4. Introduction :-
Beams are structural element subjected to transverse load in the plane of bending causing bending
moments and shear forces. Symmetrical section about z-z axis (major axis) are economical and
geometrical properties of such section are available in SP 6. The compression flange of the beam can be
laterally supported (restrained) or laterally unsupported (unrestrained) depending upon weather
restrained are provided or not. The beams are designed for maximum BM and checked for maximum SF,
local effect such as vertical buckling and crippling of webs and deflection.
Section 8/PN 52-69/IS 800:2007 shall be followed in the design of such bending members.
4
5. Type of sections :-
Beams can be of different cross sections depending on the span and loadings are shown in below.
5
6. Functional classification of flexural members
6
Flexural members are labeled in different names, such as
Purlins – Member carrying load from sheeting of roof truss
Girt – Member carrying load from side sheeting
Joist – Light member carrying load from floor in a building
Girder – Mostly member carrying heavy loads
Lintel – Member carrying load over window, door opening
Stringer – Mostly member in bridges aligned in the direction of traffic.
Spandrel – Member on periphery
7. LATERALLY SUPPORTED BEAMS
7
Conditions to qualify as laterally supported beam
Section is symmetric about yy axis
Loading is in plane containing yy axis
For sections unsymmetrical about yy axis load acts through shear center
None of its elements i.e. flange and web should buckle until a desired limit state is
achieved
15. 15
Limit states for beams
• Limit state of flexure
• Limit state of shear
• Limit state of bearing
• Limit state of serviceability
16. Modes of failure of Beams:-
A beam transversely loaded in its own plane can attain its full capacity (Plastic moment) only if local and
lateral instabilities are prevented. Local buckling of beam can be due to web crippling and web buckling.
They are avoided by proper dimensioning of the bearing plate and through secondary design checks.
1. Shear strength of Beams Ref. Cl. No. 8.4/PN 59/IS 800:2007
Vd = Design shear strength of web
γm0
= =
Vn Av fyw
√3 γm0
16
18. 3. Local Failures of Flanges :- Ref. Table. No. 2/PN 18/IS 800:2007
The local failure of flanges reduces the plastic moment capacity of the section due to buckling and is avoided
by limiting the outstand to thickness ratios as given in table 2 of IS 800 : 2007.
18
19. 3. Local Failures of Web :- Ref. Table. No. 2/PN 18/IS 800:2007
The web of a beam is thin and can fail locally at supports or where concentrated loads are acting. There are
two types of web failure
1. Web Buckling
2. Web Crippling
19
20. Web Buckling :
20
Q. State and explain the terms with neat sketch :Web buckling Dec. 2014, Dec. 2015.
The web of the beam is thin and can buckle under reaction and concentrated loads with the web behaving
like a short column fixed at the flanges.
For safety against web buckling, the resisting force shall be greater than the reaction or the concentrated load
is dispersed in to web at 450 as shown in fig.
Fwb = (b1 + n1) tw fcd ≥ Reaction, R
Fwb = Resisting force
tw = Thickness of web
fcd = Design compressive stress in web
b1 = Width of bearing plate
n1 = Width of dispersion
For concentrated loads, the dispersion is on both
side and the resisting force can be expressed as,
Fwd = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P
R
R must be less than fcd(b1+n1)tw
21. Web Crippling :
Q. State and explain the term with neat sketch :Web crippling Dec. 2014, Dec. 2015 May 2017.
Web crippling causes local crushing failure of web due to large bearing stress under reaction at supports or
concentration loads. This occurs due to stress concentration loads. Web crippling is the crushing failure of the
metal at the junction of flange and web. Web crippling causes local buckling of web at the junction of web and
flange. For safety against web crippling, the resisting Force shall be greater than the reaction or the
concentrated load. It will be assumed that the
reaction or concentrated load is dispersed in
to the web with a slope of 1 in 2.5 as shown in fig.
Fwc = Resisting force
tw = Thickness of web
fyw = Yield stress in web
b1 = Width of bearing plate
n2 = Width of dispersion
Fwc =
(b1 + n2) tw fyw
γm0
≥ Reaction,
R
Fwc
=
(b1 + 2 n2) tw fyw
γm0
≥ Concentrated load, P
R
R
R must be less than (b1 + n2) tw fyw / γm0
21
22. Laterally supported beam using single rolled steel section with and without flange plate:
Q. Explain laterally supported beam with suitable sketch. Dec. 2015, Dec. 2016, May 2017.
Beam subjected to BM develop compressive and tensile force and the flange subjected to compressive
forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of
beam. The lateral bending of beams depends on the effective span between the restrains, minimum moment of
inertial and its presence reduces the plastic moment capacity of the section.
Beams where lateral buckling of the compression flange are prevented are called laterally restrained
beams. Such continuous lateral supports are provided in two ways
i. The compression flange is connected to an
RC slab throughout by shear connectors
ii. External lateral supports are provided at closer
interval to the compression flange so that it as
good continuous lateral support Ref cl no 8.2.1
Design of such laterally supported beams are
carried out using clauses 8.2.1.2, 8.2.1.3,8.2.1.5,
8.4.8.4.1,8.4.1.1,8.4.2.1, and 5.6.1.
22
23. Design steps for laterally supported beams
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
for a given loading and support condition
2. Selection of cross section
Find
p
Z required =
M.γm0
fy
Take βb = 1 for assuming plastic or compact section
Using Annex H IS 800 : 2007 plastic properties of section. Select required section.
Or
e
Z required =
𝑍𝑝
23
𝑠ℎ𝑎𝑝𝑒 factor (𝑠)
Shape factor
i. I and channel section – 1.11 to 1.15
ii. Rectangular section – 1.5
iii. Circular section – 1.69
iv. Angle and T – section – 1.8
v. Square or diamond section – 2
∴ Zp provided > Z p required
24. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Using Table no 2 / PN 18 / IS 800 : 2007
4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V ∴ ok & Safe
5. Design bending strength (Md )
i. If V 𝑘 0.6 Vd or V < 0.6 Vd
Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007
then Md = <
βb .fy ZP 1.2 Ze .fy
.
γm0 γm0
........for simply supported beams
<
1.5 ZP . fy
………….for cantilever beams
γm0
βb = 1 for Plastic and compact section
Where
24
25. ii. If V > 0.6 Vd
then Md = Mdv
Where Md = Design bending strength under high shear.
𝜙 = Md - β (Md - Mfd
γm0
) ≤
1.2 Ze .fy
………For plastic and compact section
Ze .fy
………For semi compact section
γm0
If Md > M ∴ ok & Safe
25
26. 6. Checks
a) Web buckling Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007
Fwb = (b1 + n1) tw fcd ≥ Reaction, R
Fwd = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P ∴ ok & Safe
Where fcd = is calculated using slenderness ratio
0.7 d
r𝑦
= 2.425 ×
d
t𝑤
and table 9 c
b) Web Crippling Ref. cl no 8.7.4 / PN 67 / IS 800:2007
Fwc =
(b1 + n2) tw fyw
γm0
≥ Reaction,
R
Fwc =
(b1 + 2 n2) tw fyw
γm0
c) Deflection Limit
26
≥ Concentrated load, P ∴ ok & Safe
Ref. Table no 6 / PN 31 / IS 800:2007
∴ ok & Safe
δ actual < δ Limiting
27. Design of laterally supported beams
Example 1 A Simply supported beam of effective span 4 m carries a factored point load of 350 kN mid span.
The section is laterally supported throughout the span. Design cross section using I-section.
SPPU-Dec. 2010, 10 marks
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
Maximum Shear force
V = Reaction = W
= 175 kN-m
2
Maximum Bending Moment
M = WL
4
= 350 kN-m
2. Selection of cross section
p
Z required =
M.γm0
fy
=
350 ×10 6
×1.10
250
= 1540 × 103 mm3
Or
𝑍
Ze required = 𝑝
=
𝑠ℎ𝑎𝑝𝑒 factor (𝑠)
Select an ISLB 500 @ 75 kg/m
1540 ×10 3
27
1.14
= 1350.8 × 103 mm3
(Using annex ‘H’ IS 800:2007)
d= 500 mm, bf = 180 mm, tf = 14.1 mm, tw = 9.2 mm,
Zp = 1773.7 × 103 mm3 > Zp required , Ze = 1545. 2 × 103 mm3
Izz = 38549 × 104 mm4 ,
28. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 180 /2
14.1
= 6.38 < 9.4 ∈
∴ Flange is plastic
Web = b
tw
=
h1
tw
= 430.2
9.2
= 46.76 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
28
29. 4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V
=
Av fy
√3 γm0
=
500 × 9.2 × 250
√3 × 1.10
= 603.3 kN > V = 175 kN-m
∴ ok & Safe
29
30. 5. Design bending strength (Md ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007
i. If V 𝑘 0.6 Vd or V < 0.6 Vd
0.6 Vd = 361.8 kN
∴ V < 0.6 Vd ∴ Md = .
βb .fy ZP
γm0
Md =
βb .fy ZP
.
γm0
=
1×1773.7 ×10 3
× 250
1.10
=403.11 kN –m > M = 350 kN-m
∴ ok & Safe
30
31. 6. Checks
a) Web buckling
fwb = (b1 + n1) tw fcd ≥ Reaction, R
Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007
fwb = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P
Where fcd = is calculated using slenderness ratio
0.7 d
r𝑦
= 2.425 ×
d
t𝑤 9.2
= 2.425 × 500
=131.78
Buckling Class ‘c’
By interpolation
cd
f = 74.3 − 74.3−66.2
× (131.78 − 130)
140−130
= 72.85 N/mm2
fcd
At Concentrated load
∴ fwb = (180 + 2 × 250 ) ×9.2 ×72.85
= 455.74 kN > P = 350 kN
At Reaction ‘R’
∴ fwb = (180 + 250 ) ×9.2 ×72.85
= 288.19 kN > R = 175 kN
∴ ok & Safe
∴ ok & Safe
31
32. b) Web Crippling Ref. cl no 8.7.4 / PN 67 / IS 800:2007
Fwc
At concentrated load ‘W’
=
(b1 + 2 n2) tw fyw
γm0
=
(180 +2 ×2.5 ×14.1 × 9.2 ×250
1.10
> Concentrated load, P = 350 kN
= 523.77 kN
c) Deflection Limit
∴ ok & Safe
Ref. Table no 6 / PN 31 / IS 800:2007
working load = 350
1.5
= 233.33 kN
δ actual
WL3
=
48 EI
=
233.33 ×103
×40003
48 ×2×105
× 38549 × 10 4
= 4.03 mm
δ Limiting
= span
300
= 4000
300
32
= 13.33 mm
∴ δ actual δ Limiting
< ∴ ok & Safe
Provide an ISLB 500 @ 75 kg/m beam. … … … … … … … … … … … … … … … … . . A n s
33. Design of laterally supported beams
Example 2 Design a laterally supported simply supported beam of 7 m effective span. It carries a load 0f 250
kN which is uniformly distributed load over the whole span. In addition the beam carries a point load of 100 kN
at mid span. SPPU-Dec. 2010, 10 marks
Solution :
Given : UDL = 250 kN, UDL = 250
= 35.71 kN/m
7
Point load at centre = 100 kN
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
Maximum Shear force
V = Reaction = 1.5 [ 35.71× 7
+ 100
] = 262.48 kN
2 2
Maximum Bending Moment
2
M = 1.5 [ 35.71 × 7
8 4
+ 100 × 7
] = 590.81 kN-m
2. Selection of cross section
p
Z required =
M.γm0
fy
=
590.81 ×10 6
×1.10
33
250
= 2598.55 × 103 mm3
Select an ISLB 600 @ 99.5 kg/m (Using annex ‘H’ IS 800:2007)
d= 600 mm, bf = 210 mm, tf = 15.5 mm, tw = 10.5 mm,
Zp = 2798.56 × 103 mm3 > Zp required , Ze = 2428.9 × 103 mm3
Izz = 72867.6 × 104 mm4 ,
34. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 210 /2
15.5
= 6.77 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 520
10.5
= 49.52 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
34
35. 4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V
=
Av fy
√3 γm0
=
600 × 10.5 × 250
√3 × 1.10
= 826.66 kN > V = 262.48 kN-m
∴ ok & Safe
35
36. 5. Design bending strength (Md ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007
i. If V 𝑘 0.6 Vd or V < 0.6 Vd
0.6 Vd = 495.99 kN
∴ V < 0.6 Vd ∴ Md = .
βb .fy ZP
γm0
Md =
βb .fy ZP
.
γm0
=
1×2798.56 ×10 3
× 250
1.10
= 636.04 kN –m > M = 590.81 kN-m
∴ ok & Safe
36
37. 6. Checks
a) Web buckling
fwb = (b1 + n1) tw fcd ≥ Reaction, R
Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007
fwd = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P
Where fcd = is calculated using slenderness ratio
0.7 d
r𝑦
= 2.425 ×
d
t𝑤 10.5
= 2.425 × 600
=138.57
Buckling Class ‘c’
By interpolation
cd
f = 74.3 − 74.3−66.2
× (138.57 − 130)
140−130
= 67.35 N/mm2
fcd
At Reaction ‘R’
∴ fwb = (210 + 300) ×10.5 × 67.35
= 361.46 kN > R = 262.48 kN
∴ ok & Safe
37
38. b) Web Crippling
At concentrated load ‘W’
Ref. cl no 8.7.4 / PN 67 / IS 800:2007
Fwc
=
(b1 + 2 n2) tw fyw
γm0
=
(210 +2 ×2.5 ×15.5 × 10.5 ×250
1.10
> R = 262.48 kN
= 593.60 kN
c) Deflection Limit
∴ ok & Safe
Ref. Table no 6 / PN 31 / IS 800:2007
δ actual
=
WL3
+ 5 WL4
48 EI 348 EI
= 12.56 mm
= span
= 7000
δ Limiting 300 300
38
= 23.33 mm
∴ δ actual δ Limiting
< ∴ ok & Safe
Provide an ISLB 600 @ 99.5 kg/m beam. … … … … … … … … … … … … … … … … . . A n s
39. Design of laterally supported beams
Example 3 An ISLB 600 @ 99.5 kg/m has been used a simply supported beam over 7.2 m span. Determine the
safe uniformly distributed load ‘W’ so that the beam can carry in flexure. Assuming compressive flange is
restrained throughout the span against lateral buckling and Fe 410 steel.
SPPU-Dec. 2010, 10 marks
Solution :
1. Sectional properties
39
ISLB 600 @ 99.5 kg/m (Using annex ‘H’ IS 800:2007)
d= 600 mm, bf = 210 mm, tf = 15.5 mm,
Zp = 2798.56 × 103 mm3,
tw = 10.5 mm,
Ze = 2428.9 × 103 mm3, h1 = 520.2 mm
40. 2. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 210 /2
14.1
= 6.77 < 9.4 ∈
∴ Flange is plastic
Web = b
tw
=
h1
tw
= 520.2
10.5
= 49.54 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
40
41. 3. Maximum Bending Moment M (factored)
M =
WL2
=
8
1.5 × W × 72002
8
= 9.72 W
4. Bending Strength of section
Md = P y .
γm0
=
βb . Z f 1× 2798.56 ×10 3
× 250
1.10
= 636.03 kN-m
5. UDL ‘W’ equating Md and factored ‘M’
9.72 W = 636.03 kN-m
W = 65.43 kN/m
Now, self weight of section
=
99.5 × 9.81
41
1000
= 0.976 kN/m
Hence, UDL W = 65.43 – 0.976 = 64.45 kN/m
∴ W = 64.45 kN/m … … … … … … … … … A n s
42. Design of laterally supported beams
Example 4 Design a laterally supported beam of effective span 6 m for the following data :
Maximum bending moment M = 150 kN-m
Maximum Shear force V = 210 kN.
Grade of steel Fe 410
Solution :
Given :
SPPU-Dec. 2011, 13 marks, May 2014, 15 marks.
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
Maximum Shear force = 210 kN
Maximum Bending Moment = 150 kN-m
2. Selection of cross section
p
Z required =
M.γm0
fy
=
150 ×10 6
×1.10
42
250
= 660 × 103 mm3
Select an ISLB 350 @ 49.5 kg/m (Using annex ‘H’ IS 800:2007)
d= 350 mm, bf = 165 mm, tf = 11.4 mm, tw = 7.4 mm, h1 = 288.3 mm ,
Zp = 851.11 × 103 mm3 > Zp required.
43. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 165/2
11.4
= 7.23 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 288.3
7.4
= 38.95 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
43
44. 4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V
=
Av fy
√3 γm0
=
350 × 7.4 × 250
√3 × 1.10
= 339.84 kN > V = 262.48 kN-m
∴ ok & Safe
44
46. Laterally Unsupported Beam :-
Q. Explain laterally Unsupported beam with suitable sketch. Dec. 2016, May 2017.
Beam subjected to BM develop compressive and tensile force and the flange subjected to compressive
forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of
beam. The lateral bending of beams depends on the effective span between the restrains, minimum moment of
inertial and its presence reduces the plastic moment capacity of the section.
The value of Mcr can be calculated using the equation given in cl. 8.2.2.1 / PN 54 / IS 800 : 2007
The design of bending compressive strength can be calculated using a set of equation as specified in cl 8.2.2
(Table 13 a and 13 b / PN 54 to 57 / IS 800:2007.
The design of laterally unsupported consist of selecting a section based on the plastic sectional modulus
and checking for its shear capacity & deflection.
46
47. Modes of Failure of Beam :-
Q. Explain modes of failure of beam with suitable sketches. Dec. 2016.
i. Bending Failure : Bending failure may be done due to fracture of tension flange and due to crashing of
compression flange.
ii. Shear failure : It occurs when buckling of web of beam near location of high shear failure.
iii. Torsional buckling: Due to combined moment on the beam buckle about both axis called torsional
buckling.
iv. Lateral buckling : due to large span of beams deflection takes place beyond limits.
v. Local failure of Web : due to heavy shear in some regions of beam web fails in crippling or buckling.
47
48. Design steps for laterally Unsupported beams
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
for a given loading and support condition
2. Selection of cross section
Find
p
Z required =
Md
βb . fbd
Take βb = 1 for assuming plastic or compact section
Assume, fcd = 120 to 140 N/mm2 for I section
Using Annex H IS 800 : 2007 plastic properties of section. Select required section.
Or
e
Z required =
Zp
48
shape factor (s)
Shape factor
i. I and channel section – 1.11 to 1.15
ii. Rectangular section – 1.5
iii. Circular section – 1.69
iv. Angle and T – section – 1.8
v. Square or diamond section – 2
∴ Zp provided > Z p required
49. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Using Table no 2 / PN 18 / IS 800 : 2007
4. Effective length of beam Ref. 8.3.1 / Table No 15 / PN 58 / IS 800 : 2007
Depending upon support condition, LLT calculated using Table – 15.
5. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
49
√3 γm0
Vd = = > V
∴ ok & Safe
6. Design bending strength (Md )
Md = Zp . βb . fbd
Where, Zp = plastic sectional moduli
fbd = design bending compressive stress
Md > M
∴ ok & Safe
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
50. 6. Design bending strength (Md ) Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
7. Check : Deflection Limit
50
Ref. Table no 6 / PN 31 / IS 800:2007
<
δ actual δ Limiting
∴ ok & Safe
51. Examples on Laterally Unsupported Beams
Example 1 Design a suitable I section for and simply supported beam of span 5 m carrying a dead load of 20
kN/m and imposed load of 40 kN/m. The beam is laterally unsupported throughout the span. Take fy = 250 MPa.
SPPU- Dec. 2010, 15 Marks
Solution : Factored Load = 1.5 ×(20 + 40) = 90 kN/m
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
i. Maximum shear force :
V = WL
= 90 × 5000
= 225 kN
2 2
ii. Maximum bending force:
8 2
2 2
M = WL
= 90 × 5000
= 281.25 kN-m
2. Selection of cross section
Assuming fbd = 130 N/mm2
p
Z required =
Md
βb . fbd
6
= 281.25 ×10
130
51
= 2163.46 × 103 mm3
Select ISWB 500 @ 95.2 kg/m
d= 500 mm, bf = 250 mm, tf = 14.7 mm,
Zp = 2351.35 × 103 mm3 > Zp required,
tw = 9.9 mm, rmin = 49.6 mm,
= 52290.9 × 104 mm4
Ixx
52. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 250/2
14.7
= 8.50 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 431
9.9
= 43.53 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
52
53. 4. Effective length of beam
Effective length KL = 5000 mm.
Ref. 8.3.1 / Table No 15 / PN 58 / IS 800 : 2007
As end are restrained against torsion but compression flange is laterally unsupported.
5. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
Vd = =
γm0 √3 γ
Av fy
m0
= 649.51 kN > V = 225 kN
=
500×9.9×250
√3 ×1.1
∴ ok & Safe
6. Design bending strength (Md )
Md = Zp . βb . fbd
Where, Zp = plastic sectional moduli
fbd = design bending compressive stress
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
KL = 5000
rmin
49.6
500
53
14.7
= 100.86, h = = 34.01
tf
54. From Table no14 / PN 57 / IS 800 :2007
For
kL
rmin
= 100.86,
h
t𝐹
= 34.01
By interpolation
35−30
A = 270.9 − 270.9 −257.7
× (34.01 − 30) = 260.31 N/mm2
35−30
B = 231.1 − 231.1 −219.3
× (34.01 − 30)
cr, b
f = 260.31 − 260.31 −221.56
× (100.86 − 100)
= 221.56 N/mm2
= 256.97 N/mm2
110−100
= 256.97 N/mm2
fcr, b
54
55. Now, for rolled section αLT = 0.21
fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 152 +(
300−250
163.6 −152.3
)× (256.57 − 250)
fbd = 153.48 N/mm2
Md= Zp . βb . fbd = 1 × 2351.35 ×103 × 153.48
= 360.88 kN-m > M = 281.25
∴ ok & Safe
55
56. 7) Check : Deflection Limit Ref. Table no 6 / PN 31 / IS 800:2007
δ actual
5 W L4
=
384 EI
=
5 ×60 ×50004
384× 2×105
× 52290.9 ×104
= 4.66 mm
δ Limiting =
span
300
= 5000
300
56
= 16.67 mm
<
δ actual δ Limiting
∴ ok & Safe
Hence, Provide ISWB 500 @ 95.2 kg/m. … … … … … … … … … … … … … … … … … A n s .
57. Examples on Laterally Unsupported Beams
Example 2 A floor beam in a building has a span of 6 m. it is simply supported over supports and carries a
uniformly distributed load of 40 kN/m, inclusive of self weight. Design the beam if the compression flange is
unrestrained throughout the span against lateral buckling and Fe 410 steel. SPPU- Dec. 2011, 15 Mark
Solution : Factored Load = 1.5 × 40 = 60 kN/m
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
i. Maximum shear force :
V = WL
= 60 × 6000
= 180 kN
2 2
ii. Maximum bending force:
8 2
2 2
M = WL
= 60 ×6000
= 270 kN-m
2. Selection of cross section
Assuming fbd = 130 N/mm2
p
Z required =
Md
βb . fbd
6
= 281.25 ×10
130
57
= 2163.46 × 103 mm3
Select ISWB 500 @ 95.2 kg/m
d= 500 mm, bf = 250 mm, tf = 14.7 mm,
Zp = 2351.35 × 103 mm3 > Zp required,
tw = 9.9 mm, rmin = 49.6 mm,
= 52290.9 × 104 mm4
Ixx
58. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 250/2
14.7
= 8.50 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 431
9.9
= 43.53 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
58
59. 4. Effective length of beam
Effective length KL = 6000 mm.
Ref. 8.3.1 / Table No 15 / PN 58 / IS 800 : 2007
As end are restrained against torsion but compression flange is laterally unsupported.
5. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
Vd = =
γm0 √3 γ
Av fy
m0
= 649.51 kN > V = 225 kN
=
500×9.9×250
√3 ×1.1
∴ ok & Safe
6. Design bending strength (Md )
Md = Zp . βb . fbd
Where, Zp = plastic sectional moduli
fbd = design bending compressive stress
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
KL = 6000
rmin
49.6
500
59
14.7
= 120.96, h = = 34.01
tf
60. From Table no14 / PN 57 / IS 800 :2007
For
kL
rmin
= 120.96 ,
h
t𝐹
= 34.01
By interpolation
A = 202.4 −
35−30
202.4 −190.1
× (34.01 − 30) = 192.53 N/mm2
35−30
B = 179.0 − 179.0 −167.1
× (34.01 − 30)
cr, b
f = 192.53 − 192.53 −169.45
× (120.96 − 120)
= 169.45 N/mm2
= 190.31 N/mm2
130−120
= 190.31 N/mm2
fcr, b
60
61. Now, for rolled section αLT = 0.21
fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 106.8 +( 134.1 −106.8
) × (190.31 − 150)
200−150
128.80 N/mm2
fbd =
Md= Zp . βb . fbd = 1 × 2351.35 ×103 × 128.80
= 302.85 kN-m > M = 281.25
∴ ok & Safe
61
62. 7) Check : Deflection Limit Ref. Table no 6 / PN 31 / IS 800:2007
δ actual =
5 W L4
384 EI
=
5 ×40 ×60004
384× 2×105
× 52290.9 ×104
= 6.45 mm
δ Limiting =
span
300
= 6000
300
62
= 20 mm
<
δ actual δ Limiting
∴ ok & Safe
Hence, Provide ISWB 500 @ 95.2 kg/m. … … … … … … … … … … … … … … … … … A n s .
63. Examples on Laterally Unsupported Beams
Example 3 Determine the safe uniformly load that the beam ISLB 600 @ 99.5 kg/m has been used as a simply
supported over 7.2 m span. The compression flange of beam is not restrained against lateral buckling. At the
ends beam is fully restrained in torsion but both the flange are free to warp at the ends.
SPPU- Dec. 2012, 15 Marks, May 2016, Dec. 2016, 16 Marks
Solution :
1. Properties of ISLB 600 @ 99.5 kg/m
A = 126.69 ×102 mm2, D = 600 mm,
63
bf = 210 mm,
tw = 10.5 mm,
h1= 520.2 mm,
Ze = 2428.9 × 103 mm3
tf = 15.5 mm,
rmin = 37.9 mm,
Zp = 2798.56 × 103 mm3,
Ixx = 52290.9 × 104 mm4
64. 2. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 210/2
15.5
= 6.77 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 520.2
10.5
= 49.54 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
64
66. Now, for rolled section αLT = 0.21
for fcr, b = 92.17 N/mm2 fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 63.6 +( 70.5 −63.6
) × (89.50 − 80)
90−80
70.15 N/mm2
fbd =
Md= Zp . βb . fbd = 1 × 2798.56 ×103 × 70.15
= 196.32 kN-m
66
67. 4) Safe uniformly distributed load (w)
M = Md
8
wL2
= 196.32
w = 30.30
1.5
67
Safe UDL w = 30.30
= 20.2 kN/m Including self weight.
w = 20.2 kN … … … … … … … … … … … … … … … … … A n s .
68. Examples on Laterally Unsupported Beams
Example 4 A simply supported beam of effective spam 8 m carries uniformly distributed load w kN/m
throughout the span. The compression flange is laterally unsupported throughout the span. Determine intensity
of uniformly distributed load ‘w’ so that ISMB 400 @ 61.6 kg/m provided for beam can carry safely.
SPPU- May 2017, 10 Marks
Solution :
1. Properties of ISMB 400 @ 61.6 kg/m
A = 7846 mm2, d= 400 mm,
68
bf = 140 mm,
tw = 8.9 mm,
h1= 334.4 mm,
tf = 16 mm,
rmin = 28.2 mm,
Zp = 1176.18 × 103 mm3,
69. 2. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 140/2
15.5
= 4.375 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 334.4
8.9
= 37.57 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
69
70. 3. Design bending strength (Md )
Md = Zp . βb . fbd
From Table no14 / PN 57 / IS 800 :2007
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
= 283.68,
KL = 8000 h = 400
rmin 28.2 tf 16
= 25.
By interpolation
cr, b 290−280
f = 74.7 − 74.7 −71.8
× (283.68 − 280) = 73.63 N/mm2
fcr, b = 73..63 N/mm2
70
71. Now, for rolled section αLT = 0.21
for fcr, b = 73.63 N/mm2 fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 56.8 +( 63..6 −56.8
) × (73.63 − 70)
80−70
59.27 N/mm2
fbd =
Md= Zp . βb . fbd = 1 × 1176.18 ×103 × 59.27
= 69.71 kN-m
71
72. 4) Safe uniformly distributed load (w)
M = Md
8
wL2
= 69.71
w = 8.71
1.5
72
Safe UDL w = 8.71
= 5.80 kN/m Including self weight.
w = 5.80 kN … … … … … … … … … … … … … … … … … A n s .
