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CIE6001 Advanced Structural Analysis and Design Assignment

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Questions:

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

Answer:

Sway mechanism

H(h1) = Mpc + Mpc

At B      At B          At C

H =  

    h1

B C D  

A

h1

Q1b)

Beam mechanism

VL1  = Mpb + Mpb + Mpb

             At B         At C                       At D

     VL1       = 2Mpb + Mpb + Mpc

V =   

L

Q1c)

  1. Combined sway

W1 = Hh1 + VL1

        = 2Mpb + Mpb + Mpc + Mpc + Mpc - Mpc - Mpb

H(h1) + VL1   = Mpb + Mpb + Mpc  + Mpc

Substituting

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

5H + 2V   =

  1. Interaction diagram

sway diagram

    h1

B C D  

A

h1  

L1  

  • Equilibrium

         D         Mpc

E HE

 about D = 0

3HE – Mpc = 0

HE =

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

2VA + 5HA – Mc = 0

Mc = 200 KN

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

H = 256 KN

V = 300 KN

Mpc = 120 KN

Mpb = 240 KN

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

 About C = 0

2*300 + 5*256 – Mc 0

Mc = 1880 KNM

 About C = 0

3HE – 120 = 0

HE = 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

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

RA = 735

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

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

MB = 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 Pb = 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

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

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

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

IXX =  (20 * 3003)/12 = 4.5 * 107 mm4

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

rXX =  = 70.71 mm

l/rXX = (0.7 *800)/75 = 7.9

Pc = 148.2 N/mm2

Fc = (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

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

rXX =  = 89.12 mm

l/rXX = (0.7 *800)/89.12 = 6.3

Pc = 151.6 N/mm2

Fc = (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

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

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

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

MK = 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 Pb = 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

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

fbc = (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

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

IXX =  (20 * 3303)/12 = 6.0 * 107 mm4

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

rXX =  = 67.94 mm

l/rXX = (0.7 *2400)/67.94 = 24.7

Pc = 142.4 N/mm2

fc = (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

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

rXX =  = 132.3 mm

l/rXX = (0.7 *2400)/132.3 = 12.7

Pc = 147.2 N/mm2

Fc = (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

    Properties for plate girders that includes moment of inertia, area, modulus of section and radii of gyration are calculated from the first principle.

Maximum outstand for flange plate are always provided in BS 449, like for example

Compression flange = 16t

Tension flange = 20t

t is the thickness

shear stress of the web

will be determined by the formula

fq =

Allowable shear stress will depenf on the value of d/t and stiffeners spacing

 If the ratio of d/t exceed 85 vertical stiffeners will be required at a distance which will not exceed 1.5d

The thickness should be more than 1/180

This problem has been solved.


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