Advanced analysis of structures ppt1.pdf

ppt

1. Sanjivani Rural Education Society's GD
Sanjivani College of Engineering, Kopargaon 423603. U
Department of Civil Engineering ANJIVANI
Course Title -Advance Analysis of Structures
T. Y.BTech Civil, Semester -II(CE313)

Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method

By,
Prof. Santosh. R. Nawale
(Assistant Professor)

2. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

O Determine the approximate values of moment, shear and Axial forces
In member of frame loaded and supported as shown in figure using
Cantilever Method of Analysis. Area of each exterior columns is one
half of the area of the Interior Columns.

3. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

Q Assume point of contra flexure at centre of Beam and columns.

4. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

1) Determination of the CG of the Frame:
O Exterior Column= A; Interior Column=2A.
Q To find out CG of the frame. Taking moment of the area of columns at leftmost
column axis.
O A1 x 0+A2 x 4.54+A3 x 10.5=(A1+A2+A3)X
2A x4.5 +A x10.5=(A+2A+A) X
19.5A=4A X ; X =4.875m.
DO Consider Upper storey and Releasing frame from nodes 3,4,5.

5. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

H3+H4+H5=25
Taking Moment @ Point P
EM @p=0.

OAs per Assumption: Axial forces in column is proportional to its distance from
centroid and areas of the column group at that level.
Ratio=Stress/ Distance from C.G.

Va x 4.875 +Va x0.375 +Vsx5.625=(H +Hi+Hs)x1.5.
sd 2 e +0.154 x Va x0.375 +1.154 x Vs x5.625=
X19.
LL Vi VE
4.875 0.375 5.625 4.875 Vs + 0.058 Va +6.491 Vs =37.5

0375% say |] 11.424 Va =37.5.

Va=05x4.875 3

_5.625V. Vs =3.283 ; Va=0.506 ;Vs=3.789.

= 3=
Ve 4.875 11 54V,

6. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

V1=3.283 2 3.789= V2
Hi2007% 225m H 3m | =

— — _ I
Hz 75732H2 15m
1.5m
.—

=
Hs
Va=0.506

EF, =0.
T+ve& d-ve)
V,-3.283-0.506=0
LV, =3.789KN.
EM © EM:

Clockwise Ve & Anticlockwise +Ve Clockwise -Ve & Anticlockwise +Ve

:3.283x2.25-H.x1.5=0 ..3.283x2.25-H,x1.543.789x3=0

“A, =4.925KN. H,=12.503KN.

EF 0. EF.

(> t+ve e -ve) (> +ve & © -ve)
-,—H,4.925+25=0 2. -H,+20.076-12.503=0

-.H,20.076KN. -.H,57.573KN.

7. Unit-V = Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

Q Consider lower storey and Releasing frame from nodes 8,9,10.
Hs+Hs+Hu=25+50

Taking Moment @ Point P

EM @p=0.

Vex 4.875 +V5 x0.375
+V5x5.625=(Hs +Ho+H10)x 4.5

<+— re | LT
| Ho Hoy Hao 1 -50 x3.
Vo Vo IP Vio OAs per Assumption: Axial forces
L 4.875m eel este = in column is proportional to its
ie ‚625m —

distance from centroid
and areas of the column group at that level. Ratio=Stress/Distance from C.G.

Va x 4.875 +Vo x0.375 +V5x5.625=(Ho +Ho+Hio)x 4.5
A= 2h À -50 x3

4.875 0.375 5.625 Va x 4.875 +0.154 x Va x0.375 41.154 x Va x5.625=

Ve__05Vo _ Vio 337.50-150.

sé 2 5.625 |! 4.875 Vs + 0.058 Vs +6.491 Vo =187,50,
a Ea =

Vo=0.5%4.895 =0-154Vg|| 11.424 Va =187.50.

5.625, Ve =16,42 ; Vo=0,154x16.42=2,53;

B=1.154V,
Vio= 1.154 x 16.42=18.95.

10% 4875 8

8. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

VE Y
G sal" heey 225m Hd |
4 -

= —
DSKN sm Hi L5m fh
: += ee

| Hs | He
1323283 Va=0.506

V5=3.283 V4 =0.506 Vs=3.789
H3=4.925 H4=12.503 Hs-7.573
—=

fv. state) ism f" u am 45m
—_ 2.25m 3m =$ m

7 t D
45.065=Hs 32,54=Hr 225m

Ho 8
18,95 = Vio

Vo=16.42

9. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

Va3=3.283 Va =0.506 Vs 23.789
_Hs=4.925 fis Ha=12.503 H5=7.573
1 ism tv 13.182 Vs ism San he 15.164 v| 15m |
2.25m =
45.065= He 7 Te 32.542 H7 2.25
ur His
18.95 = eal

SFY =O(T +ve& +-ve)
—13.14+0.506+V,-2,53=0
"V,=15.164KN.

EMap Ÿ

Clockwise Ve & Anticlockwise +Ve
*,13,14x2,25-Hox2.25
15.164x3-12.503x1.5=0
,Hs=25.023KN.

EF 0. +ve & e -ve)
45.064 H,-25.023+12.503=0
.H,=32.54KN.

Eh.

[> +ve & < -ve)

:.3,283+V, -1642=0
"V,=13.14KN.

32.54-H „+7.573=0
|-H,,740.113KN.

Mar”
“lockwise -Ve & Anticlockwise +V
.H,x2.254.925x1.5
+13.14x2.25=0
..H,=9.86KN.

EF =0(> +ve & — -ve)
:.-H,+50-9.86+4.925=0
H,=45.065KN.

10. Unit-V - Approximate Analysis of Multistoried Frames.
Topic- 1) Numerical on Analysis of frame by Cantilever method.

UBending Moment Diagram

3.283 3.789

11. + THANK YOU