Chapter 8. Beam-Columns – Design of Steel Structures

8

Beam-Columns

8.1 Introduction

In a practical situation, it is very rare to find an axially loaded column. Usually, columns are subjected to both an axial load and bending moments and are known as beam-columns. Some of the common situations in which a member acts as a beam-column are described below.

  1. A member may be subjected to both an axial compression and transverse loads (Figure 8.1(a))
  2. The axial load may not be acting through the centroid of the cross-section i.e., at an eccentricity ‘e’ from the centroid (Figure 8.1(b))
  3. The axial load not be acting on the column cross-section but transferring through a bracket or a seat (Figure 8.1(c))
  4. The columns are part of rigidly jointed frames due to which bending moments are transferred from the beams. The frames may be sway frames or non-sway frames. In a non-sway frame, the relative lateral displacement between the ends of a column is prevented. One of the ways to prevent this is by providing bracings as shown in Figure 8.1(d). In a sway frame, the relative lateral displacement between the ends of a column is not prevented and hence the frame is free to sway as shown in Figure 8.1(e).

 

 

Figure 8.1 Types of beam-columns

 

A beam-column may bend / buckle in a single curvature or in a reverse curvature depending on the loading and support conditions as shown in Figure 8.2. In a beam-column bent in single curvature, the end moments are in opposite direction whereas in a beam-column bent in double (reverse) curvature, the end moments are in the same direction. The bending moments may be acting in a single plane, i.e. uniaxial bending or in two planes causing biaxial bending in addition to axial compression (Figure 8.3).

 

 

Figure 8.2 Single and double curvatures

 

 

Figure 8.3 A beam-column associated with biaxial bending

8.2 Analysis of Beam-columns

Consider the beam-column shown in Figure 8.4. The beam-column is subjected to buckling due to an axial compression and bending due to the bending moment. That is, there exists an interaction between buckling caused by the axial compression and bending caused by the bending moment. Due to this interaction, the deflection and bending moment are amplified as shown in Figure 8.4.

 

 

Figure 8.4 A beam-column subjected to transverse load

 

The maximum bending moment due to the combined action of the axial compression and the bending moments is given by

 

Mmax = M0 + max      (8.1)

 

where M0 is the maximum primary bending moment and Pδmax is the additional bending moment caused by buckling.

The maximum bending moment Mmax may be expressed in the form

 

 

where α = P / Pcr

in which

l is the effective length of the member under an axial compression and E is Young's modulus and I is the moment of inertia of the cross-section. Cm is a coefficient which accounts for the non-uniformity of the primary bending moment along the length of the member. For a simply supported beam-column loaded with a uniformly distributed load, Cm is practically equal to 1.0.

In the case of a beam-column with end moments, as shown in Figure 8.5, the maximum bending moment due to bending as well as buckling may be again expressed in the form

 

 

Figure 8.5 A beam-column subjected to end moments

 

 

where M2 is the larger end moment and α = P / Pcr

Again, Cm is a coefficient which accounts for the non-uniformity of the primary bending moment along the length of the member. Cm may be expressed as

 

 

where M1 is the smaller end moment and M2 is the larger end moment. M1/M2 is negative when the member is bent into a single curvature as shown in Figure 8.5(a). M1/M2 is positive when the member is bent into reverse curvature as shown in Figure 8.5(b). If the bending moment is uniform along the length of the beam-column (M1 = − M2 = M0), Cm = 1.0.

The just concluded discussion of beam-columns is applicable to members having ends which are not allowed to move in a transverse direction. In the case of the vertical members of an unbraced frame, a side sway may take place as shown in Figure 8.6. Δ0 is the relative displacement of the column ends due to bending only whereas Δ includes the additional side sway caused by the buckling action of the axial load P. The additional side sway and bending moment due to the buckling action of the axial load P is known as PΔ effect. This effect is taken into account by choosing the appropriate value for Cm.

 

 

Figure 8.6 A beam-column with a side sway

8.3 Modes of Failure

In general, a beam-column may fail due to yielding or due to buckling. The different possible failures of a beam column are outlined below.

8.3.1 Failure Due to Yielding

A beam-column may fail due to yielding locally at a critical section due to axial compression and bending. This type failure may take place in short beam-columns and at supports when the beam-column bends into a reverse curvature. That is, yielding occurs when a resultant stress due to an axial compression and bending reaches the yield stress.

 

 

where

A = area of the cross-section of the member

Zp = plastic section modulus

ƒy = yield stress

8.3.2 Failure Associated with Flexural Buckling and Bending

This type of failure occurs in the case of members subjected to larger compression and single curvature bending about the minor axis of a section. It may also occur in members of medium length subjected to an axial compression and a single curvature bending about the major axis. That is, this type of failure occurs when

 

 

Pcr = elastic critical load

Mp = plastic moment capacity of the section

8.3.3 Failure Associated with Flexural and Lateral Torsional Buckling

This type of failure occurs in long members subjected to axial compression and bending about the major axis when the member is free to undergo a lateral torsional buckling i.e. when there is no lateral support along the minor axis of a member. This type of failure occurs when

 

 

where Mcr = the elastic critical moment for lateral torsional buckling

8.4 Design Specifications as per IS 800:2007

8.4.1 Design for Yielding

The design of a beam-column against local failure due to yielding may be carried out using the following interaction equation

 

 

where

N = factored axial compressive force (= P)

Nd = design strength in compression due to yielding = Ag ƒy /γm0

My = factored moment acting about the minor axis of a section

Mz = factored moment acting about the major axis of a section

Mdy = design strength in bending about the minor axis due to yielding

Mdz = design strength in bending about the major axis due to yielding

8.4.2 Design for Buckling

8.4.2.1 Axial Compression and Uniaxial Bending About Major Axis

Members subjected to an axial compression and bending about the major axis of a section should be designed using the following interaction formulae.

 

 

 

The first equation accounts for the flexural buckling due to axial compression and the bending about the major axis of a section. The second equation accounts for the flexural buckling about the minor axis of a section due to axial compression and lateral torsional buckling. In the above equations,

 

   P = factored axial compressive force
Pdz = design strength in compression due to the buckling about the major axis of a section
Pdy = design strength in compression due to the buckling about the minor axis of a section
Mz = maximum factored bending moment about the major axis of a section
Mdz = design bending strength about the major axis of a section considering lateral torsional buckling
Cmz = equivalent uniform bending moment factor
= 0.6 − 0.4ψ ≥ 0.4 (for a member not subjected to transverse loads)

 

where ψ is the ratio of the smaller bending moment Mz1 to the larger bending moment Mz2 at the ends of a member. If side sway takes place, the value of Cmz should be taken as 0.9.

 

Kz = factor to account for the magnification of the bending moment Mz due to buckling caused by an axial compression (Pδ and PΔ effects)
= 1 + (λz – 0.2)nz ≤ (1 + 0.8nz)

 

where

λz = non-dimensional effective slenderness ratio = in which

 

nz = P/Pdz

KLT = factor to account for the magnification of the bending moment Mz due to lateral torsional buckling

 

 

where

 

ny = P/Pdy

CmLT = equivalent uniform bending factor for lateral torsional buckling

     = 0.6 − 0.4ψ ≤ 0.4 (for a member not subjected to transverse loads)

 

in which ψ is the ratio of the smaller bending moment Mz1 to the larger bending moment Mz2 in a segment of a beam-column between lateral supports. If there are no intermediate lateral supports, ψ is the ratio of the smaller bending moment Mz1 to the larger bending moment Mz2 at the ends of a beam-column. ƒcr, b is the extreme fibre elastic critical stress given by Equation (5.26).

It may be noted that KLT includes the effect of the non-uniformity of the bending moment by the inclusion of CmLT in its expression.

8.4.2.2 Axial Compression and Uniaxial Bending About Minor Axis

Members subjected to axial compression and bending about the minor axis of a section should be designed using the following interaction equation

 

 

where

My = maximum factored bending moment about the minor axis of a section

Mdy = design bending strength about the minor axis of a section

Cmy = equivalent uniform bending moment factor

   = 0.6 – 0.4ψ ≥ 0.4 (for a member not subjected to transverse loads)

 

in which ψ is the ratio of the smaller bending moment My1 to the larger bending moment My2 at the ends of a member. If intermediate lateral supports are provided, ψ is the ratio of the smaller bending moment My1 to the larger bending moment My2 in a segment of a beam-column between lateral supports. If side sway takes place, the value of Cmy should be taken as 0.9.

 

Ky = factor to account for the magnification of the bending moment My due to buckling caused by an axial compression (Pδ and PΔ effects )

     = 1 + (λy – 0.2)ny ≤ (1 + 0.8ny)

where λy = non-dimensional effective slenderness ratio = in which

    and     ny = P/Pdy

8.4.2.3 Axial Compression and biaxial Bending About the Major and Minor Axes

Members subjected to axial compression and bending about both the major and the minor axes of a section may be designed using the following interaction equations.

 

 

 

Equation (8.12) accounts for the flexural buckling about the minor axis of a section due to the axial load, the bending about the minor axis of the section and the lateral torsional buckling due to the bending about the major axis of the section. Equation (8.13) accounts for the flexural buckling about the major axis of a section due to an axial compression, 60% of the effect of the bending about the minor axis of the section and the bending about the major axis of the section.

Example 8.1

Design a beam-column carrying compression of 400 kN at an eccentricity of 125 mm along the minor axis. Assume that the ends of the column are hinged with an unsupported length of 5 m. The grade of the steel is E250.

 

The factored load = 1.5 × 400 = 600 kN

A trail section for the beam-column is obtained by considering only an axial load of the magnitude of twice the actual axial load. Let ƒcd = 120 MPa.

The required area of the section =

 

 

Figure 8.7

 

From Appendix A, HB 400 @ 82.2 Kg/m may be tried.

A = 10,466 mm2, bf = 250 mm, tf = 12.7 mm, tw = 10.6 mm, ry = 51.6 mm, rz = 166.1 mm, Zpz = 1,626.36 cm3.

 

   where ε = 1.0

 

∴ From Table 1.7, the section is compact.

 

h/bf = 400/250 = 1.6 > 1.2

 

∴             From Table 4.3, the buckling class is ‘a’ for the z-axis and ‘b’ for the y-axis.

 

Since the ends are hinged, the effective length KL = 5,000 mm

The factored bending moments about the z-axis, Mz1 = –Mz2 = 600 × 0.125 = 75 kNm

 

 

 

From Table 4.4, for the buckling class ‘a’, fcdz = 220 MPa

 

 

From Table 4.4, for the buckling class ‘b’, fcdy = 123 MPa

The design strengths in compression about the z and y axes are

 

Pdz = A fcdz = 10,466 × 220 = 2,303 kN
Pdy = A fcdy = 10,466 × 123 = 1,285 kN

 

From Table 5.1, assuming that the ends are fully restrained against torsion but not for warping and loading condition is normal, the length for lateral torsional buckling, LLT = 5,000 mm.

 

hf = 400 − 12.7 = 387 mm

 

Extreme fibre stress corresponding to the lateral torsional buckling is

 

 

From Table 13(a) of IS 800, for ƒy = 250 MPa, α = 0.21, ƒcr,b = 284 MPa, ƒbd = 156 MPa

The design bending strength, Mdz = βb Zpz fbd = 1.0 × 1,626.36 × 103 × 156 = 254 kNm

Since Mz is uniform along the length of the member, Cmz = CmLT = 1.0

 

 

nz = P/Pdz = 600/2,303 = 0.26 and ny = P/Pdy = 600/1,285 = 0.47
Kz = 1 + (0.34 − 0.2) × 0.26 = 1.04 < (1 + 0.8 × 0.26) = 1.21

 

 

Check for yielding

 

Nd = Afy/γm0 = 10,466 × 250/1.1 = 2,379kN

 

Check for buckling

 

 

∴ HB 400 may be provided or a smaller section may be tried.

Example 8.2

Design a beam-column of unsupported length 4 m to carry an axial compression of 500 kN and end moments of 50 kNm and 100 kNm which bend the member into a reverse curvature. Assume that the ends of the member are rigidly connected to beams and are prevented from having a side sway. The grade of the steel is E250.

The factored axial compression, P = 1.5 × 500 = 750 kN

The factored end moments, Mz1 = 1.5 × 50 = 75 kNm and Mz2 = 1.5 × 100 = 150 kNm

A trail section for the beam-column is obtained by considering only an axial load of the magnitude of twice the actual axial load. Let ƒcd = 180 MPa.

The required area of the section =

From Appendix A, SC 220 section may be tried. Its properties are

A = 8,980 mm2, rz = 93.5 mm, ry = 49 mm, bf = 220 mm, tf = 16 mm, tw = 9.5 mm

 

h/bf = 220/220 = 1.0 < 1.2

 

∴ From Table 4.3, the buckling class is ‘b’ for the z-axis and ‘c’ for the y-axis.

 

 

∴ From Table 1.7, the section is plastic.

The correct value of the effective length factor K is to be obtained from the Annexure D of IS800. However, in this problem, it is assumed as 0.7 in both the xz and the xy planes.

 

 

 

Figure 8.8

 

From Table 4.4, for the buckling class ‘b’,fcdz = 216 MPa

 

 

From Table 4.4, for the buckling class ‘c’, fcdy = 172.5 MPa

The design strengths in compression about the z and the y axes are

 

Pdz = A fcdz = 8,980 × 216 = 1,940 kN
Pdy = A fcdy = 8,980 × 172.5 = 1,549 kN

 

From Table 5.1, assuming that the ends are fully restrained against torsion and warping and the loading condition is normal, the length for the lateral torsional buckling,

 

LLT = 0.7 × 4,000 = 2,800 mm.
hf = 220 − 16 = 204 mm

 

The extreme fibre stress corresponding to the lateral torsional buckling is

 

 

From Table 13(a) of IS 800, for ƒy = 250 MPa, α = 0.21, ƒcr,b = 941.5 MPa, fbd = 206 MPa

 

 

The design bending strength, Mdz = βb Zpz fbd = 1.0 × 802 × 103 × 206 = 165.5 kNm

 

 

Since there are no lateral supports other than at the ends, CmLT = Cmz = 0.4

 

 

nz = P/Pdz = 750/1,940 = 0.39 and ny = P/Pdy = 750/1,549 = 0.48
Kz = 1 + (0.34 − 0.2) × 0.39 = 1.05 < (1 + 0.8 × 0.39) = 1.31

 

 

Check for yielding

 

Nd = Afy/γm0 = 8,980 × 250/1.1 = 2,041 kN

 

Check for buckling

 

 

Since two of the interaction equations are not satisfied, a higher section SC250 may be tried.

From Appendix A, A = 10,900 mm2, rz = 107 mm, ry = 54.6 mm, bf = 250 mm, tf = 17 mm, tw = 10 mm

 

h/bf = 250/220 = 1.0 < 1.2

 

∴ From Table 4.3, the buckling class is ‘b’ for the z-axis and ‘c’ for the y-axis.

 

 

 

Figure 8.9

 

 

∴ From Table 1.7, the section is plastic.

 

 

From Table 4.4, for the buckling class ‘b’, fcdz = 219 MPa

 

 

From Table 4.4, for the buckling class ‘c’, fcdy = 181 MPa

The design strengths in compression about the z and y axes are

 

Pdz = A fcdz = 10,900 × 219 = 2,387 kN
Pdy = A fcdy = 10,900 × 181 = 1,973 kN
hf = 250 – 17 = 233 mm

 

The extreme fibre stress corresponding to the lateral torsional buckling is

 

 

From Table 13(a) of IS 800, for fy = 250 MPa, α = 0.21, fcr,b = 1,073 MPa, fbd = 210 MPa

 

 

The design bending strength, Mdz = βb Zpz fbd = 1.0 × 1,107 × 103 × 210 = 233 kNm

 

 

nz = P/Pdz = 750/2,387 = 0.31 and ny = P/Pdy = 750/1,973 = 0.38
Kz = 1 + (0.3 − 0.2) × 0.31 = 1.03 < (1 + 0.8 × 0.31) = 1.25

 

 

Check for yielding

 

Nd = A fy/γm0 = 10,900 × 250/1.1 = 2,477 kN

 

 

Check for buckling

 

 

∴                           SC 250 section may be provided.

Example 8.3

Design a beam-column for the following data:

Pu = 500 kN, Muz1 = −67 kNm, Muz2 = 100 kNm, Muy1 = 30 kNm, Muy2 = 50 kNm

Unsupported length = 5 m. The grade of the steel is E250.

Side sway is prevented and the effective length factor, K = 0.7.

 

As the beam-column is subjected to a biaxial bending and axial compression, the trail section may be obtained by considering only an axial load of magnitude of three times the actual axial load. Let ƒcd = 120 MPa.

The required area of the section =

 

From Appendix A, HB 450 @ 92.5 kg/m may be tried.

A = 11, 789 mm2, bf = 250 mm, tf = 13.7 mm, tw = 11.3 mm, rz = 185 mm, ry = 50.8 mm and Zpz = 2,030.95 cm3.

 

h/bf = 450/250 = 1.8 >1.2 and tf = 13.7 mm < 40 mm.

 

From Table 4.3, the buckling class is ‘a’ for the z-axis and ‘b’ for the y-axis.

 

 

 

Figure 8.10

 

From Table 1.7, the section is plastic.

 

 

From Table 4.4, for the buckling class ‘a’, fcdz = 226 MPa

 

 

From Table 4.4, for the buckling class ‘b’, fcdz = 167.5 MPa

 

Pdz = 11,789 × 226 = 2,664 kN
Pdy = 11,789 × 167.5 = 1,975 kN
LLT = 0.7 × 5,000 = 3,500 mm
hf = 450 − 13.7 = 436.3 mm

 

 

From Table 13(a) of IS800, α = 0.21, fcr,b = 508.5 MPa, ƒbdz = 189.4 MPa

 

fbdy = 250/1.1 = 227.3 MPa
Mdz = βb Zpz fbd = 1.0 × 2,030.95 × 103 × 189.4 = 385 kNm

 

 

Mdy = 1.0 × 442 × 103 × 227.3 = 100.5 kNm

 

 

nz = P/Pdz = 500/2,664 = 0.19 and ny = P/Pdy = 500/1,975 = 0.25
Ky = 1 + (0.776 − 0.2) × 0.25 = 1.144 < (1 + 0.8 × 0.25)= 1.2
Kz = 1 + (0.213 − 0.2) × 0.19 = 1.0 < (1 + 0.8 × 0.19) = 1.15

 

 

Cmz = 0.6 − 0.4 ψz = 0.6 − 0.4 × (−0.67) = 0.87
Cmy = 0.6 − 0.4 ψy = 0.6 − 0.4 × 0.6 = 0.36 < 0.4
∴     Cmy = 0.4

 

Since there are no lateral supports other than at the ends, CmLT = Cmz = 0.87

 

 

Mz = Muz2 = 100 kNm and My = Muy2 = 50 kNm

Check for yielding

 

Nd = 11,789 × 250/1.1 = 2,679 kN

 

Check for buckling

 

 

Hence, HB 450 @ 92.5 kg/m is all right.

Problems

For the following problems, consider the grade of steel as E250

  1. Design a beam-column subjected to a factored axial load of 400 kN and a factored bending moment of 60 kNm at the fixed end. The other end of the beam-column is hinged. The unsupported length of the beam-column is 4 m.
  2. Re-do Example 8.1 if one end is fixed, and the other end is free.
  3. Re-do Example 8.2 if side sway is not prevented. Assume that the effective length factor (K) is 1.3.
  4. Re-do Example 8.3 if Mz1 = + 67 kNm. Assume that the side sway is prevented. The other data remains the same.
  5. Re-do Example 8.3 if Mz1 = + 67 kNm. Assume that the side sway is not prevented in both the planes (xz and yz). The other data remains the same. The effective length factor K may be taken as 1.2 for both the planes.
  6. Design an I section for the beam-column in Figure 8.11. Assume that no lateral support is provided along the length of the beam except at supports.
  7. Re-do problem 8.6 if, instead of a uniformly distributed load, a concentrated load of 50 kN acts at mid-span.

 

 

Figure 8.11