base pl pinned tad3 p0
TRANSCRIPT
-
7/31/2019 Base Pl Pinned Tad3 p0
1/15
TITLE: J.K.CEMENT WORKS, (FUJAIRAH) FZC, UAE.
1750 TPD WHITE / 2800 TPD GREY CEMENT PLAT
COAL MILL ROOF BASE PLATE REV.NO. P0
Doc No:SPADE-12-JK-CAL-101 Date: 30/12/2011
Base Plate : BP1
Staad Joint Numbers :17
Joint Load
Support Reactions Comb
Maximum Tension Fyt = -1325.935 kN 1 112
Maximum Compression Fyc = 1770.388 kN 1 101
Maximum shear Fx Fx = -87.766 kN 1 101
Maximum shear Fz Fz = -102.932 kN 1 104
Input Load factor for Force 1
Anchor Bolt design
Number of Anchor Bolts N = 8
Dia of Anchor Bolt d = 40 mm
Actual Force in each Bolt Pt,act -165.7 kN
Tensile Area of each = 949.9 mm2
Allowable Tensile stress = 120 N/mm2
Tensile Cap of each Bolt Pt,cap 113.988 kN
Tensile Stress Ratio -1.454
Check for Shear
Max shear FTotal 135.3 kN
Shear Force in each Bolt Vbolt 16.91 kN
Shear Stress in each Bolt tv 13.456 N / mm2
Allowable Shear Stress in Bolt tva 80 N / mm2
Stress Ratio
-
7/31/2019 Base Pl Pinned Tad3 p0
2/15
Thickness of Base Plate t
a) Design Considering Panel Fixed on four edges
For purpose of thickness calculations , base pressure considered is
qdes 5.00 N/mm2
Dist bet Fixed Edges a 125 mm
Dist bet Fixed & Free Edges b 125
a/b 1.00From Table-26 , Case-8a of Roarks
For a/b = 1.00
b1 0.321
Yield Stress fy 250 N/mm2
Allowable Bending Stress (0.66 fy) = 165 N/mm2
Thickness of Base-plate required from Bearing stress considerations = t1
t1 = sqrt{ (b1*q*b2)/0.66fy} t1 = 12.328367 mm
b) Design Considering cantilever portion of Base plate
Max Outstand Lc 50
Moment Mc = q (Lc)2
/ 2 Mc 6250 N-mm per unit widthThickness required t2 = 15.075567 mm
t2 =sqrt 6Mc/ (0.66*fy*b) , b=1
c) Design Considering projection of base plate as per IS:800
As per Clause 5.4.3 , IS:800-1984 ,page-44
a = greater projection of plate beyond column 50 mm
b = lesser projection of plate beyond column 35 mm
t3 = sqrt{ 3 x q x ( a2
- b2/4) / 0.66 fy }
t3 = 14.122033
d) Design Considering Tension in Anchor Bolt
Thickness of washer Plate twasher 12 mm
Assumed Thickness of BP t4 16 mm
Calculation of Pressure due to Tension of bolt
Loaded area = ( Nutsize + 2xThk of washer plate + 2xThk of baseplate )
Assuming 12 mm thick washer plate and size of nut is 1.6 times dia of bolt.
-
7/31/2019 Base Pl Pinned Tad3 p0
3/15
= ( 1.6 x 40 + 2 x 12 + 2 x 16) = 120 mm
Designing for Full Tensile Capacity of Bolt = 113.988 kN
Actual Pressure = ( 113.988 x 1000 ) / ( 120x 120)= 7.92 N/mm2
pact = 7.92 N/mm< qdes 5.00 N/mm
Base plate already designed for Bearing pressure of 5 N/mm2.
Hence 16 mm thick Base plate is OK from Bolt tension consideration
Greater of t1 to t4 t = 16
Provide 500 x 500 x 16 Thick Base-Plate
-
7/31/2019 Base Pl Pinned Tad3 p0
4/15
TITLE: J.K.CEMENT WORKS, (FUJAIRAH) FZC, UAE.
1750 TPD WHITE / 2800 TPD GREY CEMENT PLAT
COAL MILL ROOF BASE PLATE REV.NO. P0
Doc No:SPADE-12-JK-CAL-101 Date: 30/12/2011
Base Plate : BP1
Staad Joint Numbers :17
Joint Load
Support Reactions Comb
Maximum Tension Fyt = -993.245 kN 1 112
Maximum Compression Fyc = 1411.288 kN 1 140
Maximum shear Fx Fx = -82.213 kN 1 140
Maximum shear Fz Fz = -95.558 kN 1 108
Input Load factor for Force 1
Anchor Bolt design
Number of Anchor Bolts N = 8
Dia of Anchor Bolt d = 36 mm
Actual Force in each Bolt Pt,act -124.2 kN
Tensile Area of each = 816.7 mm2
Allowable Tensile stress = 120 N/mm2
Tensile Cap of each Bolt Pt,cap 98.004 kN
Tensile Stress Ratio -1.267
Check for Shear
Max shear FTotal 126.1 kN
Shear Force in each Bolt Vbolt 15.76 kN
Shear Stress in each Bolt tv 15.480 N / mm2
Allowable Shear Stress in Bolt tva 80 N / mm2
Stress Ratio
-
7/31/2019 Base Pl Pinned Tad3 p0
5/15
Thickness of Base Plate t
a) Design Considering Panel Fixed on four edges
For purpose of thickness calculations , base pressure considered is
qdes 5.00 N/mm2
Dist bet Fixed Edges a 125 mm
Dist bet Fixed & Free Edges b 125
a/b 1.00From Table-26 , Case-8a of Roarks
For a/b = 1.00
b1 0.321
Yield Stress fy 250 N/mm2
Allowable Bending Stress (0.66 fy) = 165 N/mm2
Thickness of Base-plate required from Bearing stress considerations = t1
t1 = sqrt{ (b1*q*b2)/0.66fy} t1 = 12.328367 mm
b) Design Considering cantilever portion of Base plate
Max Outstand Lc 50
Moment Mc = q (Lc)2
/ 2 Mc 6250 N-mm per unit widthThickness required t2 = 15.075567 mm
t2 =sqrt 6Mc/ (0.66*fy*b) , b=1
c) Design Considering projection of base plate as per IS:800
As per Clause 5.4.3 , IS:800-1984 ,page-44
a = greater projection of plate beyond column 50 mm
b = lesser projection of plate beyond column 35 mm
t3 = sqrt{ 3 x q x ( a2
- b2/4) / 0.66 fy }
t3 = 14.122033
d) Design Considering Tension in Anchor Bolt
Thickness of washer Plate twasher 12 mm
Assumed Thickness of BP t4 16 mm
Calculation of Pressure due to Tension of bolt
Loaded area = ( Nutsize + 2xThk of washer plate + 2xThk of baseplate )
Assuming 12 mm thick washer plate and size of nut is 1.6 times dia of bolt.
-
7/31/2019 Base Pl Pinned Tad3 p0
6/15
= ( 1.6 x 36 + 2 x 12 + 2 x 16) = 113.6 mm
Designing for Full Tensile Capacity of Bolt = 98.004 kN
Actual Pressure = ( 98.004 x 1000 ) / ( 113.6x 113.6)= 7.59 N/mm2
pact = 7.59 N/mm< qdes 5.00 N/mm
Base plate already designed for Bearing pressure of 5 N/mm2.
Hence 16 mm thick Base plate is OK from Bolt tension consideration
Greater of t1 to t4 t = 16
Provide 500 x 500 x 16 Thick Base-Plate
-
7/31/2019 Base Pl Pinned Tad3 p0
7/15
TITLE: J.K.CEMENT WORKS, (FUJAIRAH) FZC, UAE.
1750 TPD WHITE / 2800 TPD GREY CEMENT PLAT
COAL MILL ROOF BASE PLATE REV.NO. P0
Doc No:SPADE-12-JK-CAL-101 Date: 30/12/2011
Base Plate : BP1
Staad Joint Numbers :17
Joint Load
Support Reactions Comb
Maximum Tension Fyt = -576.38 kN 1 112
Maximum Compression Fyc = 983.439 kN 1 140
Maximum shear Fx Fx = -69.974 kN 1 140
Maximum shear Fz Fz = -67.156 kN 1 116
Input Load factor for Force 1
Anchor Bolt design
Number of Anchor Bolts N = 4
Dia of Anchor Bolt d = 36 mm
Actual Force in each Bolt Pt,act -144.1 kN
Tensile Area of each = 816.7 mm2
Allowable Tensile stress = 120 N/mm2
Tensile Cap of each Bolt Pt,cap 98.004 kN
Tensile Stress Ratio -1.470
Check for Shear
Max shear FTotal 97.0 kN
Shear Force in each Bolt Vbolt 24.25 kN
Shear Stress in each Bolt tv 23.821 N / mm2
Allowable Shear Stress in Bolt tva 80 N / mm2
Stress Ratio
-
7/31/2019 Base Pl Pinned Tad3 p0
8/15
Thickness of Base Plate t
a) Design Considering Panel Fixed on four edges
For purpose of thickness calculations , base pressure considered is
qdes 5.00 N/mm2
Dist bet Fixed Edges a 125 mm
Dist bet Fixed & Free Edges b 125
a/b 1.00From Table-26 , Case-8a of Roarks
For a/b = 1.00
b1 0.321
Yield Stress fy 250 N/mm2
Allowable Bending Stress (0.66 fy) = 165 N/mm2
Thickness of Base-plate required from Bearing stress considerations = t1
t1 = sqrt{ (b1*q*b2)/0.66fy} t1 = 12.328367 mm
b) Design Considering cantilever portion of Base plate
Max Outstand Lc 50
Moment Mc = q (Lc)2
/ 2 Mc 6250 N-mm per unit widthThickness required t2 = 15.075567 mm
t2 =sqrt 6Mc/ (0.66*fy*b) , b=1
c) Design Considering projection of base plate as per IS:800
As per Clause 5.4.3 , IS:800-1984 ,page-44
a = greater projection of plate beyond column 50 mm
b = lesser projection of plate beyond column 35 mm
t3 = sqrt{ 3 x q x ( a2
- b2/4) / 0.66 fy }
t3 = 14.122033
d) Design Considering Tension in Anchor Bolt
Thickness of washer Plate twasher 12 mm
Assumed Thickness of BP t4 16 mm
Calculation of Pressure due to Tension of bolt
Loaded area = ( Nutsize + 2xThk of washer plate + 2xThk of baseplate )
Assuming 12 mm thick washer plate and size of nut is 1.6 times dia of bolt.
-
7/31/2019 Base Pl Pinned Tad3 p0
9/15
= ( 1.6 x 36 + 2 x 12 + 2 x 16) = 113.6 mm
Designing for Full Tensile Capacity of Bolt = 98.004 kN
Actual Pressure = ( 98.004 x 1000 ) / ( 113.6x 113.6)= 7.59 N/mm2
pact = 7.59 N/mm< qdes 5.00 N/mm
Base plate already designed for Bearing pressure of 5 N/mm2.
Hence 16 mm thick Base plate is OK from Bolt tension consideration
Greater of t1 to t4 t = 16
Provide 500 x 500 x 16 Thick Base-Plate
-
7/31/2019 Base Pl Pinned Tad3 p0
10/15
TITLE: J.K.CEMENT WORKS, (FUJAIRAH) FZC, UAE.
1750 TPD WHITE / 2800 TPD GREY CEMENT PLAT
COAL MILL ROOF BASE PLATE REV.NO. P0
Doc No:SPADE-12-JK-CAL-101 Date: 30/12/2011
Base Plate : BP1
Staad Joint Numbers :17
Joint Load
Support Reactions Comb
Maximum Tension Fyt = -38.323 kN 1 112
Maximum Compression Fyc = 193.146 kN 1 140
Maximum shear Fx Fx = -25.829 kN 1 140
Maximum shear Fz Fz = -44.521 kN 1 116
Input Load factor for Force 1
Anchor Bolt design
Number of Anchor Bolts N = 2
Dia of Anchor Bolt d = 25 mm
Actual Force in each Bolt Pt,act -19.2 kN
Tensile Area of each = 352.5 mm2
Allowable Tensile stress = 120 N/mm2
Tensile Cap of each Bolt Pt,cap 42.3 kN
Tensile Stress Ratio -0.453
Check for Shear
Max shear FTotal 51.5 kN
Shear Force in each Bolt Vbolt 25.74 kN
Shear Stress in each Bolt tv 52.428 N / mm2
Allowable Shear Stress in Bolt tva 80 N / mm2
Stress Ratio
-
7/31/2019 Base Pl Pinned Tad3 p0
11/15
Thickness of Base Plate t
a) Design Considering Panel Fixed on four edges
For purpose of thickness calculations , base pressure considered is
qdes 5.00 N/mm2
Dist bet Fixed Edges a 125 mm
Dist bet Fixed & Free Edges b 125
a/b 1.00From Table-26 , Case-8a of Roarks
For a/b = 1.00
b1 0.321
Yield Stress fy 250 N/mm2
Allowable Bending Stress (0.66 fy) = 165 N/mm2
Thickness of Base-plate required from Bearing stress considerations = t1
t1 = sqrt{ (b1*q*b2)/0.66fy} t1 = 12.328367 mm
b) Design Considering cantilever portion of Base plate
Max Outstand Lc 50
Moment Mc = q (Lc)2
/ 2 Mc 6250 N-mm per unit widthThickness required t2 = 15.075567 mm
t2 =sqrt 6Mc/ (0.66*fy*b) , b=1
c) Design Considering projection of base plate as per IS:800
As per Clause 5.4.3 , IS:800-1984 ,page-44
a = greater projection of plate beyond column 50 mm
b = lesser projection of plate beyond column 35 mm
t3 = sqrt{ 3 x q x ( a2
- b2/4) / 0.66 fy }
t3 = 14.122033
d) Design Considering Tension in Anchor Bolt
Thickness of washer Plate twasher 12 mm
Assumed Thickness of BP t4 16 mm
Calculation of Pressure due to Tension of bolt
Loaded area = ( Nutsize + 2xThk of washer plate + 2xThk of baseplate )
Assuming 12 mm thick washer plate and size of nut is 1.6 times dia of bolt.
-
7/31/2019 Base Pl Pinned Tad3 p0
12/15
= ( 1.6 x 25 + 2 x 12 + 2 x 16) = 96 mm
Designing for Full Tensile Capacity of Bolt = 42.3 kN
Actual Pressure = ( 42.3 x 1000 ) / ( 96x 96)= 4.59 N/mm2
pact = 4.59 N/mm< qdes 5.00 N/mm
Base plate already designed for Bearing pressure of 5 N/mm2.
Hence 16 mm thick Base plate is OK from Bolt tension consideration
Greater of t1 to t4 t = 16
Provide 425 x 425 x 16 Thick Base-Plate
-
7/31/2019 Base Pl Pinned Tad3 p0
13/15
Node L/C Force-X k Force-Y k Force-Z kN Node
Maximum Tension 1 101 21.494 54.265 0.108 15Maximum Compression 1 101 21.494 54.265 0.108 15
Maximum shear Fx 1 112 -10.463 -36.678 -9.008 15
Maximum shear Fz 1 104 5.394 -0.262 -9.02 15
Node L/C Force-X k Force-Y k Force-Z kN
17 105 -2.614 -6.481 9.02
-
7/31/2019 Base Pl Pinned Tad3 p0
14/15
L/C Force-X k Force-Y k Force-Z kN Node L/C Force-X k Force-Y k Force-Z kN
101 -21.494 51.569 0.108 3 101 18.715 50.479 -0.108101 -21.494 51.569 0.108 3 101 18.715 50.479 -0.108
112 10.463 -37.432 -9.009 3 113 -10.463 -36.678 9.008
104 -5.395 -2.958 -9.02 3 105 2.615 -4.048 9.02
-
7/31/2019 Base Pl Pinned Tad3 p0
15/15
Node L/C Force-X k Force-Y k Force-Z kN
17 101 -18.714 48.046 -0.10817 101 -18.714 48.046 -0.108
17 113 10.463 -37.432 9.009
17 105 -2.614 -6.481 9.02