1. LRFD-Steel Design 1

2. Chapter 6

3. Such members are called beam-column.
6.1 INTRODUCTION
Most beams and columns are subjected to some degree of both bending and axial load especially in statically indeterminate structures.
Many columns can be treated as pure compression members with negligible error.
For many structural members, there will be a significant amount of both bending moment and axial load.
Such members are called beam-column.
Consider the rigid frame shown in the Figure:
For the given loading condition, 3

4. The horizontal member AB must not only support the vertical uniform load but must also assist the vertical members in resisting the concentrated load P1.
Member CD is a more critical case, because it must resist the load P1 + P2 without any assistance from the vertical members.
The reason is that the bracing members, prevents sidesway in the lower story. (in the direction of P, ED will be in tension and CF will be slack) 4

5. Member CD must transmit the load P1 + P2 from C to D.
The vertical members of this frame must also be treated as beam-column.
In addition, at A and B, B.M. are transmitted from the horizontal member through the rigid joints.
This is also occur at C and D and is true in any rigid frame.
Most columns in rigid frame are actually beam-columns, and the effects of bending should not be ignored.
Another example of beam-column can sometimes be found in roof trusses if purlins are placed between the joints of the top chord. 5

6. 6.2 INTRODUCTION FORMULAS
The inequality Equation may be written in the following form:
If both bending and axial compression are acting the interaction formula would be
Where
Pu is the factored axial compressive load.
Фc Pn is the compressive design strength.
Mu is the factored bending moment.
Фc Mn is the flexural design strength.

7. For biaxial bending, there will be two bending ratios:
Two formulas are given in the specification:
One for small axial load and one for large axial load.

8. Example 6.1
Determine whether the member shown in the Figure satisfies the appropriate AISC Specification interaction equation if the bending is about the strong axis.
Solution:
From the column load tables:
Фc Pn = 382 kips.
Since bending is about the strong axis,
Фb Mn for Cb can be obtained from the beam
Design chart in Part 5 of the Manual.
For Lb = 17 ft, Фb Mn = 200 ft.kips. For the end condition and loading of this problem, Cb = 1.32.

9. Фb Mn = Cb * 200 = 1.32 * 200 = ft-kips.
This moment is larger than Фb Mp = 227 ft-kips (also from Manual)
So the design moment must be limited to Фb Mp. Therefore,
Фb Mn = 227 ft-kips.
Max. B.M. occurs at midheight, so Mu = 25*17/4 = ft-kips.
Determine which interaction equation controls:
This member satisfies the AISC Specification.

10. The presence of the axial load produces secondary moment.
6.3 MOMENT AMPLIFICATION
The presence of the axial load produces secondary moment.
The total moment =
The second term may be neglected if P is small.
Because the total deflection cannot be found directly, this problem is nonlinear, and without knowing the deflection, we cannot compute the moment.
Ordinary structural analysis methods that do not take the displaced into account are referred to as first-order methods.
Iterative numerical techniques, called second-order methods, can be used to find the deflection and secondary moments.

11. These method are usually implemented with a computer program.
Most current design codes permit the use of either a second-order analysis or the moment amplification method.
This method entails computing the maximum B.M. resulting from flexural loading by a first-order analysis, then multiplying it by a moment amplification factor to account for the secondary moment.
Where, M0 is the unamplified maximum moment.
Pe is the Euler buckling load = and Pu is factored load.

12. As we describe later, the exact form of the AISC moment amplification factor can be slightly different.
Example 6.2. Compute the amplification factor for the beam-column of example 6.1.
This represents a 12 % increase in B.M.
Mmax = 1.12 * = 119 ft-kips

13. 6.4 WEB LOCAL BUCKLING IN BEAM-COLUMNS
The determination of the design moment requires that the cross section be checked for compactness.
The web is compact for all tabulated shapes if there is no axial load.
If λ ≤ λp, the shape is compact
If λp ≤ λ ≤ λr, the shape is noncompact; and
If λr ≤ λ, the shape is slender
AISC B5, Table 5.1, prescribes the following limits: .

14. For any value of
Where Py = Ag Fy
Because Py is variable, compactness of the web cannot be checked and tabulated.
Some rolled shapes satisfy the worst case limit of
Shapes listed in the column load tables in Part 4 of the Manual that do not satisfy this criterion are marked, and these shapes need to be checked for compactness of the web.
Shapes whose flanges are not compact are also marked

15. Example 6.3
A W12 x 58 of A992 steel is subjected to a bending moment and a factored axial load of 300 kips. Check the web for compactness.
The upper limit is
From the dimension and properties tables λ = h/tw = 27.0 < λp
The web is therefore compact

16. 6.5 BRACED VERSUS UNBRACED FRAMES
Two amplification factors are used in LRFD
One to account for amplification resulting from the member deflection and the other to account for the effect of sway when the member is part of unbraced frame. The following Figure illustrates these two components. .
In Figure a, the member is restrained against sidesway, and the max. secondary moment is Pδ.
If the frame is unbraced, there is an additional component of the secondary moment, shown in Figure b, and the max value of it is PΔ, which represents an amplification of the end moment.

17. The amplified moment to be used in design is:
Mu = B1 Mnt + B2 Mlt
Where
Mnt = maximum moment assuming that no sidesway occurs.
Mlt = maximum moment caused by sidesway, = 0.0 in the actuall braced frame.
B1 = amplification factor for the moment occurring in the member when it is braced against sidesway.
B2 = amplification factor for the moment resulting from sidesway.
The following sections explain the evaluation of B1 and B2 .

18. 6.6 MEMBERS IN BRACED FRAMES
The amplification factor given in section 6.3 was derived for a member braced against sidesway.
The following Figure shows a member of this type
Maximum moment amplification occurs at the center, where the deflection is largest.
For equal end moment, the moment is constant throughout the length of the member,
So the maximum primary moment also occurs at the center.
If the end moments are not equal, the maximum primary and secondary moments will occur near each other.

19. If the end moments produce reverse-curvature bending as shown.
Here the max. primary moment is at one of the ends, and the max. amplification occurs between the ends.

20. The final form of the amplification factor is
Therefore, the max. moment in a beam-column depends on the distribution of bending moment within the member.
This distribution is accounted for by a factor, Cm, applied to the amplification factor given in section 6.3.
The amplification factor given in section 6.3 was derived for the worst case, so Cm will never be greater than 1.0.
The final form of the amplification factor is
Note:
When computing pe1, use KL/r for the axis of bending and an effective length factor K less than or equal to 1.0 (braced condition)

21. Evaluation of Cm
The factor Cm applies only to the braced condition.
There are two categories of members:
Members with transverse loads applied between it’s ends.
Members with no transverse loads.

22. If there are no transverse loads acting on the member,
M1/M2 is the ratio of the bending moments at the ends of the member. (M1 is the smallest value and M2 is the largest one).
The ratio is positive for member bent in reverse curvature and negative for single curvature bending as shown in the Figure.

23. For transversely loaded members, Cm can be taken as 0
For transversely loaded members, Cm can be taken as 0.85 if the ends are restrained against rotation (fixed) and 1.0 if the ends are unrestrained against rotation (pinned).
End restrained will usually result from the stiffness of members connected to the beam-column.
Although the actual end condition may lie between full fixity and a frictionless pin, use of one of the two values given here will give satisfactory results.

24. Example 6.5
The horizontal beam-column shows in the Figure is subjected to the service live loads shown. This member is laterally braced at its ends, and bending is about the x-axis. Check for compliance with the AISC Specification.
Solution
The factored load = Pu =1.6*28=44.8 kips, wu=12*0035=0.042 kips
The maximum moment is
The member is braced against end translation, so Mlt =0.0

25. Compute the moment amplification factor
The member braced against sidesway, transversely loaded, and with unrestrained ends, Cm can be taken as 1.0.
The amplification factor is:
For the axis of bending

28. Because the beam weight is very small in relation to the concentrated live load, Cb may be taken from Figure 5.15c as This value results in a design moment of

29. This moment is greater than the plastic moment = 0. 90. 1735/12 =130
This moment is greater than the plastic moment = 0.90*1735/12 =130.1 ft-kips, so the design strength must be limited to this value.
Check the interaction formula:
Please read the remaining examples, 6.4 and 6.6 from the text book.

30. 6.7 MEMBERS IN UNBRACED FRAMES
The amplification factor given in section 6.3 was derived for a member braced against sidesway.
The following Figure shows a member of this type