Chapter 2 Example 1 approximate analysis of frames subjected to lateral loads

Dept. of Civil & Urban Eng., iOTec-HU.

Hawassa University

Example 1 Analyse the building frame shown using the three methods i.e Portal, Cantilever and Factor method.

Fig. A Solution a) Portal Method

∑

⇒

⇒

⇒

Thus: Column ID HE GF

shear 5KN 10KN 5KN

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 1

Dept. of Civil & Urban Eng., iOTec-HU.

Column ID HE GF

Hawassa University

shear 5KN 10KN 5KN

Moment 5*2=10KNm 10*2=20KNm 5*2=10KNm

Joint I: ⇒

Joint H: Joint G:

Girder IH HG

Moment 10KNm 10KNm

Span/2 2m 2.5m

Shear 10/2=5KN 10/2.5=4KN

Joint I: Joint H: Joint G:

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 2

Dept. of Civil & Urban Eng., iOTec-HU.

∑ ⇒

Hawassa University

⇒ ⇒

Thus Column AD BE CF

shear 15KN 30KN 15KN

Column AD BE CF

shear 15KN 30KN 15KN

Moment 15*3=45KNm 30*3=90KNm 15*3=45KNm

Joint D: Joint E:

⇒

Joint F:

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 3

Dept. of Civil & Urban Eng., iOTec-HU. Girder DE EF

Moment 55KNm 55KNm

Hawassa University Span/2 2m 2.5m

Shear 55/2=27.5KN 55/2.5=22KN

I

H

D

G

F

E

Joint F:

C A

B

b) Cantilever Method

Taking area of moments about column DI: ∑

In cantilever method, it is assumed that column axial stresses are propositional to the horizontal distance from the center of gravity of the columns in the storey. From similarity of triangles

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 4

Dept. of Civil & Urban Eng., iOTec-HU.

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The axial forces would be thus:

Taking moments about point m: ⇒ ⇒

I

Joint I: VIH = FDI = 3.85KN

H

D

G

F

E

Joint H: VHG = VIH +FHE = 3.85+1.05=4.9KN Joint G: VGH = FFG = 4.9KN A

Girder IH HG

Shear 3.85KN 4.9KN

Span/2 2m 2.5m

B

C

Moment 3.85*2= KNm 4.9*2.5= KNm

At each joint sum of column moments must be equal to sum of Girder moments. Joint I: MID = MIH = 7.7KNm ___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 5

Dept. of Civil & Urban Eng., iOTec-HU.

Hawassa University

Joint H: MHE = MIH +MHG = 7.7+12.25=19.95KNm Joint G: MFG = MHG = 12.25KNm

Column ID HE FG

Moment 7.7KNm 19.95KNm 12.25KNm

height/2 2m 2m 2m

Shear 7.7/2=3.85KN 19.95/2= KN 12.25/2=6.13KN

Taking area of moments about column DI:

∑

In cantilever method, it is assumed that column axial stresses are propositional to the horizontal distance from the center of gravity of the columns in the storey. From similarity of triangles

The axial forces would be thus:

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 6

Dept. of Civil & Urban Eng., iOTec-HU.

Hawassa University

Taking moments about point n: ⇒ ⇒

Joint D: VDE = FAD- FDI = 25.08-3.85=21.23KN Joint F: VEF = FCF – FGF = 31.92-4.9=27.02KN

Girder DE EF

Shear 21.23KN 27.02KN

Span/2 2m 2.5m

Moment 21.23*2=42.46KNm 27.02*2.5=67.55KNm

At each joint sum of column moments must be equal to sum of Girder moments. Joint D: MAD = MDE - MDI = 42.46-7.7=34.76KNm Joint E: MBE = MDE +MEF- MFH = 42.46+67.55-19.95=90.06KNm Joint F: MCF = MEF – MGF = 67.55-12.25=55.3KNm

Column AD BE CF

Moment 34.76KNm 90.06KNm 55.3KNm

height/2 3m 3m 3m

Shear 34.76/3=11.59KN 90.06/3=30.02KN 55.3/3=18.43KN

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 7

Dept. of Civil & Urban Eng., iOTec-HU.

Hawassa University

c) Factor Method

, Column factor=1-g Joint A B C

Girder Factor(g) 0 0 0

Column factor(c) 1 1 1

D

0.5

E

0.67

F

0.5

G

0.67

H

0.8

I

0.67

1 1 1 1 1 1 1 1 1 1 1 1

0.92 0.83 1.13 1.07 0.92 0.83 1 1.25 1.17 1.33 1 1.25

14.04

51.43

Column Moment (M)

0.92 0.83 1.13 1.07 0.92 0.83 1 1.25 1.17 1.33 1 1.25

Storey Constant (A)

Column moment factor (C)

0.25 0.33 0.33 0.4 0.25 0.33 0.5 0.25 0.5 0.33 0.5 0.25

Relative stiffness (k)

0.67 0.5 0.8 0.67 0.67 0.5 0.5 1 0.67 1 0.5 1

Half values of opposite end

ID DI HE EH GF FG DA AD EB BE FC CF

column(3) + column(4)

Ground Story

column Factor(c)

Top Story

Column

(2) Column moment factors

12.92 11.65 15.87 15.02 12.92 11.65 51.43 64.29 60.17 68.4 51.43 64.29

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 8

Dept. of Civil & Urban Eng., iOTec-HU.

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(3) Story Constant For top story Total horizontal shear of top story=20KN

Height of top story=4m

Summation of column moment factors for top story ∑

( )

∑

For bottom story Total horizontal shear of bottom story =20+40=60KN

Height of bottom story=6m

Summation of column moment factors for bottom story ∑

( )

∑

(4) Moments at the ends of Columns:

Joint Constant (B)

Girder Moment (M)

Half values of opposite end 0.1 0.17 0.17 0.1 0.17 0.25 0.25 0.17

Girder moment factor (G)

F

0.33 0.2 0.2 0.33 0.5 0.33 0.33 0.5

Relative stiffness (k)

G D E

IH HI HG GH DE ED EF FE

column(3) + column(4)

H

Girder Factor(g)

I

Girder

Joint

(5) Girder Moment factors:

0.43 0.37 0.37 0.43 0.67 0.58 0.58 0.67

2 2 2 2 2 2 2 2

0.86 0.74 0.74 0.86 1.34 1.16 1.16 1.34

15.02

12.92 7.93 7.93 12.92 63.07 37.6 37.6

10.72 15.02 47.07 32.41 47.07

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 9

Dept. of Civil & Urban Eng., iOTec-HU.

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(6) Joint Constants:

(6) Girder Moment The girder moments are calculated by the formula shown below and are given in the table above.

___________________________________________________________________________ Advanced Structural Design (CEng 5721) Chapter 2 Examples 10