epartment of Civil Engineering Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces Volume 5 P H T H E S I S BYG TU
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces
1 Preface This report is prepared as a partial fulfilment of the requirements for obtaining the Ph.. degree at the Technical Universit of enmark. The work has been carried out at the epartment of Structural Engineering and Materials, Technical Universit of enmark (BYG TU), under the supervision of Professor, dr. techn. M. P. Nielsen. I would like to thank m supervisor for valuable advise, inspiration and man rewarding discussions and criticism to this work. Thanks are also due to m co-supervisor M. Sc. Ph.. Bent Steen Andreasen, RAMBØLL, M. Sc. Ph..-student Karsten Findsen, BYG TU, M. Sc. Ph..-student Lars Z. Hansen, BYG TU, M. Sc. Ph..-student Thomas Hansen, BYG TU and M. Sc. Ph..-student João Luís Garcia omingues ias Costa, BYG TU for their engagement and criticism to the present work and m Ph..-project in general. Finall I would like to thank m wife and famil for their encouragement and support. Lngb, August 004-1 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces - -
Summar This paper treats the behaviour of concrete slabs subjected to lateral load and compressive aial force. The assumptions regarding the material behaviour are described in [14] and this paper demonstrates how to incorporate the stiffnesses found in [14] in the determination of slab behaviour. The subject is approached from a practical point of view. Thus, the solutions should be fairl simple without being too far from the eact ones. Generall the behaviour is determined from an estimation of the deflection form and the use of the work equation. This means that the solutions are upper bound solutions and the equilibrium equations are not necessaril fulfilled. Slabs subjected to transverse load, aial load or a combination, and with different support conditions are treated theoreticall. It is demonstrated how to determine the relation between aial force, transverse load and deflection. The report is limited to compressive aial force onl. One test series has been used to demonstrate the practical use of the method and good agreement has been found. - 3 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 3 Resume enne rapport behandler emnet betonplader påvirket af trknormalkræfter og tværlast. Antagelser vedrørende materialeopførelse er beskrevet i rapporten Stiffness of concrete slabs, se [14], og nærværende rapport demonstrer hvordan man kan anvende disse stivheder til bestemmelse af pladens opførsel. Emnet behandles ud fra en praktisk snsvinkel. et er tilstræbes at beskrive den overordnede opførsel af pladen på en simpel måde uden at være for langt fra eksakte løsninger. Beregningsmetoden er baseret på et udbøjningsskøn samt anvendelse af energiligningen. ette betder at der er tale om en øverværdiløsning og at ligevægtsbetingelserne ikke nødvendigvis er opfldt. Plader påvirket med normalkræfter, tværlast eller en kombination af disse er behandlet teoretisk og de fundne resultater er eksemplificeret med forskellige pladetper og forskellige armeringsplaceringer. Rapporten behandler kun trknormalkræfter. Til verificering af beregningsmetoden er der foretaget en sammenligning mellem teori og eksperimenter for en forsøgsserie. er blev her fundet god overensstemmelse. - 4 -
4 Table of contents 1 Preface...1 Summar...3 3 Resume...4 4 Table of contents...5 5 Notation...7 6 Introduction...9 7 Theor...11 7.1 GENERAL EQUATIONS...11 7.1.1 Constitutive equations...11 7.1. Compatibilit conditions...16 7.1.3 Equilibrium equations...17 7.1.4 Boundar conditions...17 7. EFLECTIONS OF SLABS WITH NO AXIAL FORCE...19 7..1 Estimation of deflections using Raleigh s principle...19 7.. Beam eample...19 7..3 Rectangular slab simpl supported at all sides...1 7..4 Rectangular slab simpl supported at three sides and one free edge...3 7..5 Rectangular slab fied at all sides...7 7..6 Rectangular slab with two adjacent sides fied and two sides free...8 7..7 Rectangular slab with two adjacent sides simpl supported and two sides free 9 7..8 Eemplification of deflection calculations...30 7.3 STABILITY OF SLABS LOAE WITH AXIAL FORCE...3 7.3.1 Rectangular slab simpl supported at all sides...33 7.3. Rectangular slab simpl supported at three sides and one free edge...34 7.3.3 Rectangular slab fied at all sides...35 7.3.4 Rectangular slab with two adjacent sides fied and two sides free...35 7.3.5 Rectangular slab with two adjacent sides simpl supported and two sides free 36 7.3.6 Eemplification of stabilit calculations...36-5 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 7.4 EFLECTIONS OF SLABS WITH AXIAL FORCE...40 7.4.1 Rectangular slab simpl supported at four sides loaded with an aial force in one direction...40 8 Theor compared with tests...49 9 Conclusion...58 10 Literature...59 11 Appendi...61 11.1 ESCRIPTION OF COMPUTER PROGRAM...61-6 -
5 Notation The most commonl used smbols are listed below. Eceptions from the list ma appear, but this will then be noted in the tet in connection with the actual smbol. Geometr L Length of a beam L,L Length of a slab in the - and - direction, respectivel h epth of a cross-section h c h c h c h c istance from the bottom face to the centre of the bottom reinforcement istance from the top face to the centre of the top reinforcement istance from the bottom face to the centre of the bottom reinforcement in the - and - direction, respectivel h c h c istance from the top face to the centre of the top reinforcement in the - and - direction, respectivel d Effective depth of the cross-section, meaning the distance from the top the slab to the centre of the reinforcement. A Area of a cross-section A c A s A s A s A s Area of a concrete cross-section Area of reinforcement per unit length close to the bottom face Area of reinforcement per unit length close to the top face Area of reinforcement per unit length close to the bottom face in the - and - direction, respectivel A s A s Area of reinforcement per unit length close to the top face in the - and - direction, respectivel 0 Compression depth,, z Cartesian coordinates Phsics σ σ c σ c, σ c Normal stress Normal stress in concrete Normal stress in concrete in the - and - direction, respectivel. - 7 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces f c φ nφ φ, φ φ, φ E E s E c n, n, n n m, m m q,q Compressive strength of the concrete. Reinforcement ratio egree of reinforcement Reinforcement ratio in the -direction for the lower and upper reinforcement, respectivel. Reinforcement ratio in the -direction for the lower and upper reinforcement, respectivel. Modulus of elasticit Modulus of elasticit for the reinforcement Modulus of elasticit for the concrete Ratio between the modulus of elasticit for the reinforcement and the modulus of elasticit for the concrete Bending stiffness for the slab in the and -direction, respectivel Torsional stiffness for the slab Aial load per unit length in the -and -direction, respectivel. Positive as compression. Shear load per unit length Bending moment per unit length in the - and -direction, respectivel Torsional moment per unit length Transverse shear load per unit length in the - and -direction, respectivel m,, n m m Applied bending moment per unit length at the edge in the n-, - and - direction, respectivel q, q, q Applied transverse shear load per unit length at the edge in the n-, - and - n p u κ,κ κ direction, respectivel Transverse load Transverse deflection Curvature in the and -direction, respectivel Torsional curvature - 8 -
6 Introduction This paper demonstrates how to incorporate the stiffnesses found in [14] in determination of slab behaviour. It is assumed that the reader is familiar with [14] and epressions and definitions from this report is used here without further eplanation. The first section describes the constitutive equations, the compatibilit conditions and the boundar conditions used in this investigation. The net section treats slabs subjected to transverse load onl. The method used in this investigation is first described in general and then followed b five determinations of the slab stiffnesses for some of the most common support conditions of rectangular slabs. Two reinforcement arrangements, one with the reinforcement placed in the centre and one where it is placed close to the faces, are used for eemplifications. Similar approach is used in the following section that treats slabs subjected to aial force onl. The stabilit loads for the five slab cases mentioned above are determined and the same two reinforcement arrangements are used for eemplification. The general approach used for a combination of lateral load and aial force is described in the net section. A rectangular slab, simpl supported at all sides with aial force in two directions, is used for demonstrating the method. In this section some considerations regarding the importance of the loading histor of the slab are made as well. As a verification of the method, test results from a series with square slabs subjected to aial force in one direction are compared with calculations. Good agreement is found. In the final section conclusions are made. - 9 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces - 10 -
7 Theor 7.1 General equations 7.1.1 Constitutive equations The constitutive equations assumed in this investigation are: u m = κ = u m = κ = u m = κ = (7.1.1) The bending stiffness and the torsional stiffness found in [14] are used. Onl a short eplanation of the calculations is given here. For a thorough treatment the reader is referred to [14]. In [14] it is shown that the stiffness ma be calculated as a function of the degree of cracking. The degree of cracking is the ratio between the cracking moment and the applied moment. The cracking moment is defined as the transition point between the cracked and the uncracked state. For pure bending the cracking moment ma be calculated b: m crack, 1 1 hc hc 1 3 hc' hc' nh + + nφ + + nφ ' 1 h h h h = 1 hc hc' + nφ + nφ'1 h h where n is the aial force per unit length, n is the ratio between the modulus of elasticit for the reinforcement and the modulus of elasticit for the concrete (E s /E c ), h is the thickness of the slab, h c is the distance from the bottom face to the lower laer of reinforcement, h c is the distance from the top face to the upper laer of reinforcement, φ is the reinforcement ratio for the lower reinforcement laer and φ is the reinforcement ratio for the upper reinforcement laer both based on the total section. (7.1.) - 11 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces In the special case with onl one laer of reinforcement placed in the centre of the slab we get: m crack, 1 nh = 61+ nφ It was shown that besides the phsical properties of the slab the bending stiffness onl depends on the n h/m ratio. The limits for n h/m corresponding to 0 =h and 0 =0 becomes: 1 h h ' + nφ + nφ'1 c c nh ( 0 = h) h h = m 1 1 hc hc 1 3 hc' hc' + + nφ + + nφ ' 1 h h h h (7.1.3) (7.1.4) hc hc' 1 ' nh φ ( 0 0) h + φ h = = m hc hc hc' hc' φ 1 3 + + φ' + h h h h Notice that n and n are positive as compression. If the n h/m ratio is higher than the limit for 0 =h the stiffness of the slab will be the uncracked stiffness given b: (7.1.5) ( 0 ) 1 = h hc 1 h c 1 3 hc hc = + nφ nφ 3 + + hec 1 h h h h (7.1.6) If the n h/m ratio is lower than the limit for 0 =0 the stiffness equals the stiffness of the reinforcement onl given b: ( 0= 0) 1 hc 3 h c hc 1 hc = nφ nφ 3 + hec h + h h (7.1.7) h In Figure 7.1 and Figure 7. the stiffnesses as a function of the degree of cracking are shown for two tpes of slabs. The first tpe is valid for slabs with reinforcement in the centre and the second tpe is valid for the reinforcement placed smmetricall about the centre, close to the faces. These are the two tpes generall used in this investigation. - 1 -
0.09 0.08 0.07 0.06 /(E c h 3 ) 0.05 0.04 0.03 0.0 0.01 nφ=0.0 h c /h= 0.5 nφ=5 h c /h= 0.5 nφ=0 h c /h= 0.5 nφ=0.05 h c /h= 0.5 0 - -1.5-1 -0.5 0 0.5 1 m crack /m Figure 7.1 Bending stiffness as a function of the cracking moment over moment ratio. h c =½h. One laer of reinforcement. 6 4 nφ= nφ = 0 h c /h= 0.5 /(E c h 3 ) 0.08 0.06 0.04 nφ=0. nφ = 0. h c /h= nφ=5 nφ = 5 h 0.0 c /h= nφ= nφ = h c /h= nφ=0.05 nφ = 0.05 h 0 c /h= -0.8-0.6 - -0. 0 0. 0.6 0.8 1 m crack /m Figure 7. Bending stiffness as a function of the cracking moment over moment ratio. h c = h c =h. Two laers of reinforcement. - 13 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces In [14] it was found that the torsional stiffness dependenc on the degree of cracking is the same as the bending stiffness dependenc on the degree of cracking for a slab with the reinforcement placed in the centre. This means that the torsional stiffness is simpl found b moving the reinforcement to the centre of the slab and then calculate the bending stiffness. This procedure is illustrated in Figure 7.3. A s A s A s= AsAs A s n n m A s n n m m z z m m he, cracked, cracked m m = 3 3 c hec Figure 7.3. The relation between torsional- and bending stiffness for an isotropic slab. This means that diagrams for bending stiffnesses ma be used if the degree of cracking for torsion is replaced b the degree of cracking for bending. The degree of cracking for torsion ma be calculated from the cracking moment given b: m, crack ( ) = sign n ( 1+ φ )( 1+ φ ) h n n n n ( + φn+ φn+ φnφn) 61 4 It should be noted that this ratio ma onl be used for aial forces of the same sign which is assumed throughout the report if nothing else is stated. (7.1.8) - 14 -
Furthermore, the cracking moment does not have an meaning if one of the aial forces is zero. In this case the cracking moment would alwas be zero meaning that the torsional stiffness is independent of the aial force, which is of course not correct. However, this problem will be ignored in this report and we shall assume that the cracking moment concept ma also be used if one of the aial forces is zero. This means that the torsional stiffness for a slab loaded b aial force in one direction will be the same as for a slab without an aial load. It should be noted that the influence of bending moments on the torsional stiffness, and the influence of the torsional moment on the bending stiffnesses have not been investigated neither in [14] nor here. Thus, the stiffnesses are assumed to be independent of each other. 7.1.1.1 Eemplification of the stiffnesses As an eample we consider a slab section with the reinforcement placed close to the face. This is named a case arrangement. For a section without an aial force the degree of cracking is zero and the stiffnesses therefore are onl depending on the degree of reinforcement. Assuming a degree of isotropic reinforcement of nφ=nφ = we find, using Figure 7., that the bending stiffness is: = 0.05 (7.1.9) 3 3 Eh c Eh c The torsional stiffness is found from Figure 7.1 using a fictitious degree of reinforcement of nφ=0. which takes into account both laers of reinforcement. Thus the torsional stiffness becomes: 0.017 3 Eh (7.1.10) c If we consider a slab section with onl one laer of isotropic reinforcement placed in the centre, named a case 1 arrangement, having a degree of reinforcement of nφ=0., we find that in this case the stiffness becomes: = = 0.017 (7.1.11) 3 3 3 Eh c Eh c Eh c As seen the position of the reinforcement is decisive for the ratio of the bending and torsional stiffness. If the reinforcement is placed in the centre, the bending and torsional stiffnesses are the same and if the reinforcement is placed close to the faces the torsional stiffness is onl about 1/3 of the bending stiffness. This conclusion is onl valid for cases without aial force. If aial force is applied, the ratio between the stiffnesses changes as a function of the aial force. - 15 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces If we consider the uncracked state the stiffnesses in case 1 (reinforcement in the centre nφ=0.) becomes: = = 0.083 (7.1.1) 3 3 3 Eh Eh Eh c c c In case (isotropic reinforcement close to the faces, nφ=nφ =, hc/h=) we have: 3 c Eh Eh 0.083 = 15 3 3 c Eh c = = 1.4 (7.1.13) If the bending stiffness in the -direction is calculated for uncracked concrete and if the bending stiffness in the -direction as well as the torsional stiffness are calculated for cracked concrete, we get: In case 1: Eh 3 c 0.083 = 0.017 3 3 c Eh c Eh = 4.9 = 4.9 (7.1.14) In case : Eh 3 c 3 c Eh 3 c Eh 15 0.083 0.017 = 1.4 = 6.7 (7.1.15) These cases are used for eemplification later in this report. 7.1. Compatibilit conditions The compatibilit conditions for slabs are: κ κ = κ κ = (7.1.16) - 16 -
If curvatures are determined from a deflection function these condition will be identicall fulfilled. A derivation ma be found in [3]. 7.1.3 Equilibrium equations The equilibrium equations for a slab element are, see [3]: m m + q = 0 m m + q = 0 q q + + p = 0 (7.1.17) Eliminating the shear forces we get: m m m + + = p (7.1.18) If this equation is fulfilled we have a staticall admissible moment field. If we introduce the constitutive equations given b formula (7.1.1) into the equilibrium equation we get: 4 4 4 u u + u 4 + = p (7.1.19) 4 If the bending stiffnesses are equal to the torsional stiffness we get well known differential equation: 4 4 4 u u u p + + = 4 4 This is called the Lagranges equation. 7.1.4 Boundar conditions In this paper the edge conditions are defined in Figure 7.4 (n normal to the edge). (7.1.0) - 17 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces Free edge Geometrical conditions: Statical conditions: m n = 0 q = 0 n Simpl supported edge u = 0 m n = 0 Fied edge u = 0 du = 0 dn Figure 7.4efinition of edge conditions The statical boundar conditions are defined using the notation of Figure 7.5. s t m n q n n m nt Figure 7.5 Forces along an edge. The statical boundar conditions, also called the Kirchhoff boundar conditions, are in general (see [3]): m q n n = mn m nt m + = qn + s s nt (7.1.1) If the edge is a straight line parallel sa to the t-ais we ma insert formulas (7.1.1) and (7.1.17) into (7.1.1). Then: u mn = n (7.1.) n For a boundar parallel to the -ais we find: 3 3 mnt u u qn + = n nt s n n t 3 (7.1.3) - 18 -
m q m Figure 7.6 Forces along an edge. u m = 3 3 m u u q + = 3 (7.1.4) (7.1.5) 7. eflections of slabs with no aial force 7..1 Estimation of deflections using Raleigh s principle If we estimate the deflected form of a slab, we ma use the energ equation to find an estimate of the deflections. This of course means that we do not necessaril satisf the equilibrium equations since the sectional forces ma not correspond to the load. This use of the energ equation is sometimes termed Raleigh s principle. The energ equation gives a relation between the load and the parameters of the deflected form. The accurac of the estimated deflection form is of course decisive for the accurac of the method. In general, it is not necessar to choose a deflection form that fulfills all boundar conditions. Nevertheless, it is obvious that the most correct solution is found when both statical and geometrical boundar conditions are fulfilled. In this paper solutions, which fulfill the geometrical boundar conditions, are used. The statical boundar conditions are not fulfilled in general. The statical boundar conditions require the fulfillment of the Kirchhoff boundar conditions, formula (7.1.5). Since the ratio between the bending stiffness and the torsional stiffness is different from slab to slab it is not possible to give a general solution. 7.. Beam eample In this section the method is illustrated for a beam, see Figure 7.7. - 19 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces u p L Figure 7.7 Beam We assume a deflection in the form of a parabola given b: The curvature is: The work equation becomes: ma u ( ) 4u L- ma = (7..1) L κ = du 8uma d = L (7..) 1 L 1 L pud = 0 M κ d 0 1 L 1 L pud = 0 0 κ d 1 1 8uma p umal= L 3 L uma p = 96 L 4 or u = 1 96 pl It is evident that the assumed deflection does not fulfil the statical boundar conditions and the equilibrium equation. The deviation between the eact solution (5/384 instead of 1/96) and this solution is 0%. 4 (7..3) If we want a deflection form that fulfils the statical boundar conditions we ma take: u u sin = ma π L The work equation gives: (7..4) - 0 -
5 π u p= 4 L or u ma 5 ma 4 4 pl = π In this case the deviation is less than 1%. A deviation of this order is acceptable for all practical calculations. 4 (7..5) 7..3 Rectangular slab simpl supported at all sides In this case we consider a rectangular slab simpl supported at all sides and loaded with a uniforml distributed load p, see Figure 7.8. L L Figure 7.8Rectangular slab simpl supported at all sides We estimate a deflection function given b: u = u ma The deflected form is illustrated in Figure 7.9. L L π + π + sin sin L L (7..6) - 1 -
0.7 Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 0.5 0. 0. 0. 0.5 0.6 0.7 0. 0.8 0.9 0.5 0.6 0.7 0. 0.6 0.5 0. /L 0-0.6 0.8 0.9 0.9 0.8 0.7-0. - - 0. 0.5 0.6 0. 0.7 0.5-0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.9eflection over maimum deflection. 0.8 0.6 0. 0.5 0. It is seen that the geometrical boundar conditions are fulfilled: u = u = u = u = 0 (7..7) L L L L = = = = In this case we should have a bending moment of zero along the edges in order to fulfil the statical boundar conditions. These are fulfilled if: Using the energ or work equation we get: κ = κ = κ = κ = 0 (7..8) L L L L = = = = L L L L L L = L L + + 1 1 ( ) pudd M κ M κ M κ dd L L L L L L = L L + + 1 1 ( ) pudd κ κ κ dd pu L L 1 u π = + + 4 ma ma 4 4 3 3 π 8 L L 6 u L maπ p = 16 + L ( L L L L ) + L L 4 4 4 4 L L (7..9) - -
7..4 Rectangular slab simpl supported at three sides and one free edge We now consider a rectangular slab simpl supported at three sides, one free edge and loaded with a uniforml distributed load p,see Figure 7.10. L L Figure 7.10Rectangular slab simpl supported at three sides In this case, we estimate a deflection function given b: u = u ma The deflected form is illustrated in Figure 7.11 L L π + π + sin sin L L (7..10) 0.5 0. 0. 0.5 0. 0.6 0.7 0.5 0.8 0. 0.6 0.7 0.9 /L 0 0. 0.6 - -0. - - 0. 0.5-0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.11 eflection over maimum deflection. 0.7 0. 0.8 0.6 0.5 0.7 0.9 0. 0.8 0.6 0.5-3 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces It is seen that the geometrical boundar conditions are fulfilled: u = u = u = 0 u L L L = = = L = = u ma sin ( + L ) L π In this case the statical boundar conditions are not fulfilled since we have: κ = κ = κ = 0 κ The moment m (=½L) becomes: m L L L = = = L = L = u π = 4 ma sin L ( + L ) L π ( + L) (7..11) (7..1) π u ma sin L π = (7..13) 4 L which means that the boundar condition for m is not fulfilled. Further the Kirchhoff boundar conditions (7.1.5) along the free edge are not fulfilled. Using the work equation we get: L L L L L L L L 1 1 ( ) pudd = M κ + M κ + M κ dd 6 umaπ p = 56 L + 16L + 8 L L 4 4 4 4 L L (7..14) We now assume another deflection form given b: u = u ma L L π + + sin L L (7..15) - 4 -
The deflected form is illustrated in Figure 7.1. 0.5 0. 0. 0. 0.5 0. 0.6 0.5 0.7 0.8 0.9 /L 0 0.6 0.7-0.8-0. - - 0. -0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.1 eflection over maimum deflection. 0. 0.5 0.6 0. 0.5 0.7 It appears that the statical boundar condition (7.1.5) is still not fulfilled. However, this deflection form leads to no bending moments along the edges and is therefore more accurate. In this case we find the following relation between p and u ma. 3 maπ 6L +π L 4 6 L L u p = (7..16) Notice that in this case the bending stiffness in the - direction has no influence. As a third possibilit consider a deflection given b, see Figure 7.13: u = u ma L L π + Arctan 3 + 3sin sin 4 L L ( ) 3 (7..17) - 5 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 0.5 0. 0.5 0. 0.7 0.8 0. 0.6 0.9 /L 0 0.5 - -0. 0.6 0.7 0.8-0. 0.5 - -0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.13eflection over maimum deflection. 0. Still the statical boundar conditions are not fulfilled in general but the bending moment m at the free edge is zero. We find the following relation between p and u ma. ( ) π ( ( )) π ( ( )) 4 5 4 4 ( Arc ( )) L ( Arc ( )) L 4 4 L L ( + ) 4 4 4 4 405π Arc tan L 7 L + 1458π tan 16 tan u 891 tan + 3645 tan ma p = 51 4 3 9 p u 3 Arc L L Arc L L 4 4 ( 46,6L + 79,4L L + 13, 4L ) ma 4 4 L L If we compare this solution with the previous solution we notice that the last solution involves the bending stiffness in the direction. Whether this is relevant or not depends on the ratio between the side lengths. If L is ver large compared to L it is logical that the bending stiffness in the - direction will have a small effect since the slab is unable to carr an substantial load in that direction. The other etreme is when L is ver large compared to L and in this case the slab will of course be able to carr a substantial part of the load in the direction. - 6 - (7..18)
7..5 Rectangular slab fied at all sides In this case we consider a rectangular slab fied at all sides and loaded with a uniforml distributed load p, see Figure 7.14. L L Figure 7.14Rectangular slab fied at all sides We estimate a deflection function given b, see Figure 7.15: u = u ma L L π + π + sin sin L L (7..19) 0.5 0. 0. 0.5 0.6 0.8 0. 0.9 0.5 0.7 0.6 0. /L 0 - -0. - - 0. 0.5 0. 0.7 0.7 0.6 0.8 0.9 0.5 0.7 0. 0.8 0.6 0.5 0. -0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.15 eflection over maimum deflection - 7 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces In this case it ma be shown that both the geometrical and statical boundar conditions are fulfilled. Using the work equation we get: 3L + 3L + L L p = u (7..0) 4 4 4 maπ 4 4 L L 7..6 Rectangular slab with two adjacent sides fied and two sides free In this case we consider a rectangular slab fied at two sides net to each other and loaded with a uniforml distributed load p, see Figure 7.16. L L Figure 7.16Rectangular slab fied at two sides and with two sides free. In this case, we ma estimate a deflection function given b, see Figure 7.15: u = u ma L L π + π + 4sin sin 4L 4L (7..1) - 8 -
0.5 0. 0.5 0.8 0.9 0.7 0.6 0. 0.5 0. /L 0 - -0. 0. - - -0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.17 eflection over maimum deflection It ma be shown that the geometrical boundar condition are fulfilled. The statical boundar condition (7.1.5) is not fulfilled. Using the work equation we get: 4 4 5, L + 5, L + 3,1L L p uma (7..) 4 4 L L 7..7 Rectangular slab with two adjacent sides simpl supported and two sides free In this case we consider a rectangular slab simpl supported at two sides net to each other and loaded with a uniforml distributed load p see Figure 7.18. L L Figure 7.18Rectangular slab simpl supported at two sides, two sides free. - 9 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces In this case, we ma estimate a deflection function given b, see Figure 7.19: u = u ma 3 3 L ( ) ( ) L + Arctan Arctan 7 + sin sin 8 L L (7..3) 0.5 0. 0.5 0.6 0.8 0.7 0.9 0. 0. /L 0 - -0. - - -0.5-0.5 - - -0. - 0 0. 0.5 /L Figure 7.19 eflection over maimum deflection It ma be shown that the geometrical boundar conditions are fulfilled. The statical boundar condition (7.1.5) is not fulfilled. Using the work equation we get: 1L + 1L + 5.4 L L p = u (7..4) 4 4 ma 4 4 L L 7..8 Eemplification of deflection calculations In order to evaluate the influence of the torsional stiffness and the influence the position of the reinforcement we show the result of some eamples. In section 7.1.1.1 it was demonstrated that if the reinforcement is placed close to the faces the torsional stiffness is of the order one third of the bending stiffness and if the reinforcement is placed in the centre the torsional stiffness is the same as the bending stiffness. - 30 -
If we consider a square slab and assume that the reinforcement is isotropic we ma compare the maimum deflection for a given load using the formulas above. L /L =1, Isotropic reinforcement Formula Case 1. Reinforcement placed in the centre. ( = = ) Case. Reinforcement placed close to the faces. (3 = = ) L (7..9) u ma p 4 4 L u 160. ma = p = 4 L p p 1.5 case1 case L L L (7..14) L (7..16) (7..18) u p = 93.9 p = ma 4 L u 8.0 ma 4 L u p = 139, 4 ma 4 L (7..0) uma p 779.3 4 L u p = 73.9 ma 4 L u p = 61.3 ma 4 L u p = 86.5 ma 4 L u 649.4 ma = p = 4 L 1.3 1.3 1.6 1. L L (7..) uma p 33,5 4 L u 18.1 ma = p = 4 L 1.9 L L (7..4) uma p 73. 4 L u 38.3 ma = p = 4 L 1.9 L Table 7.1Relation between load and deflection for different slabs. Table 7.1 demonstrates the effect of a change in the ratio between the bending stiffness and the torsional stiffness. In is clear that the effect is most pronounced for the slabs with two free adjacent sides. This is to be epected since the slab carries the load mainl in torsion. - 31 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces Also a slab simpl supported at all four sides is quite affected whereas the slab fied at all four sides is much less affected, which is due to the fact that the torsional moments are small. For the slab simpl supported at 3 sides and one free, 3 formulas for different deflection forms are used as described in section 7..4. Formula (7..14) is a deflection form that leads to bending moments along the free edge and thus is does not lead to a particularl good solution. However, it is seen that even though this deflection form is not accurate the result does not deviate that much from the other two solutions where bending moments along the free edge are zero. As for the two other solutions the difference is here whether the deflection varies linearl or sinusoidal in the -direction. It is seen that for the present ratio between the side lengths a linear variation leads to a lower stiffness. Since we are dealing with upper bound solutions the solution with the lower stiffness is to be preferred. If the ratio between the side lengths, L /L, is more than approimatel 3, the sinusoidal deflection form will lead to a better result. 7.3 Stabilit of slabs loaded with aial force If a slab is subjected to aial forces in one or two directions the maimum load ma be governed b instabilit. If the slab is subjected to aial forces (compressive) in both directions the slab initiall will be uncracked, and the stiffness ma be calculated for uncracked concrete. If the reinforcement is placed in the centre and the bending stiffnesses are the same as the torsional stiffness, the calculations ma be carried out according to the standard theor of elastic stabilit. In this case the solutions described in for eample [3] or [15] ma be used directl. The nonlinear stress-strain relation for concrete ma be taken into account b letting the modulus of elasticit depend on the normal force, cf. the standard theor for concrete columns. In all other cases the load at instabilit must be found in a wa that takes into account the difference in the stiffnesses. In this report the use Bran s equation is described. Bran s equation for a slab is (see[1],[3]): - 3 -
1 u u u 0= dd + + (7.3.1) 1 u u u u + n n n dd + + Here n and n are positive for tensile forces. The sign convention for n follows the coordinate sstem in the usual wa. The method is to estimate a deflection form and then use Bran s equation to find the load at instabilit. The method is demonstrated in the net section. 7.3.1 Rectangular slab simpl supported at all sides In this case we consider a rectangular slab simpl supported at all sides. The slab is loaded b a uniforml distributed load p and there ma be aial forces in one or two directions, see Figure 7.0. n n L L Figure 7.0Rectangular slab simpl supported at all sides We estimate the same deflection function as in the case with no aial forces: u = u ma Inserting into Bran s equation we get: L L π + π + sin sin L L L + L + L L L L n = π L L 4 4 4 4 n π (7.3.) (7.3.3) Similar calculations are made without an further comments for other slabs in the following sections. - 33 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 7.3. Rectangular slab simpl supported at three sides and one free edge n n L Figure 7.1Rectangular slab simpl supported at three sides L eflection function: Bran s equation: n u = u ma L L π + + sin L L π 3L + π L L nl 4 3 L (7.3.4) = (7.3.5) eflection function: Bran s equation: u = u ma L L π + Arctan 3 + 3sin sin 4 L L ( ) 3 4 4 4 4 L L (7.3.6) 3.34L + 11.58L + 9.87 L L 1.17L L n n = (7.3.7) - 34 -
7.3.3 Rectangular slab fied at all sides n n L Figure 7.Rectangular slab fied at all sides L eflection function: Bran s equation: u = u ma L L π + π + sin sin L L 4 (7.3.8) n 4 4 4 1L + 1L + 4L L 3L L π n π = (7.3.9) 3 L L 7.3.4 Rectangular slab with two adjacent sides fied and two sides free n n L Figure 7.3Rectangular slab fied at two sides, two sides free L eflection function: Bran s equation: u = u ma L L π + π + 4sin sin 4L 4L (7.3.10) - 35 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 4 4 4 5.44L L L L n +.47L +.47L n = (7.3.11) 4 L L 7.3.5 Rectangular slab with two adjacent sides simpl supported and two sides free n n L L Figure 7.4Rectangular slab simpl supported at two sides, two sides free eflection function: u = u Bran s equation: ma 3 3 L ( ) ( ) L + Arctan Arctan 7 + sin sin 8 L L 4 4 4 4 L L (7.3.1) 8.41 L L L L n + 3.34 L + 3.34 L n = (7.3.13) 7.3.6 Eemplification of stabilit calculations Again we evaluate the influence of the torsional stiffness and the influence the position of the reinforcement through some eamples. We consider a square slab subjected to the same aial force in both directions and assume that the reinforcement is isotropic. The stiffnesses in this situation are given b formulas (7.1.1) and (7.1.13). Using the formulas from the previous sections we get the results in Table 7.. - 36 -
L /L =1, Isotropic reinforcement, Formula Case 1. Reinforcement Case. Reinforcement n n case, 1 case, n /n =1 placed in the placed close to the centre. faces. Formula Formula (7.1.1) (7.1.13) ( = = ) (1.4 = = ) n L n (7.3.3) n = 14.8 n = 13.4 L L 1.1 L n L n (7.3.5) n = 9.9 L n = 9. L 1.1 L (7.3.7) n = 11.4 L n 1 = L 1.1 n L n (7.3.9) n 46.1 = n = 44. L L 1.0 L L n n (7.3.11) n = 5. n = 4.4 L L 1. L L n n (7.3.13) n = 7.5 n 6.3 = L L 1. L Table 7.Load at instabilit for different cases and slabs. Table 7. demonstrates the effect of a change in the ratio between the bending stiffness and the torsional stiffness. As epected the effect is most pronounced for slabs with two free adjacent sides. It is seen that if the slab is subjected to compressive aial force in two directions, the position of the reinforcement does not influence the stabilit load significantl. - 37 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces It should be noted that for the slab simpl supported at three sides and with one free edge the lower value of the aial force at instabilit should be used since we are dealing with upper bound solutions. We now consider a square slab subjected to aial force in one direction onl and assume that the reinforcement is isotropic. The stiffnesses in this situation are given b formulas (7.1.14) and (7.1.15). Using the formulas from the previous sections we get the results shown in Table 7.3. - 38 -
L /L =1, Isotropic reinforcement, n =0 L n (7.3.3) Formula Case 1. Reinforcement placed in the centre. Formula (7.1.14) ( =4.9 = 4.9 ) n Case. Reinforcement placed close to the faces. Formula (7.1.15) ( =1.4 = 6.7 ) = 13.9 n = 18.4 L L n n case, 1 case, 0.8 L L n (7.3.5) n =.6 L n = 7.5 L L n L (7.3.7) (7.3.5) (7.3.7) n = 6.6 L ( =4.9 = 4.9 ) n = 34.5 L n = 11. L ( =1.4 = 6.7 ) n = 33.9 L 0.6 1.0 L n = 14.3 L n = 15.4 L 0.9 L n (7.3.9) n = 50. n = 69.6 L L 0.7 L L n (7.3.11) n 4.1 = n 5.1 = L L 0.8 L L n (7.3.13) n = 5.7 n = 7.0 L L 0.8 L Table 7.3Load at instabilit for different cases and slabs. - 39 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces In these cases neither the ratio between bending stiffnesses and the torsional stiffness nor the ratio between the two bending stiffnesses is the same. It is seen that the change in the ratio between the bending stiffnesses gives the most dominant effect. As epected, the slabs with two sides free are not as affected b a difference between the bending stiffnesses as the other ones. 7.4 eflections of slabs with aial force If a slab is subjected to both aial forces and transverse load both tpes of loads must be taken into account in the work equation. The work equation becomes: 1 1 u u u pudd dd = + + 1 u u u u n n n dd + + Notice that in this equation n and n are positive for tension. For such slabs there are some further complications. First of all, the stiffnesses change as a function of the loads and as a function of the position in the slab. The stiffnesses are functions of the ratio between the aial forces and the moments. Since the moments change throughout the slab the stiffnesses will also change throughout the slab. For most practical purposes a reasonable estimate of the behaviour of the slab is sufficient. Therefore we will demonstrate how to carr out the calculations using onl one set of stiffnesses throughout the slab equal to the lower values of the actual stiffnesses. The procedure is demonstrated in the following section. 7.4.1 Rectangular slab simpl supported at four sides loaded with an aial force in one direction In this section we consider a slab simpl supported at all sides. The slab is loaded b an aial force in one direction onl and a uniforml distributed transverse load, see Figure 7.5. (7.4.1) - 40 -
n L L Figure 7.5Rectangular slab simpl supported at all sides. Uniform load and aial force in one direction. The deflection form is given b formula (7.3.). The work equation, formula (7.4.1), leads to the following relation between the aial force, maimum deflection and transverse load: 1 π L + π L + π L L L L n p = u (7.4.) 16 4 4 4 4 maπ 4 4 L L We have, as mentioned above, assumed that the stiffnesses are the same throughout the slab. This assumption is correct regarding the bending stiffness in the -direction and it is also correct for the torsional stiffness if we use the constitutive equations described previousl in section 7.1.1. However, the assumption is not correct regarding the bending stiffness in the -direction. This bending stiffness is estimated b using the stiffness at the point of maimum bending moment in the -direction. This point is determined b: dκ = 0 = 0 d π u κ dκ = = 0 = 0 d ma,ma L Thus the bending stiffness in the centre of the slab is used throughout the slab. Formula (7.4.3) ma also be used to determine the relation between bending moment, stiffness and maimum deflection. This is: L m (7.4.3) π uma,ma,ma =,ma ma = (7.4.4) L π,ma m u The calculation of an interaction diagram between the aial force and the transverse load ma now be done in the following wa: For a given aial load a maimum bending moment in the -direction is assumed, and b using the constitutive equations described in section 7.1.1 the stiffnesses are determined. - 41 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces Then the maimum deflection ma be determined b formula (7.4.4). Knowing the maimum deflection, formula (7.4.) ma then be used to determine the transverse load. B varing the bending moment for different constant aial forces diagrams like those shown in Figure 7.6 and Figure 7.7 are obtained. Increasing aial force Zero aial force u ma Figure 7.6 Bending stiffness as a function of maimum deflection for different aial loads. - 4 -
Zero aial force p Increasing aial force u ma Figure 7.7 Transverse load as a function of maimum deflection for different aial loads. Using Figure 7.7 the maimum transverse load ma be found for a given aial force as illustrated in Figure 7.8. It is evident that if the aial force is sufficientl low the load continues to increase as the maimum deflection increases. In such cases, we do not have failure b instabilit and these points are therefore left out in Figure 7.8. We ma still use the formulas to determine the relation between aial force, maimum deflection and transverse load. The failure will be a material failure. In such case, the load carring capacit ma be determined b introducing maimum stresses. A short description of the procedure for ma be seen in List 1. - 43 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 1. The following data for the slab are loaded: f c, φ, φ, h, h c, L, L and E s.. A loop is started where the aial force in -direction is increased in each step. a. A loop is started where the bending moment in -direction is increased in each step. i. n=e s /E c is calculated. ii. and are calculated using n and Figure 7.1 iii. The maimum deflection is determined according to formula (7.4.4). iv. The transverse load is determined according to formula (7.4.). v. If the maimum stress eceeds the maimum admissible stress the solution for this deflection and transverse load is disregarded. b. The maimum deflection versus transverse load is plotted and the maimum transverse load and the corresponding load is determined for a given aial force. 3. The aial load as a function of the maimum transverse load is plotted. List 1. escription of the calculation procedure. n p Figure 7.8 Aial force as a function of transverse load. - 44 -
7.4.1.1 Influence of the loading procedure Tests made b MacGregor, [6], indicate that the loading procedure influences the load carring capacit. This can not be eplained b the present method. Indeed, we ma show that the loading procedure is without influence if the modulus of elasticit is kept constant. In Figure 7.9 the solid line shows how the slab behaves if it is loaded with a transverse load first and then with a lateral load. The dotted line shown the opposite loading procedure. Both failure b instabilit and b material failure are covered. It is seen that both loading procedures lead to the same result. p Zero aial force n p p Increasing aial force n u ma Figure 7.9 Transverse load as a function of maimum deflection for different aial loads. Such are the facts when the modulus of elasticit remains constant. However, it is well known that this is actuall not true for concrete because of its nonlinear behaviour. To take this into account we ma assume that the modulus of elasticit is a function of the maimum stress, meaning that a higher stress leads to a lower modulus of elasticit. As an eample we might assume a modulus of elasticit of the form: 1 σ c Ec = 1000 fc 1 fc,ma (7.4.5) - 45 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces Here σ c,ma is the maimum compressive stress and f c is the compressive strength of the concrete. An assumption of this kind ma lead to quite different behaviour of the slab as illustrated in Figure 7.30. The red lines illustrate the behaviour of the slab if it is loaded b a relativel large transverse load. The solid line is valid when the transverse load is applied first and the dotted line is valid if the aial load is applied first. It is seen that in this case the two loading procedures do not lead to the same aial load even though the transverse load is the same. p Zero aial force n p p Increasing aial force n u ma Figure 7.30 Transverse load as a function of maimum deflection for different aial loads. The black lines illustrate the behaviour if the transverse load is relativel low. In this case both procedures result in the same transverse and aial load. Nevertheless, attention should be given to the fact that if the transverse load is applied first the maimum deflection will decrease and then increase again. Such behaviour will of course influence the behaviour of the concrete since some of the concrete that was cracked starts to be compressed. This fact is not taken into consideration in this investigation. The results are illustrated b some data from tests, [11]. Figure 7.31 and Figure 7.3 show how two different levels of aial load for the same level of transverse load ma be reached. - 46 -
Figure 7.33 illustrates how the deflection of the slab ma decrease as the aial load is increased. 700 600 N [kn/m] 500 400 300 00 100 u [mm] 0 0 10 0 30 40 50 60 70 30 5 q [kn/m ] 0 15 10 5 u [mm] 0 0 10 0 30 40 50 60 70 Figure 7.31. The loading curves for slab no 6 in [11]. - 47 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 100 1000 N [kn/m] 800 600 400 00 u [mm] 0 0 5 10 15 0 5 30 35 40 45 30 5 q [kn/m ] 0 15 10 5 u [mm] 0 0 5 10 15 0 5 30 35 40 45 Figure 7.3. The loading curves for slab no 16 in [11]. 100 1000 N [kn/m] 800 600 400 00 u [mm] 0 0 5 10 15 0 5 30 35 40 30 5 q [kn/m ] 0 15 10 5 u [mm] 0 0 5 10 15 0 5 30 35 40 Figure 7.33 The loading curves for slab no 4 in [11] - 48 -
8 Theor compared with tests In this section we demonstrate how the methods described in the previous sections ma be used to calculate the load carring capacit for an actual slab that has been tested. The test series was carried out b Larz Z. Hansen and the author, see[11]. In this investigation a square slab loaded with aial force in the -direction was eamined for different combinations of aial force and transverse load. The main data and results are given in Table 8.1 and Table 8.. No f c E c f Y h Laer A s h c ρ 0 l h c A s ρ 0 l [MPa] [MPa] [MPa] [mm] [mm /m] [mm] [ ] [mm] [mm] [mm /m] [ ] [mm] 1 6,5 14495 593 63,4 1 53,599 35 0,0083 000 5 53,599 0,0086 000 56,0 17545 593 61,38 1 53,599 35 0,0085 000 5 53,599 0,00853 000 3 60,4 18081 593 61,66 1 53,599 35 0,0085 000 5 53,599 0,00849 000 4 59,5 1745 593 6,03 1 53,599 35 0,0084 000 5 53,599 0,00844 000 5 58,8 1766 593 61,63 1 53,599 35 0,0085 000 5 53,599 0,0085 000 6 64,6 18688 593 61,37 1 53,599 35 0,0085 000 5 53,599 0,00853 000 7 64,0 18466 593 61,6 1 53,599 35 0,0085 000 5 53,599 0,00855 000 8 61,4 17718 593 60,99 1 53,599 35 0,0086 000 5 53,599 0,00858 000 9 66,7 18744 593 61,56 1 53,599 35 0,0085 000 5 53,599 0,00851 000 16 66,7 19394 593 61,48 1 53,599 35 0,0085 000 5 53,599 0,0085 000 Table 8.1. The data for the reinforced concrete slabs, see [11] for further details. It should be noted that h c in this case is the distance from the top of the slab to the reinforcement No q N u Notes Table 8.. Results from tests. [kn/m ] [kn/m] [mm] 1 5,3 0,0 1,6 Test slab no failure 49,8 0,0 6 Test slab no failure 3 74,5 0,0 78 Material failure 4 1,5 1084,1 9 Rig failure 5 33, 46,9 53 Stabilit failure 6 5,1 653,3 46 Stabilit failure 7 41,5 436,0 61 Stabilit failure 8 16,7 800,0 4 Stabilit failure 9 8,5 1103,4 17 Material failure 16 5,1 1030,1 37 Material failure - 49 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces The calculations in this section will be based on the following data for the slab: f c =6MPa, L =L =000mm, h=6mm,h c =31mm, φ =φ =0.0085 (one laer of reinforcement). It should be noted that in this case h c is the distance from the top of the slab to the reinforcement. Also it should be noted that in Table 8. the comment material failure means material failure at the place where the aial force was applied, i.e. a local failure. The moduli of elasticit are: E s = 5 10MPa 1000 fc E0c = min 3 f 51000 c 4 fc + 13 fc k = 0.8 400 fc in MPa E 0c σ ma Ec = 1 k 1 k fc ( ) σ f where σ ma and σ min is the maimum and minimum compressive strength in the section, respectivel. Furthermore it is assumed that material failure will take place when the maimum stress in the concrete eceeds 1.5f c. These assumptions are basicall those given in the anish Code of Practice S411 and represent a simple wa of taking the nonlinear behaviour of the concrete into account. Maimum and minimum stresses should strictl speaking be taken as the maimum and minimum principal stresses. However, determination of the principal stresses is not possible on the basis of the present constitutive equations. The reason is that we have assumed that the stiffnesses are independent of each other, which means that the depth of the compression zone when dealing with the torsional stiffness ma not be the same as the depth of the compression zone when dealing with the bending stiffnesses. Thus it is not possible to make an correct calculation of the principal stresses. Instead it is assumed that the actions in the -direction are the dominating ones so that the -direction ma be considered one of the principal directions. Hence the maimum and minimum stresses in formula (7.4.6) are the maimum and minimum stresses in the -direction. A description of the computer program ma be found in Appendi 11.1. Results of the calculations ma be seen in Figure 8.1 to Figure 8.3 min c (7.4.6) - 50 -
3500 3000 500 N [kn] 000 1500 1000 500 0 0 0.01 0.0 0.03 0.04 0.05 0.06 0.07 0.08 p [MPa] Figure 8.1 Aial force as a function of lateral load.* marks failure b instabilit,+ marks rig failure or material failure and o marks slabs that have not been loaded to failure. 3500 3000 500 N [kn] 000 1500 1000 500 0 3.1 3.105 3.11 3.115 E c [MPa] 3.1 3.15 3.13 10 4 Figure 8. Aial force as a function of modulus of elasticit. - 51 -
Behaviour of Concrete Slabs Subjected to Transverse Load and Compressive Aial Forces 3500 3000 500 N [kn] 000 1500 1000 500 0 35 40 45 50 55 60 65 70 75 80 σ c [MPa] Figure 8.3 Aial force as a function of maimum stresses. - 5 -