BEAM COLUMNS - Weebly

Prof. Dr. Zahid Ahmad Siddiqi

BEAM COLUMNS

Beam columns are structural members that are

subjected to a combination of bending and axial

stresses.

The structural behaviour resembles

simultaneously to that of a beam and a column.

Majority of the steel building frames have

columns that carry sizable bending moments in

addition to the usual compressive loads.


FLOW CHART FOR DESIGN OF

BEAM-COLUMNS

Known Data: Pu, M

ntx, Mltx , M

nty, Mlty, Kx

Lx, K

yL

y

Mr

= Mu

= Mnt

+ Mlt for the first trial

Calculate Mr

both in the x and y directions

Peq

= Pr

+ Mrx

(mx) + M

ry(m

y)

Assume an approximate magnification of

15% for the moments only.


Select section as a simple column depending

upon the following criteria:

1. Asel

≈ Areq

2. Minimum weight

3. Connecting leg width b > bmin

4. Depth of W-section ≤ 360 mm

mx (for first trial) = 8.5 − 0.7K1xLx

my (for first trial) = 17 − 1.4K1yLy

mx = 10 − 14(d / 1000)2 − 0.7K1xLx

my = 20 − 28(d / 1000)2 − 1.4K1yLy


The column selection tables may also be

employed to select the section using the values of

Peq

and KyL

y.

See rx/r

yfrom column selection table for selected

section

Calculate (KyL

y)

eq=

yx

xx

rr

LK

Re-enter the table for greater of KyL

yand (K

yL

y)

eq

and revise to obtain suitable section for the load

Peq

.


Find new values of m for subsequent trials.

Select a new section and repeat until values

of load capacities, Peq, and m are stabilized.

Peq

= Pr

+ Mrx

(mx) + M

ry(m

y)

Select a new section and repeat until values of

load capacities, Peq

and m are stabilized.


Calculate Cmx

and Cmy

for no sway conditions

Calculate , , and x

xx

r

LK1

y

yy

r

LK1

R = maximum of the above values

Check for maximum slenderness ratio: R ≤ 200

x

xx

r

LK2

y

yy

r

LK2


Axial strength of trial section:

Calculate φcF

crcorresponding to the R-value or

directly read it from the table in Reference-1 and

evaluate the compression capacity by multiplying

with the area of cross-section.

Pc = φcP

n= φcFcr

Ag

/ 1000

Calculate Euler buckling strength (Pe1

)x, (P

e1)y

, (Pe2

)x

and (Pe2

)y

for both lt and nt cases.

Pe1 = π2 EI / (K1 L)2 / 1000 (kN)


Calculate no-sway moment magnification factors

B1x

= ≥ 1.0 : B1y

= ≥ 1.0

Note: Pr

in the above formulas is the actual

factored axial load and not Peq

.

Calculate B2x

and B2y

.

B2

=

( )xe

r

mx

P

P

C

1

1 α− ( )ye

r

my

P

P

C

1

1 α−where α = 1.0 for LRFD procedure.

2

1

1

e

nt

P

P

∑∑− α


Calculate design moments

Mrx

= Mux

= B1x

Mntx

+ B2x

Mltx

Mry

= Muy

= B1y

Mnty

+ B2y

Mlty

Bending strength of the trial section:φbM

ny= φ

bF

yZ

y/ 106 (kN-m)

There are no chances of lateral buckling because

the lateral direction for y-axis bending is the

stronger direction.


Check conditions of compact section as a beam.

Find Lp

and Lr

from column table and check

against Lbx

.

Calculate φbM

nxas for a beam using L

bx, L

p, L

r

and beam selection tables. Use Cb

= 1.0 in the

expressions.

Calculate to see which interaction

equation is applicable.c

r

P

P


Check interaction equations:

≤ 1.0For ≥ 0.2

≤ 1.0For < 0.2

Get the value of Left Side of equation (LS) up to

2nd decimal place, truncating the 3rd decimal digit,

which should not be more than 1.00.

This means that LS can be as high as 1.0099 but

not 1.01.

c

r

P

P

c

r

P

P

++

cy

ry

cx

rx

c

r

M

M

M

M

P

P

9

8

++

cy

ry

cx

rx

c

r

M

M

M

M

P

P

2


Values of LS between 0.9 and 1.0 → Very economical design

Values of LS between 0.8 and 0.9 → Economical design

Values of LS between 0.7 and 0.8 → May be acceptable,

but better to try

an economical section

Values of LS lesser than 0.7 → Revise by selecting

a lighter section

Values of LS greater than 1.0099 → Select a stronger section

Check shear strength, which may usually be

omitted in hot rolled W sections because of very

high available strength.

Write the final solution using standard designation.


Example 5.1: Design the columns in a single-

bay multi-storey unbraced frame shown in Figure

5.6, where P is the load from the top stories.

Ratio of moment of inertia of beams with respect

to columns may be assumed as shown in the

figure. Approximate analyses results are also

provided in Figures 5.7 and 5.8. Assume that

sway is not allowed in the y-direction.

Solution:

Total Factored Loads

1. Load Combination 1, Gravity Load

Combination (1.2 D + 1.6 L)


8.5 m

II

PP

wI I

H

6.0 m

6.0 m

1.4 I

Figure 5.6. Frame And Loading For Example 5.1.


Pu

= 1.2(1025) + 1.6(410) = 1886 kN

wu

= 1.2(7.3) + 1.6(22.0) = 43.96 kN/m

P = 1025 kN dead load

= 410 kN live load

w = 7.3 kN/m dead load

= 22.0 kN/m live load

H = 345 kN wind load


1886kN1886kN

43.96kN/m

75.8 kN-m

37.9 kN-m

227 kN-m

Figure 5.7. Partial Gravity Load Analysis Results.


2. Load Combination 2, Wind Load

Combination (1.2D+0.5L+1.3W)

Pu

= 1.2(1025) + 0.5(410) = 1435 kN

Hu

= 1.3(345) = 448.5 kN

wu

= 1.2(7.3)+0.5(22.0) = 19.76 kN/m

Value of Kx

Gtop

= = = 2.02( )( ) beamsforLI

columnsforLI

∑∑ ( )

5.84.1

62

I

I

Gbotton

= 1.0 for sway columns


1435kN1435kN

19.76kN/m

51.2 kN-m

25.6 kN-m

( No Sway Part )

93.7 kN-m

138 kN138 kN

448.5kN

586 kN-m

759 kN-m

∆oh

( Sway Part )

586 kN-m

759 kN-m

Figure 5.8. Partial Lateral Load Analysis Results.


Kx

= 1.0 for braced frame

Kx

= 1.45 for unbraced conditions

Value of Ky:

No data of connected elements is given for y-

direction and hence the approximate value may

conservatively be assumed for no sway in this

direction.

Ky

= 1.0


Here, design is made for the wind combination and

check is then made for the gravity combination.

Design for Combination 2:

Pr

= 1435 + 138 + 19.76 × 8.5/2

= 1656.7 kN

According to AISC, max. moments for different

types of loading (nt or lt case), acting at different

locations or of different signs, are to be added

magnitude-wise in any combination.

The Right column is critical for the axial load.


Mntx

= 51.2 kN-m

Mltx = 759 kN-m

Mnty

= Mlty = 0

K2x

Lx= 1.45 × 6 = 8.7 m for lt-case

K1x

Lx= 1.00 × 6 = 6.00 m for nt-case

K1y

Ly= 1.00 × 6 = 6.00 m

Peq

= Pu

+ 1.15 Mux

(m)

= 1656.7 + 1.15 (51.2 + 759.0) (4.3)

= 5663 kN

mx (for first trial) = 8.5 − 0.7K1xLx

= 8.5 − 0.7 × 6 = 4.3

Assume 15% magnification of moments.


Using column tables of Reference 1 for this Peq

and KyL

y= 6.0 m;

Trial Section = W360 × 262

Peq

= Pu

+ 1.15 Mux

(m)

= 1656.7 + 1.15 (51.2 + 759.0) (3.99)

= 5374 kN

Revised mx = 10 − 14(d / 1000)2 − 0.7K1xLx

= 10 − 14 × 0.362 − 0.7 × 6

= 3.99


Trial Section-1: W360 × 237

A = 30,100 mm2

rx

= 162 mm, ry

= 102 mm

rx/r

y= 1.60

Ix

= 79,100 × 104 mm4

M1

/ M2

is positive because of reverse curvature

Cmx

= 0.6 – 0.4 2

1

M

M

= 0.6 – 0.4 = 0.4

2.51

6.25

= = 37.04 ( for nt case)x

xx

r

LK1

162

10000.6 ×

yx

xx

rr

LK

/

2

60.1

7.8(K

yL

y)

eq= = = 5.44 m (not critical)


= = 53.70 (for lt-case)x

xx

r

LK2

162

10007.8 ×

= = 58.82y

yy

r

LK1

102

10000.6 ×

R ≈ 59 < 200 OK

φcF

cr= 187.09 MPa

Pc

= φcF

crA

g= = 5,631 kN

1000

100,3009.187 ×

for nt-case

Pe1x

= π2 EI / (K1x

L)2

= 10006000

10100,79000,2002

42

××××π

= 43,371 kN


B1x

= =

= 0.42 ∴ B1x

= 1.0

Pe2x

= π2 EI / (K2x

L)2

= 10008700

10100,79000,2002

42

××××π

= 20,628 kN for lt-case

xer

mx

PP

C

,11 α− 371,437.165611

4.0

×−

ΣPnt

= 1435 × 2 + 19.75 × 8.5 = 3038 kN

ΣPe2,x

= 2 × 20,628 = 41,256 kN


B2x

= = = 1.08

Mrx

= B1x

Mntx

+ B2x

Mltx

= 1.0 (51.20) + 1.08(759.00)

= 870.9 kN-m

From column selection table:

Lp

= 5.06 m, Lr

= 25.43 m

xe

nt

P

P

,2

1

1

∑∑− α

256,41

30380.11

1

×−

Pr

= Pnt

+ B2

Plt

= 1518.98 + 1.08(138)

= 1668.02kN


Check conditions of compact section:

= 6.5 < λp

= 10.8 OKf

f

t

b

2

= = 0.245yb

u

P

P

φ ( ) 100,301000/2509.0

7.1656

××For web, λ

p= 3.4233.28.31 ≥

−

yb

u

P

P

φ= 66.3 for A36 steel

= 15.3 < λp

OKwt

h


Lb

= 6.00m > Lp, bending strength is to be

calculated using the inelastic LTB formula.

Mp

= 250 × 4700 × 103 / 106 = 1175.0 kN-m

φbM

p= 0.9 × 1175 = 1057.5 kN-m

Mr

= 0.7 × Fy× S

x/ 106

= 0.7 × 250 × 4160 × 103 / 106

= 728.0 kN-m

BF = = 21.94 kN

−−=−

−06.543.25

7281175

pr

rp

LL

MM


Mcx = φb[Mp − BF(Lb − Lp)]

= 0.9 [1175 − 21.94(6.0 − 5.06)]

= 1038.9 kN-m

Check Interaction Equation:

c

r

P

P

631,5

02.1668= = 0.296 > 0.2

+cx

rx

c

r

M

M

P

P

9

8= 0.296 +

9.1038

92.870

9

8

= 1.041 > 1.00 NG


Trial Section-2: W360 × 262

A = 33,400 mm2

rx = 163 mm, ry = 102 mm

rx/ry = 1.60

Ix = 89,100 × 104 mm4

(KyL

y)eq

= yx

xx

rr

LK

/

2

= 60.1

7.8= 5.44 m (not critical)


Cmx = 0.4 (as before)

x

xx

r

LK1

163

10000.6 ×= = 36.81 ( for nt case)

x

xx

r

LK 2

163

10007.8 ×= = 53.37 ( for lt case)

y

yy

r

LK1

102

10000.6 ×= = 58.82

R ≈ 59 < 200 OK

φcFcr = 187.09 MPa


Pc

= φcP

n= φ

cF

crA

g

= 1000

400,3309.187 ×= 6,248 kN

Pe1x

= π2 EI / (K1x

L)2

= 10006000

10100,89000,2002

42

××××π

= 48,854 kN for nt-case

Pe2x

= π2 EI / (K2x

L)2

= 10008700

10100,89000,2002

42

××××π

= 23,236 kN for lt-case


xer

mx

PP

C

,11 α−854,487.165611

4.0

×−

B1x

=

= = 0.41 ∴ B1x

= 1.0

ΣPnt = 1435 × 2 + 19.75 × 8.5

= 3038 kN

ΣPe2,x = 2 × 23,236 = 46,472 kN

B2x

=

xe

nt

P

P

,2

1

1

∑∑− α =

472,46

30380.11

1

×− = 1.07


Mrx = B1x Mntx + B2x Mltx

= 1.0 (51.20) + 1.07(759.00)

= 863.33 kN-m

Pr = Pnt + B2 Plt

= 1518.98 + 1.07(138)

= 1666.64 kN

From column selection table:

Lp = 5.08 m, Lr = 30.44 m


Check conditions of compact section:

f

f

t

b

2= 6.0 < λ

p= 10.8 OK

yb

u

P

P

φ ( ) 400,331000/2509.0

7.1656

××= = 0.220

For web,

λp

= 3.4233.28.31 ≥

−

yb

u

P

P

φ for A36 steel

= 67.1

wt

h= 13.7 < λ

pOK


Lb = 6.00m > Lp, bending strength is to be

calculated using the inelastic LTB formula.

Mp = 250 × 5240 × 103 / 106

= 1310.0 kN-m

φbMp = 0.9 × 1310 = 1179 kN-m

Mr = 0.7 × Fy × Sx / 106

= 0.7 × 250 × 4600 × 103 / 106

= 805.0 kN-m


BF =

−−=−

−08.544.30

8051310

pr

rp

LL

MM

= 19.91 kN

Mcx

= φb[M

p− BF(L

b− L

p)]

= 0.9 [1310 − 19.91(6.0 − 5.08)]

= 1162.5 kN-m

c

r

P

P

248,6

64.1666= = 0.267 > 0.2



+cx

rx

c

r

M

M

P

P

9

8

5.1162

33.863

9

8= 0.267 +

= 0.927 < 1.00 OK

Section Selected For Wind

Combination: W360 × 262

Check for Combination 1:

Pr = Pu = 1886 + 43.96 × 8.5/2

= 2073 kN

Mntx = 75.8 kN-m


Cmx

= 0.4 same as before

xer

mx

PP

C

,11 α− 854,48207311

4.0

×−B1x

= =

= 0.42 ∴ B1x

= 1.0

Mrx = B1x × Mntx = 75.8 kN-m

c

r

P

P

248,6

2073= = 0.332 > 0.2



+cx

rx

c

r

M

M

P

P

9

8

5.1162

8.75

9

8= 0.332 +

= 0.39 < 1.00 OK

Final Selection: W360 × 262

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