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

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

1/15

ZND332_A-AMT-Td-0002-Rev.

Page. 1

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


2/15


Page. 2

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


3/15


Page. 3

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


4/15


Page. 4

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


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


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)



5/15


Page. 5

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


6/15


Page. 6

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


7/15


Page. 7

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


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

Compression = 80.06kNs


8/15


Page. 8


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;


9/15


Page. 9

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,


10/15


Page. 10

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


11/15


Page. 11

A) Tension


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


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


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


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


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


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!


12/15


Page. 12

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.


13/15


14/15


Page. 14

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


15/15


Page. 15

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.

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

Documents