Building Frames - 4 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : Building Frames - 4 Civil Engineering (CE) Notes | EduRev

 Page 1


Example 36.3 
Estimate approximate column reactions, beam and column moments using 
cantilever method of the frame shown in Fig. 36.8a. The columns are assumed to 
have equal cross sectional areas. 
 
Solution: 
This problem is already solved by portal method. The center of gravity of all 
column passes through centre column. 
 
()
m 5
10 5 0
=
+ +
+ +
= =
?
?
A A A
A A A
A
xA
x (from left column) 
 
 
 
                                                                             
                                                         
Page 2


Example 36.3 
Estimate approximate column reactions, beam and column moments using 
cantilever method of the frame shown in Fig. 36.8a. The columns are assumed to 
have equal cross sectional areas. 
 
Solution: 
This problem is already solved by portal method. The center of gravity of all 
column passes through centre column. 
 
()
m 5
10 5 0
=
+ +
+ +
= =
?
?
A A A
A A A
A
xA
x (from left column) 
 
 
 
                                                                             
                                                         
 
 
Taking a section through first storey hinges gives us the free body diagram as 
shown in Fig. 36.8b. Now the column left of C.G. must be subjected to 
tension and one on the right is subjected to compression.  
. .e i CB
From the third assumption, 
 
y y
y y
O M
A
O
A
M
- = ?
×
- =
× 5 5
 
 
Taking moment about O of all forces gives, 
 
() () ? = ? =
= × - ×
kN 3 ; kN 3
0 10 5 . 1 20
y y
y
O M
M
 
 
Taking moment about R of all forces left ofR ,  
 
 
                                                                              
                                                         
Page 3


Example 36.3 
Estimate approximate column reactions, beam and column moments using 
cantilever method of the frame shown in Fig. 36.8a. The columns are assumed to 
have equal cross sectional areas. 
 
Solution: 
This problem is already solved by portal method. The center of gravity of all 
column passes through centre column. 
 
()
m 5
10 5 0
=
+ +
+ +
= =
?
?
A A A
A A A
A
xA
x (from left column) 
 
 
 
                                                                             
                                                         
 
 
Taking a section through first storey hinges gives us the free body diagram as 
shown in Fig. 36.8b. Now the column left of C.G. must be subjected to 
tension and one on the right is subjected to compression.  
. .e i CB
From the third assumption, 
 
y y
y y
O M
A
O
A
M
- = ?
×
- =
× 5 5
 
 
Taking moment about O of all forces gives, 
 
() () ? = ? =
= × - ×
kN 3 ; kN 3
0 10 5 . 1 20
y y
y
O M
M
 
 
Taking moment about R of all forces left ofR ,  
 
 
                                                                              
                                                         
()
1.5 3 2.5 0
5kN
M
M
V
V
×-× =
=?
 
 
 
Taking moment of all forces right of about , S S
 
. kN 5 0 5 . 2 3 5 . 1 = ? = × - ×
O O
V V 
0 20 0 = - + + =
? O N M X
V V V F 
 
. kN 10 =
N
V 
 
Moments  
 
kN.m 5 . 7 5 . 1 5 = × =
CB
M 
 
kN.m 5 . 7 - =
CF
M 
 
kN.m 5 . 7
kN.m 5 . 7
kN.m 5 . 7
kN.m 5 . 7
kN.m 15
- =
=
- =
- =
=
IF
IH
FI
FC
FE
M
M
M
M
M
 
 
 
Tae a section through hinges (ref. Fig. 36.8c). Since the center of gravity 
passes through centre column the axial force in that column is zero. 
L K J , ,
 
                                                                              
                                                         
Page 4


Example 36.3 
Estimate approximate column reactions, beam and column moments using 
cantilever method of the frame shown in Fig. 36.8a. The columns are assumed to 
have equal cross sectional areas. 
 
Solution: 
This problem is already solved by portal method. The center of gravity of all 
column passes through centre column. 
 
()
m 5
10 5 0
=
+ +
+ +
= =
?
?
A A A
A A A
A
xA
x (from left column) 
 
 
 
                                                                             
                                                         
 
 
Taking a section through first storey hinges gives us the free body diagram as 
shown in Fig. 36.8b. Now the column left of C.G. must be subjected to 
tension and one on the right is subjected to compression.  
. .e i CB
From the third assumption, 
 
y y
y y
O M
A
O
A
M
- = ?
×
- =
× 5 5
 
 
Taking moment about O of all forces gives, 
 
() () ? = ? =
= × - ×
kN 3 ; kN 3
0 10 5 . 1 20
y y
y
O M
M
 
 
Taking moment about R of all forces left ofR ,  
 
 
                                                                              
                                                         
()
1.5 3 2.5 0
5kN
M
M
V
V
×-× =
=?
 
 
 
Taking moment of all forces right of about , S S
 
. kN 5 0 5 . 2 3 5 . 1 = ? = × - ×
O O
V V 
0 20 0 = - + + =
? O N M X
V V V F 
 
. kN 10 =
N
V 
 
Moments  
 
kN.m 5 . 7 5 . 1 5 = × =
CB
M 
 
kN.m 5 . 7 - =
CF
M 
 
kN.m 5 . 7
kN.m 5 . 7
kN.m 5 . 7
kN.m 5 . 7
kN.m 15
- =
=
- =
- =
=
IF
IH
FI
FC
FE
M
M
M
M
M
 
 
 
Tae a section through hinges (ref. Fig. 36.8c). Since the center of gravity 
passes through centre column the axial force in that column is zero. 
L K J , ,
 
                                                                              
                                                         
 
 
Taking moment about hingeL , can be evaluated. Thus, 
y
J
 
0 10 10 3 5 . 1 40 3 20 = × - × + × + ×
y
J 
 
( ) ( ) ? = ? = kN 15 ; kN 15
y y
L J 
 
  
Taking moment of all forces left of  about gives, P P
 
0 5 . 1 5 . 2 15 5 . 2 3 5 . 1 5 = × + × - × + ×
j
V 
 
() ? = kN 15
J
V 
 
Similarly taking moment of all forces right of Q aboutQ gives, 
 
0 5 . 1 5 . 2 15 5 . 2 3 5 . 1 5 = × + × - × + ×
L
V 
 
( ) ? = kN 15
L
V 
 
06
30 kN.
XJKL
K
FVVV
V
=++-
=
?
00=
 
 
                                                                             
                                                         
Page 5


Example 36.3 
Estimate approximate column reactions, beam and column moments using 
cantilever method of the frame shown in Fig. 36.8a. The columns are assumed to 
have equal cross sectional areas. 
 
Solution: 
This problem is already solved by portal method. The center of gravity of all 
column passes through centre column. 
 
()
m 5
10 5 0
=
+ +
+ +
= =
?
?
A A A
A A A
A
xA
x (from left column) 
 
 
 
                                                                             
                                                         
 
 
Taking a section through first storey hinges gives us the free body diagram as 
shown in Fig. 36.8b. Now the column left of C.G. must be subjected to 
tension and one on the right is subjected to compression.  
. .e i CB
From the third assumption, 
 
y y
y y
O M
A
O
A
M
- = ?
×
- =
× 5 5
 
 
Taking moment about O of all forces gives, 
 
() () ? = ? =
= × - ×
kN 3 ; kN 3
0 10 5 . 1 20
y y
y
O M
M
 
 
Taking moment about R of all forces left ofR ,  
 
 
                                                                              
                                                         
()
1.5 3 2.5 0
5kN
M
M
V
V
×-× =
=?
 
 
 
Taking moment of all forces right of about , S S
 
. kN 5 0 5 . 2 3 5 . 1 = ? = × - ×
O O
V V 
0 20 0 = - + + =
? O N M X
V V V F 
 
. kN 10 =
N
V 
 
Moments  
 
kN.m 5 . 7 5 . 1 5 = × =
CB
M 
 
kN.m 5 . 7 - =
CF
M 
 
kN.m 5 . 7
kN.m 5 . 7
kN.m 5 . 7
kN.m 5 . 7
kN.m 15
- =
=
- =
- =
=
IF
IH
FI
FC
FE
M
M
M
M
M
 
 
 
Tae a section through hinges (ref. Fig. 36.8c). Since the center of gravity 
passes through centre column the axial force in that column is zero. 
L K J , ,
 
                                                                              
                                                         
 
 
Taking moment about hingeL , can be evaluated. Thus, 
y
J
 
0 10 10 3 5 . 1 40 3 20 = × - × + × + ×
y
J 
 
( ) ( ) ? = ? = kN 15 ; kN 15
y y
L J 
 
  
Taking moment of all forces left of  about gives, P P
 
0 5 . 1 5 . 2 15 5 . 2 3 5 . 1 5 = × + × - × + ×
j
V 
 
() ? = kN 15
J
V 
 
Similarly taking moment of all forces right of Q aboutQ gives, 
 
0 5 . 1 5 . 2 15 5 . 2 3 5 . 1 5 = × + × - × + ×
L
V 
 
( ) ? = kN 15
L
V 
 
06
30 kN.
XJKL
K
FVVV
V
=++-
=
?
00=
 
 
                                                                             
                                                         
Moments 
 
5 1.5 7.5 kN.m ; 15 1.5 22.5 kN.m
BC BA
MM =× = = × = 
 
30 kN.m
BE
M =- 
 
10 1.5 15 kN.m ; 30 1.5 45 kN.m
EF ED
MM =× = = × = 
 
30 kN.m 30 kN.m
EB EH
MM =- =- 
 
5 1.5 7.5 kN.m ; 15 1.5 22.5 kN.m
HI HG
MM =× = = × = 
 
30 kN.m
HE
M =- 
 
 
Summary 
In this lesson, the building frames are analysed by approximate methods. 
Towards this end, the given indeterminate building fame is reduced into a 
determinate structure by suitable assumptions. The analysis of building frames to 
vertical loads was discussed in section 36.2.  In section 36.3, analysis of building 
frame to horizontal loads is discussed. Two different methods are used to 
analyse building frames to horizontal loads: portal and cantilever method. Typical 
numerical problems are solved to illustrate the procedure. 
 
                                                                             
                                                         
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