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Transcript of 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
Page 1
SS 10 mmØ threaded rod
Tributary area
Tributary area
Page 2
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
Page 3
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
Page 4
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.
Page 5
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
Page 6
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
Page 7
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
Page 8
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
Page 9