The document discusses the design of compression members according to IS 800:2007. It defines compression members as structural members subjected to axial compression/compressive forces. Their design is governed by strength and buckling. The two main types are columns and struts. Common cross-section shapes used include channels, angles, and hollow sections. The effective length of a member depends on its end conditions. Slenderness ratio is a parameter that affects the load carrying capacity, with higher ratios resulting in lower capacity. Design involves checking the member for short or long classification, buckling curve classification, and calculating the design compressive strength. Examples are included to demonstrate the design process.

1. Design of compression members
As per IS 800 : 2007

2. Compression members
• Structural Members subjected to axial
compression/compressive forces
• Design governed by strength and buckling
• Two types of compression member :
1. Column
2. Strut

3. • Sections used for compression member
 In theory numerous shapes can be used for columns
to resist given loads.
 However, from practical point of view, the number of
possible solutions is severely limited by section
availability, connection problems, and type of
structure in which the section is to be used.

6. Effective length of compression
member
• Column End Condition And Effective Length
1.Both end hinged.
2.Both end fixed.
3.One end fixed and other hinged.
4.One end fixed and other free.

7. Effective length

8. • Slenderness ratio ( λ ) :
Slenderness Ratio =
effective length of column/Minimum radius of gyration
λ = le/r
If λ is more , its load carrying capacity will be less.

9. • Short compression member
1. L/r ≤ 88.85 for Fy = 250 Mpa
2. Failure stress equal to yield stress
3. No buckling
• Long compression member
1. They will buckle elastically
2. Axial buckling stress is below proportional limit.

10. Compression Member Failure
10
 There are three basic types of column failures.
 One, a compressive material failure( very short and
fat).
 Two, a buckling failure,(very long and skinny).
 Three, a combination of both compressive and
buckling failures.(length and width of a column is in
between a short and fat and long and skinny column).

11. Compression Member Failure
11
 Local Buckling This occurs when some part or
parts of x-section of a column are so thin that
they buckle locally in compression before other
modes of buckling can occur

12. Compression Member Failure
12
 Flexural Buckling (also called Euler Buckling) is the
primary type of buckling members are
subjected to bending or flexure when they
become unstable

13. Compression Member Failure
• Squashing :
When the length of column is relatively small and
column is stocky and its component plates are
prevented from local buckling, then column will be
able to attain its full strength before failure.

14. Cross section classification

15. Design compressive strength

16. Clause 7.3.2

17. IS 800: 2007 Clause 7.1.2.1

18. Column buckling curves
• Classification of different sections under
different buckling class a, b,c and d are given
in Table 10 of IS 800: 2007 ( page 44).
• The stress reduction factor χ , and the design
compressive stress fcd, for different buckling
class, yield stress and effective slenderness
ratio is given in table 8 ( page 37)
• Table 9( page 40) shows the design
compressive stress, fcd for different buckling
class a to d.

19. • The curve corresponding to different buckling class are presented in
non-dimensional form as shown in the figure below. Using this curve
one can find the value of fcd ( design compressive stress)
corresponding to non- dimensional effective slenderness ratio λ (
page 35)

21. Angle struts
• Single angle strut: ( IS : 800 cl. 7.5.1)
The compression in single angle struts may be
transfered either concentrically to its centroid through
end gusset or eccentrically by connecting one of its
legs to a gusset or adjacent member.
1. Concentric loading
2. Loaded through one leg √ k1 + k2 + ʎ2
vv + k3 ʎ2
ɸ

22. • Double angle struts
1. Connected back to back on opposite sides of G.P.
The effective length kL in plane of end
gusset shall be taken as between 0.7 and 0.85 times
the distance between intersections.
2. Connected back to back to one side of
G.P.(cl.7.5.2.2)
The outstanding legs shall be connected by
tack bolting or tack welding spaced at a distance not
exceeding 600 mm. (cl. 10.2.5.5)

23. • Determine compressive strength of a single ISA 100×100×8
mm @ 12.1 kg/m with length of member 2.5 m . Ends of the
members are hinged. Assume that the load is applied
concentrically to the angle. Fy = 250 MPa

24. • For ISA 100×100×8 mm
a = 1539 mm2 ( Steel table )
rmin = rvv = 19.5 mm
• Section Classification :
b/t = 100/8 = 12.5 < 15.7 ε
d/t = 100/8 = 12.5 < 15.7 ε ( IS : 800 , T – 2 , P.18)
(b+d)/ t = 200/8 = 25 ≤ 25 ε where ε = (250/fy )0.5
= 1
• Since all three criteria is satisfied , the section is semi
compact.
• For the semi compact section, gross area is taken as
effective area. Ae = 1539 mm2

25. • Effective length : ( IS :800 , T-11 , P.45 )
Both end pinned,
Effective length = kL
= 1 × 2500 = 2500 mm
• As the member is concentrically loaded, the design
compressive strength is calculated as per cl. 7.1.2 of code
IS 800 – 2007.
• Euler’s buckling stress : ( cl. 7.1.2 P. 34 )
fcc = п2 E / ( kL / r )2
= п2 × 2 × 105 / (2500/19.5)2
= 120.09 N/ mm2

26. • Non dimensional effective slenderness ratio :
( T.10 P.44 )
λ = √ fy / fcc = √ (250/ 120.09) = 1.44
α = Imperfection factor
= 0.49
( For a single ISA , buckling class is ‘c’ ) ( from T- 7 , α =
0.49)
ɸ = 0.5[ 1 + α (λ - 2) + λ 2 ]
= 1.8406

27. • Stress reduction factor χ :
χ = 1 / [ɸ + (ɸ2 - λ 2 )0.5 ]
= 1 / [ 1.8406 + (1.84062 – 1.442 )0.5 ]
= 0.335
• Design compressive stress ( f cd ) :
f cd = (fy / ɣmo ) / [ɸ + (ɸ2 - λ 2 )0.5 ]
= χ . fy / ɣmo
= 0.335 × 250/1.10
= 76.13 N/mm2

28. • Design compressive strength ( cl. 7.1.2):
Pd = Ae . Fcd
= 1539 × 76.13
= 117.16 kN

29. • Design a double angle discontinuous strut to carry a
factored load of 200 kN. The length of member is 3 m
between intersections. The two angles are placed back to
back on same side of gusset plate. Assume Fe 410 steel
with fy 250 Mpa.
Angles are connected to the same side of gusset
plate, as per IS : 800 cl.7.5.2.2 the member is designed as
two single ISA.
Load for single ISA = 200/2 = 100 kN
Effective length = c/c length of member = 3000 mm
Assume slenderness ratio kL/r = 120
Fcd = 83.7 N/mm2 ( T- 9(c) , P.42)

30. • Ag required = P / fcd
= 100×103 / 83.7
= 1195 mm2
• Choose single ISA 90×90×8 mm
a = 1379 mm2
rmin = rvv = 17.5 mm
kL/r = 3000/17.5
= 171.42 < 180 ........OK
• Section Classification :
b/t = 90/8 = 11.5 < 15.7 ε
d/t = 90/8 = 11.5 < 15.7 ε ( T – 2 , P.18)
(b+d)/ t = 180/8 = 22.5 ≤ 25 ε where ε = (250/fy )0.5
= 1
Since all three conditions are satisfied the section is
semi compact.

31. • Ae = 1379 mm2
• Assume two bolts at each end with fixed end condition.
k1 = 0.20
k2 = 0.35 IS : 800 , T – 12 , P. 48
k3 = 20
λ vv = ( L / rvv ) / [ ε (√ п2 E / 250) ]
= (3000 / 17.5 ) / [1 √ (п2 2 × 105 / 250 )]
= 1.929
λ ɸ =[ ( b1 + b2 ) / 2t] / [ ε √ п2 E / 250 ]
= ( 90 + 90/2 × 8 ) / 88.86
= 0.1266

32. • λ e = √ k1 + k2 + λ 2
vv + k3 λ 2
ɸ
= √ 0.20 + 0.35(1.929)2 + 20(0.1266)2
= 1.35
• Imperfaction factor = α = 0.49 ( T- 7 , P.35)
ɸ = 0.5 [ 1+ α (λ - 0.1) + λ 2 ]
= 1.693
• Stress reduction factor :
χ = 1 / [ɸ + (ɸ2 - λ 2 )0.5 ]
= 1 / 1.693 + (1.6932 – 1.352 )0.5
= 0.368
• Design compressive strength : (cl. 7.1.2.1 , P.34)
fcd = χ . fy / γmo
= 0.368 × 250/1.10
= 83.64 N/mm2

33. • Design compressive strength ( Pd ) :
Pd = Ae fcd
= 1379 × 83.64
= 115.34 kN
For 2 – ISA ,
Pd = 2× 115.34
= 230.67 > 200 kN.........SAFE
Provide 2 – ISA, 90 × 90 × 8 mm on same side of the gusset
plate.

34. • Determine the design axial load on the column section of ISMB
350 having height 3 m , hinged at both ends. Fy = 250 Mpa
Properties of ISMB 350 ( steel table )
A = 6671 mm2
bf = 140 mm , b = 70 mm
tf = 14.2 mm
tw = 8.1 mm
h = 350 mm
R = 14.0 mm
rzz = 142.9 mm
ryy = 28.4 mm

35. • Section classification :
ε = (250/fy )0.5
= 1
Flange , b / tf = 70 / 14.2 = 4.93 < 9.4 ε ......plastic
Web , d / tw = [ 350 – 2(14.2+14.0)] / 8.1 = 36.42 < 42 ε
.....semi compact
Flange and web both are fully effective.
Ae = A = 6671 mm2
• Effective length : ( IS : 800 , T-11 , P.45 )
(both end hinged)
kL = 1.0 L = 3000 mm

36. • Buckling curve classification :
rmin = ryy = 28.4 mm
Failure is by buckling about minor axis y-y ,
h / bf = 350 / 140 = 2.5 > 1.2 ( T-10 , P.44 )
t f = 14.2 mm < 40 mm
For failure about y-y axis use buckling curve ‘ b ‘ .
kL / ryy = 3000/28.4 = 105.63 ( T- 9(B) , P.41 )
fy = 250 Mpa P.41
10 14 (difference)
5.63 ? ( 7.882 ) fcd = 118 – 7.882
= 110.12 N / mm2

37. Pd = Ae × fcd
= 6671 × 110.12
= 734.61 kN

38. • Design a steel column to carry factored axial load of
1500 kN. The length of column is 3.6 m and hinged at
both ends. Assume fy = 250 Mpa.
P = 1500 kN
Assume fcd = 0.6y
= 0.6 × 250
= 150 N/mm2
Ag = P / fcd = 1500 × 103 / 150 = 100 cm2

39. • Choose ISHB 400@ 77.4 kg/ m ,
A= 9866 mm2
h = 400 mm
bf = 250 mm , b = 125 mm
tf = 12.7 mm
tw = 9.1 mm
R = 14 mm
rzz = 168.7 mm
ryy = 52.3 mm
• Section classification :
where ε = (250/fy )0.5 ( T – 2 , P.18)
= 1
Flange , b/tf = 125/12.7 = 9.84 < 10.5 ε
web , d/tw = [ 400 – 2 (12.7+14)] / 9.1
= 38.08 < 42 ε.......semi compact

40. • Full section is fully effective.
• Ae = 9866 mm2
• Effective length : ( IS : 800 , T – 11 , P.45 )
column hinged at both ends
kL = 1 . 0 × L
= 3600 mm
• Buckling curve classification : ( IS : 800 , T – 10 , P.44)
h / bf = 400 / 250 = 1.6 > 1.2
tf = 12.7 < 40
• For buckling about y-y axis use curve – b.
Since ryy is minimum failure is by buckling about
y-y axis.

41. • ( kL / ryy ) = 3600 / 52.6 = 68.44 < 180......OK
Fy = 250 Mpa ( IS : 800 , T – 9(b) , P.41)
10 15 (difference)
8.44 ? (12.66)
Fcd = 181 – 12.66 = 168.34 N/mm2
• Design compressive strength ,
Pd = Ae × fcd
= 9866 × 168.34
= 1660.84 > 1500 kN..........SAFE
Provide ISHB 400 as a column.

42. THANK YOU