Portal Method :- Introduction:- Used for the analysis of building frames subjected to lateral loading. More Appropriate to low rise building frames. (i.e. Height is less than width of frame) Assumption:- An point of Contra flexure is located at centre of each column and beam. The Horizontal shear is divided among all the columns on the basis that each interior columns takes twice as much as exterior column.

1. Sanjivani Rural Education Society's
Sanjivani College of Engineering, Kopargaon 423603.
Department of Civil Engineering
Course Title –Advance Analysis of Structures
T. Y.BTech Civil, Semester –II(CE313)
Unit-V – Approximate Analysis of Multistoried Frames.
Topic- Numerical on Analysis of frame by Portal method
Unit-V – Approximate Analysis of Multistoried Frames.
Topic- Numerical on Analysis of frame by Portal method
By,
Prof. Santosh. R. Nawale
(Assistant Professor)

2. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
 Determine the approximate values of moment, shear and Axial forces
In member of frame loaded and supported as shown in figure using
Portal Method of Analysis.

3. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
 Assume point of contra flexure at centre of Beam and columns.

4. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
 Consider Upper storey and Releasing frame from nodes 4,5,6,7.
 The Horizontal shear is divided among all the columns on the basis that each
interior columns takes twice as much as exterior column.



F =0.
X
H+2H+2H+H=30
H=5KN.

5. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
( +ve & -ve)
-
F =0.
X
H 5 +30=0
 


( +ve & -ve)
F =0.
X
H +25-10=0
 


1
1
1
1
+Ve & Anticlockwise -Ve
-
H 5 +30=0
H = 25KN.
M =0.
@I
V 1.75+5 1.5=0
V =4.28KN.
Clockwise
 



  




F =0.
Y
4
4
( +ve& -ve)
+4.28
=4.28
-V =0
V KN.
2
2
2
2
+Ve & Anticlockwise -Ve
H +25-10=0
H =15KN.
M =0.
@J
4.28 1.75 V 1.75 10 1.5=0
V =4.29KN.
Clockwise
 



     




F =0.
Y
5
5
( +ve& -ve)
+4.29-4.28
=0.01
V =0
V KN.

6. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
( +ve & -ve)
F =0.
X
15 10 H =0
 

    
F =0.
Y
( +ve& -ve)
-4.28
V =0
V KN.
3
3
3
3
+Ve & Anticlockwise -Ve
15 10 H =0
H =5KN.
M =0.
@K
4.29 1.75 V 1.75 10 1.5=0
V =4.28KN.
Clockwise
 
  


     




F =0.
Y
6
6
( +ve& -ve)
-4.29+4.28
=0.01
-V =0
V KN.
7
7
 


F =0.
Y
( +ve& -ve)
-4.28
=4.28
V =0
V KN.

7. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
 Consider lower storey and Releasing frame from nodes 11,12,13,14.
 The Horizontal shear is divided among all the columns on the basis that each
interior columns takes twice as much as exterior column.
( v e ; - v e )
  




F = 0 .
X
H + 2 H + 2 H + H = 2 5 + 5 0
6 H = 9 0 .
H = 1 5 K N .

8. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
( +ve & -ve)
=
F =0.
X
60 15 5 H 0
 

   
( +ve & -ve)
F =0.
X
50 10 30 H =0
 

   
8
8
8
8
+Ve & Anticlockwise -Ve
=
60 15 5 H 0
H = 50KN.
M =0.
@E
15 1.5 V 1.75+5 1.5=0
V = 17.14KN.
Clockwise
 
   


    




F =0.
Y
11
11
( +ve& -ve)
+4.28+17.14
=21.42
-V =0
V KN.
9
9
9
9
+Ve & Anticlockwise -Ve
50 10 30 H =0
H =30KN.
M =0.
@F
17.14 1.75 V 1.75 10 1.5 30 1.5=0
V = 17.14KN.
0.01 17.14
Clockwise
 
   


       


  

F =0.
Y
12
12
( +ve& -ve)
+17.14
=0.01
V =0
V KN.

9. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
( +ve & -ve)
=
F =0.
X
30 10 30 H 0
 

    F =0.
Y
( +ve& -ve)
10
10
10
10
+Ve & Anticlockwise -Ve
=
30 10 30 H 0
H = 10KN.
M =0.
@G
17.14 1.75 30 1.5 V 1.75+10 1.5=0
V = 17.14KN.
17.14
Clockwise
 
   


      


 

F =0.
Y
13
13
( +ve& -ve)
-0.01+17.14
=0.01
V =0
V KN.
17.14
 

 

F =0.
Y
14
14
( +ve& -ve)
-4.28
=21.42
V =0
V KN.

10. Unit-V – Approximate Analysis of Multistoried Frames.
Topic- 2) Numerical on Analysis of frame by Portal method.
Bending Moment Diagram

11. THANK YOU