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7/26/2019 Calculation Note-Construction Case Structure

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DESIGN CALCULATION ON CONSTRUCTION CASESTRUCTURE

Project: STEEL CASE STRUCTUREDate: April 12, 2015

Author: Dr. NGUYEN HUU THANH

Table of Content

1. Geometry Structure Modelling ....................................................................................... 2

2. Material properties ............................................................................................................. 3

3. Section properties .............................................................................................................. 3

4. Section Assignment ........................................................................................................... 5

5. Boundary condition ........................................................................................................... 7

6. Loading condition .............................................................................................................. 8

7. Analysis results ................................................................................................................ 10

8. Design Check for Strength capacity and stability ........................................................... 14

9. Comments and Conclusions ............................................................................................ 16


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1.

Geometry Structure Modelling

3D model

Elevations and plan views


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2.

Material properties

The structure is in steel material with the properties shown in below Table

MaterialWeight density

(T/m3)

Elastic modulus

E (MPa)

Yield stress

(MPa)Note

1. Milk steel 7.85 200000 250

2. Rebar 7.85 200000 250

3. Concrete 2.4 24000 30

3.

Section propertiesThe steel section is pipe shape. The section properties are shown in the following Table andFigures in unit of cm.

TABLE: Frame Section Properties 01 - General

Section Name Material Shape t3 tw Area Torsonst !33 !"" AS" AS3

Tex Tex Tex !" !" !"2 !"4 !"4 !"4 !"2 !"2

#I#E 2$%2 ST&2500 #i'e 2.$ 0.2 1.5$ 2.4$ 1.24 1.24 0.$( 0.$(

#I#E 4(%2.5 ST&2500 #i'e 4.( 0.25 3.)5 1(.* (.( (.( 1.*3 1.*3

#I#E )0%3 ST&2500 #i'e ) 0.3 5.3$ 43.$) 21.** 21.** 2.)( 2.)(

TABLE: Frame Section Properties 01 - General

SectionName Material Shape t3 tw S33 S"" #33 #"" $33 $""

Tex Tex Tex !" !" !"3 !"3 !"3 !"3 !" !"

#I#E 2$%2 ST&2500 #i'e 2.$ 0.2 0.(1 0.(1 1.25 1.25 0.**)$ 0.**)$

#I#E 4(%2.5 ST&2500 #i'e 4.( 0.25 4.04 4.04 5.41 5.41 1.)4)4 1.)4)4

#I#E )0%3 ST&2500 #i'e ) 0.3 $.2( $.2( (.$) (.$) 2.01* 2.01*


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4.

Section Assignment

All structural members are assigned with steel pip section.

Section assignment to steel members


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Section assignment


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5.

Boundary condition

Boundary condition describing wheels is applied to the model as presented in following

figures. U=0 means the displacement is not constrained, U=1 means the constrained.

Rolling constrain

Ux=0, Uy=0, Uz=1

Rolling constrain

Ux=0, Uy=1, Uz=0

Rolling constrain

Ux=0, Uy=1, Uz=0


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Boundary condition

6. Loading condition

Three load cases are created and analyzed

Dead Load (self-weight) (DL): the weight of structure calculated internally in the software.

Live Load 1 (LL1): Describe one worker with tools and equipment stand at center of thedeck. The load is concentrate force of magnitude of 1kN.

Live Load 2 (LL2): Two people standing on outer half of the chair. The load is uniform load

on half of the deck area with magnitude of 5.0KN/m2. Total load on the deck will be 5.0 x0.4 x 1 = 2kN. The applied load is shown in following Figures.

Rolling constrain

Ux=0, Uy=0, Uz=1

Rolling constrainUx=0, Uy=1, Uz=0

Rolling constrain

Ux=0, Uy=1, Uz=0


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Live load 1 (LL1), Concentrate Load P=1kN

Live load 2 (LL2), Uniform load on half deck area, q=5kN/m2


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7.

Analysis results

DISPLACEMENT - DEFORMATION

Displacement of Dead Load case, mm

Displacement of LL1 Load case, mm


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Displacement of LL2 Load case, mm

INTERNAL FORCES

M33 bending moment diagram of Dead Load case, kN.m


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Bending moment M33 of Load Combination, kN.m

Axial force N of Load Combination, kN


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D/C ratio strength design check result


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9.

Comments and Conclusions

- The structure of the case will work in safety if the design of structure and member

section follows the above input and analysis data.

- Designer is requested to revise the drawing in accordance with this analysis report.


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Design result of member with section Pipe 60x3

TABLE: Steel %esi&n 1 - S'mmar( %ata - A!S-L$F%)3

Frame %esi&nSect %esi&nT(pe Stat's $atio $atioT(pe om*o Location ErrMs&

Tex Tex Tex Tex U+i,e-- Tex Tex " Tex

1( #I#E)0%3 e" N e--e- 0.02134) # CO3 0.5 N e--e-

20 #I#E)0%3 e" N e--e- 0.02134) # CO3 0 N e--e-

21 #I#E)0%3 e" N e--e- 0.023($1 # CO3 0.5 N e--e-

22 #I#E)0%3 e" N e--e- 0.023($1 # CO3 0 N e--e-

24 #I#E)0%3 e" N e--e- 0.00*)2$ # CO3 0 N e--e-

25 #I#E)0%3 e" N e--e- 0.00*)2$ # CO3 0 N e--e-

2) #I#E)0%3 e" N e--e- 0.00*)2$ # CO3 0.1 N e--e-

2$ #I#E)0%3 e" N e--e- 0.00*)2$ # CO3 0.1 N e--e-

2* #I#E)0%3 C,"+ N e--e- 0.00($44 # CO3 0 N e--e-

2( #I#E)0%3 C,"+ N e--e- 0.00($44 # CO3 0 N e--e-

4) #I#E)0%3 C,"+ N e--e- 0.)035)5 # CO3 0.* N e--e-

4* #I#E)0%3 C,"+ N e--e- 0.4($003 # CO3 0.* N e--e-

50 #I#E)0%3 C,"+ N e--e- 0.115(*5 # CO3 0.4 N e--e-

51 #I#E)0%3 C,"+ N e--e- 0.1$(1(( # CO3 0 N e--e-

52 #I#E)0%3 C,"+ N e--e- 0.1*43$3 # CO3 0.4 N e--e-

5( #I#E)0%3 C,"+ N e--e- 0.)035)5 # CO3 0.* N e--e-

)0 #I#E)0%3 C,"+ N e--e- 0.4($003 # CO3 0.* N e--e-

)2 #I#E)0%3 C,"+ N e--e- 0.115(*5 # CO3 0.4 N e--e-

)3 #I#E)0%3 C,"+ N e--e- 0.1$(1(( # CO3 0 N e--e-

)4 #I#E)0%3 C,"+ N e--e- 0.1*43$3 # CO3 0.4 N e--e-

*1 #I#E)0%3 e" N e--e- 0.)01(04 # CO3 0 N e--e-

*2 #I#E)0%3 e" N e--e- 0.445$*3 # CO3 0.35 N e--e-

*3 #I#E)0%3 e" N e--e- 0.)01(04 # CO3 0 N e--e-

*4 #I#E)0%3 e" N e--e- 0.445$*3 # CO3 0.35 N e--e-


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Design result of member with section Pipe 27x2

TABLE: Steel %esi&n 1 - S'mmar( %ata - A!S-L$F%)3

Frame %esi&nSect %esi&nT(pe Stat's $atio $atioT(pe om*o Location ErrMs&

Tex Tex Tex Tex U+i,e-- Tex Tex " Tex

2 #I#E2$%2 r!e N e--e- 0.4$3((1 # CO3 0 N e--e-

3 #I#E2$%2 e" N e--e- 0.23221 # CO3 0 N e--e-

4 #I#E2$%2 e" Oer-re--ed 0.($$*$2 # CO3 0 N e--e-

5 #I#E2$%2 e" N e--e- 0.$142$1 # CO3 0.2 N e--e-

) #I#E2$%2 e" N e--e- 0.$0321) # CO3 0 N e--e-

$ #I#E2$%2 C,"+ Oer-re--ed 1.10$2(5 # CO3 0 N e--e-

( #I#E2$%2 C,"+ N e--e- 0.2*12$ # CO3 0 N e--e-

11 #I#E2$%2 r!e N e--e- 0.4$3((1 # CO3 0 N e--e-

12 #I#E2$%2 e" N e--e- 0.23221 # CO3 0 N e--e-

13 #I#E2$%2 e" Oer-re--ed 0.($$*$2 # CO3 0 N e--e-

14 #I#E2$%2 e" N e--e- 0.$142$1 # CO3 0.2 N e--e-

15 #I#E2$%2 e" N e--e- 0.$0321) # CO3 0 N e--e-

1) #I#E2$%2 C,"+ Oer-re--ed 1.10$2(5 # CO3 0 N e--e-

1$ #I#E2$%2 C,"+ N e--e- 0.2*12$ # CO3 0 N e--e-

30 #I#E2$%2 C,"+ N e--e- 0.51)$$* # CO3 0 N e--e-

31 #I#E2$%2 C,"+ N e--e- 0.53()45 # CO3 0.4 N e--e-

32 #I#E2$%2 C,"+ Oer-re--ed 1.02(03( # CO3 0 N e--e-

33 #I#E2$%2 C,"+ N e--e- 0.51)$$* # CO3 0 N e--e-

34 #I#E2$%2 C,"+ N e--e- 0.53()45 # CO3 0.4 N e--e-

35 #I#E2$%2 C,"+ Oer-re--ed 1.02(03( # CO3 0 N e--e-

3* #I#E2$%2 e" N e--e- 0.0205)5 # CO3 0.5 N e--e-

3( #I#E2$%2 e" N e--e- 0.0205)5 # CO3 0 N e--e-

40 #I#E2$%2 C,"+ N e--e- 0.0($$3$ # CO3 0.3 N e--e-

41 #I#E2$%2 C,"+ N e--e- 0.245(2 # CO3 0.4 N e--e-

$$ #I#E2$%2 e" N e--e- 0.020*1( # CO3 0.5 N e--e-

$* #I#E2$%2 e" N e--e- 0.0440*2 # CO3 0.5 N e--e-

*$ #I#E2$%2 e" N e--e- 0.2(2(2$ # CO3 0 N e--e-

** #I#E2$%2 e" N e--e- 0.0)$111 # CO3 0.4 N e--e-*( #I#E2$%2 e" Oer-re--ed 1.111)52 # CO3 0 N e--e-

(0 #I#E2$%2 e" N e--e- 0.224)55 # CO3 0.4 N e--e-

(1 #I#E2$%2 C,"+ N e--e- 0.1144)4 # CO3 0 N e--e-

(2 #I#E2$%2 C,"+ N e--e- 0.1(5)42 # CO3 0 N e--e-

(3 #I#E2$%2 C,"+ N e--e- 0.221$0) # CO3 0.4 N e--e-

(4 #I#E2$%2 e" N e--e- 0.)13*(1 # CO3 0 N e--e-

(5 #I#E2$%2 e" N e--e- 0.3014$( # CO3 0.4 N e--e-

() #I#E2$%2 e" N e--e- 0.4(23)) # CO3 0 N e--e-

($ #I#E2$%2 e" N e--e- 0.21242( # CO3 0.4 N e--e-


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145 #I#E2$%2 C,"+ N e--e- 0.10$) # CO3 0.4 N e--e-

14) #I#E2$%2 C,"+ N e--e- 0.1($*45 # CO3 0.4 N e--e-

14$ #I#E2$%2 e" N e--e- 0.01))22 # CO3 0 N e--e-

14* #I#E2$%2 e" N e--e- 0.01))22 # CO3 0.5 N e--e-

14( #I#E2$%2 e" N e--e- 0.1*$44* # CO3 0 N e--e-

150 #I#E2$%2 e" N e--e- 0.23)*$4 # CO3 0.35 N e--e-