Grc Column

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PROJECT : OSSAIMI APPARTMENTS AND BEACH RESORT LOCATION : JUMEIRAH, DUBAI, UAE SUBJECT : GRC COLUMN FIXING CALCULATION - rev 11 DATE : 27-07-2015 ITEM NO. 1 GRC COLUMN - ELEVATION - SECTION Page 1

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DESIGN OF GRC COLUM

Transcript of Grc Column

Page 1: Grc Column

PROJECT : OSSAIMI APPARTMENTS AND BEACH RESORT

LOCATION : JUMEIRAH, DUBAI, UAE

SUBJECT : GRC COLUMN FIXING CALCULATION - rev 11

DATE : 27-07-2015

ITEM NO.

1 GRC COLUMN - ELEVATION - SECTION

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SS 10 mmØ threaded rod

Tributary area

Tributary area

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Design for the bolt fixing support for the grc crown

detail SS 10 mmØ rod steel bracket 50x50x6mm

r

Compute for the Weight of GRC

Given :

GRC Unit Wt. = 2000 Kg/M³

For Steel :

E = 200000 MPa

Steel Unit Wt. = 7850 Kg/M³

Compute for the section area of the crown:

As = 73880 mm² diam Ø = 2400 mm

0.07388 M²

Vol = As (π Ø) For threaded rod:

0.07388(3.141592654 x 2.4) Wt .2 = 1/2 Wt.1

Vol = 0.557 M³ wt.2 = 2732.5 N

Check for Shear on bolt

Compute for GRC Wt. fv = Wt.2

As

Wt.1 = Unit Wt. x Vol 2732.5

Wt.1 = 2000 ( 0.557 ) 78.54

1114.084 Kg x 9.81 fv = 34.79 N/mm²

10929.17 N ( divided by 2) as per tributary area

Wt.1 = 5464.583

say Wt.1 = 5465 N Allow. Fv = 0.4Fy = 0.4(250)

Allow. Fv = 100 N/mm²

OK

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5465 N

Check for the strength of the

Steel Bracket 724 mm section of

630 mm M 50 x 50 x 6mm angle

Section Area (50 x 50 x 6)

Atot = 564 mm²

178 A1 = 300 mm²

555 84 A2 = 264 mm²

A1

s y1 = 3

y2 y2 = 28

Compute for the Moment

M = GRC Wt. ( 84 ) A2

M = 5465 (84) by area moment method, s =

M = 459060 N-mm Atot (s) = A1 y1 + A2 y2

564 s = 300( 3 ) + 264( 28 )

Check for the Bending s = 8292

564

fb = M c but c = 50 - s s = 14.70

I c = 35.30

I = bh³ + A a²

fb = 459060 x 35.3 12

137624 but a1 = s - y1137624 but a1 = s - y1

fb = 117.74 N/mm² 11.70

a2 = y2 - s

Allow. Fb = 0.66 Fy since Fy < 65 Ksi 13.30

I1 = (50)(6³) + 300(11.7²)

allow. Fb = 0.66(250) = 165 N/mm² 12

41981.94 mm4

I2 = (6)(44³) + 264(13.3²)

Check for Deflection 12

95642.02 mm4

def = Pa²(3 L - a) a = 84

6 E I a² = 7056 I = I1 + I2= 137624 mm4

def = (5465)(7056)(3(178)-84)

6(200000) 137624

1.74E+10

1.65E+11

def = 0.105 mm OK

Allow. Def = L/240 = 178(240) = 0.742 mm

trib. Area = 563884 mm²

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5465 N

Y

X

at joint b 5465 N

fba

b

fdb

Sum of forces along Y-axis = 0 = fdb(250/555) - 5465

fdb = 5465(555)

250

12132.3 N (compression)

Sum of forces along X axis = 0 = 12132.3(546/555) - fba

a b c

d

Sum of forces along X axis = 0 = 12132.3(546/555) - fba

fba = 11935.56 N ( tension)

For compression, check for the slenderness of member fdb

KL/r = where K = 1

L = 555 mm

Compute for r:

r = I since KL/r < Cc use the allow. Fa

A 1 - 0.5(KL/r)² x Fy

r = 137624 Fa = Cc²

564 5 + 3(KL/r) - 1(KL/r)³

r = 15.62 3 8(Cc) 8(Cc)³

KL/r = 35.53

Check for Cc Fa = 240.01 (KL/r)/Cc= 0.282732

Cc = 2π²(E) 1.770

Fy (KL/r)²/Cc²= 0.079938

Cc = 2π²(200000) Fa = 135.61 N/mm² (allowable)

250 (KL/r)³/Cc³= 0.022601

Cc = 125.66 fa = P

A

Note : see Mungo Fixing Design Calc fa = 5465 fa = 9.69 N/mm²

for the anchor bolts 564 OK, less than allow.

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At the Crown SS threaded rod 10mm diam.

Check for the Wind Load

Wp = 1.3 Kpa

Fw = Wp ( Ta )

Ta = 563884 mm²

0.563884 M²

Fw = 1.3(0.563884)

0.73305 KN

Fw = 733.05 N

Say Fw = 735 N

For Shear A area of sect. = 78.54 mm²

fv = P sum of forces along x-axis

A Fx = 0 = Fw - 3r

245 r = Fw

78.54 3

fv = 3.12 N/mm² 735

OK, less than allowable 3

r = 245 N the reaction isr = 245 N the reaction is

small

Allow. Fv = 0.4 Fy = 0.4(250) = 100 N/mm²

At the column

Wp = 2.09 Kpa

Fw = Wp ( Ta )

Ta = 1124680 mm²

1.12468 M²

Fw = 2.09(1.12468)

2.35058 KN

Fw = 2350.58 N

Say Fw = 2350 N 1100

Solve for r:

Sum of forces along y = 2r - Fw = 0

2r = 2350 N

r = 1175 N less than 10KN, it is safe

1.175 KN

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For the GRC Column Tributary Area

2995

1515

Tributary height = 2995 mm

Diameter Ø = 1100 mm

Average thickness= 30 mmAverage thickness= 30 mm

Area = Arch length x height

851 x 2995

2548745 mm²

Compute for the Volume

Vol = Area x ave thick

Vol = (2548745 x 30 )

Vol = 76462350 mm³

0.076462 M³

Compute for the Wt.

Wt. = Unit wt x Volume

Wt. = 2000 x 0.076462

152.92 Kg x 9.81

1500.191 N Check for Shear on the bolt SS 10mm Ø:

say Wt. = 1500 N fv = r/A = 750

r = Wt./2 78.54

r = 1500 /2 = 750 N fv = 9.55 N/mm² OK, less than allow.

Allow. Fv = 0.4Fy = 0.4(250) = 100 N/mm²

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Tributary height = 2950

Diameter Ø = 1100

Average thickness= 30

Area = Arch length x height

743 x 2950

2191850 mm²

Compute for the Volume

Vol = Area x ave thick

Vol = (2191850 x 30 )

Vol = 65755500 mm³

0.065756 M³

Compute for the Wt.

Wt. = Unit wt x Volume

Wt. = 2000 x 0.065756

131.51 Kg x 9.81

1290.123 N

say Wt. = 1290 N

Check for Shear on the bolt SS 10mm Ø:

r = Wt./2 fv = r/A = 645

r = 1290 /2 = 645 N 78.54

fv = 8.21 N/mm² OK, less than allow.

Allow. Fv = 0.4Fy = 0.4(250) = 100 N/mm²

The design of 10 mmØ rod is OK

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x

30 mm

average thickness

compute for the centroid of the section

A1 = 839(30) 25170 x1 = 419.5

A2 = 100(30) 3000 x2 = 824

A3 = 350(30) 10500 x3 = 1014

A4 = 70(30) 2100 x4 = 1174

A5 = 396(30) 11880 x5 = 1031

A6 = 432(30) 12960 x6 = 678A6 = 432(30) 12960 x6 = 678

A7 = 545(30) 16350 x7 = 272.5

81960

Solve for the centroid of the section or dist x

by Area moment method:

81960 x = 25170(419.5) + 3000(824) + 10500(1014) + 2100(1174) + 11880(1031)+

12960(678) + 16350(272.5)

x = 629.99

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