Structural Analysis and Design Project of Ls

December 11, 2018 | Author: kvranapratap | Category: Truss, Framing (Construction), Column, Bending, Beam (Structure)
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1

STRUCTURAL STEEL DESIGN

PLANS AND DRAWINGS  A.  VICINITY MAP 12 12 356565333

LOCATION: PUERTO PRINSESA, PALAWAN, PHILIPPINES

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

B. DETAILS OF FLOOR PLAN/ FRAMING PLAN

C1

C2

C3

C5

C4

B1

S1

5.375m.

S3

S2

S4

B4 B5

B8

B6

B7

C6 B2

S5

5.375m. C8

C7

S6

S7

C9

C10

S8

C11

C12

B3 1.5m.

S10

S9

S11

S12

B9 C13

C14 10.75m.

C15 10.75m.

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

C17

C16 10.75m.

10.75m.

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STRUCTURAL STEEL DESIGN

B. DETAILS OF FLOOR PLAN/ FRAMING PLAN

C1

C2

C3

C5

C4

B1

S1

5.375m.

S3

S2

S4

B4 B5

B8

B6

B7

C6 B2

S5

5.375m. C8

C7

S6

S7

C9

C10

S8

C11

C12

B3 1.5m.

S10

S9

S11

S12

B9 C13

C14 10.75m.

C15 10.75m.

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

C17

C16 10.75m.

10.75m.

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STRUCTURAL STEEL DESIGN

GROUND FLOOR PLAN

SECOND FLOOR PLAN

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

C. ELEVATIONS

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

D. STRUCTURAL DETAILS AND SPECIFICATIONS

Design code: NSCP ASD sdsds 2232 2322 Strength: LC channel purlins: Fy= 170 Mpa Columns, Beams and Girders (A36 wide flanges and A36 angle bars): Fy= 248 MPa Tierods and Sagrods: Fy=248 Mpa E60xx: Fu= 415Mpa  A325 fasterners: Fv= 145Mpa wwewewe  A36 gusset plate: Fy= 248 Mpa

Concrete pedestal: F’c= 21 Mpa Square Base plate: Fy= 248 Mpa Computation of Forces, Moments (Criticals): use of  STAAD PRO 2007 and by manual Roofing: corrugated galvanized iron (g.i.): gage 22 Slab Weight

Concrete

150 mm thick 

Frame Partitions

Wood or steel studs, 13 mm. gypsum board each side

0.38 kPa

Ceiling Load

Gypsum board + mechanical duct allowance

1.2 + 0.2 = 1.4 kPa

Flooring

Concrete fill finish + hardwood flooring

0.345 + 0.19 = 0.535 kPa

Frame walls

Window, glass, frames and sash + wood sheating

0.38+ 0.855 = 1.235 kPa

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

STRUCTURAL ANALYSIS AND DESIGN  A. BASIS OF DESIGN

Student Number: 2007101109 Code: 101109 Z=1  Y=0 X=2 W=6  V=0 U=2 Type of Occupancy: Commercial Building (Office) Location of the Project: Puerto Prinsesa, Palawan, Philippines Type of Truss: Pratt Truss Span of Truss, ST = 10.75 meters  Angle of Inclination (Truss), θ = 22.10° Bay Distance, L = 10.75 meters

  1.5

L

L

L

L

Concrete slab thickness = 150mm. Beams and columns are made of wide-flange sections

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

`

  1.5m ROOF FRAMING PLAN LC (channel) purlins with sagrads and tierods

Roofing: corrugated galvanized iron (g.i.): gage 22 HEIGHT OF COLUMN, H

H1= 3.0m H2= 2.75m.

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

Roof  height

H2

0.6

H1

0.6 LEFT SIDE ELEVATION

c = -0.01

F D

H J

B

A

L C

E

G

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

I

K Page 8

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STRUCTURAL STEEL DESIGN

B. LOADING COMPUTATIONS Galvanized iron roofing gage 22 (w = 2.4 lbs/ft2)

x = ST / division of truss = 10.75 / 6 = 1.792 m.

S = x / cosө = 1.792 / cos(22.1 ) = 1.934 m. h = (tanө)*(ST/2) = 2.183 m. a. Dead Loads, DL Galvanized iron roofing gage 22 (w = 2.4 lbs/ft2)

RL= (2.4

)(  )(9.81 )(  ) (0.895m) = 102.86 N/m 2

b. Live Loads, LL  

Tributary Area, AT = 10.75 x 10.75 = 115.5625 mm2 RLL= (0.6) (0.895) = 0.537 KN/m = 537 N/m

c. Wind Loads, WL

2.183

2.7

0.6

3.0

0.6 WL= (c) (q) H= 0.6+ 3.0 + 0.6 + 2.7 +2.183 = 9.133 meters

 ) = 29.962 ft 

Height Zone = (9.133 m) ( C=

1.3 sin  

 0.5  1.3 sin 22.10   0.5  0.010908 q= 20 lb/ft ( zone III )

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

c = -0.01

 

 ) (9.81 m/s ) (  ) (0.967m) = 9.26 N/m      = 0.5 ( ) () (9.81 m/s ) () (0.967m) = 463.08 N/m

WLWINDWARD = 0.010908 (

WLLEEWARD

2

)(

2

2

2

C. DESIGN OF PURLINS, SAGRODS AND TIERODS

Design of Purlins Critical load = (Suction is not included)

[Getting the data from Association of Structural Engineers of the Philippines (ASEP) Steel Manual] Data Given: Fy = 170 Mpa LC 225 x 90 x 25 x 4.5

Weight, w (kg/m)  Area, A (mm2)

15.81 2015

Section Modulus about X, S x (x 103 mm3)

150.8

Section Modulus about Y, S y (x 103 mm3)

30.3

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

 According on what we learn in the previous topic of this subject,

f bx  Fbx



f by 

 1.00

F by

 

This is the formula in computing the adequacy of purlins… Where: f bx & f by = actual bending stress along X and Y axis respectively Fbx & Fby = allowable bending stress along X and Y axis respectively The section is said to economical if the interaction expression falls under the range 0.8



f bx  Fbx



f by  F by  

 0.9 .

 Assume that the purlins have compact sections. If it is compact, Fbx = 0.66Fy and Fby = 0.75Fy where Fy = 170 MPa

Computing for the total weight, wT Self weight= (15.81

) (9.81 m/s ) = 155.0961 N/m 2

wT  

 DL   LL

WT = [(155.0961+537+102.86)= 794.9561 N/m  After solving the total weight, solve for the total weight components, W x and W y

Wx = WT cos22.1˚ = 794.9561 N/m (cos22.1˚) Wx = 736.5496 N/m Wy = WT sin22.1˚ = 794.0561 N/m (sin22.1˚) Wy = 299.0818 N/m

GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

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STRUCTURAL STEEL DESIGN

Solving for moment about X and Y, M x and M y

   Mx =

   My= 

Mx =

My =

Mx = 10.64 kN-m

My = 1.08 kN-m

Solving actual stress along X and Y, f bx and f by f bx = f bx =

 

f by = f by =

f bx =70.56 MPa

 

f by =40 MPa

Solving allowable stress about X and Y, Fbx and Fby Fbx = 0.66 Fy

Fby = 0.75 Fy

Fbx = 0.66 (170)

Fby = 0.75 (170)

Fbx = 112.2 MPa

Fby = 127.5 MPa

Solving for the interaction expression,

f bx  f by   = Fbx F by  

f bx  Fbx



f by  F by  

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