Chapter 2-analysis of statically determinate structures

Chapter 2
Analysis of Statically
Determinate Structures

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

Support connection
• Pin support and pin connection

• Fixed support & fixed connection

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

Chapter 2-analysis of statically determinate structures

Support for coplanar structures

Idealized Structure

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
 
   
 

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

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.

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
 
   
 

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.

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

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

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

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

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

degree4
2
10
th



n
r

• Stability
Partial Constraints
caseloadingwith thisunstablesatisfiedbenotwill0  xF

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

• 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

Classify The structure Stable or Unstable
3
1
3 3(1) Stable
r
n
no special cases


 

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

Unsable)3(37
3
7



n
r

Application of
Equilibrium Equation

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

How Important is the Free-Body Diagram?

Example 1
Determine the Reactions

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










Example 2

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

Example 3

 
 
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










Example 4

 
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












Example 5

 
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













Example 6

 
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
)(















Example 7

 
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














Problem 1

Problem 2

Problem 3

Chapter 2-analysis of statically determinate structures

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