beam exampleproblems

Section Moment Capacity: Example Problem No.1

What is the section moment capacity of 530UB92.4 Grade 300 beams about the major

principal axis (x)?

tw = 10.2 mm fyw = 320 MPa

tf = 15.6 mm fyf = 300 MPa

Section capacity (For the failure mode of “local buckling and yielding”)

Flange e = (bf - tw)/2tf)250

f y = 6.37

250

300 = 6.98 < ep of 9. Compact flange

Web e = (d1/tw)250

f y = 49.2

250

320= 55.7 < ep of 82 Compact web

Section is therefore compact.

Zex = Lesser of Sx or 1.5Zx = Lesser of 2370 x 103 or 1.5 x 2080 x 10

3 = 2370 x 10

3 mm

3

Msx = fy Zex = 0.9 x 300 x 2370 x 103 = 640 kNm

In this case of compact flange and web (compact section), the capacity is not affected by local

buckling, and is equal to the full plastic moment capacity Mp.

Note: Since the web and flange thicknesses are different (15.6 and 10.2 mm), their yield

stresses are not the same (320 and 300 MPa) despite being the same steel grade of 300. In the

plate slenderness e calculations, a nominal yield stress value of 300 could have been used to

simplify the calculations, however, actual values of 300 and 320 have been used here. The use

of actual values makes very little difference to the final answers. For the section capacity

calculations, the minimum yield stress value of 300 MPa has been used (conservative).

Example Problem No.2

What is the section moment capacity of 200UC52.2 Grade 300 beams about the minor

principal axis (y)?

tw = 8 mm fyw = 320 MPa

tf = 12.5 mm fyf = 300 MPa



f y = 7.84

250

300= 8.59 < ep of 9. Compact flange

No need to check the web as it is on the neutral axis.

Section is therefore compact.

Zey = Lesser of Sy or 1.5Zy = Lesser of 264 x 103 or 1.5 x 174 x 10

3 = 261 x 10

3 mm

3

Msy = fy Zey = 0.9 x 300 x 261 x 103 = 70.5 kNm

In this case of compact section, the capacity is not affected by local buckling, and is equal to

the full plastic moment capacity Mp. Note that Sy > 1.5Zy, so 1.5Zy is used.


Figure 4.4 shows a lightly welded box-girder made of Grade 350 steel. What is its section

moment capacity about the major principal axis (x)?


tf = 12 mm fyf = 360 MPa

Figure 4.4 Welded Box-girder in Example Problem No.3


Flange e = (bf/tf)250

f y = (360/12)

250

360 = 36 > ep of 30, but < ey of 40.

Non-compact flange

Web e = (d1/tw)250

f y = [(400-2x12)/10]

250

360 = 45 < ep of 82 Compact web

Section is therefore non-compact.

Effective section modulus Ze will be between Z and Zc (Lesser of S or 1.5 Z )

Ze = Z +( Zc - Z )

ey e

ey ep

Elastic section modulus

Z = [2x10x(400-2x12)3/12 + 2x400x12(400/2-12/2)

2]/200 = 2.250x10

6 mm

3

Plastic section modulus

S = 2x10x(400-2x12)2/4 + 2x400x12(400/2-12/2) = 2.569x10

6 mm

3

Zc = Lesser of S or 1.5 Z = 2.569x106 mm

3

Zex = 2.250x106

+ (2.569-2.250)x106x

)3040(

)3640(

= 2.377x10

6 mm

3

(Note that 2.250x106 corresponds to ey of 40 whereas 2.569x10

6 corresponds to the ep of 30)

Design section moment capacity

Msx = fy Zex = 0.9 x 360 x 2.377 x 106 = 770 kNm

In this case of compact web and non-compact flange (non-compact section), the capacity is

affected by inelastic local buckling of the compression flange, and therefore will not reach

the full plastic moment capacity Mp.

360 mm 400 mm


Figure 4.5 shows a lightly welded I-section beam made of Grade 350 steel. What is its section


tw = tf =10 mm fy = 360 MPa

Figure 4.5 Welded I-section in Example Problem No.4



f y = [(400-10)/2x10]

250

360 = 23.4 > ey of 15. Slender flange

Web e = (d1/tw)250

f y = [(400-2x10)/10]

250

360= 45.6 < ep of 82 Compact web

Section is therefore slender.

Elastic section modulus

Z = [10x(400-2x10)3/12 + 2x400x10(400/2-10/2)

2]/200 = 1.750x10

6 mm

3

Effective section modulus Ze = Z

ey

e

for flat plate element in uniform compression.

Zex = 1.750x106 x 15/23.4 = 1.122x10

6 mm

3


Msx = fy Zex = 0.9 x 360 x 1.122 x 106 = 363.5 kNm

In the above method, the section modulus component of the web also gets reduced despite

being compact. Therefore a more accurate Zex can be obtained as shown next if the reduction

is only applied to the section modulus component of the slender flange.

Zex = [10x(400-2x10)3/12 + 2x400x10(400/2-10/2)

2x(15/23.4)]/200 = 1.204x10

6 mm

3

Msx = fy Zex = 0.9 x 360 x 1.204 x 106 = 390.1 kNm

This method gives about 7% more in bending capacity.

In this case of compact web and slender flange (slender section), the capacity is affected by

elastic local buckling of flanges, and therefore will not even reach the first yield moment

capacity My.

400 mm

400 mm


Figure 4.6 shows a lightly welded plate girder made of Grade 350 steel. What is its section




Figure 4.6 Welded Plate Girder in Example Problem No.5



f y = [(300-10)/2x25]

250

340 = 6.76 < ep of 8. Compact flange

Web e = (d1/tw)250

f y = [(1500-2x 25)/10]

250

360 = 174 > ey of 115 Slender web.

Section is therefore slender, governed by web element, ie. Section slenderness s = 174

Ignoring the contribution of web to the section modulus

Zex = [2x300x25x(1500/2-25/2)2]/750 = 10.878x10

6 mm

3

Msx = fy Zex = 0.9 x 340 x 10.878x106 = 3329 kNm

The AS 4100 design rules do not address this case of slender web with both edges supported.

Instead it gives the following rule for flat plate element with maximum compression at

unsupported edge and tension at the other edge (including zero tension).

Effective section modulus Ze = Z (

ey

e

)2

Applying this to the section modulus of the web only,

Zex = [{10x(1500-2x25)3/12} x (115/174)

2 + 2x300x25(1500/2-25/2)

2]/750 =12.358x10

6 mm

3

WEB componet FLANGE component


Msx = fy Zex = 0.9 x 340 x 12.358 x 106 = 3780 kNm

This method gives about 13.5% more in capacity.

In this case of compact flange and slender web (slender section), the capacity is affected by

elastic local buckling of webs, and therefore will not even reach the first yield moment

capacity My.

300 mm

1500 mm


Figure 4.7 shows a welded plate girder made of Grade 350 steel. What is its section moment

capacity about the major principal axis (x)?



Figure 4.7 Welded Plate Girder in Example Problem No.6



f y = [(450-10)/2x15]

250

350 = 17.4 > ey of 15.

Slender flange

Web e = (d1/tw)250

f y = [(1500-2x15)/10]

250

360= 176.4 > ey of 115

Slender web.

Section is therefore slender and its slenderness will be that of web as (176.4/115) >

(17.4/15). Section slenderness s = 176.4

Ignoring the contribution of web to the section modulus and allowing for the slender flange in

uniform compression, ie. Ze = Z

ey

e

for flat plate element in uniform compression;

Zex = {[2x450x15x(1500/2-15/2)2]/750} x (15/17.4) = 8.555x10

6 mm

3

Msx = fy Zex = 0.9 x 350 x 8.555x106 = 2695 kNm

The AS 4100 design rules do not address this case of slender web with both edges supported.

Instead it gives the following rule for flat plate element with maximum compression at

unsupported edge and tension at the other edge (including zero tension).


ey

e

)2

Applying this to the section modulus of the web only,

Zex = [{10x(1500-2x15)3/12}x(115/176.4)

2+{2x450x15x(1500/2-15/2)

2}x(15/17.4)]/750

= 10.055x106 mm

3 WEB componet FLANGE component

450 mm

1500 mm


Msx = fy Zex = 0.9 x 350 x 10.055 x 106 = 3519 kNm

This method gives about 30.5% more in capacity.

In this case of slender web and flange (slender section), the capacity is affected by elastic

local buckling of both web and flange, and therefore will be much less than the first yield

moment capacity My.


How do you calculate the section moment capacity of an I-section beam about the minor

principal axis (y) if its flanges are slender?.

The AS 4100 design rules address this case of slender flanges with maximum compression at

unsupported edge and zero tension at the other edge.


ey

e

)2

Zey = Zflange (

ey

e

)2 where ey = 25 for hot-rolled sections.

The section modulus component of web, Zweb, is negligible.


What is the section moment capacity of a Grade 350 1500 x 10 CHS?

t = 10 mm fyf = 360 MPa do = 1500 mm


s = t

d o 250

f y = (1500/10) x (360/250) = 216 > ey of 120.

Slender section

Ze = Lesser of Z(sy/s) and Z(2sy/s)2 = Lesser of Z x (120/216) or Z x (2x120/216)

2

= Lesser of 0.745 Z or 1.235 Z = 0.745 Z

Ze = 0.745 [{(15004-1480

4)/64}/750] = 12.903x10

6 mm

3

Ms = fy Ze = 0.9 x 360 x 12.903x106 = 4181 kNm

Member Capacity: Example Problem No.1

Figure 4.14 shows the roof structure of a service station building. Determine the size of a

suitable UB section for the main beams B if the wind uplift pressure on the roof is 2.5 kPa.

Beam

B

2m

Column A

Elevation

1.5 m

Plan View

Figure 4.14. Service Station Building for Example Problem 1

Using the roofing and purlin manufacturer’s charts,

The required roof sheeting: 0.42 mm Monoclad with an internal span of 1700 mm and

end spans of 1350 mm.

the required C-section purlin to be used for the given wind uplift pressure is C300-30.

11 m 2 m

2 m

11 m 2 m

Column A Purlins

Main Beams B

1.5 m

8 m

Metal sheeting Purlins

Load Evaluation for Beam Design

For each main beam, the uniformly distributed design load due to

wind uplift load = 2.5 x 5.5 x 15/15 = 13.75 kN/m

self-weight of sheeting = 4.28 kg/m2 x (9.81/1000) x 5.5 = 0.231 kN/m

self-weight of 10 purlins = (12.72 kg/m x 5.5 x 10 / 15) x (9.81/1000) = 0.458 kN/m

Assume beam self-weight to be 80 kg/m = 0.78 kN/m

Total self-weight of roof allowing for connections = (0.231 + 0.458 + 0.78) 1.1 = 1.616 kN/m

w kN/m

Figure 4.15 Loading diagram for Main Beams B

Maximum bending moment Mmax = 7.5 w x 5.5 – 7.5 w x 3.75 = 13.125 w kNm

Wind Uplift Case (dead load plus wind load combination), Design load w = 0.9 x (1.616) – 13.75 = - 12.30 kN/m (net wind uplift)

Design moment M* = 13.125 x 12.30 = 161.4 kNm

Effective length factors:

Segment length is 11 m and its ends can be assumed to be fully restrained (F).

Since purlins are located on the tension flange, they are considered not to provide any lateral

restraint to the main beam, hence the Full span segment has to be considered (ie. 11 m)

kt = 1; load applied on the sheeting, ie. on the tension flange, kl = 1; kr = 1

Effective length le = 1 x 1 x 1 x 11 m = 11 m

Try 610UB101, Zex = 2900 x 103

Msx = 0.9 x 300 x 2900 x 103 = 783 kNm (Msx = 870 kNm)

For le = 11 m, the reference buckling moment Mo = 223 kNm from the table given at the end

of this problem. Alternatively, it can be calculated using appropriate section properties of

610UB101 and an effective length of 11 m in the buckling formula given earlier in this

section.

s = 0.6 [ )223

870(3)

223

870( 2 ] = 0.220

Moment distribution factor m = 1.13 for a bending moment distribution due to a uniformly

distributed load on a simply supported beam (ignoring the effect due to overhangs). Mbx = m s Msx = 1.13 x 0.220 x 783 = 194.7 kNm > M* of 161.4 kNm

Hence the choice of 610UB101 is satisfactory.

Check the self-weight of beam 610UB101 = 101 kg/m = 0.99 kN/m

Revised self-weight of roof = (0.231 + 0.458 + 0.99) x 1.1 = 1.85 kN/m

Revised design moment = 13.125 x [0.9 x 1.85 – 13.75] = 158.6 kNm, OK.

11m 2 m 2 m

Now check for Gravity Load Case (Dead load plus live load combination)

For each main beam, the uniformly distributed design load due to

Live load Q = 0.25 kPa x 5.5 = 1.375 kN/m

Dead load G = 1.85 kN/m

Design load w = 1.2 x 1.85 + 1.5 x 1.375 = 4.28 kN/m

Design moment M* = 13.125 x 4.28 = 56.2 kNm

Effective length factors:

In this case purlins are fixed to the compression flange of the beams and hence will improve

its buckling strength. They will also divide the main span into many small segments.

Segment length l is equal to purlin spacing = 1.7 m and its ends can be assumed to be fully

restrained (F) – see AISC publication.

kt = 1;

load is applied on the sheeting, ie. on the compression flange, so usually kl could have been

1.4, but the load is transferred via the purlin at the segment end, so kl = 1.0

kr = 1

Effective length le = 1 x 1 x 1 x 1.7 = 1.7 m

For le = 1.7 m, the reference buckling moment Mo = 6338 kNm

s = 0.6 [ )6338

870(3)

6338

870( 2 ] = 0.960

Moment distribution factor m can be taken as 1 for a short segment of length 1.7 m.

Mbx = m s Msx = 1 x 0.960 x 783 = 752 kNm > M* of 56.2 kNm

Hence the choice of 610UB101 is satisfactory, but over-design; the section is too big.

This is not an economical design; reduce effective length for wind uplift case by providing

fly braces at every purlin (spaced at 1.7 m).

Effective length le = 1 x 1 x 1 x 1.7 m = 1.7 m

Moment distribution factor m can be taken as 1 for a short segment of length 1.7 m.

Using ASI design capacity tables, required section is 310UB46.2.

Its design capacity is m x 176 = 176 kNm > M* of 158.6 kNm OK

For gravity loads, effective length le = 1 x 1.0 x 1 x 1.7 m = 1.7 m

As before, m = 1, Mbx = 176 kNm for the chosen 310UB46.2. This is > M* of 56.2 kNm.

OK

Use 310UB46.2

Example Problem No. 2

Design the beam ABCD shown in Figure 2. The loads shown include load factors. The

connection and support conditions at A, B and C are as per the AISC publication (Types 1, 11

and 8, respectively).

Type 11

A 3 m B 3 m C 3 m D

Type 1 Type 8

Figure 4.16. Beam ABCD for Example Problem 2

Reaction at A: 6R = 400 x 3 – 100 x 3 R = 150 kN

Design moments M* at B = 450 kNm, at C = 300 kNm

Assuming an m value of 1.5, M* for AISC design capacity tables = 450/1.5 = 300 kNm

Effective length = 3 m, Required beam = 610UB101

Ms of 610UB101 = 782 kNm

Segment AB BC CD

End Restraints PF FF FU

Length l (m) 3 3 3

kt 1+ 3)

6.102

8.14(

3000

572

x

= 1.065

1 1

kl 1 1 2

kr 1 1 1

le (m) 3 x 1.065 = 3.2 3 x 1 = 3 3 x 2 = 6

Mo (kNm) 1792 1993 570

s 0.788 0.810 0.470

m 1.75 m=300/450 = 0.67,

so m = 2.5

1.25

Mb

= m s Ms

1079>Ms of 782

so 782

1584>Ms of 782

so 782

459

M* (kNm) 450 450 300

Conclusion OK, but overdesign OK, but overdesign OK, but overdesign

400 kN

100 kN

Bending moment

beam exampleproblems

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