Week 3 Challenge

CHALLENGE 3

Aim: The main aim of this challenge is to gain the knowledge about the concept of design of a compression members and the various parameters that are to be taken into consideration while designing of a compression members. And to determine the sizes that are required for the compression member to taken up the load is also learnt.

Introduction: A structural member loaded axially in compression is generally called a compression member. Vertical compression members in buildings are called columns, posts or stanchions. A compression member in roof trusses is called struts and in a crane is called a boom. Columns which are short are subjected to crushing and behave like members under pure compression. Columns which are long tend to buckle out of the plane of the load axis.

A member which carries an axial compression is known as a compression member. It is also called a strut. A compression member may be called – short or long depending on the slenderness ratio (KL/r), where KL is the effective length and r is the least radius of gyration of the cross section. For small values of KL/r, a compression member undergoes only simple compression. Such a compression member is known as a short column. For large values of KL/r, a compression member undergoes buckling i.e., side way deflection. Such a compression member is known as a long column.

Classification of columns based on their lengths and their behaviour:

• Short columns

Short columns are very short a compression members. Slenderness ratio of such columns is very low. The failure of such columns occurs by yielding and hence stresses at failure are yield stresses. No buckling is observed in such columns.

• Slender or long compression members

The strength of any compression member decreases with the increase in its length. Compression members with high slenderness ratio and which satisfy all the conditions of Euler’s Formula for buckling are called slender or long compression members. These columns will fail by elastic buckling. The stresses induced during failure (buckling) are well below the yield limit and lie in the elastic zone. So, the failure occurs elastically.

• Intermediate columns

A column under an axial load has some fibres yielded some fibres in the elastic limit are known as intermediate columns. These compression members would fail both by yielding and buckling. The failure would fall under the ‘inelastic’ category. Hence, Euler’s formula is not applicable for such columns.

Possible failure modes of a compression member (axially loaded):

• Local buckling

Failure occurs by buckling or deflection of one or more parts of the member, for example: flange or web of an I-section. No overall deflection is observed in this kind of buckling.

• Squashing

Squashing occurs in relatively small length columns. It occurs by yielding of a cross section of the column.

• Overall flexural buckling

In this mode, failure of the member occurs by excessive deflection in the plane of weaker principal axis.

• Torsional and flexural torsional buckling

Torsional buckling failure occurs by twisting of the column about shear centre in the longitudinal axis. A combination of flexural and torsional buckling is called flexural- torsional buckling.

Lacings and battens for built-up compression members:

In a built-up section, the different components are connected together so that they act as a single column. Lacing is generally preferred in case of eccentric loads. Battening is normally used for axially loaded columns and in sections where the components are not far apart. Flat bars are generally used for lacing. Angles, channels and tubular sections are also used for lacing of very heavily columns. Plates are used for battens.

A lacing system should generally conform to the following requirements:

• The compression member comprising two main components laced and tied should, where practicable, have a radius of gyration about the axis perpendicular to the plane of lacing not less than the radius of gyration at right angles to that axis.
• The lacing system should not be varied throughout the length of the strut as far as practicable.
• Cross (except tie plates) should not be provided along the length of the column with lacing system, unless all forces resulting from deformation of column members are calculated and provided for in the lacing and its fastening.
• The single-laced systems on opposite sides of the main components should preferably be in the same direction so that one system is the shadow of the other.
• Laced compression members should be provided with tie plates at the ends of the lacing system and at points where the lacing system are interrupted. The tie plates should be designed by the same method as followed for battens.

Battens:Compression members composed of two main components battened should preferably have these components of the same cross-section and symmetrically disposed about their X – X axis. The battens should be placed opposite to each-other at each end of the member and at points where the member is stayed in its length, and should as far as practicable, be spaced and proportioned uniformly throughout. The effective length of columns should be increased by 10 percent.

Design of Compression Member

1. Effective length factor determination problem

Height of column is 10 m and it is effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction). Bottom of column is fixed and top of column is Free.

Calculate effective length factor of columns in following cases

Effective length factor =1.0L

=1*5 = 5m

  Effective length factor =2.0L

=2*5 = 10m

Effective length factor =1.2L

=1.2*5=6m

2. Design axial strength determination of a column with its end conditions

Calculate the Axial compressive capacity of Column made up of hot rolled ISMB 300 with flange cover plate of 200x12 mm on each flange having length of beam as 3 m. Assume the bottom of the column is Fixed and Top as Roller (Rotation Fixed, translation free).  Yield strength of material fy= 250 N/mm2.

Given: ISMB 300, length of beam = 3m, yield strength of material fy= 250mpa                       

For ISMB 300  

A=58.60cm2  

Ixx =8990cm4

 Iyy=486cm4

bf=140mm 

D=300 

tf=13.1mm  

tw=7.7mm

rxx=12.37cm4

ryy=2.84cm4

Cover plates - 200x12(Top and Bottom)

Determination of radii of gyration for the compound section

Area of compound section = 5860mm2+2x(200x12)mm2

 =(5860+4800)mm2=10660mm2

Izz for plates = 2x(200x12^3/12 +200x12x156^2)=11687x10^4

Total Izz of beam and plates= (11687x10^4+8990x10^4)=20677x10^4mm4

Hence Rzz =√I/A

=(20677x10^4/10660)

=139.24

Iyy of ISMB 300 = 486x 10^4

 Iyy of two plates= 2x  (12x200^2/12)=1600x10^4

Total Iyy of beam and plates=(486+1600)x10^4 =2086x10^4mm4

 ryy=  `sqrt(2084 x 10^4/10660) = 44.24mm

Buckling curve classification  from table 10 using curve b   tw<40mm

Since ryy is small the buckling will be about the YY axis

Effective length=1.2L(support is fixed and other end is roller)

Slenderness  ratio λ= (1.2x 3 x 10^3)/44.24`=81.4mm<180

From table 9c of the code for fy=250      λ=81.4

fcd = 134.2

Hence design strength =134.2x 10660/1000= 1430.5 KN

The axial strength of the column is 1430.5KN

3. Design circular tube strut member for certain load

Design 2 m long tubular strut member to carry factored axial load of 400 kN. Assume the ends of the member are simply supported and the yield strength fy=210 Mpa. Also find Equivalent Square hollow sections.

Given: Length of strut = 2 m

Factored axial load = 400 KN

Yield strength of strut fy= 210n/mm2

Assume the ends of member simply supported

Area required= 400x10^3/210=1904.76mm2

For the given area a square hollow section of 100x100x6, Area of square section =2163mm2

For a circular hollow section of 125 outer bore dia=139.7 thickness=4.5, Area of circular section =1910mm2

From table 10 both sections class a

Effective length =1.0L =2.0m(both sections)

For square section Slenderness ratio λ=2x10^3/r min=2x10^3/37.9=52.77<180

For circular section Slenderness ratio λ=2x10^3/r min=2x10^3/47.8=41.84<180

From table 9c of the code for fy=210

For square section fcd=174.1N/mm2

Hence, the design strength= 174.1x21.63=376.58KN

For circular section fcd=180 N/mm2

Hence, the design strength= 180x  19.10=343.8KN

4. Design of laced column

Design a single laced column made up of two channels back to back of 10 m height to carry factored axial load of 1000kN. The lacings are bolted. Design of bolted connection not required.

Given: Height = 10M

Axial load= 1000 KN

Design of column P=1000x10^3

L=10m

Assume  design stress of 125 N/mm2(0.5 fy)

Required area =1000x10^3/125=8000mm2

Select two ISMC 300 channel back to back

properties A=91.28cm2  

rxx=118.1mm 

ryy=38.7mm

tf=13.6 mm 

cyy=23.6 mm

tw=7.6 mm  

Ixx=12725.2 cm4

When the radius of gyration of yy axis is increased in such a way that it is equal to radius of gyration about zz axis. This can be achieved by spacing the section in such a way that rzz becomes rmin. So first checking the safety of the section and then workout the required spacing between the two channels.

L/rzz=10x10^3/118.1=84.67

L/r of the built up section should be taken as 1.05x84.67=88.90

fcd taken from code for given rzz=120N/mm2

= 9128 x 120=1095 kn> 1000kn.

Hence the section is safe.

Spacing of channels=12725.2x104 =2 (IYY + A(S/2 +cyy)2)

=2 x(310.8 + 4564(s/2 +23.6)2

solving the above equation   S =183.1mm

Lacing system of the built-up section using single lacing system with the inclination of lacing bar gauge length g=50 mm for 90 mm flange

Spacing of lacing bars Lo= 2x(183+50+50) cot 45

=2x  283 x 1=566 mm

Lo/ryy should be <0.7xL/r of whole column

566/26=21.76 < 0.7 x 88.98=62.3 

Hence safe

Maximum shear =(2.5/100)x 1000x10^3=25000 KN

Transverse shear in each panel =V/N=25000/2=12500 N

Compressive force in the lacing bar =(V/N) cosec45

= 12500 x1.414

=17675 N    

5. Design of battened column

Design a battened column made up of two channels back to back of 5 m height to carry factored axial load of 2000kN. The Battens are welded. Design of weld not required. Use Hot rolled channel sections. 

Given: Height = 5m

Axial load= 2000 kN

Design of column    P   =2000x10^3N

L=5m

Assume  design stress of 150 N/mm2(0.6 fy)

Required area =2000x10^3/150=13333mm2

Select two ISMC 350 channel back to back

 properties  A=53.66 x2=10732cm2  

rxx=136.6mm

 ryy=28.3mm

bf=100       

 tf=13.5 mm 

cyy=24.40 mm

tw=8.1 mm  

Ixx=12725.2 cm4

when the radius of gyration of yy axis is increased in such a way that it is equel to radius of gyration about zz axis .this can be achieved by spcing the section in such away that rzz becomes rmin. So first checking the safety of the section and then workout the required spacing between the two channels.

L/rzz= 5x10^3/136.6 =36.60

k=1 fy=250n/mm2 using Table 9c of code.

fcd taken from code for given rzz=36.60 

Fcd=202.4/mm2

Capacity of built up column    = 10732 x 202.4 =2172 KN >2000 KN   

Hence safe.

Spacing of channels =  2 x 20016 x104 =2 (IYY + A(S/2 +cyy)2)

=2 x(430.6 x10^4 + 5366(s/2 +24.4)2

solving the above equation   S =333.4 mm

spacing of battens  Lo/ryy should be less than 0.7x L/ryy

Lo /ryy < 0.7 x36.6

Lo < 0.7 x36.6x28.3 =725 mm

Also, Lo/ryy should be  less than 50

Lo/ryy < 50       

 Lo < 50x ryy =50x28.3 =1415 mm

Hence provide batten at spacing 725 mm

Size of end battens

Overall depth of batten = 334 + 2X CYY

=334 +2 X24.4= 383MM

Provide 60 mm overlap of batten on channel flange for welding

Length of batten=334 x 2 x 60=454mm=460mm

Thickness of battens =1/50x 460=9.2 mm

Provide batten size =460 x 390 x 9.2 mm

Size of intermediate battens

Overall depth = 3/4 x 390 = 293 mm

Provide 460 x300x 9.2 mm.              

Conclusion:

From the project carried out it can be concluded that the concept of the design procedure of the compression member and the various reasons for the failure of the compression member is learnt. The method for the determination of the size of the compression member is also learnt. The concept of the design of the laced and the batten pattern is also learnt.

Vijayalakshmi S H