Structural Analysis and Design Project of Ls

PLANS AND DRAWINGS PLANS AND DRAWINGS

 A.

 A.  VICINITY MAP 1 VICINITY MAP 12 3565653332 356565333

LOCATION: PUERTO PRINSESA, PALAWAN, PHILIPPINES LOCATION: PUERTO PRINSESA, PALAWAN, PHILIPPINES

B.

B. DETAILS OF FLOOR PLAN/ FRAMING PLANDETAILS OF FLOOR PLAN/ FRAMING PLAN

5.375m. 5.375m. 5.375m. 5.375m. 1.5m. 1.5m. 10.75m. 10.75m. 10.75m. 10.75m. 10.75m. 10.75m. 10.75m.10.75m. B1 B1 B2 B2 B3 B3 B9 B9 B4 B4 B5 B5 B6B6 B7 B7 B8 B8 S1 S2 S1 S2 S3 S3 S4S4 S5 S5 S6S6 _{S7 S7 S8S8} S9 S9 S10 S10 S11S11 S12S12 C1 C1 C2C2 C3C3 C4 C4 C5C5 C6 C6 C7 C7 C8 C8 C13 C13 C9 C9 C14 C14 C10 C10 C15 C15 C11 C11 C16 C16 C12 C12 C17 C17

B.

B. DETAILS OF FLOOR PLAN/ FRAMING PLANDETAILS OF FLOOR PLAN/ FRAMING PLAN

5.375m. 5.375m. 5.375m. 5.375m. 1.5m. 1.5m. 10.75m. 10.75m. 10.75m. 10.75m. 10.75m. 10.75m. 10.75m.10.75m. B1 B1 B2 B2 B3 B3 B9 B9 B4 B4 B5 B5 B6B6 B7 B7 B8 B8 S1 S2 S1 S2 S3 S3 S4S4 S5 S5 S6S6 _{S7 S7 S8S8} S9 S9 S10 S10 S11S11 S12S12 C1 C1 C2C2 C3C3 C4 C4 C5C5 C6 C6 C7 C7 C8 C8 C13 C13 C9 C9 C14 C14 C10 C10 C15 C15 C11 C11 C16 C16 C12 C12 C17 C17

GROUND FLOOR PLAN GROUND FLOOR PLAN

SECOND FLOOR PLAN SECOND FLOOR PLAN

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 +

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

`

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.





1.5m

LEFT SIDE ELEVATION Roof  height H2 0.6 H1 0.6 c = -0.01 A B D F H J L C E G I K

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



)(

 



)2_{(0.895m) =102.86 N/m} b. Live Loads, LL  Tributary Area, A_T = 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

WL_WINDWARD= 0.010908 (





) (

 



) (9.81 m/s2) (

 



)2(0.967m) =9.26 N/m WL_LEEWARD = 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

 According on what we learn in the previous topic of this subject, 1.00 by  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 F_bx & F_by= 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 bx  by  0.9 bx by   f  f  F F 







.

 Assume that the purlins have compact sections.

If it is compact, F_bx = 0.66F_y and F_by= 0.75F_ywhere F_y = 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 xand 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˚)

Solving for moment about X and Y, M xand M y Mx =



My =



_

Mx =





My=





Mx =10.64 kN-m My =1.08 kN-m

Solving actual stress along X and Y, f bxand 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, Fbxand 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, bx  by  bx by   f  f  F



F  <1.0 by  bx  bx by   f  f  F



F  =





+





=0.909< 1.0 OK!

.: the purlins is adequateand economical.

Design of Sagrods TSAGRODS =



WyL TSAGRODS =



(299.08) (10.75) TSAGRODS =2.01 KN T = F_T A_g 2.01 x 10^3 = 0.6 (248) [



dsagrods2] x7

dsagrods = 10.97mm, use dsagrods= 12 mm

Design og Tierods

0

H 

F 





TTierods = Tsagrods cos22.1˚

TTierods = N

2.17 x 10^3 = 0.6 (248) [

_

dtierods2]

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)

C = (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

Weight,w (kg/m) 44.28

 Area,A(mm2) 5238

Radius of Gyration about X, r x

(mm) 29.95

Radius of Gyration about Y, r y

Web Members (Both Tension & 

Compression)

L 100 x 100 x 14

Orientation

Weight,w (kg/m) 20.56

 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 6@1.79m 3Rx 3Rx 3Rx 4Rx 4Rx 4Rx 3@ 1.93m. 4Rx Ɵ_=22.1 2.18m. 14Ry 14Ry

SAMPLE COMPUTATION OF FORCES IN CRITICAL MEMBERS

JOINT A

TABLE OF ANALSYSIS

TOP 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

22.1

O

Ɵ

22.1

O 171.67 KN

3R 

X

14R 





 AB sin22.1o + 171669.621 –3R xcos22.1o – 14R y sin22.1 = 0

 AB = -420.172 kN (T)  AB = 420.172 kN (C)

 AC + AB cos22.1o + 3R _x sin22.1o –14R _ycos22.1o = 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 mm2look 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

200



r 

kL

3 3 ) ( 8 ) / ( ) ( 8 ) / ( 3 3 5 Cc r  k l  Cc r  k l   FS _{  }

 Fy

 E 

Cc

2

2

 





 

 



 

 

















 Fs

 Fy

Cc

r 

kl 

 Fa

₂ 2

)

(

2

)

/

(

1

 A

 P 

 Fa



 Fy

 Ft 



0

.

6

 A

 P 

 Fa



P critical = -420.172 KN

 A=2822.98 mm2look for the manual for most efficient section  A/2=1411.49 mm2for 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 mm2look for the manual for most efficient section  A/2 =454.802 mm2for 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  k L 3 3 ) ( 8 ) / ( ) ( 8 ) / ( 3 3 5 Cc r  k l  Cc r  k l   FS _{  }

 Fy

 E 

Cc

2

2

 





 

 



 

 

















 Fs

 Fy

Cc

r 

kl 

 Fa

₂ 2

)

(

2

)

/

(

1

 A

 P 

 Fa



 Fy

 Ft 



0

.

6

P= 171669.621 P= 171669.621

H=3m. solving for Required Area:

 A=



_{}



= 1550.639 mm2

 A=1550.639 mm2look for the manual for most efficient section  A/2=775.319 mm2for 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 kL 0

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 

 



KnowingkL 0 r   , 5 3 FS



Then,

 

248 1 5 3 148.8 a a MPa F  F MPa      _{ }     

 A

 P 

 Ft 



H=3m. H=3m.

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 sectionW 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( x103mm ) 128 S Y ( x103mm ) 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

solving for 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

 Fy

 E 

Cc

2

2

 



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





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 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

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 (x103mm3) 5260

Section Modulus about Y, S Y (x103mm3) 1885

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

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

Depth , d 382.27 Thickness 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

)

( L

ST 

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 x103mm3 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



0

.

66

Sx

 Mx

 fb



 Fy tf   bf   170 2



 Fy tw d  1680 

 Fy

 Fb



0

.

66

Sx

 Mx

 fb



 Fb

 fb



 L

 L

wL

 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

Section Modulus about X,S X (x103mm3) 15355

Section Modulus about Y, S Y (x103mm3) 3097

Radius of Gyration about X, r x(mm) 259.08

Radius of Gyration about Y, r y (mm) 83.06

Depth , d 660.91 Thickness of web , tw 43.94 Base of flange, bf  340.49 Thickness of flange, tf  79.5 Moment of Inertia, Ix 5078 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 BEAM Self Weight, SW

W 14 x 61

Orientation

Weight,w (kg/m) 90.83

 Area,A(mm2) 11548

Section Modulus about X,S X (x103mm3) 1510

Section Modulus about Y, S Y (x103mm3) 351

Radius of Gyration about X, r x(mm) 151.88

Radius of Gyration about Y, r y (mm) 62.1

Depth , d 352.8 Thickness of web , tw 9.5 Base of flange, bf  253.9 Thickness of flange, tf  16.4 Moment of Inertia, Ix 266.388 Moment of Inertia, Iy 44.537  Fy tw d  1680 

 Fy

 Fb



0

.

66

 Fb

 fb



 L

 L

wL

 Ra



(

/

2

)

Sx

 PL

wST 

 fb



 

 



 

 





8

4

2  Fy tf   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 km ( + + eight of slab)(_T) 4 =



9.475



(93.91) 4(10.75) = 20.693 km (+eight of slab)(T) 4 =



3.55+3.525



(93.91) 4(10.75) =15.4515 km

 Assume the section is compact. Solving for thereactions

 1 + 2



1+2



+ c2 + 61a

̅

1 ₁ + 62b

̅

2 ₂ = 6( h1 ₁+ h2 ₂)

Consider that there isno settlement in the reactions, Ma= Me = 0, and

6₁a

̅

₁ ₁ = 6₃a

̅

₃ ₃



6₃b

̅

₃ ₃



w3 4 = 15.4515(10.75)3 4 =4798.8375N-m2 6₂a

̅

₂ 2 =62b

̅

2 2 =64b

̅

4 4



w3 4 = 20.693(10.75)3 4 =6426.7123N-m 2 Where L1 = L2 = L3 = 10. 75m L L L L 15.4515 KN/m 20.693 KN/m 15.4515 KN/m 20.693 KN/m

Considering the first and second span 2



1+2



+ c2 + 61a

̅

1 ₁ + 62b

̅

2 ₂ =



2



21.5



+ c(10.75) + 4798.8375+6426.7123 = 0 43_  10.75_ = 11225.2498

Considering the second and third span _₂+ 2_c



₂+₃



+ _₃ + 62a

̅

2 2 + 63b

̅

3 3 = 0 _



10.75



+ 2_c(21.5) + _(10.75) + 4798.8375+6426.7123 = 0 10.75_  43_ 10.75_= 11225.2498

Considering the third and fourth span _c₃+ 2_



₃+₄



+ 63a

̅

3 3 + 64b

̅

4 4 = 0 c



10.75



+ 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

S_x =

223.76 x 106



= 148.19  Solving the allowable bending stress, Fb

Fb= 0.66Fy

Fb= 0.66 (248) = 163.68 MPa

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 Fa_column = 0.60Fa Facolumn= 0.60 (148.8MPa) Facolumn= 89.28 MPa Fa_column=   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 (x103mm3) 298

Section Modulus about Y, S Y (x103mm3) 61

Radius of Gyration about X, r x(mm) 88.65

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

Depth , d 210.31 Thickness of web , tw 6.35 Base of flange, bf  133.86 Thickness of flange, tf  10.16 Moment of Inertia, Ix 31 Moment of Inertia, Iy 4

Reference: Association of Structural Engineers of the Philippines (ASEP) Steel Manual

Checking the adequacy of the section:

 r = 1.0(3000) 32 =93.75 Cc =

 

22_ Fy =

 

22_(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 

Sincef 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.

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 F_p

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

Solving column load, Pc

c=313639 N+9.81 m s2



31 g m



(3.0 m) _=314551.33 

Solving allowable bearing stress of concrete, Fp

Fp=0.35

(

f c

)

=0.35(21 a)

Solving the required area of the base plate, A_plate  _plate=c Fp =314551.33 N 7.35 =42796.1 mm 2

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 p f _y

Where x is larger between m& n, andf pis 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 p f y =2(171.456mm)

 

1.55 248 =27.11, say t=30 mm

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.

Fv= _max  v  _v=max F_v N

[

 4(25mm) 2

]

=420.172(1000) 145 N =5.9 say 6

 Analyzing bearing P failure conditions.

Fp= _max  _v  p= max F_p N



25(10)



=420.172(1000) 0.4(400) N =10.5 say 11

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.

 Analyzingshear V failure conditions.

Fv=  _max  _v  v= max F_v N

[

 4(28mm) 2

]

=313.639(1000) 145 N = 3.51 say 4

 Analyzingbearing P failure conditions.

F_p= max  p  _p=max Fp N



(28)(10)



=313.639(1000) 0.4(400) N = 7

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

 Analyzingshear V failure conditions.

Fv=  _max  _v  v= max F_v .707(8mm)()=313



639(1000) 0.3(415) L =445.4 say 450 mm.