chapter 2 – strip method for slabs

Chapter 2 – Strip Method for Slabs

Dr.-Ing. Girma ZerayohannesDr.-Ing. Adil Zekaria

2

Chapter 2- Strip Method for Slabs

• 2.1 Introduction• Different methods of analysis are allowed by

EBCS-2• One of these is plastic methods• Strip method is a plastic method of analysis

Dr.-Ing. Girma Zerayohannes

Dr.-Ing. Girma Zerayohannes 3


• The upper bound theorem of the theory of plasticity is presented in chapter 1. The YL method of slab analysis is an upper bound approach to determining the capacity of the slab

• Disadvantages:• An upper bound analysis if in error will be on the

unsafe side. The actual carrying capacity will be less than, or at best equal to the capacity predicted, which is a cause for concern.



• When applying this method it is necessary to assume that the distribution of reinforcement is known over the whole slab.

• a tool for review.• Can be used for design only in an iterative

sense, i.e., trail design until a satisfactory amount is found



• These circumstances motivated Hillerborg (1956) to develop what is known as strip method for slab design

• In contrast to yield line analysis, the strip method is a lower bound approach, based on the satisfaction of equilibrium requirements every where in the slab



• By the strip method, a moment field is first determined that fulfills equilibrium requirements, after which the reinforcements of the slab at each point is designed for this moment field



• Lower Bound Theorem: If a distribution of moments can be found that satisfies both equilibrium and boundary conditions for a given external loading, and if the yield moment capacity of the slab is nowhere exceeded, then the given external loading will represent a lower bound of the true carrying capacity



• Advantages:• The strip method gives results on the safe

side, which is certainly preferable in practice• The strip method is a design method, by which

the needed reinforcement can be calculated



• 4.2 Basic Principles• The governing equilibrium equation for a small

slab element having sides dx and dy is:



Figure 1



wyx

mym

xm xyyx

2

2

2

2

2

2



• Where w = the external load per unit area• mx, my = Bending Moments per unit width in

the x and y directions and• mxy = the twisting moment



• So according to the lower bound theorem, any combination of mx, my, and mxy that satisfies the equilibrium at all points in the slab and that meets boundary conditions is a valid solution, provided that the reinforcement is placed to carry these moments



• The basis for the simple strip method is that the torsional moment is chosen equal to zero; no load is assumed to be resisted by the twisting strength of the slab mxy = 0

• The equilibrium equation then reduces to:

wym

xm yx

2

2

2

2



• This equation can be split conveniently into 2 parts, representing twist less beam strip action.

wkym

kwxm

y

x

)1(2

2

2

2



• Where the proportion of load taken by the strips is k in the x-direction and (1-k) in the y-direction (concept of load dispersion)

• In many regions in slabs, the value k will be either 0 or 1, i.e., load is dispersed by strips in x or in y direction

• In other regions, it may be reasonable to assume that the load is divided equally in the two directions, i.e. k=0.5



• 2.3 Choice of load distribution• Theoretically, the load w can be divided

arbitrarily b/n x and y directions.• Different divisions will, of course, lead to

different patterns of reinforcement, and all will not be equally appropriate.



• The desired goal is to arrive at an arrangement of steel that is safe and economical and will avoid problems at service load level associated with excessive cracking or deflections.

• In general, the designer may be guided by his knowledge of the general distribution of elastic moments.



• To see an example of the strip method and to illustrate the choices open to the designer, consider the square, simply supported slab shown below, with side length a and a uniformly distributed factored load w per unit area.

• The simplest load distribution is obtained by setting k=0.5 over the entire slab, as shown in Figure 2.



Figure 2



• The load on all strips in each direction is thus w/2 ( with k=0.5), as illustrated by the load dispersion arrows

• This gives maximum design moments mx = my = wa2/16, implying a constant curvature for all strips in the x-direction at mid-span corresponding to a constant moment wa2/16 across the width of the slab (see fig. 2)



• The same applies for y-direction strips• It is recognized however that the curvatures in the

strips (say x-direction strips) near the supports, for such a slab, are less than near mid-span.

• If the slab were reinforced according to this solution, extensive redistribution of moments would be required, certainly accompanied by much cracking in the highly stressed regions near the middle of the slab



• So what we need is a type of load distribution (dispersion) which can give a moment distribution that gives rise to greater curvatures in strips near the middle of the slab and less near the ends

• Try the alternative, more reasonable distribution shown in Figure 3 next slide.



Figure 3



• Here the regions of different load dispersion separated by the dash-doted discontinuity lines follow the diagonals, and all of the load on any region is carried in the direction giving the shortest distance to the nearest support.

• k=0 or 1 in the different regions



• The lateral distribution of moments shown in Fig (3) would theoretically require a continuously variable bar spacing impractical

• A practical solution would be to reinforce for the average moment over a certain width, approximating the actual lateral variation in Fig. (4) in a stepwise manner.



• Hillerborg notes that this is not strictly in accordance with the equilibrium theory and that the design is no longer certainly on the safe side, but other conservative assumptions, e.g., neglect of membrane strength in the slab or strain hardening of the reinforcement, would compensate for the slight reduction in safety margin



• A third alternative is with discontinuity lines parallel to the edges.

• Here again the division is made so that the load is carried to the nearest support, as before, but load near the diagonals is divided with one-half taken in each direction.

• Thus k is given the values 0 or 1 along the middle edges and 0.5 in the corners and center of the slab



Figure 4



• Two different strip loadings are now identified, strip along A-A and along B-B.

• This design leads to practical arrangement, one with constant spacing through the center strip of width a/2 and a wider spacing through the outer strips, where the elastic curvatures and moments are known to be less.



• The averaging of moments necessitated in the second solution is avoided here, and the 3rd (Fig. 4) solution is fully consistent with the equilibrium theory.

• The three examples also illustrate the simple way in which moments in the slab can be found by strip method, based on familiar beam analysis.



• It is important to note too that the load on the supporting beams is easily found because it can be computed from the end reactions of the slab-beam strips in all cases.


Chapter 2- Strip Method for Slabs• 2.4 Rectangular slabs with simple support• Discontinuity lines parallel to the edges as shown in

the figure• In the x-direction:• Side strips: mx = w/2×b/4×b/8 = wb2/64• Middle strips: mx = w×b/4×b/8 = wb2/32• In the y-direction• Side strips: my = wb2/64• Middle strips: my = wb2/8

Figure 5 Rectangular slab with discontinuity lines originating at the corners.




Figure 6 Rectangular slab with discontinuity lines parallel to the edges



• Design the rectangular slab using the strip method for slabs

• Use a=6.0 m, b= 4.5 m, t = 150 mm, C-25 concrete and S-300 reinforcing steel.

• Compare the results with the solution using the coefficients in EBCS-2

• Take variable load q = 3.0 kN/m2

• Floor finish-30 mm screed and 20mm thick marble



• 2.5 Fixed Edges and Continuity• Up to now we have dealt with positive

moments in strips, where a large amount of flexibility in assigning loads to the various regions of the slab was provided

• This same flexibility extends to the assignment of moments b/n negative and positive bending sections of slabs that are fixed or continuous over their supported edges



• Some attention should be paid to elastic moment ratios to avoid problems with cracking and deflection at service loads

• Figure 7 (next slide) shows a uniformly loaded rectangular slab having two adjacent edges fixed and the other two edges simply supported

• Let us consider slab strips with one end fixed and one end simply supported as shown in Fig. 7



A A

B

B

Figure 7



• In designing by strip method, slab strips carrying loads only near supports and unloaded in the central region are encountered

• It is convenient if the unloaded region is subject to a constant moment (and zero shear) because this simplifies the selection of positive reinforcement



• The discontinuity lines are shifted to account for the greater stiffness of the strips with fixed ends (i.e. bigger reaction at the fixed support)

• Their location is defined by a coefficient , with a value less than 0.5, so that the edge strips have widths greater and less than b/4 at the fixed and simple end respectively



• For a BMD for x-direction middle strips (section A-A) with constant moment over the unloaded part, the following maximum moments are achieved

• and

8422

22

2 wbbbwbwmxf

2)21(

84)1(

2)1(

222 wbwbbbwmxs



• The first term is the “cantilever” moment at the left end

• So the negative moment at a support plus the span moment = the “cantilever” moment

• Now the ratio of negative to positive moments in the x-direction middle strip is:



• mxs/mxf = 1-2/2

• Hillerborg notes that as a general rule for fixed edges, the support moment should be about 1.5 to 2.5 times the span moment in the same strip.

• For mxs/mxf =2.0 = 0.366• Determine moment in the x-direction edge

strips They are half middle strip values



• Determine moments in the y-direction middle strip

• It is reasonable to choose the same ratio b/n support and span moments in the y-direction as in the x-direction.

• To achieve this, choose the distance from the right support to maximum moment section as b



• myf = wb(b)- wb(b/2)= 2(wb2/2)• The cantilever span = (1-)b• mys =w(1-)b.(1-)b/2 = (1-2)(wb2/2)• So the ratio of negative to positive moment is

as before mys/myf = 1-2/2

• Determine moment in the y-direction edge strips

• myf = w(b)2/16



• Cantilever moment=(w/2)(1-)(b/2).(1-)(b/4)• mys=(1-)(wb2/16) 1/8 of y-direction

middle strip• With the above expressions, all the design

moments for the slab can be found once a suitable value for is chosen



• 0.350.39 give corresponding ratios of Negative to positive moments from 2.45 to 1.45

• 2.6 Unsupported Edges• The real power of the strip method becomes

evident when dealing with nonstandard problems, such as with unsupported edge, slabs with holes, or slabs with reentrant corners (L-shaped)



• For a slab with one edge unsupported, a reasonable basis for analysis by the simple strip method is that a strip along the unsupported edge takes a greater load per unit area than the actual load acting, i.e., that the strip along the unsupported edge acts as a support for the strips at right angles.



• Such strips have been referred to as “strong bands”.

• A strong band is, in effect, an integral beam, usually having the same total depth as the remainder of the slab but containing a concentration of reinforcement.

• The strip may be made deeper than the rest of the slab to increase its carrying capacity, but this will not usually be necessary



• Consider the rectangular slab carrying a uniformly distributed ultimate load w with fixed edges along three side and no support along one short side, shown in Figure 8.



Figure 8



• The following are observed:• Discontinuity lines are chosen as shown• The load on a unit middle strip in the x–

direction, includes the downward load in the region adjacent to the fixed left edge and an upward reaction kw in the region adjacent to the free edge



• M about the left end (with moments +ve clockwise and with the unknown support moment mxs acting clockwise)

• mxs+wb2/32-(kwb/4)(a-b/8)=0 • k=(1+32mxs /wb2)/(8(a/b)-1)• k will be known once mxs is selected

• Selection of mxs will depend on the shape of the slab.



• If a is large relative to b, the strong band in the y-direction at the edge will be relatively stiff, and the moment at the left support of the x-direction strips will approach the elastic value for a propped cantilever



• If the slab is nearly square, the deflection of the strong band would tend to increase the support moment; a value about half the free cantilever moment might be selected

• With mxs selected and k calculated from the above equation, the max span moment is determined



• mxf = (kwb2/32)((8a/b)-3+k)• Determine moments in the x direction edge

strips They are one-half those in middle strip• In the y direction middle strip, the cantilever

moment is wb2/8• Adopting a ratio of support to span moment of

2 results in support and span moments, respectively, of



• myf + mys = myf + 2myf = wb2/8• myf = wb2/24 and mys = wb2/12• Determine moments in y-direction strip adjacent

to the fixed edge• It is one-eighth the middle strip values (check)• In the y-dir strip along the free edge, moments

can, with slight conservatism, be made equal to (1+k) times y-dir middle strip values



• 2.7 Slabs with Holes• Slabs with small openings can usually be

designed as if there were no openings, replacing the interrupted steel with bands of rebar of equivalent area on either side of the opening in each direction.

• Smaller dimensions are those needed to accommodate heating, plumbing and ventilating risers, etc.

60


• Larger size holes are required by stairways and elevator shafts

• Slabs with larger openings must be treated more rigorously

• The strip method offers a rational and safe basis for design in such cases. Integral load-carrying beams (strong bands) are provided along the edges of the opening, usually having the same depth as the remainder of the slab but with extra reinforcement, t pick up the load from the affected regions and transmit it to the supports

Dr.-Ing. Girma Zerayohannes



• In general, these integral beams should be chosen so as to carry the loads most directly to the supported edges of the slab.

• The width of the strong bands should be selected so that the steel ratios are at or below the maximum for beams



• Example: Rectangular slab with central opening

• Figure shows a 5m×8m slab with fixed supports along 4 sides. A central opening 1.2m×2.4m must be accommodated. Estimated slab thickness is 200 mm. The slab is to carry a uniformly distributed factored load of 15kN.m2 including self weight.



• Devise an appropriate system of strong bands to reinforce the opening, and determine moments to be resisted at all critical sections of the slab

• Discontinuity lines for the basic slab (w/o hole) are first chosen and the moments determined which are used as a guide in selecting moments for the actual slab with hole



• Edge strips are defined having width equal to 5/4 = 1.25m

• In the central region, 100% of the load is assigned to the y direction

• Moments of the basic case w/o hole will be calculated and later used as a guide in selecting moments for the actual slab with hole. A ratio of support to span moments of 2.0 will be used generally



Moments for the slab w/o is: x direction middle strips:• Cantilever mx = wb2/32 = 552/32=11.72

kNm/m• Negative mxs = 11.722/3 = 7.81 kNm/m

• positive mxf = 11.721/3 = 3.91 kNm/m• X direction edge strips are ½ middle strips



Y direction middle strips• Cantilever my = wb2/8= 1552/8=46.88

kNm/m• negative mys = 46.882/3=31.25 kNm/m

• positive myf = 46.881/3=15.63 kNm/m

• Y direction edge strips are 1/8 middle strip values



• Because of the hole, certain strips lack support at one end. To support them, 0.3m wide strong bands will be provided in the x direction at the long edges of the hole and 0.6 m wide strong bands in the y direction on each side of the hole.

• The y dir bands will provide for the reactions of the x dir bands.



• With the distribution of loads shown in figure, strip reactions and moments are found as follows:

• Strip A-A: Assuming propped cantilever action with the restraint moment along the slab edge taken as 31.25

• 31.25+w1(0.3)(1.75)-15(1.6)2/2 = 0

• w1=-7.95 kN/m



• The negative value indicates that the cantilever strips are serving as supports for strip DD, and in turn for the strong bands in the y-direction, which is hardly a reasonable assumption.

• Hillerborg notes that the restraint moment should stay as close to the basic value” w/o w1 being negative

• w1=0 (cantilever alone)



• Note: with w1 = 0 chosen -kw = 0 k=0 loading on the strong band = (1+k)w = w = 15 KN/m2

• Now mys = 15(1.6)2/2 = -19.2 kNm/m• Strip B-B:• mxs = 7.81 kNm/m (basic value) 7.81+w2(0.6)

(2.5)-15(1.25)2/2 = 0 w2=2.61 kN/m



• Because of the positive reaction by the strong band the load dispersed in the y direction must be greater than 15 kN/m2

• Determine k• -kw = -15k = -2.61 kN/m2 k=0.174• load dispersed in strong band in y-dir in the

middle: (1+k)w = (1.174)15=17.61 kN/m2



• load dispersed in strong band in y-dir near the edge: (1+k/2)w = (1.087)15=16.31 kN/m2

• Determine max span moment:• Shear is zero at: 15x = 0.6(2.61) x = 0.1m• mxf = 0.6(2.61)(0.3+0.95+0.1)-15(0.1)2/2 = 2.04

kNm/m



• Strip C-C:• Negative and positive moments and the

reaction to be provided by strip C-C, are al one-half the corresponding values for strip B-B.

• Strip D-D:• The 0.3m wide strip carries 15kN/m2 in the x-

direction with reactions provided by the strong bands E-E (loading 150.3 = 4.5 kN/m)



• (we0.6)2 = 4.52.4 we= 9.0 kN/m (over a 0.3 m wide strip)

• mxf = 9(0.6)(1.2+0.3)-4.5(1.2)/2 = 8.1 – 3.24 = 4.86 kNm/(0.3m width)

• Strip E-E:• Direct load dispersed (1+k)w and (1+k/2)w are

17.61 and 16.31 kN/m2 respectively



• Reactions from strong bands D-D is 9.0 kN/m over 0.3 m width or 9/0.3 = 30 kN/m2 300.6 = 18 kN/m over 0.6 m wide strip

• 17.610.6 = 10.566 kN/m over 0.6 m wide strip

• 16.310.6 = 9.786 kN/m over 0.6 m wide strip



• Determine moments:• Cantilever 9.786(2.5)2/2+(10.566-9.786)

(1.25)(1.25+0.625)+18(0.3)(1.6+0.15) = 41.86 kNm (per 0.6 m width)

• Negative: mys = 41.86(2/3) = 27.911 kNm (per 0.6 m width)

• positive: myf = 41.86(1/3) = 13.95 kNm (per 0.6 m width)



• Reference• Nilson and Winter, 14th edition or newer

chapter 2 – strip method for slabs

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