6.
0 TORSIONAL STRESSES IN I-SHAPED SECTIONS
  6.1 PURE TORSION (SAINT VENANT TORSION) - Fibers are twisted
          where                           Shear Modulus of Elasticity
  6.2 WARPING TORSION - Extreme fibers translates laterally in opposite direction
                       Figure 6.2.1 Common Torsional Loading
                  Figure 6.2.2 Torsion of an I-Shaped Section
From figure 6.2.2, we define the displacement of a point at the flange uf to be a
function of the twist angle   .
Consequently
                                         eqn 1
Consider upper flange and recalling Beam Curvature Relation
Differentiating wrt x, we obtain
Substitute in (1)
Define Warping Torsion
Let
                              >> warping torsional constant
Then
The torsional moment is the sum of the rotational part and the lateral bending part
   Recall
So that
Rewriting and dividing by ECw
The above equation is a non-homogenous PDE
Let                           and solve for the homogenous solution
Assume a solution in the form
Then
The above equation requires that
                                 or
Thus
                                                 >> exponential form
Note that
for the particular solution since Mz is in general some function of z, the form of the
solution may be written as
  such that
  where the terms on the left side must be paired with the terms on the right side. Rarely
  will the function f1(z) be required to contain higher than second degree terms.
  Study Example 8.5.1
6.3 DERIVATION OF PDE FOR LATERAL TORSIONAL BUCKLING (ELATIC THEORY)
                I-SHAPED BEAM IN SLIGHTLY BUCKLED POSITION
The rotated primed axes is given by the transformation matrix equivalent to the matrix
of direction cosines.
The curvature equations in x, y and z axes are
Differentiating wrt to z
Substituting in the above equation the Moment -Curvature Relation
We get
Or rearranging
                                                                >> PDE for the
                                                                Angle of Twist
Note : See Salmon and Johnson for the complete solution of above PDE, page 437.
6.4 ELASTIC LATERAL TORSIONAL BUCKLING MOMENT Mcr
     (LATERALLY UNSUPPORTED)
  LATERAL SUPPORT
    Two categories :
       1. Continous lateral support by embedment of compression flange.
       2. Lateral support intervals. (Laterally Braced)
  STRENGTH OF I-SHAPED BEAMS UNDER UNIFORM MOMENT
    Failure Modes even if Mu<Mp
       1. Local buckling of flange in compression
       2. Local buckling of web in flexural compression
       3. Lateral torsional buckling
  FOUR CATEGORIES OF BEHAVIOUR
       1. Plastic moment strength Mp is achieved with large deformation (rotation
       capacity)
       2. Inelastic behaviour where Mp is achieved but with little rotation capacity.
       3. Inelastic behaviour Mr, the moment above which residual stresses cause
       inelastic behaviour to begin is reached or exceeded.
       4. Elastic behaviour in which moment strength Mcr is controlled by inelastic
       buckling.
  ELASTIC LATERLA TORSIONAL BUCKLING MOMENT Mcr
       Recall PDE for Elastic Lateral Torsional Buckling
       The above equation may be written in this form
             where
  The characteristic roots are
  Substitute boundary conditions for simple support noting that for simple supports may
  not twist but are free to warp.
                                                   >> end twist are zero
                                                   >> end torque are zero
  Solving the above problem using Boundary Value Method, the final
  solution becomes
                                                               >> basis for LRFD Eqns
6.5 LOAD AND RESISTANCE FACTOR DESIGN - I SHAPED BEAMS SUBJECTED TO
STRONG AXIS OF BENDING
     The strength requirement according AISC-B3.3 may be stated as
           where
  Whether or not plastic moment strength is reached, failure will be on one of the
  following modes.
                                  Figure 6.5.1 Beam Behaviour
Case 1 : Plastic moment is reached (Mn=Mp) along with large plastic rotation capacity
(R>=3 in Figure 6.5.1)
  Note :
    1. The section must be compact to prevent local buckling.
    2. Lateral bracing must be provided where Lb<Lpd
       Figure 6.5.2 Deformation requirements for developing plastic strengths
        Note: You may use M2 instead of Mp (AISC-App 1.7), because M2=Mp
        under case 1.
Case 2. Plastic Moment is reached (Mn=Mp) but with relatively little rotation capacity (R<3
in Figure 6.5.1
        Note :
          1. The section must be compact to prevent local buckling.
          2. Lateral bracing must be provided where Lb<Lp
Case 3. Lateral torsional buckling of compact sections may occur in the inelastic range
(Mp>Mn>=Mr)
        Note :
              1. The section must be compact to prevent local buckling.
              2. Lateral bracing must be provided where Lp<Lb<Lr
              3. The nominal strength Mn is a linear function of the lateral torsional
          buckling strength, thus taking Mr=0.7Fy*Sx
        where
        Note: The length Lr is obtained by equating
                                         and
        and solving for L which yields AISC F2.2b
        where
        Note: rts may be approximated as
Case 4. General limit state where nominal moment strength Mn occurs in the inelastic
range (Mp>Mn>=Mr)
        Note :
              1. The condition is very uncommon for rolled shapes.
              2. Lateral bracing must be provided where Lp<Lb<Lr
              3. The nominal strength Mn is a linear function of the lateral torsional
          buckling strength, thus taking Mr=0.7Fy*Sx
              4. The flanges are within the non compact limits (web local buckling
          excluded)
              5. Lateral bracing must be provided where Lp<Lb<Lr
              6. The nominal strength Mn is a linear function of the lateral torsional
          buckling strength, thus taking Mr=0.7Fy*Sx
  The strength is the lower value obtained from..
                                                             >> non compact case
     and
                                                             >> inelastic LTB
     Other parameters within the equation for inelastic LTB are the same for case 3.
Case 5. General limit state where nominal moment strength Mn equals the elastic
buckling strength Mcr. Mn<Mr, where Mr=0.7FySx
        Note :
              1.        for flange and web (non compact range)
              2.        for slender elements
        Using properties rts and ho defined previously
        and the nominal strength is therefore
6.6 MOMENT GRADIENT Cb
  The moment gradient Cb is a modification factor for non-uniform bending moment
  variation for a beam segment laterally unbraced except at segment ends.
    Old Formula 1961
    where M1 and M2 are end moments such that M1<M2. M1/M2 is negative for
    single curvature bending.
    New Formula 1993 (Kirby and Nethercot) - Rm term added for mono
    symmetric section AISC 2005. Note Cb has a minimum value of 1.0 and a
    maximum value of 3.0
    where
                           >> For doubly symmetric sections and for singly
                           symmetric sections bending in single curvature
             Note : Rm is reduce when section is singly symmetric in reverse
             curvature bending. AISC F-1. This formula is more accurate when
             the unbraced segment has a non-linear moment variation.
6.7 PLOT OF BEAM MOMENT STRENGTH CURVE
    Given a W 18x97 using ASTM A992 with Fy=50ksi
    Material and Sectional Properties
    LTB Properties
                                                    >> Formula 1
                                                    >> Formula 2
      Iterate on Lb
Figure 6.7.1 Beam Moment Strength Curve using Formula 1
         Figure 6.7.2 Beam Moment Strength Curve using Formula 2
Problem Set No. 3
1. Determine the product of inertia and locate the shear center for the following steel
shapes.
2. Study Example 9.9.1 and 9.9.2. Discuss the procedure on how to undergo design of
beams in general.
3. Create a flow chart or a program on how to design a beam considering both
requirements on compactness and lateral support requirement and consider the effect
of moment gradient factor Cb.You may want to use the example above in checking
accuracy the results.
6.8 DETERMINATION OF MOMENT GRADIENT - MODIFICATION FACTOR Cb
EXAMPLE : Determine the moment gradient factor Cb for a simply supported beam with
a uniform load only. Lateral braces located at the ends and quarter points.
  The moment equation may be expressed as a function of its distance x from the left
  support.
  Assume values               and                     >> units are consistent
                                                  >> Moment equation
  The plot of the Moment diagram is a parabola.
  Segment 1-2 and 4-5
  1961 Formula
  1993 Formula
Segment 2-3 and 3-4
1961 Formula
1993 Formula
*** Try solving a simple beam with point load at the midspan and braced laterally at
the ends and at quarter points.
*** Try solving a fixed end beam with uniform loading and braced laterally at the
ends and at quarter points.