Chapter 8 formwork Bách Khoa

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Đây là tài liệu của các bạn sinh viện hiện tại đang học tại Đại học Bách Khoa TP HCM. Đồng thời cũng là giáo án của giảng viên tại Đại học Bách Khoa. Nó sẽ rất hữu ích cho công việc học tập của các Bạn. Chúc Bạn thành công. Chapter Concrete formwork Construction methods and management ©2010 Đỗ Thị Xuân Lan , GVC Ths Chapter Concrete formwork 8.1 General requirements for formwork 8.2 Classification of formwork 8.3 Typical formwork (footing form, wall form, column form, beam and slab form) 8.4 Shores and scaffolding 8.5 Concrete form design 8.6 Formwork inspection and acceptance ©2010 Đỗ Thị Xuân Lan , GVC Ths Causes of failures of formwork • Improper or inadequate shoring • Inadequate bracing of members • Lack of control of rate of concrete placement • Improper vibration or consolidation of concrete • Improper or inadequate connections • Improper or inadequate bearing details source: Peurifoy Oberlender, 1996, tr.156 ©2010 Đỗ Thị Xuân Lan , GVC Ths Causes of failures of formwork • • • • • • Premature stripping of formwork Errors in placement of reshoring Improper or lack of design of formwork Inadequate strength of form material Failure to follow codes and standards Modifications of vendor-supplied equipment • Negligence of workers or supervisors source: Peurifoy Oberlender, 1996, tr.156 ©2010 Đỗ Thị Xuân Lan , GVC Ths Chapter Concrete formwork GENERAL REQUIREMENTS FOR FORMWORK ©2010 Đỗ Thị Xuân Lan , GVC Ths Formwork/concrete life cycle source: Hanna, 1999 ©2010 Đỗ Thị Xuân Lan , GVC Ths Objectives in building and designing formwork • Quality— must be designed and built with sufficient stiffness and accuracy so the size, shape, position, and finish of the cast concrete are attained within the required tolerances • Safety— must be built with sufficient strength and factors of safety so they are capable of supporting all dead and live loads without collapse or danger to workers and to the concrete structure • Economy— must be built efficiently, minimizing time and cost in the construction process ©2010 Đỗ Thị Xuân Lan , GVC Ths Principles in building and designing formwork Each component of the forms must be able to support its load from two points of view: • Strength: based on the physical properties of the material used • Serviceability: the ability of the selected sections to resist the anticipated loads without exceeding deflection limits ©2010 Đỗ Thị Xuân Lan , GVC Ths General requirements for formwork • Be strong enough to withstand the pressure of plastic concrete and to maintain their shape during the concrete placing operation • Be tight enough to prevent wet concrete from leaking through joints and causing unsightly fins and ridges • Be simple to build • Be easy to handle on the job ©2010 Đỗ Thị Xuân Lan , GVC Ths 10 SLAB FORM DESIGN Select wale spacing: consider the stud as a uniformly loaded beam supporting a strip of design load 304mm w = (0.304m) x (1) x (30kN/m2) = 9.12 kN/m Bending: 100 Fb S 12 l= ( ) 1000 w 100 (12480)(0.5019 x105 ) ( ) = 926mm l= 1000 9.12 ©2010 Đỗ Thị Xuân Lan , GVC Ths 131 WALL FORM DESIGN • Shear 1.11 FV A + 2d 1000 w 1.11 (827)(3387) l= + 2(88.9) = 519mm 1000 9.12 l= • Deflection 73.8 EI 1/3 l= ( ) 1000 w 73.8 (11.7 x106 )(2.23x106 ) 1/3 l= ( ) = 1048mm 1000 9.12 • Shear governs in this case and the maximum wale span is 519mm We will select a 406mm wale spacing for modular©2010 units Đỗ Thị Xuân Lan , GVC Ths 132 WALL FORM DESIGN Select tie spacing: Based on a wale spacing of 406mm 406 x9.12 = 12.18kN / m w= 304 Bending: 100 Fb S 12 l= ( ) 1000 w 100 (12480)(2 x0.5019 x105 ) ( ) = 1014mm l= 1000 12.18 ©2010 Đỗ Thị Xuân Lan , GVC Ths 133 WALL FORM DESIGN • Shear 1.11 FV A + 2d 1000 w 1.11 (827)(2 x3.387 x103 ) l= + 2(88.9) = 688mm 1000 12.18 l= • Deflection 73.8 EI 1/3 l= ( ) 1000 w 73.8 (11.7 x106 )(2 x 2.231x106 ) 1/3 l= ( ) = 1199mm 1000 12.18 • Shear governs and the maximum stringer span is 688mm Select a 610mm©2010 tiecủaspacing forThs a modular value 134 Đỗ Thị Xuân Lan , GVC WALL FORM DESIGN • Now, we check tie load The maximum stringer span: (406)(610) P= x30 = 7.42kN ≺ 13.34kN (1000)(1000) ©2010 Đỗ Thị Xuân Lan , GVC Ths 135 WALL FORM DESIGN • Check bearing: - Stud on wales: Bearing area (A) (double wales) A= (2)(38)(38) = 2888mm2 Load P = load/m of stud x wale spacing (9.12)(406) = 3.7kN 1000 3.7 x106 f cL = = 1222kPa ≺ Fc = 3340kPa 2888 ©2010 Đỗ Thị Xuân Lan , GVC Ths 136 WALL FORM DESIGN • Check bearing: - Tie wedges on wales Bearing area (A) (double wales) A= (2)(38)(38) = 2888mm2 Load P = 7.11kN 7.11x106 f cL = = 2462kPa ≺ Fc = 3340kPa 2888 ©2010 Đỗ Thị Xuân Lan , GVC Ths 137 WALL AND COLUMN FORM DESIGN • Sheathing: 1.2x2.4m sheets of 19mm class I Plywood • Stud: 50x100mm at 304mm spacing • Wales: 2x50x100mm at 406mmm spacing • Tie: 13.34kN at 610mm interval FIGURE 13-3 Wall form, Example ©2010 Đỗ Thị Xuân Lan , GVC Ths 13-2 DESIGN OF LATERAL BRACING • Many failures of formwork have been traced to omitted or inadequately designed lateral bracing ©2010 Đỗ Thị Xuân Lan , GVC Ths Lateral Braces for Wall and Column Forms Table 13-3 Recommended minimum lateral design load for wall forms ©2010 Đỗ Thị Xuân Lan , GVC Ths Lateral Braces for Wall and Column Forms • Determine the maximum spacing of nominal 50x100-mmm lateral braces for the wall form of example 13.2 placed as shown in Figure Assume that local code wind requirements are less stringent than Table 13.3 Allowable stress values for the braces are as follows: Fc Allowable stresses (kPa) 5861 Fl E 4999 9.7x106 ©2010 Đỗ Thị Xuân Lan , GVC Ths 141 Lateral Braces for Wall and Column Forms Determine the design lateral force per unit length of form: H = 1.46kN/m Determine the length of the strut: l = (h '2 + l '2 )1/2 = (1.832 + 1.532 )1/2 = 2.38m The axial concentrated load on the strut produced by a unit length of the form: Hxhxl (1.46)(2.44)(2.38) = = 3.03kN / m P' = h ' xl ' (1.83)(1.53) ©2010 Đỗ Thị Xuân Lan , GVC Ths OH h ' l 'h' = → OH = 142 l' l l Lateral Braces for Wall and Column Forms • Determine the l/d ratio: 2380 l/d = = 62.5 > 50 38.1 • Use a single lateral support at the midpoint of each strut l=1.19m 0.3E (0.3)(9.7 x106 Fc ' = = = 2.97kPa 2 (l ) (1190 ) d 38 • As Fc’ < Ft < Fc so the value Fc’ governs the maximum allowable compressive = (38)(89)(2967) = 10.03kN 10.03m = 3.31 • The maximum strut spacing: s = 3.03 ©2010 Đỗ Thị Xuân Lan , GVC Ths 143 Lateral Braces for Slab Forms Determine the design lateral force for the slab form 152mm, 6.1m wide and 3.5m long as shown in Figure13-5 The slab is to be poured in one pour Assume concrete density is 2403kg/m3 and the formwork weighs 0.72kPa ©2010 Đỗ Thị Xuân Lan , GVC Ths FIGURE 13-5 Slab form bracing design, 144 Lateral Braces for Slab Forms Dead load: (0.152)(1)(2403)(9.8) + 0.72 = 4.3kPa 1000 For the 6.1m face, the width of the slab is 30.5m H = (0.02)(4.3)(30.5) = 2.62kN/m dl = For the 30.5m face, the width of the slab is 6.1m H = (0.02)(4.3)(6.1) = 0.52kN/m
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