Design of Welded Girder and Sway Mechanism

1. Apply analytical, quantitative, and computer methods relevant to the engineering discipline, in order to solve engineering problems

2. Application of plastic analysis methods for the analysis and design of statically indeterminate fames

3. Understand appropriate codes of practice and industry standards

Load Calculation

Sway mechanism

H(h₁) = M_pc+ M_pc

At B At B At C

H =

h₁

B C D

h₁

Q1b)

Beam mechanism

VL₁ = M_pb + M_pb + M_pb

At B At C At D

VL₁= 2M_pb + M_pb + M_pc

V =

L₁

Q1c)

Combined sway

W₁ = Hh₁ + VL₁

= 2M_pb + M_pb + M_pc + M_pc+ M_pc - M_pc- M_pb

H(h₁) + VL₁= M_pb + M_pb + M_pc + M_pc

Substituting

5*H* + V*2= 200* + 200* + 100* + 100*

5H + 2V=

Interaction diagram

sway diagram

h₁

B C D

h₁

L₁

Equilibrium

D M_pc

E H_E

about D = 0

3H_E – M_pc = 0

H_E =

_x = 0

about E = 0

-2V + 2H + 3H = 0

5 * = 2V

V = KN

Thus the moment at E, from a free body diagram of ABC

About C = 0

2V_A + 5H_A – M_c = 0

M_c = 200 KN

Since there is a plastic hinged at C of value M_pC = 200 KNM, we have the equilibrium

H = 256 KN

V = 300 KN

M_pc = 120 KN

M_pb = 240 KN

L1 = 2, L2 = 3, h2 = 3, H1= 5 M

About C = 0

2*300 + 5*256 – M_c 0

M_c = 1880 KNM

About C = 0

3H_E – 120 = 0

H_E = 40 KN

Moment at B

256 *2.5 = 640 KNm

Moment at D

1880 – 300 *3 = 980 KNm

BMD

640 KNm 120 KNm

B C D

640 KNm

A 1880 KNm

Part B

Span of the welded girder = 9 m

It carries a load of approximately 600 KN each

The loads are at a distance of 3 m from each end of the girder

The girder carries uniform distributed load of 30 KN/m, which include the self-weight of girder

Designing the girder

600 KN 30 kn/m 600 KN

A C E D B

3 m 3 m 3 m

_B = 0

R_A * 9 = 600*6 + 600*3 + 30*9 * 9/2 = 6615 KN

R_A = 735

R_B = 600 + 600 + 30 * 9 – 735 = 735 KN

M_c = 735 * 3 – 30 * 3 * 1.5 = 2160 KNm

M_B = 735 * 4.5 – 600 * 1.5 – 30 *4.5* 4.5/2 = 2103.75 KNm

The overall depth of girder = span/10 = 900 mm

Assume that the girder has 30 mm thick therefore the allowance P_b = 140 N/mm²

If we assume that the flange plate resist the bending moment then the moment will be resisted by lever arm of about 860 mm

Flange area = (2103.75 *10⁶)/(860 *140) = 1.75 * 10⁴ mm²

If we assume the flange that was used is of dimensions 350 mm by 50 mm

Assume the allowance shear stress is 100 N/mm²

Depth of web plate is 450 mm and shears 735 KN at girder support

Thickness of web plate = (735 *10³)/(800 *100) = 9.19 mm

If a thickness of 20 mm was used

Checking the bending stress

I_XX = (2 * 50 * 350 * 425²) + (20 *800³)/12 = 7.18 *10⁹ mm⁴

f_bc = (2103.75 *10⁶ *735)/ (7.18 * 10⁹) = 215.36 N/mm² (it is safe)

Stress Calculation

load bearing stress are required at the support under the point loads

The spacing should not exceed

1.5d = 1200mm

180t = 180 *20 = 3600

Stiffeners under 500 KN

Assuming the stiffeners each 150 mm by 20 mm

Bearing stress = (600 * 10³)/(2*135 * 20) = 111.11 (safe )

The area of centerline of web

I_XX = (20 * 300³)/12 = 4.5 * 10⁷ mm⁴

A = (300 * 10) + (150 * 20*2) = 9, 000 mm²

r_XX = = 70.71 mm

l/r_XX = (0.7 *800)/75 = 7.9

P_c = 148.2 N/mm²

F_c = (735 *10³)/(9000) = 81.67 N/mm²

A smaller size stiffener was used

End plate

Assume the end plate of 20 mm thick was used

Checking the bearing

Bearing stress = (735 * 10³)/(350 *20) = 105 (safe )

Checking the section acting at the strut

A = 350 * 20 + (200 *10) = 9000mm²

I_XX = (20 * 350³)/12 = 7.14 * 10⁷ mm⁴

r_XX = = 89.12 mm

l/r_XX = (0.7 *800)/89.12 = 6.3

P_c = 151.6 N/mm²

F_c = (735 *10³)/(9000) = 81.67N/mm²

Assume the weld stretch = 300 N/mm

Approximately the welded length = (735 *10³)/300 = 2450 mm

Part B (ii)

Minimum yield strength at nominal thickness 16 mm

For steel S275 = 275 N/mm²

Tensile strength between 3 mm and 16 mm = 370 to 530 Mpa

Dead load

s.w = 20 KN

point load w1d = 200 KN, w2d = 200 Kn

imposed load

udl = 40 KN/m

point load, w1d = 300 KN, w2d = 300 KN

720 KN 40 kn/m 720 KN

B E K G J

9 m 7 m 9 m

Total weight = 1.2 D.L + 1.6 L.L

Load w1d = 1.2*200 + 1.6*300 = 720 KN

Load w2d = 1.2 * 200 + 1.6* 300 = 720 KN

= 0

R_B * 25 = 720*16 + 720*9 + 40*25 * 25/2 = 1220 KN

R_J = 720 + 720 + 40 * 25 – 1220 = 1220 KN

M_E = 1220 * 9 – 40 * 9 * 4.5 = 9360 KNm

M_K = 1220 * 12.5 – 720 * 3.5 – 40 *12.5 * 12.5/2 = 9605 KNm

Girder section

The overall depth of girder = span/10 = 25000/10 = 2500 mm

Take the cover to be 40 mm thick therefore the allowance stress bending P_b = 275N/mm²

If we assume that the flange plate resist the bending moment then the moment will be resisted by lever arm of about 2460 mm

Flange area = (9605 *10⁶)/(2460 *275) = 1.42 * 10⁴ mm²

If we assume the flange that was used is of dimensions 300 mm by 50 mm

Depth of web plate is 2400 mm and shears 1220 KN at girder support

Thickness of web plate = (1220 *10³)/(2400 *100) = 5.08mm

If a thickness of 10 mm was used

Checking the bending stress

I_XX = (2 * 50 * 300 * 1225²) + (10 *2400³)/12 = 1.156 *10¹⁰ mm⁴

f_bc = (9605 *10⁶ *1200)/ (1.156 * 10¹⁰) = 997 N/mm² (it is safe)

for the web the ratio d/t = 2400/10 = 240 the for the intermediate stiffener must be provided

Design and Construction

Load bearing stress is required at the support under the point loads

The spacing should not exceed

1.5d = 3600 mm

180t = 180 *10 = 1800

From the table of allowable shear stress assume 100 N/mm²

Stiffeners under 720 KN

Try two stiffeners each 200 mm by 20 mm

Bearing stress = (720 * 10³)/(2*175 * 20) = 103 N/mm² (safe )

The area of centerline of web

I_XX = (20 * 330³)/12 = 6.0 * 10⁷ mm⁴

A = (500 * 10) + (200 * 20*2) = 13, 000 mm²

r_XX = = 67.94 mm

l/r_XX = (0.7 *2400)/67.94 = 24.7

P_c = 142.4 N/mm²

f_c = (720 *10³)/(13000) = 55.38 N/mm²

it is advisable to use smaller size stiffener

take weld strength approximately to 450 N/mm

length = (720*10³)/450 = 1600 mm

End plate

Assume the end plate of 20 mm thick was used

Checking the bearing

Bearing stress = (1220 * 10³)/(500 *20) = 22 N/mm²

Maximum stiffener = 11t = 11 * 20 = 220 mm (safe)

Checking the section acting at the strut

A = 500 * 20 + (1900 *10) = 11.9 * 10³ mm²

I_XX = (20 * 500³)/12 = 20/83 * 10⁷ mm⁴

r_XX = = 132.3 mm

l/r_XX = (0.7 *2400)/132.3 = 12.7

P_c = 147.2 N/mm²

F_c = (1220*10³)/ (11900) = 102.5 N/mm²

Try the weld strength = 500 N/mm

Approximately the welded length = (1220 *10³)/500 = 2440mm

Intermediate stiffeners

Use stiffeners of dimensions of each 100 mm by 10 mm

Maximum value = 10t = 10 *20 = 200

Moment of inertia = I = (10 *210³)/12 = 7.73 * 10⁵ mm⁴

Distance between stiffeners = 1800 mm

Required thickness of web = 1300/ 180 = 7.22 mm

For shear strength, t = 5.08

I = (1.5 *1200³ *7.35³)/(10⁴ *1800²) = 3.2 *10⁶ mm

Web to flange web

Horizontal shear per weld

= (1220 *10³ *300 *50 *625)/(1.156 *10¹⁰ *2) = 495 N/mm ( hence ok)

Design consideration

Considering the types plate girder

Plate girder construction involves use of welded steel plate which together will form an I section

stresses and loads

application of loads to the plate girder are through stanchions, floor slab and floor beams, which will be carried by the girder

the flange of the plate will carry the bending moments while the web will have a resistance to the shear force.

To avoid plate from failing and working effectively the rules from BS449 will be used to govern the with or thickness ratios of the plates to ensure no buckling takes place, the rules also will govern the positions of both the intermediate and stiffeners.

plate girder depth

the depth should always be about 1/10 of the span for average loading.

The lightly loaded girders the depth will be between 1/15 to 1/20

The flange depth is always about 1/3 of the depth

Web buckling are prevented by provision of stiffeners

plate girders deflections

Clause 15 of BS449 will be used determining the deflection requirement

permissible stresses

if the plate thickness will exceed 40 mm a lower stress must be used

the permissible bending stress to be used are provided in BS 449

bending stresses

Clause 17, 27 and 32 of BS 449 sets section area for the girders