Lecture_24

Lecture_24 - 6.4 Punching Shear 6.4.1 Punching failure A...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6.4 Punching Shear 6.4.1 Punching failure A concentrated load P on a slab causes shearing stresses on a section around the load; this effect is referred to as punching shear. Fig. 6.4-1 Punching shear failure Punching shear failure may occur in flat slab construction, column footings, pile caps and bridge decks. Fig. 6.4-2 Punching shear failure of reinforced concrete slab-column connection due to axial column load 228 6.4.2 Analysis (BS 8110: Part 1, clause 3.7.7.4) (1) Control perimeter (critical section) It is at a distance of 1.5d from the load and has square corners whether the shape of the loaded area is in square or circular. (a) Slab subjected to punching loaded area control perimeter (b) (c) Fig. 6.4-3 Control perimeter adopted by design codes (2) Punching shear stress is calculated by v= N ud (6.5) where u = control perimeter (perimeter of the critical section). In Fig. 6.4-3(b), u = 4·(c + 2 × 1.5d) . • If the punching shear stress v < vc, no shear reinforcement is required. • Checks must be undertaken to ensure: The shear stress calculated for the perimeter at the face of the loaded area vmax ≤ 0.8√fcu or 5N/mm2, whichever is the lesser. 229 Text Example 8.1: Punching shear A slab 175 mm thick, d = 145 mm, is constructed with grade 30 concrete and is reinforced with 12 mm bars at 150 mm centres one way and 10 mm bars at 200 mm centres in the other directions. Determine the maximum load that can be carried on an area, 300 × 400 mm, without exceeding the ultimate shear stress. For φ12@150, 100 As 100 × 754 = = 0.52 bd 1000 × 145 and for φ10@200, 100 As 100 × 393 = = 0.27 bd 1000 × 145 Average value of 100 As = 0.395 bd From Table 3.8 of BS 8110, vc = 0.62 N/mm2 for grade 30 concrete Punching shear perimeter = (2a + 2b +12d) = 600 + 800 + 12 × 145 = 3140 mm Maximum load = vc × perimeter × d = 0.62 × 3140 × 145 = 282 kN At the face of the loaded area, the shear stress 282 × 103 N v= = = 1.39 N/mm 2 (2a + 2b)d (600 + 800)145 which is less than 0.8√fcu and 5 N/mm2. 230 6.4.3* Design for shear reinforcement (Clauses 3.7.7.5 & 3.7.7.6) • Shear perimeters Fig. 6.4-4 Punching shear zones o The first perimeter is checked. If v ≤ vc , no further checks are required. o If v > vc , successive perimeters have to be checked until one is reached where v ≤ vc . • When vc < v < 2vc , shear reinforcement in form of links may be provided to increase the shear resistance of the punching shear resistance, where the slab thickness should be over 200 mm. (IstructE suggests that the slab thickness be at least 250 mm.) • Amount of punching shear reinforcement For cases where v ≤ 1.6vc , ∑A sv sinα ≥ (v − vc )ud 0.95 f yv (6.6) For cases where 1.6vc < v ≤ 2vc , ∑A sv * sinα ≥ 5(0.7v − vc )ud 0.95 f yv (6.7) Optional course materials for CIVL 232 231 In either case, ∑A sv sinα ≥ 0.4ud 0.95 f yv where α = angle between shear steel and the plane of the slab u = length of the outer perimeter of the zone • Zones for punching shear reinforcement Fig. 6.4-5 Zones for punching shear reinforcement o The reinforcement should be provided on at least two perimeters between the column perimeter and the first perimeter. o The first perimeter of reinforcement should − be located approximately 0.5d from the face of the column; − contain not less than 40% of Asv . o The second perimeter of reinforcement should be located at not more than 0.75d from the first. 232 o The reinforcement is distributed evenly around a perimeter and the spacing of the legs of links should not exceed 1.5d. o The spacing of perimeters of reinforcement should not exceed 0.75d. • Illustrated detailing of punching shear reinforcement (a) (b) Fig. 6.4-6 Shearhoop reinforcement system. (a) Plan view; (b) basic shearhoop Fig. 6.4-7 Reinforcement cages (North American practice) Fig. 6.4-8 Shear stud reinforcement cages 233 6.5 Stair Slabs 6.5.1 General information and types 550 mm < G + 2R < 700 mm Fig. 6.5-1 General information. (a) Dimensions; 234 Fig. 6.5-1 (cont’d) (b) detailing Fig. 6.5-2 Types of stairs • Stairways are normally sloping one-way spanning slabs. • The usual forms of stair slabs can be classified into two types: o Transverse spanning stair slab o Longitudinal spanning stair slab 235 6.5.2 Transverse spanning stair slabs The slabs span between walls, a wall and string (an edge beam) or two stringers. breadth (Effective depth = D/2) Fig. 6.5-3 Transverse panning stair slabs • The stair slab is designed as a series of beam consisting one step with a breadth b and an effective depth of d = D/2. • For a cantilevered stair, the effective depth d is taken as the mean effective depth of the section. Fig. 6.5-4 Cantilever stairs 236 6.5.3 Longitudinal spanning stair slabs (a) Plan (b) Section Fig. 6.5-5 Slab spanning between supports at top and bottom of the flight (a) (b) Fig. 6.5-6 Slab spanning around a lift well. (a) Plan; (b) section A-A. 237 (1) The supports of a stair slab may be beams, walls or landing slabs. o The effective span l is measured horizontally between the centres of the supports. o The thickness of the waist is taken as the slab thickness. o Span-depth ratio of stair slab = (1 + 15%) span/depth ratio of normal slabs. o When the staircase is built monolithically at its ends into supporting members spanning at right-angles (Fig. 6.5-6), effective span (span of flight) = l a + (l b1 + l b 2 ) / 2 where l a = clear span l b1 , l b 2 = breadths of supporting members at different ends, respectively, or 1.8 m whichever is the smaller. (2) Loading o The dead load is calculated along the slope length of the stairs; the live load is based on the plan area. o Loads common to two spans that intersect at right angles and surround an open well (or lift well) may be assumed to be divided equally between the spans. (3) The maximum moment near mid-span and over supports is taken as Wl / 10 , where W is the total load of the flight. Text Example 8.10: Design of a stair slab, pp.211-212 238 Example 6.5-1: Stair slab 1. Specification Design the side flight of a staircase surrounding an open stair well. A section through the stairs is shown in Fig. 6.5-7(a). The stair slab is supported on a beam at the top and on the landing of the flight at right angles at the bottom. The imposed loading is 5 kN/m2. The stair is built 110 mm into the side wall of the stair well. The clear width of the stairs is 1.25 m and the flight consists of eight risers at 180 mm and seven goings of 220mm with 20 mm nosing. The stair treads and landings have 15 mm granolithic finish and the underside of the stair and landing slab has 15 mm of plaster finish. The materials are grade 30 concrete and grade 460 reinforcement. (a) Fig. 6.5-7 Example stair slab. (a) Section through stairs; 239 (b) Fig. 6.5-7 (cont’d) (b) loading diagram 2. Loading and Moment Assume the waist thickness of structural concrete is 100 mm, the cover is 25 mm and the bar diameter is 10 mm. The loaded width and effective breadth of the stair slab are shown in section AA in Fig. 6.5-7(a). The effective span of the stair slab is the clear horizontal distance (1540 mm) plus the distance of the stair to the centre of the top beam (235 mm) plus one-half of the breadth of the landing (625 mm), i.e. effective span = 1540 + 235 + 625 = 2400 mm The design ultimate loading on the stairs is calculated first. (i) Landing slab The overall thickness including the top and underside finishes is 130 mm: Dead load = 1.4 × 0.13 × 24 = 4.4 kN/m2 Imposed load = 1.6 × 5 = 8.0 kN/m2 Total load on landing slab = 0.5 × (4.4 + 8.0) × 0.625 × 1.1 = 4.26 kN 240 One-half of the load on the landing slab is included for the stair slab under consideration. The loaded width is 1.1 m. (ii) Stair slab The slope length is 2.29 m and the steps project 152 mm perpendicularly to the top surface of the waist. The average thickness including finishes is 100 + 152/2 + 30 = 206 mm dead load = 1.4 × 0.206 × 2.29 × 24 × 1.1 = 17.44 kN imposed load = 1.6 × 5 × 1.78 × 1.1 = 15.66 kN total load on stair slab = 33.1 kN The dead load is calculated using the slope length while the imposed load acts on the plan length. The loaded width is 1.1 m. The total load on the span is V = {33.1(2.4−1.775/2)+4.26(2.4−1.775)/2}/2.4 = 21.42 kN 4.26 + 33.1 = 37.36 kN The maximum shear at the top support is 21.42 kN. The design moment for sagging moment near mid-span and the hogging moment over the supports is 37.36 × 2.4/10 = 8.97 kNm 3. Moment Reinforcement The effective depth d is 100 – 25 –5 = 70 mm. The effective width will be taken as the width of the stair slab, 1250 mm. 8.97 × 10 6 M = 1.46 = bd 2 1250 × 70 2 100As / bd = 0.39 241 As = 0.39 × 70 × 1250/100 = 341 mm2 Provide 8T8@180 (402 mm2). The same steel is provided in the top of the slab over both supports. The minimum area of reinforcement is 0.13 × 100 × 1000/100 = 130 mm2 Provide T8@300 (167 mm2/m) 4. Shear Resistance Shear = 21.4 kN 21.4 ×10 3 v= = 0.24 N / mm 2 70 ×1250 100 As 100 × 402 = = 0.46 bd 1250 × 70 vc = 0.79(0.46)1 / 3 (400 / 125)1 / 4 (30 / 25)1 / 3 / 1.25 = 0.69 N / mm 2 The slab is satisfactory with respect to shear. Note that a minimum value of d of 125 mm is used in the formula. 5. Deflection The slab is checked for deflection. The basic span/d ratio is 26. fs = 5 × 460 × 341.3 = 244.1N / mm 2 8 × 402 The modification factor for tension reinforcement is 0.55 + 477 − 244.1 = 1.37 120(0.9 + 1.46) Allowable span/d ratio = 26 × 1.37 × 1.15 = 41.0 Actual span/d ratio = 2400/70 = 34.3 242 Note that the stair flight with a plan length of 1540 mm occupies 64% of the span and the allowable span/d ratio can be increased by 15% (BS8110: Part1, clause 3.10.2.2). 6. Cracking For crack control the clear distance between bars is not to exceed 3d = 210 mm. The reinforcement spacing of 180 mm is satisfactory. 7. Reinforcement The reinforcement is shown below. 243 6.6 Introduction to Yield Line Analysis (Details of the theory and analysis are introduced in CIVL 521 Advanced Reinforced Concrete) • Johansen’s yield line theory is an ultimate-load theory and is based on an assumed collapse mechanism and plastic properties of underreinforced slabs. • The bands in which yielding has occurred are referred as yield lines. Fig. 6.6-1 Development of yield lines at failure • Rotation along the yield lines will occur at a constant moment ( = the ultimate moment of resistance of the section), and will absorb energy. This can be equated to the energy expended by applied loads undergoing a compatible displacement (the virtual work principle). • Basic assumptions in the yield-line analysis: (1) The yield lines divide the slab into several rigid regions which remain plane, so that all rotations take place in yield lines. (2) Yield lines are straight and end at the boundary. 244 (3) A yield line between two rigid regions must pass through the intersection of the axes of rotation of the two regions. (4) An axis of rotation lies along a line of support and passes over columns. • Yield-line patterns Fig. 6.6-2 Examples of yield-line patterns • Details of the yield-line analysis can be found in Chapter 8 of Ref. by F.K. Kong & R.H. Evans. 245 ...
View Full Document

Ask a homework question - tutors are online