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EN2400 Civil Engineering For Concrete Frame and Steel Frame 

Would you do the design in the same way next time?
• What have you learnt from the design?
• What areas have you found particularly difficult and why?
• On a number of occasions we have identified areas where we have simplified the design –what differences would you expect if you were to follow the design codes without these simplifications?

Answer:

Concrete Frame

Ref

 

 

 

 

 

 

 

 

 

 

 

 

 

From the table

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Calculations

 

Effective depth =  = 334 mm

From LUSAS data

 

 

=  = 2.90 N/mm2

 

 

Cover

Fire expose = 35 mm

Therefore,

Assume diameter of main steel = 10 mm

d = h – cover  – diameter/2

  = 180 – 35 – 10/2

= 140 mm

Assume diameter of secondary steel = 10 mm

d = h – cover  – diameter - diameter/2

  = 180 – 35 – 10 - 10/2

= 130 mm

 

At support

Mass at the support = 170.5 kNm

K =

 =

= 0.96655 > 0.167

Z = d(0.5 + )  0.95d

Z = 0.95*140 = 133 mm

As =

   =

= 2947 mm2

 

Secondary steel

 

Moment = k*bd2*Fcu = 0.167 *300*1172 * 30 = 20.6 kNm

M2


 = m – mrequire = 170.5 – 20.6 = 149.9 kNm

K =

 =

= 0.9855

Z = d(0.5 + )  0.95d

Z = 0.95*130 = 123.5 mm

As =

   =

= 2790 mm2

Tension reinforcement

Required = 2790 mm2

Provide [email protected] 250mm center (As = 3220 mm2)

 

Compression reinforcement

Required = 2947 mm2

[email protected] 100 mmcenter (As = 3140 mm2

 

At span

 

M at span  = 105.2 kNm

K =

 =

= 0.5964 > 0.167

Z = d(0.5 + )  0.95d

Z = 0.95*140 = 133 mm

As =

   =

= 1818 mm2

Moment = k*bd2*Fcu = 0.167 *300*1172 * 30 = 20.6 kNm

M2 = m – mrequire = 105.2 – 20.6 = 84.6 kNm

K =

 =

= 0.5562

Z = d(0.5 + )  0.95d

Z = 0.95*130 = 123.5 mm

As =

   =

= 1575 mm2

 

Tension reinforcement

 

Required = 1818 mm2

Provide [email protected] 150 mm center (As = 2090 mm2)

 

Compression reinforcement

 

Required = 1575 mm2

Provide [email protected] mm center (As = 1610 mm2)

 

LINKS

 

Using Wuls at supports

Take Wuls = 55.9

V = Wuls(b/2 – d) = 55.9(6000/2 – 117) = 161.2 kN

Vmax = 0.18*bd*fcu(1-fcu/250)

= 0.18 *300 *117 *30*(1-30/250)

  = 167 kN

Vmax1 = 0.69* Vmax2

  = 115.23 kN

For reference

V = 0.12(1 +

 = 0.12(1 +)

61.4 kN

 

Links are required

 

Required

Asw/s =

Where cot

Asw/s =

= 1.1247

 

Minimum

Asw/s = 0.08 *b *

        = 0.08 *300 *

    = 0.263

1.1247 > 0.263

Maximum spacing = 0.75d = 0.75*117 = 87.75 mm

For the link H8 use 100 mm2

100/1.1247 = 88.91 mm

s = 75 mm

provide

Asw/s = 100/75 = 1.333 mm > 1.1247 it ok

Provide H8 links at 75 mm center to center

 

Output

 

From LUSAS model

B = 300 mm

Depth = 180 mm

 

 

 

 

 

 

 

 

Cover = 35 mm

Main steel diameter = 10mm

 

d = 140 mm

 

Secondary steel diameter = 10 mm

 

d = 130 mm

 

 

 

 

 

 

 

Z = 133 mm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z = 123.5 mm

 

 

 

 

 

Provide [email protected] mm

 

 

 

Provide  [email protected] mm

 

 

 

 

 

 

 

 

Z = 133 mm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Provide [email protected] 150 mm

 

 

 

 

 

Provide [email protected] 125 mm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Cot

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Provide H8 links at 75 mm center to center

 

 

Steel Frame

Ref

Calculations

 

Slab span = 3.5 m

Primary beam span = 6.0 m

Beam design

Assume the self weight = 0.8 kN/m

Permanent load = 0.87 + 15.3125 + 6.3 + 1.75 = 24.2325 kN/m

 

Variable load = 14 kN/m

 

W = 1.35*24.2325 + 1.5 *14 = 53.714 kN/m

Bending moments

Mcapacity =  = 465 kNm

Mdesign =  =  = 109.05 kNm

Bending moment capacity > bending moment design

Hence ok

 

Shear

 

Thickness = 10 mm

Assume the steel ISMB 450

Av = 450 *10 = 4500 mm2

Vcapacity =

=

= 680 kN

Vdesign  = wL/2 = 24.2325*6/2 = 72.7 kN

Vcapacity > Vdesign

 

Web crushing

 

Take kv = 5.35 for the transverse stiffeners

cr.e = (kv*

 = (5.35 *3.1422E)/(12(1-0.32)(450/10)2)

Take E = 200,000

= (5.35 *3.1422*200000)/(12(1-0.32)(450/10)2)

478 N/mm2

w =

=

= 0.576 < 0.8

Therefore

B = fy/ = 275/ = 158.77 N/mm2

VD = d*t*

   = 450 *10 *158.77*10-3/1.05 = 680.44 kN

Since VD >Fv ( 680 kN) it is safe

 

Deflection

 

 

Maximum deflection

 

=  = 47.69 mm

 

47.67 > 25 mm ok

 

 

 

Primary beam design roof

 

Permanent load =  beam self weight + slab self weight + roof finishes

      = 15.3125 + 1.35 + 6.3 = 22.9625 kN/m

Variable load of roof = 2.1 kNm

W = 1.35*22.9625 + 1.5 *2.1 = 34.15 kN/m

 

Bending moments

 

Mcapacity =  = 328 kNm

Mdesign =  =  = 154 kNm

Bending moment capacity > bending moment design

Hence ok

 

Shear

 

Thickness = 10 mm

Assume the steel ISMB 400

Av = 400 *10 = 4000 mm2

Vcapacity =

=

= 605 kN

Vdesign  = wL/2 = 34.15*6/2 = 102.5 kN

Vcapacity > Vdesign

 

Web buckling

 

Take kv = 5.35 for the transverse stiffeners

cr.e = (kv*

 = (5.35 *3.1422E)/(12(1-0.32)(450/10)2)

Take E = 200,000

= (5.35 *3.1422*200000)/(12(1-0.32)(400/10)2)

605 N/mm2

w =

=

= 0.5123 < 0.8

Therefore

B = fy/ = 275/ = 158.77 N/mm2

VD = d*t*

   = 400 *10 *158.77*10-3/1.05 = 605 kN

Since VD >Fv ( 680 kN) it is safe

 

Deflection

 

 

Maximum deflection

 

=  = 67.21 mm

67.21 mm > 25 mm ok

 

 

Output

 

 

 

 

 

 

 

 

 

 

 

 

 

465 kNm > 109.05 kNm

 

 

 

 

 

 

 

 

 

 

 

 

680 kN > 72.7 kN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

680.44 kN > 680 kN

 

 

 

 

 

 

 

 

47.67 mm > 25 mm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

328 kNm > 154 kNm

 

 

 

 

 

 

 

 

 

 

 

 

 

605 kN > 102.5 kN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

67.21 mm > 25 mm

 

 

 

 

 

 

 

Reflective report on the design of concrete frame and steel frame 

The design of steel frame will first involve considering the types steel frame where the steel frame construction involves use of welded steel plate which together will form an I section, secondly, consideration of the stresses and loads that will acting on the steel frame by checking the source of the application if it is through stanchions, floor slab and floor beams, which will be carried by the girder, it should be noted that the flange of the plate will carry the bending moments while the web will have a resistance to the shear force. And in order 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. Thirdly, the girder depth will always be about a tenth of the span for average loading, whereas the lightly loaded girders the depth will be between 1/15 to 1/20 and also the flange depth is always about 1/3 of the depth. Fourthly, to determine the plate girders deflections Clause 15 of BS449 will be used, at the same point the permissible stresses are determined and counterchecked when the BS449 are used, and this will require a lower stress to be used when the thickness of the steel frame will exceed 40 mm a lower stress must be used, and finally the bending stresses will be determined as per the directive of Clause 17, 27 and 32 of BS 449 sets section area for the girders, similarly the shear stress of the web will be determined by the formula fq =

It should be noted that the allowable shear stress will depend on the value of d/t and stiffeners spacing, where the ratio of d/t will exceed 85 vertical stiffeners will be required at a distance which will not exceed 1.5d, and always the thickness should be more than 1/180. 

I have spent time to think on the procedure I used in designing the concrete frame and steel frame, and found out that it is the best and easier method that I will use more and more in future, through this method of design I have learned while dealing with a  continuous concrete beam as per task involved are usually not statistically determinate which means that more complex design techniques must be implemented in order to obtain the shear force and bending moment on it respective members, where an application of an easier method that will facilitate the calculation bending moment and shear force at the support of the continuous member will be through moment distribution techniques, where I  should be able to determine fix end moment , determine the stiffness factor and finally I will evaluate the distribution factors for each and every member that will be meeting at a joint point, further moment at the far end I will determine for each member, the procedure I will repeat it until the balance moment are negligible and finally the moment at each span is determined by summation  the determined moments at the joints.

Secondly, I have learned that  in order to determine the ultimate loads  I will first determine all the dead load through involvement in summing up all the loads which will be acting on the concrete though they are not alive, then I will also determine  imposed load by summing up all live loads acting on the concrete, at the same point the self-weight of the concrete I will also determine it, therefore  the ultimate load that I determined will  be equivalent to the sum of dead load and self-weight of the concrete which I added it up and multiplied with dead load factor and procedure i also did it  on the imposed load.

After I determined of the ultimate load, I also learned how to determine the design moment through multiplication of the ultimate load by length and dividing up with value number eight, for uniformly distributed load.

I also learned how to determine  effective depth by considering the cover, the diameter of the main link and diameter of the secondary link, the calculated effective depth is to assist in determining the ultimate moment using the formula Mu = 0.156fcubd2, which involves multiply a constant value of 0.156 by compressive strength of the concrete by the breadth and square of the effective depth determined, and if the ultimate moment will be greater than design moment the beam will singly be reinforced anf if not then it will be doubly reinforced.

I also learned that the concrete beam shear reinforcement must be checked in order to examine if it suitable, the deflection too will be check in order to avoid formations of cracks and breaking up of beams when they are subjected to the load either above or below them and finally the reinforced detailing is made.

I found out that the hardest part is when am designing the steel frame where am supposed to first  consider the types steel frame where the steel frame construction involves use of welded steel plate which together will form an I section, secondly,  am supposed to consider  the stresses and loads that will acting on the steel frame by checking the source of the application if it is through stanchions, floor slab and floor beams, which will be carried by the girder, where am suppose know that  the flange of the plate will carry the bending moments while the web will have a resistance to the shear force. And in order to avoid plate from failing and working effectively the rules from BS449 i will use it to govern the width 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. Thirdly, the hurdle part am supposed to remember that the girder depth will always be about a tenth of the span for average loading, whereas the lightly loaded girders the depth will be between 1/15 to 1/20 and also the flange depth is always about 1/3 of the depth. Fourthly, am supposed to know that  the permissible stresses are determined and counterchecked when the BS449 are used, and this will require a lower stress to be used when the thickness of the steel frame will exceed 40 mm a lower stress must be used, and finally the bending stresses am supposed to determine as per the directive of Clause 17, 27 and 32 of BS 449 sets section area for the girders, similarly the shear stress of the web will be determined by the formula fq =

The method of design I used is the accurate method since am involved in checking several characteristic of both the concrete and steel frame, ranging from deflection, web bucking, shear strength and bending moment, fro this range of characteristic there will be no error that may be achieved.   

Design, Construction, Operation 

Concrete

Activity and hazards

L *

S

= RR

Noise from vibration

5

5

25

Vehicles collisions  

2

3

6

Use of vehicles

4

1

4

Use of cement

5

5

25

Manual handling of equipment’s

3

3

9

 

Reduce  hazards

L *

S

= RR

Use of ear defenders

5

5

25

Reversing to be avoided wherever possible, if unavoidable suitable reversing aids to be fitted and the use of a trained banks-man considered

2

3

6

· Vehicles kept in good condition by adequate maintenance

· Vehicles fitted with warning lamps and all round visibility

4

1

4

· Use correct cements and do not expose skin to cement products .

· Suitable washing facilities to be available, wash cement off the skin immediately

· Inform all workers of the cement hazards at induction

5

5

25

Avoid manual handling and use mechanical aids wherever possible

3

3

9

Steel frame

Activity and hazards

L *

S

= RR

Flying sparks while welding of steel frame

4

4

16

Crane collapse   

4

3

12

Electrical fire

4

3

12

Manual handling  

3

5

15

Working at height

4

3

12

 

Reduce hazards

L *

S

= RR

Proper eye protection will be used

4

4

16

Crane to be operated by licensed person

4

3

12

Ensure all installation of electric are done properly and use of electrical guards should be used

4

3

12

Avoid manual handling and use mechanical aids wherever possible

5

5

25

Qualified persons must handle the steel work under supervision

2

4

8

This problem has been solved.


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