Module3 rajesh sir

Structural Analysis - II
Approximate Methods
Dr. Rajesh K. N.
Assistant professor in Civil EngineeringAssistant professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN

Module IIIModule III
Approximate Methods of Analysis of Multi-storey Frames
• Analysis for vertical loads - Substitute frames-Loading
conditions for maximum positive and negative bending
moments in beams and maximum bending moment in columns
• Analysis for lateral loads - Portal method–Cantilever method–y
Factor method.
2

Why approximate analysis?
• Rapid check on computer aided analysis
y pp x y
• Preliminary dimensioning before exact analysis
Advantage? • Fasterg
Disadvantage? • Results are approximateg pp
• Approximate methods are particularly useful for
multi-storey frames taller than 3 storeys.

Approximate analysis for Vertical LoadsApproximate analysis for Vertical Loads
SUBSTITUTE FRAME METHOD
• Analyse only a part of the frame – substitute frame
SUBSTITUTE FRAME METHOD
Analyse only a part of the frame substitute frame
• Carry out a two-cycle moment distribution

Substitute frame
Actual frame

• Analysis done for:y
• Beam span moments
• Beam support moments
• Column moments
• Liveload positioning for the worst condition
• For the same frame, liveload positions for maximum span
t t t d l tmoments, support moments and column moments may
be different
• For maximum moments at different points, liveload
positions may de different
positions may de different

LL positions for maximum positive span moment at Bp p p
B
Influence
D d l d Li l d
line for MB
Dead loads Live loads

LL positions for maximum negative support moment at Apos t o s o ax u egat e suppo t o e t at
A
Influence
Live loads
line for MA
Dead loads
Dead loads

LL positions for maximum column moment M1 at Cpos t o s o ax u co u o e t M1 at C
C
M1
Live loadsLive loads
Dead loads

LL positions for maximum column moment M2 at Dpos t o s o ax u co u o e t M2 at
D
M2
D
Live loadsLive loads
Dead loads

Problem 1: Total dead load is 12 kN/m. Total live load is 20ob e : N/
kN/m. Analyse the frame for midspan positive moment on BC.
4m
6 m 6 m 6 m
B C DA
4m4m
11

12+20 kN/m
B D
12 kN/m 12 kN/m
B C D
A 6 m 6 m 6 m
Fi d d t
2 2
12 6
36AB
wl
FEM kNm
− − ×
= = = − 36BAFEM kNm=
Fixed end moments
12 12
AB
2
32 6
96FEM kN
− × 96FEM kNm=96
12
BCFEM kNm= = −
36CD DCFEM FEM kNm− = =
96CBFEM kNm=
36CD DCFEM FEM kNm

Distribution factors
1 4 6
0.25
4 6 4 4 4 4
AB DC
K EI
DF DF
K K K EI EI EI
= = = =
1 2 3 4 6 4 4 4 4
AB DC
K K K EI EI EI+ + + +
4 6K EI1
1 2 3 4
4 6
0.2
4 4 4 6 4 4 4 6
BA
K EI
DF
K K K K EI EI EI EI
= = =
+ + + + + +
0.2BC CD CB BADF DF DF DF= = = =

A B C D
0.2 0.2
* * * * * * FEM
0.2 0.20.25 DFs0.25
* * * * * *
* * * * CO
Dist
* *
* * Final Moments
Dist

A B C D
0.2 0.2
-36 36 -96 96 -36 36 FEM
0.2 0.20.25 DFs0.25
9 12 12 -12 -12 -9
6 4.5 -6 6 -4.5 -6 CO
Dist
2.25 0.3 0.3 -0.3 -0.3 -2.25
-18.75 52.8 -89.7 89.7 52.8 18.75 Final Moments
Dist

B
89.7kNm 89.7kNm
32kN mA B
89.7kNm 32kN m
2
3 32 6×
Midspan positive moment on BC,
3 32 6
89.7 32 3 54.3
2 2
EM kNm
×
= − − × + × =

Problem 2: Analyse the frame for beam negative moment at B.
M t f i ti f b i 1 5 ti th t f l T t l d dMoment of inertia of beams is 1.5 times that of columns. Total dead
load is 14 kN/m and total live load is 9 kN/m.
6 m 4m4 m
3.5m.5m
B
A
m3
DC
3.5m3.5m
17
3

14+9 kN/m 14+9 kN/m
I
B D6 m
14 kN/mI
B C D
A
6 m 4 m 4 m
I
1.5I 1.5I 1.5I
Fi d d t
2 2
23 6
69AB
wl
FEM kNm
− − ×
= = = − 69BAFEM kNm=
Fixed end moments
12 12
AB
2
23 4
30.67BC CBFEM FEM kNm
×
− = = = 30.67
12
BC CBFEM FEM kNm
2
14 4
36CD DCFEM FEM kNm
×
− = = =
36
12
CD DCFEM FEM kNm

Distribution factors
( )
( )
1 4 61.5
0.304
4 6 4 3 5 4 3 5
AB
K E I
DF
K K K E EI EI
= = =
( )1 2 3 4 6 4 3.5 4 3.51.5
AB
K K K E EI EII+ + + +
1 5 6K I1
1 2 3 4
1.5 6
0.209
1.5 6 3.5 3.5 1.5 4
BA
K I
DF
K K K K I I I I
= = =
+ + + + + +
1
1 2 3 4
1.5 4
0.313
1.5 6 3.5 3.5 1.5 4
BC
K I
DF
K K K K I I I I
= = =
+ + + + + +1 2 3 4
0.284, 0.284, 0.396CB CD DCDF DF DF= = =

A B C D
0.209 0.313 0.284 0.2840.304 DFs0.3960.209 0.313
* * * * *
* * * *
FEM
Dist
0. 8 0. 8 0.396
* *
* *
CO
Dist
* * Final Moments

A B C D
0.209 0.313
-69 69 -30.67 30.67 -18.67 FEM
0.284 0.2840.304 DFs0.396
20.98 -8.01 -12 -3.41
10.49 -1.71 CO
Dist
-1.84 -2.75
69.64 -47.13 Final Moments
Dist
B 69 64kNm47 13kNm B 69.64kNm47.13kNm
Max. beam negative moment at B 69.64 kNm=

Approximate analysis for Horizontal LoadsApproximate analysis for Horizontal Loads
1 P t l th d1. Portal method
2. Cantilever method
3. Factor method
22

PORTAL METHODPORTAL METHOD
Assumptions
1. The points of contraflexure in all the members lie at their
midpoints.
2. Horizontal shear taken by each interior column is doubley
that taken by each exterior column.
Horizontal forces are assumed to act only at the joints.

CA B C DA
P1
P 2P 2P PP
P
2P
2P
2P
2P
P
P
FE G
HP2
Q 2Q 2Q Q
Q 2Q 2Q Q
J K LI
24

P
1 2 2P P P P P= + + +
1
6
P
P⇒ =

2 2P P Q Q Q Q+ = + + + 1 2P P
Q
+
⇒ =1 2 2 2P P Q Q Q Q+ = + + + 1 2
6
Q⇒ =

Problem 3: Analyse the frame using portal method.ob e 3: y g p
B C DA
m
120 kN
7 m 3.5 m 5 m
FE G
H
3.5m
180 kN
m
H
3.5m
J K LI J K LI
27

Horizontal shears:
1, 2 2For the top storey P P P P P= + + +
120
20
6
P kN⇒ = =
1 2
,
6
P P
For the bottom storey Q
+
=
120 180
50
6
kN
+
= =,
6
y Q
6

120kN 3 5 m
A
Moments:
35kNm
m
3.5 mMoments:
35kN
20kN
1.75m
35kNm 10kN
20kN
35kNm 35kNm
B
35kNm 35kNm
40kN10kN
70kNm
10kN
29

Beam moments:
B C DA
Beam moments:
35
35
35
35
35
3535 35 35
FE G
H
122.5
122.5
122.5 122.5
122.5 122.5
J K LI
30

Column moments:
B C DA 7035 kNm
70
35 kNm
Column moments:
FE G
H
35 87.5 87.5175 70 3570 175
J K LI
87.5 87.5175 175
31

Beam and Column moments:
35
B D
35 35 35
7035 70
35
35 35
35 87 5
87.5175
70 3570
175
122.5 122.5
122 5
122.5
122 535 87.5 122.5 122.5 122.5
87 5 87 5175 17587.5 87.5175 175
32

Home work
B CA40 kN B CA
5m
40 kN
5 m 7.5 m
FED
3.5
80 kN
m5m
H IG
33

CANTILEVER METHODCANTILEVER METHOD
• Frame considered as a vertical cantilever
Assumptions
1. The points of contraflexure in all the members lie at
their midpoints.
2. The direct stresses (axial stresses) in the columns are
directly proportional to their distance from thedirectly proportional to their distance from the
centroidal vertical axis of the frame.

P1
y1
y2 y3
y4
P2
A1 A2
A3 A4
Area of cross
ti
Centroidal vertical axis
of the frame
section
1 1 2 2 3 3 4 4Ad A d Ad A d
y
A A A A
+ + +
=
+ + +
To locate centroidal vertical
axis of the frame,
35
1 2 3 4A A A A+ + +

V1 V2 V3 V4
x
My
I
σ =
MM
I
is constant at a given height (of the ‘vertical cantilever’).
1 2 3 4
1 2 3 4y y y y
σ σ σ σ
= = = 31 2 431 2 4
1 2 3 4
V AV A V A V A
y y y y
⇒ = = = ( )1
1 2 3 4y y y y 1 2 3 4y y y y

1P
h m1
m2
H1 H2 H3 H4
2 B
V1 V2 V3 V4
1 1 1 2 2 3 3 4 4
2
B
h
M P Vm V m V m V m⇒ = + − −∑ ( )2
( ) ( ) 1 2 3 4, , , , .1 2From and V V V V can be found( ) ( ) 1 2 3 4, , , , .1 2From and V V V V can be found

1P
H1 H2 H3 H4
1 1 2 3 4P H H H H= + + +

Problem 4: Analyse the frame using cantilever method, if allob e : y g ,
the columns have the same area of cross section.
B C DA
m
120 kN
7 m 3.5 m 5 m
FE G
H
3.5m
180 kN
m
H
3.5m
J K LI J K LI
39

To locate centroidal vertical axis of the frame,
1 1 1 10 7 10.5 15.5A A A A
y
A A A A
× + × + × + ×
=
+ + +
To locate centroidal vertical axis of the frame,
33
8.25
4
m= =
1 1 1 1A A A A+ + + 4
120
8.25
1.25 2.25
7.25
180
Also, 31 2 431 2 4
V AV A V A V A
= = = 1 2 3 4V V V V
⇒ = = =
40
Also,
8.25 1.25 2.25 7.25
= = =
8.25 1.25 2.25 7.25
⇒

1 25 2 25 7 25V V V1 1 1
2 3 4
1.25 2.25 7.25
, ,
8.25 8.25 8.25
V V V
V V V= = =
1P
2
h m1
m2
O
H1 H2 H3 H4
2 O
V1 V2 V3 V4
1 1 1 2 2 3 3 4 4
2
O
h
M P Vm V m V m V m⇒ = + − −∑,For the top storey
1 2 3 4
3.5
120 15.5 8.5 5 0
2
V V V V⇒ × = × + × − × − ×
41

3 5 ⎛ ⎞ ⎛ ⎞1 1
1
3.5 1.25 2.25
120 15.5 8.5 5
2 8.25 8.25
V V
V ⎛ ⎞ ⎛ ⎞⇒ × = × + × − ×⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1 13.615V kN⇒ =
2
1.25 13.615
2.063 ,
8.25
2 25 13 615
V kN
×
= =
3
2.25 13.615
3.713 ,
8.25
7 25 13 615
V kN
×
= =
×
4
7.25 13.615
11.965
8.25
V kN
×
= =
: 13.615 2.063 3.713 11.965 0Check + − − =

H1 H2 H3 H4
V1 V2 V3
V4
O
V1 V2 V3 4
3 53 5⎛ ⎞
,For the bottom storey
1 2 3 4
3.53.5
120 180 15.5 8.5 5 03.5
22
OM V V V V⎛ ⎞⇒ × + × = × + × − × − ×+⎜ ⎟
⎝ ⎠
∑

3 5 1 25 2 253 5 V V⎛ ⎞ ⎛ ⎞⎛ ⎞ 1 1
1
3.5 1.25 2.253.5
120 180 15.5 8.5 53.5
2 8.25 8.252
V V
V ⎛ ⎞ ⎛ ⎞⎛ ⎞⇒ × + × = × + × − ×+⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1 61.267V kN⇒ =
1 25 61 267×
2
1.25 61.267
9.283 ,
8.25
2.25 61.267
16 709
V kN
V kN
×
= =
×
3 16.709 ,
8.25
7.25 61.267
53 841
V kN
V kN
= =
×
= =4 53.841
8.25
V kN= =
: 61 267 9 283 16 709 53 841 0Check + − − =
: 61.267 9.283 16.709 53.841 0Check + − − =

120kNMoments:
47.652kNm
A120kN 3.5 m
Moments: A
1.75m
47.652kNm 13.615kN
27.3kN
k13.615kN
45

Beam and Column moments:
29.9
B D
47.6 27.4 29.9
7547.6 57.3
B D
47.6
27.4 29.9
47 6 119 2
74.8187.8
57.3 29.975
143.2
166.8 96
96
104.7
104 747.6 119.2
74.8187.8 143.2
166.8 96 104.7
119 2 74 8187 8 143 2119.2 74.8187.8 143.2
46

BA25 kN
Home work
BA25 kN
6 m
5m
DC50 kN
3.53.5m
k
FE55 kN
4.5m
HG
47

FACTOR METHODFACTOR METHOD
• More accurate than Portal and Cantilever methodsMore accurate than Portal and Cantilever methods
• Specially useful when moments of inertia of various
b d ffmembers are different.
Basis:
• At any joint the total moment is shared by all the members
in proportion to their stiffnesses
• Half the moment gets carried over to the far endHalf the moment gets carried over to the far end

Gi d d l f tGirder and column factors:
• Relative stiffness of a member
I
k =
Girder factor at a joint
Relative stiffness of a member k
L
=
Girder factor at a joint
,k of all meeting at the joint
g
columns
=
∑
,
g
k of all members meeting at the joint
=
∑
Column factor at a joint
,
1
,
k of all meeting at the joint
c g
k of all members meeting at the joint
beams
= = −
∑
∑
, f g j∑

Moment factor for a member
,mC c k for a column= ,
,
m
m
f
G g k for a beam=
m
m
where c c half of column facor of far end
and g g half of girder facor of far end
= +
= +m
storeC sum of column moment factors for a y→∑
G sum of beam moment factors for a joint→∑

Problem 5: Analyse the frame using factor method.
H IG40 kN
ob e 5: y g
H IG
5m
40 kN
5 m 7.5 m
FED
3.5
80 kN
m5m
B CA

Total column moment above DEF 40 3.5 140kNm= × =
Total column moment above ABC 40 8 5 80 5 740kNm= × + × =Total column moment above ABC 40 8.5 80 5 740kNm× + ×

1 2 3 4 5 6 7 8 9 10 11 12
JOINT MEMBER k=I/L Ʃk FACTOR c/2 5+6 MOMENT Total Col DFB BeamJOINT MEMBER k I/L Ʃk FACTOR c/2,
g/2
from
far
end
5+6 MOMENT
FACTOR
Total
Col.
Mom, MT
Col.
Mom,
MC =
MT×
C/ƩC
DFB
=G/ƩG
Beam
Mom
= MC ×
DFBCol Beam Col Beam c
=Ʃk(be
g
=Ʃk(c
cm gm C=
cm ×
G =
gm × C/ƩC(
ams)/Ʃ
k
(
olumn
s)/Ʃk
k k
D
DA 0.2
0.686
0.29 0.5 0.79 0.158 740 99.4
DE 0.2 0.71 0.3 1.01 0.202 1 122.6
DG 0 286 0 29 0 3 0 59 0 169 140 23 2DG 0.286 0.29 0.3 0.59 0.169 140 23.2
E
ED 0.2
0.819
0.59 0.36 0.95 0.19 0.59 83.25
EH 0.286 0.41 0.27 0.68 0.194 140 26.6
EF 0.133 0.59 0.4 0.99 0.132 0.41 57.85
EB 0.2 0.41 0.5 0.91 0.182 740 114.5
FE 0 133 0 79 0 3 1 09 0 145 1 133 03
F
FE 0.133
0.772
0.79 0.3 1.09 0.145 1 133.03
FI 0.286 0.21 0.16 0.37 0.106 140 14.6
FC 0.2 0.21 0.5 0.71 0.142 740 89.3
G
GD 0.286
0.486
0.59 0.15 0.74 0.212 140 29.1
GH 0.2 0.41 0.23 0.64 0.128 1 29.1
H
HG 0.2
0.772
0.46 0.21 0.67 0.134 0.559 16.5
HE 0.286 0.54 0.21 0.75 0.215 140 29.5
HI 0.133 0.46 0.34 0.8 0.106 0.441 13.0
I
IH 0.133
0.419
0.68 0.23 0.91 0.121 1 16.58
IF 0 286 0 32 0 11 0 43 0 123 140 16 58IF 0.286 0.32 0.11 0.43 0.123 140 16.58
A AD 0.2 X 0.2 1 0 0.15 1.15 0.23 740 144.7
B BE 0.2 X 0.2 1 0 0.21 1.21 0.242 740 152.2
C CF 0.2 X 0.2 1 0 0.11 1.11 0.222 740 139.6
0 169 0 194 0 106 0 212 0 215 0 123 1 019F h C∑
, 0.169 0.194 0.106 0.212 0.215 0.123 1.019For the top storey C = + + + + + =∑
, 0.158 0.182 0.142 0.23 0.242 0.222 1.176For the bottom storey C = + + + + + =∑

Home work
B CA120 kN
6 m 6 m
4m
60 kN
FED60 kN
6m
H IG
54

SummarySummary
Approximate Methods of Analysis of Multi-storey Frames
• Analysis for vertical loads - Substitute frames-Loading
conditions for maximum positive and negative bending
moments in beams and maximum bending moment in columns
• Analysis for lateral loads - Portal method–Cantilever method–y
Factor method.

Module3 rajesh sir

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