Section3 3 PDF
Section3 3 PDF
Devdas Menon
                Introduction
                The analysis of flexural members under service loads involves the calculation of the
                following quantities.
                     a) Cracking moment.
                     b) Location of kern points.
                     c) Location of pressure line.
                The cracking moment (Mcr) is defined as the moment due to external loads at which the
                first crack occurs in a prestressed flexural member. Considering the variability in stress
                at the occurrence of the first crack, the evaluated cracking moment is an estimate.
                Nevertheless, the evaluation of cracking moment is important in the analysis of
                prestressed members.
                Based on the allowable tensile stress the prestress members are classified into three
                types as per IS:1343 - 1980. The types are explained in Section 1.2, Advantages and
                Types of Prestressing. For Type 1 (full prestressing) and Type 2 (limited prestressing)
                members, cracking is not allowed under service loads. Hence, it is imperative to check
                that the cracking moment is greater than the moment due to service loads. This is
                satisfied when the stress at the edge due to service loads is less than the modulus of
                rupture.
                The modulus of rupture is the stress at the bottom edge of a simply supported beam
                corresponding to the cracking moment (Mcr). The modulus of rupture is a measure of
                the flexural tensile strength of concrete. It is measured by testing beams under 2 point
                loading (also called 4 point loading including the reactions or middle third loading). The
                modulus of rupture (fcr) is expressed in terms of the characteristic compressive strength
                (fck) of concrete by the following equation (IS:456 - 2000). Here, fcr and fck are in N/mm2.
                The following sketch shows the internal forces and the resultant stress profile at the
                instant of cracking.
                                                         Mcr
                                              CGC
                                         yb              Pe
                                                                                     fcr
                                        Internal forces in                    Resultant stress
                                        concrete                              profile
                             Figure 3-3.1       Internal forces and resultant stress profile at cracking
                The stress at the edge can be calculated based on the stress concept as follows. The
                cracking moment (Mcr) can be evaluated by transposing the terms.
                                                     Pe Pe ey b Mcr y b
                                                 -     -        +         = fcr
                                                     A     I        I
                                                                  Mcr y b        P P ey
                                                             or,          = fcr + e + e b
                                                                    I             A    I
                                                                            f I PI
                                                             or,    Mcr = cr + e + Pe e                         (3-3.2)
                                                                             y b Ay b
                The above equation expresses Mcr in terms of the section and material properties and
                prestressing variables.
                When the resultant compression (C) is located within a specific zone of a section of a
                beam, tensile stresses are not generated. This zone is called the kern zone of a section.
                For a section symmetric about a vertical axis, the kern zone is within the levels of the
                upper and lower kern points. When the resultant compression (C) under service loads
                is located at the upper kern point, the stress at the bottom edge is zero. Similarly, when
                C at transfer of prestress is located at the bottom kern point, the stress at the upper
                edge is zero. The levels of the upper and lower kern points from CGC are denoted as kt
                and kb, respectively.
                Based on the stress concept, the stress at the bottom edge corresponding to C at the
                upper kern point, is equated to zero. The following sketch shows the location of C and
                the resultant stress profile.
                                             CGC    C   kt
                                        yb
The value of kt can be calculated by equating the stress at the bottom to zero as follows.
                                                         C Ckt y b
                                                        -  +        =0
                                                         A      I
                                                         C Ck y
                                                    or, - + t 2 b = 0
                                                         A Ar
                                                                      r2
                                                             or, kt =                                      (3-3.3)
                                                                      yb
                The above equation expresses the location of upper kern point in terms of the section
                properties. Here, r is the radius of gyration and yb is the distance of the bottom edge
                from CGC.
                Similar to the calculation of kt, the location of the bottom kern point can be calculated by
                equating the stress at the top edge to zero. The following sketch shows the location of
                C and the resultant stress profile.
                                         yt
                                              CGC   C   kb
                                                       C Ck b y t
                                                      -  +        =0
                                                       A     I
                                                       C Ck y
                                                  or, - + b 2 t = 0
                                                       A Ar
                                                                    r2                                            (3-3.4)
                                                          or, k b =
                                                                    yt
                                                                                  fcr         fcr
                                        Location of
                                                                                        Resultant stress
                                        resultant
                                                                                        profile
                                        compression
                            Figure 3-3.4       Resultant stress profile at cracking of the bottom edge
                The cracking moment can be expressed as the product of the compression and the
                lever arm.        The lever arm is the sum of the eccentricity of the CGS (e) and the
                eccentricity of the compression (ec). The later is the sum of kt and z, the shift of C
                outside the kern.
                                                   Mcr = C ( e + ec )
                                                          = C ( e + kt + z )
                                                                          fcr I                                   (3-3.5)
                                               or, Mcr = C ( e + kt ) +
                                                                           yb
                Substituting C = Pe, kt = r2/yb and r2 = I/A, the above equation becomes same as the
                previous expression of Mcr.
                                                               r 2  fcr I
                                                      Mcr =Pe  +e+
                                                               yb  yb
                                                           f I PI
                                                  or, Mcr = cr + e +Pee                                      (3-3.6)
                                                            yb Ayb
                The pressure line in a beam is the locus of the resultant compression (C) along the
                length. It is also called the thrust line or C-line. It is used to check whether C at
                transfer and under service loads is falling within the kern zone of the section. The
                eccentricity of the pressure line (ec) from CGC should be less than kb or kt to ensure C
                in the kern zone.
                The pressure line can be located from the lever arm (z) and eccentricity of CGS (e) as
                follows. The lever arm is the distance by which C shifts away from T due to the moment.
                Subtracting e from z provides the eccentricity of C (ec) with respect to CGC. The
                variation of ec along length of the beam provides the pressure line.
                                                             M
                                                           z=
                                                             C
                                                        ec = z - e                                           (3-3.7)
                A positive value of ec implies that C acts above the CGC and vice-versa. If ec is
                negative and the numerical value is greater than kb (that is |ec| > kb), C lies below the
                lower     kern     point   and   tension    is   generated    at   the   top   of   the    member.
                If ec> kt, then C lies above the upper kern point and tension is generated at the bottom
                of the member.
Pressure line CL
                                  CGC                                                         kt
                                  CGS                                                         kb
                                                              Pressure line
                                                                                  CL
                                  CGC                                                         kt
                                  CGS                                                         kb
                Limiting Zone
                For fully prestressed members (Type 1), tension is not allowed under service conditions.
                If tension is also not allowed at transfer, C always lies within the kern zone. The limiting
                zone is defined as the zone for placing the CGS of the tendons such that C always lies
                within the kern zone.
                For limited prestressed members (Type 2 and Type 3), tension is allowed at transfer
                and under service conditions. The limiting zone is defined as the zone for placing the
                CGS such that the tensile stresses in the extreme edges are within the allowable values.
                The following figure shows the limiting zone (as the shaded region) for a simply
                supported beam subjected to uniformly distributed load.
CL
                                                                             Locus of emin
                                        CGC
                                                                       Locus of emax
                                    Figure 3-3.4     Limiting zone for a simply supported beam
                The determination of limiting zone is given in Section 4.4, Design of Sections for Flexure
                (Part III).
Example 3-3.1
                For the post-tensioned beam with a flanged section as shown, the profile of the
                CGS is parabolic, with no eccentricity at the ends. The live load moment due to
                service loads at mid-span (MLL) is 648 kNm. The prestress after transfer (P0) is
                1600 kN. Assume 15% loss at service. Grade of concrete is M30.
CGC
                                                               18.0m
                                                             500
200
                                                             150
                                              1000                      Values in mm.
                                              CGS
                                                     150               200
                                                             250
                                                Cross-section at mid-span
                Compare the stresses with the following allowable stresses at transfer and at
                service.
                       For compression, fcc,all =  18.0 N/mm2
                       For tension,           fct,all = 1.5 N/mm2.
Solution
                The section is divided into three rectangles for the computation of the geometric
                properties. The centroid of each rectangle is located from the soffit.
                                                           1
                                                                         yt
                                                           +                  CGC
                                                           2
                                        900                                         Values in mm.
                                                                     y   yb
                                              500
                                                           3
                                                    100
                                        A   = A1 + A2 + A3
                                            = 240,000 mm2
                Therefore,
                                                    y b = 583.3 mm
                                                    y t = 1000.0 - 583.3
                                                         = 416.7 mm
                                                    e = y -150
                                                         = 583.3 -150
                                                         = 433.3 mm
                                                       1
                                                  I1 =   500 2003 + A1 (900 - 583.3)2
                                                      12
                                                    = 1.0361010 mm4
                                                       1
                                               I3 =       250 2003 + A3 (583.3 -100)2
                                                      12
                                                    = 1.1841010 mm4
                                                I = I1 + I2 + I3
                                                 = (1.036 + 0.336 +1.184)1010
                                                 = 2.5521010 mm4
                                                   I
                                              r2 =
                                                   A
                                                   2.5521010
                                                 =
                                                     240,000
                                                 = 1.063105 mm2
                                      r2                                                    r2
                              kt =                                                   kb =
                                      yb                                                    yt
                                    1.063105                                            1.063 105
                                  =                                                    =
                                       583.3                                                416.7
                                  = 182.2mm                                            = 255.1mm
                                                      CGC          182.2
                                                                           +
                                                                   255.1        Kern zone
CGS
Values in mm.
= 5.76 kN/m
                                              w DL L2
                                        MDL =
                                                 8
                                              5.76 18.02
                                            =
                                                     8
                                            = 233.3 kNm
                                                  fcr I PeI
                                          Mcr =        +    + Pe e
                                                  y b Ay b
                                                3.83 2.5521010 0.851600103  2.5521010
                                              =                    +
                                                      583.3               240103 583.3
                                                                  + 0.81600 103  433.3 Nmm
                                              = 167.6 + 247.9 + 554.6
                                              = 970.1 kNm
                Since the given live load moment (648.0 kNm) is less than the above value, the section
                is uncracked.
                          The moment of inertia of the gross section can be used for computation of
                stresses.
                At transfer
                              MDL
                          z=
                               C
                              233.3103                              ec = z - e
                            =
                                 1600                                   = 145.8 - 433.3
                            = 145.8 mm                                  = - 287.5mm
Since the magnitude of ec is greater than kb, there is tension at the top.
                                        CGC
                                                            +
                                          287.5                      Kern zone
                                                            x
                                                                       Location of pressure line
                                        CGS
                                                   Value in mm.
                At service
                              MDL+LL
                          z=
                                C
                                                                              ec = z - e
                              (233.3 + 648.0)103
                            =                                                     = 648.0 - 433.3
                                  0.851600
                                                                                  = 214.7mm
                            = 648.0 mm
Since the magnitude of ec is greater than kt, there is tension at the bottom.
                                                            x
                                                                       Location of pressure line
                                                  214.7
                                        CGC
                                                            +
                                                                      Kern zone
CGS
                                                     Value in mm.
                d) Calculation of stresses
                The stress is given as follows.
                                                         P Pey My
                                                  f =-        
                                                         A   I   I
                                              M
                                  CGC                           +             +             =
                                          P
                                            P0    1600 103
                                               =-
                                            A      240 103
                                               = -6.67 N/mm2
                Stress at the top fibre
                                                 = 0.84 N/mm2
                Stress at the bottom fibre
= - 17.19 N/mm2
= -10.44 N/mm2
= 1.0 N/mm2
                                                                                 Numeric values
                                                                                 in N/mm2.
                                           17.19                            1.0
                                         At transfer                     At service