesatisfiedbenotwill0  xF

42. • Improper Constraints
This can occur if all the support reactions are concurrent at a point.
0dP

43. • This can occur also when the reactive forces are all parallel
In General
r < 3n, Then the structure is Unstable
r >= 3n, Also, Unstable if member reactions are
concurrent or parallel or some of the components
form a collapsible mechanism

44. Classify The structure Stable or Unstable
3
1
3 3(1) Stable
r
n
no special cases


 
Structural Analysis IDr. Mohammed Arafa

45. Sable)2(38
2
8



casesno special
n
r
UnsableBarconcurrentarereactionsthreethe)1(33
1
3



n
r

46. Unsable)3(37
3
7



n
r

49. Application of
Equilibrium Equation
Structural Analysis IDr. Mohammed Arafa

50. Application of Equation of Equilibrium
1. If not given, establish a suitable x-y coordinate system.
2. Draw a free body diagram (FBD) of the object under
analysis.
3. Apply the three equations of equilibrium to solve for
the unknowns.
Procedure Steps

51. How Important is the Free-Body Diagram?

52. Example 1
Determine the Reactions

53. kA
A
kB
BM
kA
AF
y
y
y
yA
x
xx
4.13
05.3860sin600F
5.38
050)14()1(60cos60)10(60sin600
30
060cos600
y










54. Example 2
Determine the Reactions

55. 600
0)6(60)4(600
120
060600F
00
y
kNM
MM
kNA
A
AF
A
AA
x
x
xx









56. Example 3
Determine the Reactions

57.  
 
IbA
A
IbA
AF
IbN
NNM
y
y
x
xx
B
BBA
2700
05.133135000F
1070
05.13310
5.1331
0)10()4()5.3(35000
5
3
y
5
4
5
3
5
4










58. Example 4
Determine the Reactions

59.  
ftkA
A
AF
k.ftIb.ftM
MM
IbC
CM
y
y
xx
A
AAat
y
yRightB
.6.7
080004000F
00
7272000
0)10(80006000)35(4000
400
06000150
y
point












60. Example 5
Determine the Reactions

61.  
kNA
A
kNA
AF
kNC
CM
kNC
CM
y
y
x
xx
x
xA
y
yRightB
4.9
0)(8630F
87.9
0)(87.140
7.14
0)2(8)3(6)4(3)5.1(0
3
0)1(620
5
4
y
5
3













62. Example 6
Determine the Reactions

63.  
mkNM
MM
kNE
EF
EF
kNC
CM
kNA
AM
E
ErightD
y
y
xx
y
yLeftD
y
yLeftB
.33.5
0)4(33.5)2(80
33.5
067.64880
00
67.6
0)8(8)11(430
4
0)3(860
)(
y
)(















66. Example 7
Determine the Reactions

67.  
kNA
A
kNA
AF
kNC
CM
kNC
CM
y
y
x
xx
x
xRightB
y
yA
120
045sin9.8445sin6.2542400F
285
045cos9.8445cos6.254601801950
195
0)(9.84)5.4(60)3(240)6(0
240
0)5.4(45cos9.84)5.4(45cos9.84)5.1(45sin6.254
)5.4(45cos6.254)5.1(60)5.1(180)6(0
y
2
23














68. Problem 1
Determine the Reactions

69. Problem 2
Determine the Reactions

70. Problem 3
Determine the Reactions