znd332_a-base & top plate & anchors-rev-e (repaired)

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  • 8/2/2019 ZND332_A-Base & Top Plate & Anchors-REV-E (Repaired)

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    ZND332_A-AMT-Td-0002-Rev.

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    ZND332_A-AMT-Td-0002-Rev. E

    FAST PLANETS Co Ltd.,

    ENGINEERING DEPARTMENT

    P.0. BOX : 34620

    RIYADH 15978

    Base & Top Plate Connection Calculation(CI & 3G 2011 PROJECT)

    Site No.: ZND332_ASite Name: ZND332_A / Bldg. of Ali Hussain Al-Sageer/ Dammam

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    Preface

    General:

    This document contains the capacity calculations of base and top plate and its weldedjoints, bolted joints and chemical anchorage bolts supporting the steel frames of the

    equipment shelter.

    The document consists of drawings and calculations necessary to assess the plates and its

    connection capacity.

    Analyzed by: Checked by: Approved by:

    J.B.Vasquez J. Abo Hasson F.C.Falgui

    Telecom Engineer Structural Engineer Prin. Structural Engineer

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    A. Connection Anchorage Check to the Existing Concrete Slab & to Bottom Plate

    1. JOINT 17 (Exterior Column = 200 x 600)most Critical Joint

    A) Tension

    From STAAD Pro Output, Factored forces (Please refer to page 11/30 of STAAD Output)

    Rx = -1.65 kNs Ry = -25.82 kNs Rz = 45.06 kNs (Load No. 10)

    The orthogonality/45 degree direction of horizontal forces has been included in the load combinations of

    STAADPro Inputs (Load Nos. 13 & 14) and from the results as per Support reactions, the values of forces are

    critical. Please see page 11/30 of STAAD Pro outputs for reference.

    Tension = -25.82 kNs

    Actual tension for 1Anchor Bolt = 25.82 / 4 = 6.455 kNs

    Using UpatUKA 3 EAP chemical anchor, M20

    Shear: 27.0kN (@edge distance = 170 mm)

    (@anchor spacing = 340 mm)

    However for uniformity the size of bottom plate of 250 x 500mm is adopted.

    100

    200

    100

    100 300 100

    500

    factor 1 = ( 1 + 100 / 170) / 2 = 0.794

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    factor 2 = ( 1 + 170 / 170) / 2 = 1.00

    factor 3 = ( 1 + 100 / 340) / 2 = 0.647

    factor 4 = ( 1 + 340 / 340) / 2 = 1.00

    red. F = 0.794 x 0.647 x 1.0 0 x 1.00 x 27.0 = 13.87 kNs > 6.455 kNs OK!

    2. JOINT 17 ( Exterior Column = 200 x 600)most Critical JointB) Shear

    From STAAD Pro Output, Factored forces (Please refer to page 11/30 of STAAD Output)

    Rx = -15.65 kNs Ry = 40.74kNs Rz = 101.41 kNs (Load No.12)

    The orthogonality/45 degree dirction of horizontal forces has been included in the load combinations of

    STAADPro Inputs(Load Nos. 13 & 14) and from the results as per Support reactions, the values of forces are n

    critical. Please see page 11/30 of STAAD Pro outputs for reference.

    Total Factored Shear, V = (15.65)\2

    + 101.412

    )0.50

    = 102.61 kNs

    Considering the effect of friction. Vf

    Vf = 0.90x Fy x U, where U = 0.55

    Vf = 0.90 x40.74 x 0.55

    Vf = 20.17 kNs

    Net Shear, V n = 102.61 - 20.17 = 82.44 kNs

    Actual shear for 1Anchor Bolt = 82.44 / 4 = 20.61 kNs

    Using UpatUKA 3 EAP chemical anchor, M20

    Shear: 27.0kN (@edge distance = 200 mm)

    (@anchor spacing = 340 mm)

    However for uniformity the size of bottom plate of 250 x 500mm is adopted.

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    100

    200

    100

    100 300 100

    500

    factor 1 = ( 1 + 100 / 200) / 2 = 0.750

    factor 2 = ( 1 + 200 / 200) / 2 = 1.00

    factor 3 = ( 1 + 100 / 340) / 2 = 0.647

    factor 4 = ( 1 + 340 / 340) / 2 = 1.00

    red. F = 0.750 x 0.647 x 1.0 0 x 1.00 x 27.0 = 13.10 kN s < 20.61 kNs Not OK!

    Therefore shear key is required.

    Net Shear to be resisted by shear key = 4(20.6113.10) = 30.04 kNs

    Consider a 2- 100 mm x 200 mm x 20 mm shear plate. Since the effective area will be 50 mm that will be embedded

    into the concrete, the bearing pressure developed by the shear plate is = 30.04 x 1000 / (50 x 200) = 3.004 mPa

    The max bearing pressure as per AISC = 0.35fc = 0.35(20) = 7.0 mPa > 3.004 mPa Therefore OK!

    Check for thickness of the plate:

    M = 3.004 x 200 x 50 x (50 + 25) = 2253000 N-mm

    fb = 6M / bd2

    fb = 6 x 2253000 / 200d2

    Allow Fb = 0.75Fyd = 0.75 x 248 = 186 mPa

    d2 = 6 x 2253000 / (200) x 186

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    d = 19.06 mm

    However, the thickness of the shear plate is adopted as 20mmdue to the following facts;

    1.

    The shear plate is subjected to corrosion which is not considered in the calculations.

    2. The strength of the shear plate is reduced by the intense heat due to welding process.3. The shear plate is subjected to thermal loads due to temperature changes which is not considered in th

    calculations.

    4. The minimum thickness of plate if 6 mm thick of weld is used is from 12 -20mm as per AISC manua

    Check for Weld Capacity:

    1.Shear Plate

    P = 30.04 kNs Fu = 485 Mpa (for E-70XX) (Please refer to page 11/30 of STAAD Output)

    fd = P/A Throat = 0.707 x 6 = 4.242 mm

    The critical length of weld is 200 mm;

    A = (200) x 4.242 = 848.40 mm2

    fd = ( 30.04 x 1000 ) / 848.40 = 35.41 mPa

    F = 0.30 Fu

    F = 0.30 x 485

    F = 145.5 mPa > fd = 35.41 mPa OK!

    Therefore adopt 2-100 x 200 x 20 shear keys collinear with the longer side in all bottom plate plates

    Weld Capacity Check:1. Bottom plate to Stiffener/column connection

    P = 102.61 kNs Fu = 485 Mpa (for E-70XX)

    fd = P/A Throat = 0.707 x 6 = 4.242 mm

    From inspection , the length of weld is 175 x 3 = 525 mm

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    A = (525) x 4.242 = 2227.05 mm2

    fd = ( 102.61 x 1000 ) / 2227.05 = 46.07 mPa

    F = 0.30 Fu

    F = 0.30 x 485

    F = 145.5 mPa > fd = 46.07 mPa OK!

    1. Top PlateP = 102.61 kNs Fu = 485 Mpa (for E-70XX) (Please refer to page 11/30 of STAAD Output)

    fd = P/A Throat = 0.707 x 6 = 4.242 mm

    From inspection, the length of weld is 270 x 3 = 810 mm

    A = (810 x 4.242 = 3436.02 mm2

    fd = ( 102.61 x 1000 ) / 3436.02 = 29.86 mPa

    F = 0.30 Fu

    F = 0.30 x 485

    F = 145.5 mPa > fd = 29.86 mPa OK!

    B. Thickness of the 200 x 500 mm2

    Bottom Plate:

    JOINT 17 ( Interior Column = 200 x 600)most Critical Joint

    1. Compression

    From STAAD Pro Output, Factored forces (Please refer to page 11/30 of STAAD Output)

    Rx = -16.13 kNs Ry = 80.06 kNs Rz = 74.20 kNs

    Compression = 80.06kNs

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    However for uniformity the size of bottom plate of 200 x 500mm is adopted.

    100

    100

    200

    100

    100 300 100

    500

    The bearing pressure developed on the base plate = 80.06 x 1000/.(200 x 500)

    = 0. 814mPa

    The max bearing pressure as per AISC = 0.35fc = 0.35(20) = 7.0 mPa > 0 0.814 mPa OK!

    Considering a strip of 1 mm of base plate,

    From inspection, the length of the base plate subjected to cantilever effect, L = 100 mm

    The length of the base plate subjected to cantilever effect, L = 100mm

    M = WL2

    / 2 = 0.814 x 1002/2 = 40700 N-mm

    fb = 6M / bd2

    fb = 6 x / 1.0d2

    Allow Fb = 0.75Fyd = 0.75 x 248= 186 mPa

    d2 = 6 x 40700 / 186

    d = 11.45 mm

    However, the thickness of the base plate is adopted as 20mm due to the following facts;

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    1. The base plate is subjected to corrosion which is not considered in the calculations.

    2. The strength of the base plate is reduced by the intense heat due to welding process.

    2.

    The base plate is subjected to thermal loads due to temperature changes which is not considered in th

    calculations.

    3. The minimum thickness of plate if 6 mm thick of weld is used is from 12 -20mm as per AISC manua

    D. Thickness of the 320 x 500 Top Plate

    For the load acting on the top plate, neglecting the selfweight of the pipe pedestal, equilibrium is applied and from

    STAAD Output, the factored reaction in Y direction is Ry = 80.06 kNs (Please refer to page 11/30 of STAAD Outp

    The flange width of the main beam is 290 mm. A 320 x 500 plate is considered for the frame.

    However for uniformity the size of top plate of 320 x 500mm is adopted.

    100

    160

    320

    160

    100 300 100

    500

    For a 320 x 500 plate, the pressure is 80.06 x 1000/ (320 x 500) = 0.500 mPa

    From Inspection, the length of the cantilever = 100 mm

    Considering a 1mm strip of plate,

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    The moment, M = 0.500 x 1002/2 = 2500 N-mm

    The actual fb = 6M / bd2 = 6 x 2500 / 1d2

    The allow Fb = 0.75 x 248 = 186.0 mPa

    186 = 6(2500) / 1d2

    d = 8.98 mm

    However, the thickness of the top plate is adopted as 20mm due to the following facts;

    1. The top plate is subjected to corrosion which is not considered in the calculations.2. The strength of the top plate is reduced by the intense heat due to welding process.3. The top plate is subjected to thermal loads due to temperature changes which is not

    considered in the calculations.

    4. The minimum thickness of top plate if 6 mm thick of weld is used is from 12 -20mm asper AISC manual.

    5. The minimum thickness of the top plate shall not be less than the flange thicknessof the main beam, tf = 17.50 mm for a HE360A .

    Check Bolt Anchorage of Main Beam to Top Plate:

    1. JOINT 17 (Exterior Column = 200 x 600)most Critical JointHowever for uniformity the size of top plate of 320 x 500mm is adopted.

    100

    160

    320

    160

    100 300 100

    500

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    A) Tension

    From STAAD Pro Output, Factored forces (Please refer to page 11/30 of STAAD Output)

    Rx = -1.65 kNs Ry = -25.82 kNs Rz = 45.06 kNs (Load No. 10)

    The orthogonality/45 degree dirction of horizontal forces has been included in the load combinations of

    STAADPro Inputs (Load Nos. 13 & 14)and from the results as per Support reactions, the values of forces are n

    critical. Please see page 11/30 of STAAD Pro outputs for reference.

    Tension = -25.82 kNs

    Actual tension for 1Anchor Bolt = 25.82 / 4 = 6.455 kNs

    For M20 A325 Bolt

    Allowable Tension Stress, Ft = 303 mPa

    Actual Tensile Stress, ft = 6.455 x 1000 / (3.14 x 202/4) = 20.55 mPa < 303 mPa Therefore OK!

    3. JOINT 17 ( Exterior Column = 200 x 600)most Critical JointB) Shear

    From STAAD Pro Output, Factored forces (Please refer to page 11/30 of STAAD Output)

    Rx = -15.65 kNs Ry = 40.74kNs Rz = 101.41 kNs (Load No.12)

    The orthogonality/45 degree dirction of horizontal forces has been included in the load combinations of

    STAADPro Inputs (Load Nos. 13 & 14)and from the results as per Support reactions, the values of forces are n

    critical. Please see page 11/30 of STAAD Pro outputs for reference.

    Total Factored Shear, V = (15.65)\2 + 101.412 )0.50 = 102.61 kNs

    Actual shear for 1Anchor Bolt = 102.61 / 4 = 25.65 kNs

    For M20 A325 Bolt

    Allowable Shear Stress, Fv = 117 mPa

    Actual Shear Stress, fv = 25.65 x 1000 / (3.14 x 202/4) = 81.68 mPa < 117 mPa Therefore OK!

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    C) Bearing

    For M20 A325 Bolt

    Allowable Bearing Stress, Fb = 145 mPa

    Actual Bearing Stress, fb = 25.65 x 1000 / (20 x 17.50) (Thinner of two plates is applied)

    Actual Bearing Stress, fb = 73.28 mPa < 145 mPa Therefore OK!

    Check Short Column between Bottom Plate & Top Plate:

    2. JOINT 17 most Critical JointHowever for uniformity the size of Short Column of 200mm x 120mm x 20mm is adopted.

    100

    160

    320

    160

    100 300 100

    500

    From Inspection, the Critical Tributary Area is 150mm x 320mm

    For a 320 x 500 plate, the pressure is 80.06 x 1000/ (320 x 500) = 0.500 mPa

    Compression force acting on the short column = 0.500 x 1000\x 0.320 x 0.150 = 24 kNs

    For fixed-fixed ends, K = 0.50

    Iy = bh3/12 = 0.020 x 0.2003/12 = 1.333 x 10-05q.m.

    Ix = bh3/12 = 0.200 x 0.020

    3/12 = 1.333 x 10

    -07q.m.

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    100

    100

    200

    100

    100 300 100

    500

    From algrebra, the area of the segment of connecting circles = 3364.31mm2

    Area of Circle = D2/4 = 3.14 x 1702/4 = 22686.50 mm2

    Net Area for 4 Circles subjected to 45 degree Cone Failure, An = 4 x 22686.503 x 3364.31 = 80653.07 mm2

    For Anchor Tension Failure load;

    Tensile capacity of 4 concrete cone, Fu = 4(fc)xAn = 0.075x 0.65 x (3000) x 125 = 178001 lbs

    Fu = 79.21 kNs > ft = 25.82 kNs Therefore OK!

    Check for min edge distance; The ACI

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    Conclusion:In view of the above calculations, it is concluded that the top plate, top plate, Uma Ukat Chemical

    Anchors and M20 A325 Bolts can safely carry the loads imposed by the Penetrating Tower + shelter

    equipment including appurtenances.