7/31/2019 Base Pl Pinned Tad3 p0

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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

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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.

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= ( 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

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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.

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= ( 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

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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.

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= ( 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

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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.

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= ( 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

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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

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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

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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