structural analysis and design project of ls

7/27/2019 Structural Analysis and Design Project of Ls

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GUILLERMO, Lou Stalin Dominique D.J. - 2007101109

1STRUCTURAL STEEL DESIGN

PLANS AND DRAWINGS

A. VICINITY MAP 12 356565333

LOCATION: PUERTO PRINSESA, PALAWAN, PHILIPPINES





B. DETAILS OF FLOOR PLAN/ FRAMING PLAN

5.375m.

5.375m.

5m.

10.75m. 10.75m. 10.75m. 10.75m.

B1

B2

B3

B9

B4

B5 B6B7

B8

S1 S2 S3 S4

S5 S6S7 S8

S9 S10 S11 S12

C1 C2 C3C4

C5

C6

C7

C8

C13

C9

C14

C10

C15

C11

C16

C12

C17





GROUND FLOOR PLAN

SECOND FLOOR PLAN





C. ELEVATIONS





D. STRUCTURAL DETAILS AND SPECIFICATIONS

Design code: NSCP ASD sdsds 2232 2322222s

Strength: LC channel purlins: Fy= 170 Mpa

Columns, 222 22

Slab Weight Concrete 150 mm thick

Frame PartitionsWood or steel studs, 13 mm.

gypsum board each side0.38 kPa

Ceiling LoadGypsum board + mechanical duct

allowance1.2 + 0.2 = 1.4 kPa

FlooringConcrete fill finish + hardwood

flooring0.345 + 0.19 = 0.535 kPa

Frame walls

Window, glass, frames and sash +

wood sheating 0.38+ 0.855 = 1.235 kPa

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

Concrete slab thickness = 150mm.

Beams and columns are made of wide-flange sections

`

1.5

L L L L

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.

Roof

height

H2

0.6

H1

0.6





LEFT SIDE ELEVATION

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)

c = -0.01

A

B

D

F

H

J

L

C E G I K





RL= (2.4)(

)(9.81 )( )2(0.895m) = 102.86 N/m

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

WL= (c) (q)

H= 0.6+ 3.0 + 0.6 + 2.7 +2.183 = 9.133 meters

Height Zone = (9.133 m) (

) = 29.962 ft

C= 1.3 sin 0.5 1.3 sin 22.10 0.5 0.010908

q= 20 lb/ft ( zone III )

2.7

0.6

3.0

0.6

2.183





WLWINDWARD = 0.010908 ( ) ( ) (9.81 m/s2

) ( )2

(0.967m) = 9.26 N/m

WLLEEWARD = 0.5 ( ) ( ) (9.81 m/s2) ( ) 2 (0.967m) = 463.08 N/m

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) 15.81

Area, A (mm2) 2015

Section Modulus about X, S x (x 103

mm3)

150.8

Section Modulus about Y, S y (x 103

mm3)

30.3

c = -0.01





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

1.00by bx

bx by

f f

F F

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 0.9by bx

bx by

f f

F F .

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/s2) = 155.0961 N/m

T w 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





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 by =

f bx = 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,by bx

bx by

f f

F F <1.0

by bx

bx by

f f

F F =

+ =0.909< 1.0 OK!

.: the purlins is adequate and economical.

Design of Sagrods

TSAGRODS = WyL

TSAGRODS = (299.08) (10.75)

TSAGRODS = 2.01 KN

T = FT Ag

2.01 x 10^3 = 0.6 (248) [ dsagrods

2] x7

dsagrods = 10.97mm, use dsagrods = 12 mm

Design og Tierods

0H

F

TTierods = Tsagrods cos22.1˚

TTierods = N

2.17 x 10^3 = 0.6 (248) [ dtierods

2]

dTierods = 4.31 mm

use dTierods= 5mm





D. LOADINGS AND ANALYSIS OF TRUSS

Considering one purlin:

x =x

2

=(10.75)

2

= 3958.97

y =y

16=

(299.08)(10.75)

16=200.94

Types of Truss R x R y

Middle Full Truss

Side Full Truss

Half Truss

2R x = 2(3.95897)

= 7.91794 KN

2R y = 2(200.94)

= 401.88 KN

Ceiling Load, CL

Using suspended metal lath and gypsum plaster as ceiling (From NSCP Vol. 1,

Section 2-6, Table 204-2)

= (0.02+0.008)(14.75)(7.25)=22.243 or 22243 The Section to Be Used For Each Type of Truss Members

Type of Truss Members The Section & Its Properties

Top Chords and Bottom

Chords

L 100 x 100 x14

Orientation


Area, A (mm2) 5238

Radius of Gyration about X, r x

(mm) 29.95

Radius of Gyration about Y, r y

(mm) 29.95





Web Members

(Both Tension &

Compression)

L 100 x 100 x 14

Orientation


Area, A (mm2) 2.619

Radius of Gyration about X, r x

(mm) 29.95

Radius of Gyration about Y, r y

(mm) 29.95

Self Weight

3Rx= 11876.91 N

4Rx= 15835.88 N

14Ry= 2813.16 N

CL= 0.48 kpa (10.75 m)

CL= 5.16 KN/m = 5160 N/m 1154.18 N/m

UL= CL+SW

UL= 5160 + 1154.18

UL= 6314.18 N/m

SOLVING FOR THE FORCES OF THE MEMBERS OF THE TRUSS

3Rx

[email protected]

3Rx 3Rx

3Rx

4Rx

4Rx

4Rx

3@ 1.93m.

4Rx

Ɵ=22.1

2.18m.

14Ry14Ry





SAMPLE COMPUTATION OF FORCES IN CRITICAL MEMBERS

JOINT A

TABLE OF ANALSYSISTOP CHORD T/C FORCE (KN) LENGTH (m)

AB=JL C -420.172 1.934

BD=HJ C -420.172 1.934

DF=FH C -394.593 1.934

BOTTOM CHORD T/C FORCE 9KN) LENGTH (m)

AC=KL T 389.302 1.792

CE=IK T 365.602 1.792

EG=GI T 420.059 1.792

WEB MEMBER T/C FORCE (KN) LENGTH (m)

BC=JK C -7.934 0.728

CD=HK T 30.531 2.308

DE=HI T 77.651 1.455

EF=FI C -85.828 2.824

FG T 146.314 2.183

A AC

AB

22.1O

Ɵ

22.1O

171.67 KN 3R X

14R

AB sin22.1o

+ 171669.621 – 3R x cos22.1o

– 14R y sin22.1 = 0 AB = -420.172 kN (T)

AB = 420.172 kN (C)

AC + AB cos22.1o + 3R x sin22.1o – 14R y cos22.1o = 0

AC = 389.302 kN (T)





DESIGN OF TOP AND BOTTOM CHORD

E. DESIGN FOR TOP CHORD

L= 1.934 m.

r= 29.955

Test for buckling: Solving for FS:

64.555 < 200 OK FS= 1.842

Cc= 126.169 Fa= 117.026 Mpa

kL/r < Cc

Therefore: Intermediate column

Solving for Required Area:

A=

= 3590.412 MPa

A= 3590.412 mm2 look for the manual for most efficient section

A/2 = 1795.206 mm2 for single angle bar

Adopt the equal angle bar section L 100 x 100 x 14

F. DESIGN FOR BOTTOM CHORD

Pcritical=420.059 KN

L= 1.792 m

Ft= 148.8 MPa

Solving for Required Area

A=

= 2822.980 mm^2

200r

kL3

3

)(8

)/(

)(8

)/(3

3

5

Cc

r kl

Cc

r kl FS

Fy

E Cc

22

Fs

Fy

Cc

r kl Fa

2

2

)(2

)/(1

A

P Fa

Fy Ft 6.0

A

P Fa

P critical = -420.172 KN






A/2= 1411.49 mm2 for sngle angle bar

Adopt the Equal angle bar section L 100 x 100 x 14

G. DESIGN FOR WEB MEMBER

For compression:

Pcrit= -85.828 KN

L= 2.824m

test for buckling Solve for FS

94.267 < 152.390 OK! FS = 1.895

Cc = 126.169 Fa= 94.357 MPa

kL/r < Cc

Therefore: Intermediate Column

solving for Required Area:

A

A=

= 909.604 mm2

A = 909.604 mm2 look for the manual for most efficient section

A/2 = 454.802 mm2 for single angle bar

Adopt the Equal angle bar section L100 x 100 x 14

For Tension:

Pcritical=146.314 KN

L= 2.824m

Ft=148.8 MPa

390.152

r

kL3

3

)(8

)/(

)(8

)/(3

3

5

Cc

r kl

Cc

r kl FS

Fy

E Cc

22

Fs

Fy

Cc

r kl Fa

2

2

)(2

)/(1

A

P Fa

Fy Ft 6.0





P= 171669.621P= 171669.621

H=3m.

solving for Required Area:

A=

= 1550.639 mm2


A/2= 775.319 mm2 for single angle bar

Adopt the Equal angle bar section L100 x 100 x 14

DESIGN FOR SECOND FLOOR COLUMN

Consider the most critical column due to axial load alone and note that the height of the second

floor column H2 = 2.75 m = 2,750mm. Assume that 0kL

r

that is less that Cc making it an

intermediate column and that the column is hinged at both ends.

0 c kL C IntermediateColumn

r

Knowing 0kLr ,

5

3

FS

Then,

248

153

148.8

a

a

MPaF

F MPa

A

P Ft

H=3m.H=3m.





P= 171669.621

Fs= 1.764 Fa= 135.58 MPa

Assume 0.6 Fa

60% of Fa= 81.348 MPa

Solving for required Area

A =

A= 2110.319 mm2

look in the manual for most efficient section

Adopt wide flange section W 8 x 10

3.0 m

3

3

)(8

)/(

)(8

)/(3

3

5

Cc

r kl

Cc

r kl Fs

Fs

Fy

Cc

r kl Fa *

)(2

)/(1

2

2

A

P Fa %60





ORIENTATION W 8 x 10

Weight, W (kg/m) 15

Area, A (m2) 1910

S x ( x10

3

mm ) 128

S Y ( x103 mm ) 17

R x (mm) 81.79

R y (mm) 21.36

Reference: Fundamentals of Structural Steel Design by DIT Gillesania

Check for the Adequacy of the section

K = 1.000

solvingfor kl/r

kl/r < 200.000

33.623 < 200.000

Solving for sovling for Cc

= =33.623 < Cc E = 200000

.:intermediate column Cc= 126.169

Fs= 1.764

Fa= 135.58 MPa

= =89.879 MPa

Therefore adequate!

Fy

E

Cc

22

3

3

)(8

)/(

)(8

)/(3

3

5

Cc

r kl

Cc

r kl Fs

Fs

Fy

Cc

r kl Fa *)(2

)/(1 2

2

A

P fa

Fa fa





H. DESIGN OF BEAM & GIRDER

Beam Design

Dead Loads, DL

The following dead loads are to be considered:

Type of Dead Load Component of Dead Load Required Amount

Slab Weight Concrete 150 mm thick

Frame PartitionsWood or steel studs, 13 mm.

gypsum board each side0.38 kPa

Ceiling LoadGypsum board + mechanical duct

allowance1.2 + 0.2 = 1.4 kPa

FlooringConcrete fill finish + hardwood

flooring0.345 + 0.19 = 0.535 kPa

Frame wallsWindow, glass, frames and sash +

wood sheating0.38+ 0.855 = 1.235 kPa

Reference: National Structural Code of the Philippines (NSCP) Volume 1, Section 204 – DEAD LOADS

Live Loads, LL

According to NSCP Vol. 1 Section 205 Table 205-1, the minimum uniform live load for

commercial building (office) is 2.4 kPa.

Assume section of wide flange:Using W 12 x 230 whose properties are

W12 x 230

Orientation

Weight, w (kg/m) 342

Area, A (mm2) 43677

Section Modulus about X, S X (x103 mm3) 5260

Section Modulus about Y, S Y (x103 mm3) 1885

Radius of Gyration about X, r x (mm) 151.64

Radius of Gyration about Y, r y (mm) 84.07

Depth , d 382.27Thickness of web , tw 32.64

Base of flange, bf 327.53

Thickness of flange, tf 52.580

Moment of Inertia, Ix 1007

Moment of Inertia, Iy 309





w = total dead loads + live load + self weight of section

Dead loads = partitions + ceilings + floor finishes + floor framing

DL= 3.55 kPa

LL= 2.4 kPa

DL + LL= 5.95 kPa

ɣconc= 23.50

weight of slab= ɣconc * thickness

= 3525 N/m2

= 3525 KN/m2

DL+LL+weight of slab= 9.475 KN/m2

Tributary area= 57.781 m2

(DL+LL+wt. slab)(TA)/L = 50.928 KN/m

(self weight of section)(9.81)/1000 = 4.905 KN/m

total w= 55.833 KN/m

ANALYSIS OF BEAM

w = 55.833

L= 10.75 m.

R R

)2(4

)( LST TA





use the assume wide flange, W 12 X 230

fb= 163.68 Mpa

solving for Sx to determine the most economical section

Sx= 4790.671 x103 mm3

Note: look for the manual for the most efficient section

Adopt the wide flange section W 12 x 230

Check for the Adequacy:

Test for compactness:

3.115< 13.038

Compact!

11.712 < 128.85

Compact!

Fb= 163.68 MPa

fb = 149.075 MPa

149.075 < 163.8

.: the section is adequate

=

Ra = 291.772 KN

Fy fb 66.0

Sx

Mx fb

Fytf

bf 170

2

Fytw

d 1680

Fy Fb 66.0

Sx

Mx fb

Fb fb

L

LwL Ra

)2/(





ANALYSIS OF GIRDER

assume section of wide flange W 21 x 402

W 21 x 402

Orientation

Weight, w (kg/m) 598

Area, A (mm2) 76129





Depth , d 660.91

Thickness of web , tw 43.94




Moment of Inertia, Iy 529

Use the same loadings

DL+LL+weight of slab = 9.475 Kn/m2

Tributary area = 57.781 m2

(DL+LL+wt. slab)(TA)/ST = 50.928 Kn/m

(self weight of section)(9.81)/1000 = 5.866 KN/m

total w= 56.795 KN/m

solving for Sx to determine the most economical section

Sx= 14593.648 x 10^3 mm3

Adopt the wide flange section W 14 x 500

Sx

Mx fb





Check for the Adequacy:

Test for compactness:

2.141<10.795 Compact!

15.041 < 106.680

Compact!

Fb= 163.680 MPa

fb = 155.564 MPa

155.564 < 163.680

.:the section is adequate

=

Ra = 597.042 KN

ANALYSIS FOR CONTINUOUS BEAMSelf Weight, SW

W 14 x 61

Orientation


Area, A (mm2) 11548





Depth , d 352.8




Moment of Inertia, Ix 266.388

Moment of Inertia, Iy 44.537

Fytwd

1680

Fy Fb 66.0

Fb fb

L

LwL Ra

)2/(

Sx

PLwST

fb

48

2

Fytf

bf 170

2





self weight of section = (89)(9.81)(1/1000) = 0.87 kN/m

Dead loads = 3.55 kPa

Live load = 2.4 KPa

eight of slab = Ɣ conc x thk e ee ee e wewewewe

Weight of slab = 23.5 x 150 mm = 3.525 KN/m2

W = ∑ + LL + weight of slab = 3.55 + 2.4+ 3.525 = 9.475 KN/m2

Tributary Area, T=0.25ST(4)

4+1.5(4)()=

0.25(10.75)(4)(10.75)

4+1.5(4)(10.75)= 93.91 km

( + + eight of slab)(T)

4=

9.475(93.91)

4(10.75)= 20.693 km

(+eight of slab)(T)

4=

3.55+3.525(93.91)

4(10.75)= 15.4515 km

Assume the section is compact.

Solving for the reactions

1 + 21+2 + c2 +61a 1

1+

62b2

2= 6(

h1

1+

h2

2)

Consider that there is no settlement in the reactions, Ma = Me = 0, and

61a

1

1 =

63a

3

3 63b

3

3 w3

4 =

15.4515(10.75)3

4 =4798.8375N-m2

62a 2

2=

62b2

2=

64b4

4 w3

4=

20.693(10.75)3

4=6426.7123N-m2

Where L1 = L2 = L3 = 10. 75m

L

L

L

L

15.4515 KN/m 15.4515 KN/m20.693 KN/m20.693 KN/m





Considering the first and second span

21+2 + c2 +61a 1

1+

62b2

2=

221.5 + c(10.75) + 4798.8375+6426.7123 = 0

43 10.75 = 11225.2498

Considering the second and third span

2+ 2c2+3 + 3 +62a 2

2+

63b3

3= 0

10.75 + 2c(21.5) + (10.75) + 4798.8375+6426.7123 = 0

10.75 43 10.75= 11225.2498

Considering the third and fourth span

c3+ 23+4 + 63a 3

3+ 64b 4

4= 0

c10.75 + 2(21.5) + 4798.8375+6426.7123 = 0

10.75 43 = 11225.2498

Solving for the internal moments simultaneously.

MB= -223.76kN-m

MC= -149.17 kN-m

MD= -223.76 kN-m

Checking for adequacy:

Solving the actual bending stress, f b

f b=max

Sx=

223.76 x 106

= 148.19

Solving the allowable bending stress, Fb

Fb = 0.66Fy

Fb = 0.66 (248) = 163.68 MPa

Since f b<Fb, the section is adequate as long girder.





15.4515 KN/m

20.693 KN/m

Solving for the reactions:

SEGMENT AB:

SEGMENT BC

SEGMENT CD:

SEGMENT DE:

A B

15.4515 KN/m

B C

20.693 KN/m

C D

D E





REACTIONS:

Ay=142.504 KN Dy=313.639 KN

By=313.639 KN Ey=73.166 KN

Cy=254.897 KN

V max=313.639 KN

I. DESIGN OF COLUMNS

FIRST FLOOR COLUMN

Firsts Floor Exterior Column Design

The height of the first floor column H1 = 3 m = 3000mm. Assume that kL/r= 0

that is less that Cc making it an intermediate column and that the column is hinged at both ends.

kL/r = 0 <Cc, intermediate column!

Therefore, FS = 5/3

Fa = 0.6(Fy) = 0.6(248) = 148.8 MPa

Assume Facolumn = 0.60Fa

Facolumn = 0.60 (148.8MPa)

Facolumn = 89.28 MPa

acolumn

=

89. 28 a =313.639(1000)

A = 3512.98 mm2

3.0 m

313.639 kN





Using W 8 x 21 whose properties are

W 8 x 21

Orientation

Weight, w (kg/m) 31

Area, A (mm2) 3974

Section Modulus about X, S X (x10

3

mm

3

) 298Section Modulus about Y, S Y (x103 mm3) 61


Radius of Gyration about Y, r y (mm) 32

Depth , d 210.31





Moment of Inertia, Iy 4Reference: Association of Structural Engineers of the Philippines (ASEP) Steel Manual

Checking the adequacy of the section:

r=

1.0(3000)

32=93.75 c = 22

y= 22(200000)

248= 126.169

kL/r <Cc, therefore, intermediate column!

Solving allowable compressive stress, Fa

FS= 1.89 Fa=94.83 MPa

Solving for actual stress, fa

f a=

=

313.639(1000)

3974=78.92

Since f a<Fa, the section is adequate as second floor column.

Note:

For design purposes: for other columns including those on the second floor, use the same

section from the computation with respect to the critical load.

No internal column, all columns were on the sides (exterior) of the structure.

3

3

)(8

)/(

)(8

)/(3

3

5

Cc

r kl

Cc

r kl Fs

Fs

Fy

Cc

r kl Fa

2

2

)(2

)/(1





J. DESIGN OF BASE PLATE

Assume the base plate is square that rests on full area of concrete pedestal whose f’ c = 21

MPa (the most common).

plate=c

p

Where Pc = column load and Fp = allowable bending stress of concrete

Solving column load, Pc

c=313639 N+9.81m

s231

g

m (3.0 m)

=314551.33

Solving allowable bearing stress of concrete, Fp

p=0.35 (f c) =0.35(21 a)

p=7.35





Solving the required area of the base plate, Aplate

plate=c

p=

314551.33 N

7.35=42796.1 mm2

Knowing the base plate is square,

=N=

plate

=N=√ 42796.1

==206.872 mm Say use 210mm x 210mm

However, it is too small to support the section, thus, use 450mm x 450mm

Solving for thickness, t

t=2x f pf y

Where x is larger between m & n, and f p is the actual bearing stress

Solving the actual bearing stress, f p

f p=c

actual=

314551.33

(450)2

=1.55

Solving m & n

m=N - 0.95d

2=

450-0.95(210.31)

2=125.103 mm

n= - 0.8bf

2= 450-0.80(133.86)

2=171.456 mm

Then, x = 171.456 mm

Hence,

t=2x f pf y

=2(171.456mm) 1.55

248=27.11, say t=30 mm

Henceforth, use 450mm x 450mm x 30 mm A36 base plate





K. DESIGN OF CONNECTION

Truss Member Connection (Riveted Connection)

Considering the member that carries the highest value of internal force.

TYPE OF

MEMBER MEMBER

INTERNAL

FORCE

(N)

LENGTH

(m)

CROSS

SECTIONAL

AREA

(mm2)

ACTUAL

STRESS

(MPa)

Top chord AB 420172 1.934 5238 80.22

Using 10mm A36 gusset plate, 25

mm ø A325 fasteners (whose Fv=145MPa)

Analyzing shear V failure conditions.

v=max

v

v=max

v

N [4

(25mm)2] =

420.172(1000)

145

N =5.9 say 6

Analyzing bearing P failure conditions.

p=max

v

p=max

p

N 25(10)= 420.172(1000)

0.4(400)

N =10.5 say 11

Therefore, use 11 – 25mm ø A325 fasteners to resist both shear and bearing failure





Beam and Girder Connection (Riveted Connection)

Consider the shear force V of 313.639 kN (the maximum shear that occurs in the short

girders), use 28 mm ø A325 fasteners (whose Fv=145MPa). Assume that the connecting angle is

10mm.


v= max

v

v=max

v

N [4

(28mm)2] =

313.639(1000)

145

N = 3.51 say 4

Analyzing bearing P failure conditions.

p= max

p

p=max

p

N (28)(10)= 313.639(1000)

0.4(400)

N = 7

Therefore, use 7 – 28mm ø A325 fasteners to resist both shear and bearing failure





Frame and Support Connection (Wielded Connection)

Using E60XX electrode whose Fu = 415MPa, we consider the shear force V of 313.639kN (the

maximum shear that occurs in the short girders). The wield thickness is assumed to be 8mm


v= max

v

v=

maxv

.707(8mm)()=313639(1000)

0.3(415)

L =445.4 say 450 mm.

Therefore, use 450 mm long 8mm thick wield to resist shear failure.




L. APPENDICES

-ASEP Steel Handbook 2004

-NSCP (National Structural Code of the Philippines)

-Steel review book (Fundamentals of Structural Steel Design)

By: DIT Gillesania-STAAD PRO 2007

-AUTOCADD 2009

-Google Earth and Google Maps

-Progress Reports

M. ACKNOWLEDGMENTS

I’m gld tht I hd finished my projet on time even though it is quite rush in

the making, I thank all those who had help me on the times of working of the project

matter. First of all, I thank God for the guidance, wisdom and knowledge he blessedto me. Second, I thank our beloved professor in CE134P (steel and timber design),

Engr. Edgardo Cruz for all what he had taught to us regarding the subject matter

including his personal experience. Third, I thank my fellow steelmates because they

help me in times of problem regarding the project matter, they explain what is not

clear to me and they share their work, it help a lot, it acts like a reference. Fourth,

Manuel Trinidad, Jr. and his family, because they welcome us, they let us work our

project on their house, they fed us and even let us use their electricity, phone,

internet connection and other more, all that is necessary to use for the project,

special thanks to the father of Manuel for guiding us and teaching us to use STAAD, it

became handy for getting the forces on the truss, frame and beam. Fifth, I thank myfriend, Christiane Bernardo for helping me to create my floor plan using autocadd.

Our experience regarding the making of this project will be an unforgettable one due

to what we experience, it is the first time for me and some of my classmates to have

not just one day overnight of working having very less time for sleep and even almost

none but because of it we learned many lessons and we learned to budget time and

srifie our leisure time for work nd mny more. I’m relly very gld nd very

thankful that I have finished my project just in time.

structural analysis and design project of ls

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