TOWER CRNE MAST ANCHORAGE TIE DESIGN

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1 TOWER CRNE MST NCHORGE TIE DESIGN Sujit Halder P&M Department ITD Cementation India Limited bstract: One variant is piled base cast in fixing angle type with top slewing tower crane which needs mast anchorage tie to ascertain stability beyond working manufacturer specified freestanding mast height. Major challenge arises at the time of fixing ties with the existing building or structure as shape, location, distance, load bearing capacity of the structure may differ from project to project. Consulting manufacturer service department may not be possible all the time. ttempt is made to find an approximate value of the forces to be encountered by the ties in service condition and designing the ties. Keyword - Tower Crane, Mast nchorage Tie, In service condition, Propped Cantilever, Designing the ties. 1. INTRODUCTION mong different types of lifting gear tower cranes are most popular and most effective for high rise buildings, chimneys, towers, power plants etc. Space requirement is less. Loads can be lifted at a greater height with wide coverage area. Selection of tower crane type depends on site conditions and requirements. In Service Condition: Self weight of the crane Lifted load Wind load Dynamic effects for hoisting, slewing and trolleying. Out of Service Condition: Self weight of the crane Wind loading 2. CRNE SPECIFICTION Maximum lifting capacity up to 14.7m radius : 6 MT Maximum lifting capacity up to 50m radius : 1.3 MT Maximum lifting speed : 18.8 m/min Free standing mast height : 48 m First mast anchorage tie from base : 27 m Slewing speed (max.) : 0.7 r/min Slewing torque (max.) : 120 N.m Trolleying speed : 15 m/min Maximum possible mast height : m Maximum jib length : 50 m Considering a dynamic load co-efficient: FINDING CG OF THE CRNE IN SERVICE CONDITION Counter clockwise moments: S. No. Item description Weight of the Lever from true geometric Moment (Kg. m) component (Kg) vertical axis of the mast (m) 1 Counter jib allast Total counter clockwise moments Clockwise moments: S. No. Item description Weight of the component (Kg) Lever from true geometric vertical axis of the mast (m) Moment (Kg. m) 1 Jib foot section Jib section Jib section Jib section , IJIRE- ll Rights Reserved Page -10

2 5 Jib section Jib section Jib nose Jib trolley (13 m radius) Tower head, ladder and monorail 10 Lifted load 6000 X Total clockwise moments CG (service condition) = = = m (clockwise) Total Weight Load eccentricity, e = m 4. FINDING CG OF THE CRNE OUT OF SERVICE CONDITION Counter clockwise moments: S. No. Item description Weight of the Lever from true geometric Moment (Kg. m) component (Kg) vertical axis of the mast (m) 1 Counter jib allast Total counter clockwise moments Clockwise moments: S. No. Item description Weight of the component (Kg) Lever from true geometric vertical axis of the mast (m) Moment (Kg. m) 1 Jib foot section Jib section Jib section Jib section Jib section Jib section Jib nose Jib trolley (13 m radius) Tower head, ladder and monorail Total clockwise moments CG (out of service condition) = = = 1.46 m (anti-clockwise) Total Weight Load eccentricity = e = 1.46 m 5. WIND EFFECT Wind load in KN, F = p C f is the solid surface area perpendicular to the wind direction. C f is the force coefficient, depends on the shape of the element facing wind. P is wind pressure in KN/m 2 calculated using the formula, P = x 10-3 x v 2 s. Where, v s is the calculated wind speed in m/s. In service wind speed varies from 14 to 28 m/s. Out of service storm wind speed varies between 36 to 46 m/s depending upon height above ground level. Force coefficient for square lattice tower with flat sided sections given by, C f = 1.7 x (1 + η) Shielding factor, η = 0.43 Force coefficient, C f = 1.7 x ( ) = Wind facing frontal area of the tower mast = 33 m , IJIRE- ll Rights Reserved Page -11

3 In service wind force on the tower crane mast: For 20m/s wind speed (Normal crane in service wind speed), P = x 10-3 x v 2 s = x 10-3 x 20 2 = 0.25 KN/m 2 Maximum in service wind force, F = x p x C f = 33 x 0.25 x = 20 KN Out of service wind force on the tower crane mast: For 46 m/s wind speed (Storm wind speed), P = x 10-3 x v 2 s = x 10-3 x 46 2 = KN/m 2 Maximum in service wind force, F = x p x C f = 33 x x = 104 KN Pressure intensity increases from bottom to top. For simplicity of calculation let us assume that it is uniform throughout the height. Pressure point is the mid -point of the mast. 6. MOMENT OF INERTI OF MST CROSS SECTION Cross section of mast main cord: 160 x 15 3 Moment of Inertia about XX = YY = [{ + (160 x 15) x } x 4] x Maximum ending Moment of mast without wind, P M = P. e sec l ( ) EI + [{ + (145 x 15) x } x 4] 12 = m 4 7. FORCES ON MST TIE IN SERVICE CONDITION , IJIRE- ll Rights Reserved Page -12

4 29323 x 9.81 = x 9.81 x 2.17 x sec {48 } 2.1 x 10 5 x 1000 x 1000 x = x 9.81 x 2.17 x sec (0.5493) = x 9.81 x 2.17 x sec (31.47º) = Nm M x l² Deflection at 48 m height = 2 EI x 48 2 = 2 x 2.1 x 10 5 x1000 x 1000 x = m = 383 mm Deflection of mast due to 20 KN horizontal wind force on the mast acting at an height of 24 m( i.e., middle of the mast), 5x20000 x 48 3 δ = 48 x 2.1 x 10 5 x 1000 x 1000 x = m = 105 mm Total maximum deflection in service condition = = 488 mm s per manufacturer recommendation, first mast tie to be added at 27 m from base to resist deflection of the mast at top. Let us assume tie will exert a force P which deflects the mast to its vertical position, P x 27 3 Opposing deflection = 3 x 2.1 x 10 5 x 1000 x 1000 x P x X (48 27) 2 x 2.1 x 10 5 x 1000 x 1000 x = x 10-6 x P x 10-6 x P = x 10-6 x P Net deflection of 48 m height = 0 Therefore, x 10-6 x P = Or, P = N Or, P = 7688 Kg 8. FORCES ON MST TIE OUT OF SERVICE CONDITION Maximum ending Moment of mast without wind, P M = P. e sec l ( ) EI x 9.81 = x 9.81 x 1.46 x sec {48 } 2.1 x 10 5 x 1000 x 1000 x = x 9.81 x 1.46 x sec (0.4835) = x 9.81 x 1.46 x sec (27.7º) = Nm M x l² Deflection at 48 m height = 2 EI x 48 2 = 2 x 2.1 x 10 5 x1000 x 1000 x = m = 193 mm Deflection of mast due to 104 KN horizontal wind force on the mast acting at an height of 24 m (i.e., middle of the mast) , IJIRE- ll Rights Reserved Page -13

5 5 x104 x 1000 x 48 3 δ = 48 x 2.1 x 10 5 x 1000 x 1000 x = m = 545 mm Total maximum deflection in service condition = = 738 mm s per manufacturer recommendation, first mast ties to be added at 27 m from base to resist deflection of the mast at top. Let us assume tie will exert a force P which deflects the mast to its vertical position, P x 27 3 Opposing deflection = 3 x 2.1 x 10 5 x 1000 x 1000 x P x X (48 27) 2 x 2.1 x 10 5 x 1000 x 1000 x = x 10-6 x P x 10-6 x P = x 10-6 x P Net deflection of 48 m height = 0 Therefore, x 10-6 x P = Or, P = N Or, P = Kg 9. FINDING FORCE CTING ON LEGS Force acting from the geometrical axis of the mast at anchorage tie level and its effect on each leg has to be found out resolving reactions at each leg support. Condition 1: N force acting in the perpendicular direction from tie support wall and away from the wall: N C D E mm 2500 mm 5000 mm R R C = E = 6236 mm DE = 300 mm D = 6708 mm Taking moments about, R x 5 = x 2.5 Or, R = N , IJIRE- ll Rights Reserved Page -14

6 R = N Free body diagram at joint, C N 74⁰ Force on C = E = = N (Tensile) Sin 74⁰ Horizontal force at joint = x cos74⁰ = N Free body diagram at joint, D N 63⁰ Force on D = = N (Tensile) Sin 63⁰ Horizontal force at joint = x cos63⁰ = N Condition 2: N force acting from the mast axis and parallel to the wall / supporting structure: N C D E mm 2500 mm R 5000 mm , IJIRE- ll Rights Reserved Page -15 R

7 Taking moments about, R x 5 = x 6.8 Or, R = N Taking moments about, R x 5 = x 6.8 Or, R = N Free body diagram at joint, N C 74⁰ Force on C = = N (Compressive) Sin 74⁰ Horizontal force at joint = x cos74⁰ = N Free body diagram at joint, N D 63⁰ Force on D = = N (Compressive) Sin 63⁰ Horizontal force at joint = x cos63⁰ = N Condition 3: N force acting from the mast axis and at an angle 45⁰with the wall N C D E mm 2500 mm 5000 mm R , IJIRE- ll Rights Reserved Page -16 R

8 Net force = N Resolving along vertically and horizontally, Vertical component = N Horizontal component = N Taking moments about, R x x 2.5 = x 6.8 Or, R = N R = N Free body diagram at joint, C 69355N 74⁰ Force on C = = N (Tensile) Sin 74⁰ Horizontal force at joint = x cos74⁰ = N Free body diagram at joint, N D 63⁰ Force on D = = N (Tensile) Sin 63⁰ Horizontal force at joint = x cos63⁰ = N Force Table: Component 0⁰ 45⁰ 90⁰ C N (Tensile) N (Tensile) N (Tensile) D N (Tensile) N (Tensile) N (Tensile) E N (Compressive) N (Compressive) N (Tensile) 10. CONCLUSION 1. Once forces on the legs are determined sections, pins, tie length etc. can be selected with factor of safety as per standard code of practice. 2. Mast should be tied in perfect vertical position to facilitate jacking and climbing. 3. Legs are considered to be on perfect level i.e., at an angle 90⁰ with the vertical mast. References: [1].Tower Crane Stability, by Hilary Skinner, Tim Watson, ob Dunkley, Paul lackmore. [2]. [3].Indian Code of Practice for Cranes Wind Load ssessment (IS 14467: 1997). [4].Validation of Use of FEM (NSYS) for Structural nalysis of Tower Crane Jib an Static and Dynamic nalysis of Tower Crane Jib Using NSYS [ISSN: ],by jinkya Karpe, Sainath Karpe, jaikumar Chawrai, Prof. Sachin Ranjan Vankar. [5].Strength of Materials, by S. Ramamurtam, ISN X, 6 th edition , IJIRE- ll Rights Reserved Page -17

(Please write your Roll No. immediately) Mid Term Examination. Regular (Feb 2017)

(Please write your Roll No. immediately) Roll No... Mid Term Examination Regular (Feb 2017) B.Tech-II sem Sub-Engineering Mechanics Paper code- ETME-110 Time-1.5 hours Max Marks-30 Note: 1. Q.No. 1 is