Solutions-DOS-II Unit-I

1.
Design a suitable section of a carrying a live and dead load of 50 kN/m over
a simply supported span of 7 m. Carrey out the usual checks.
Sol: Assume the beam is laterally supported thought the compression flange (i.e.
compression is restrained against lateral buckling).
. .
Assuming the wall thickness as 300mm on either side, leff=7 + + = 7.3 m

Assume the self-weight of the beam = 1.0 kN/m
Total udl on the beam = 51.0 kN/m

∗ .
Max. BM= = = 333.0625 kN-m = 333.0625 * 106 N-mm
∗ .
Max. SF= = = 182.5 kN = 182.5 * 103 N

Design:
.
Z required = , (
, = 0.66fy = 0.66*250 = 165 MPa (N/mm2)

. ∗
= 2018560.6 mm3 = 2018.56 cm3.

Consider ISMB 550 @ 1.03 kN/m
Section Properties: D = 550 mm (over all depth)
t (thickness of the web) = 11.2 mm

Ixx= 64893.6 cm4= 64893.6*104 mm4, Zxx = 2359.8 cm3
Check for shear:

. ∗!
τav (average shear stress in the section) = = = 29.63 N/mm2 < (0.4fy
∗ ∗.
= 100 N/mm2, i.e. maximum allowable shear stress), Hence the section is safe in
shear.
Check for deflection:

.∗!
Maximum allowable deflection, "#,$% = = = 22.46mm

) ∗! ( .∗! ))
"#,&%' = * = * * = 14.53 mm <
( *+ ( ! .∗. ∗( /.∗)
"#,$% (19.38 mm) , Hence the section is safe against deflection
2. A steel beam in a building has a span of 6.0m. It is simply supported over

supports and carries a uniform load of 40.0 kN/m inclusive self-weight.
Design the beam (fy = 250MPa) if
(i) The compression flange is restricted against lateral buckling.
(ii) The lateral support for the compression flange is provided at the ends
(unrestrained). Perform the check for shear and deflection.
Sol: Span = 6.0 m
Assuming 300mm supports on either side of the beam, effective span leff = 6.0
. .
+ + = 6.30 m

Total udl on the beam w= 40.0 kN/m

(∗.
Maximum BM= = = 198.45 kN-m = 198.45*106 N-mm (@ centre of the
span)
(∗.
Maximum SF = = = 126 kN = 126*103 N (@supports)

Design of the beam:

(i) (Compression flange is laterally supported thought the compression
flange)
.
Z required = , ( , = 0.66fy = 0.66*250 = 165 MPa (N/mm2)

/ .( ∗
= 1202727.27 mm3 = 1202.727 cm3.

From steel tables select a section whose Z is greater than the Z required
Consider ISMB450 @ 0.724 kN/m

Z provided = 1350.7 cm3, Ixx= 30390.8 cm4, D = 450 mm and t (thickness of
the web) = 9.4mm
Check for shear:

∗!
τav (average shear stress in the section) = = = 29.79 N/mm2 < (0.4fy =
∗ ( ∗/.(
100 N/mm2, i.e. maximum allowable shear stress), Hence the section is safe in
shear.

.∗!
Maximum allowable deflection, "#,$% = = = 19.38 mm

) (∗! ( .∗! ))
"#,&%' = * = * * = 13.49 mm <
( *+ ( ! .∗. ∗/. ∗)
"#,$% (19.38 mm) , Hence the section is safe against deflection.
(ii) (Compression flange is laterally supported only at the supports i.e.

laterally unsupported thought the compression flange)
.
Z required = , ( = 100 to 110 is to be considered)

Let = 110 N/mm2

/ .( ∗
Z required = = 1804090.90 mm3 = 1840.09 cm3 (consider a section whose Z

value is about 20% more than the Zrequired)
Consider ISMB 550 @ 1.03 kN/m

Section Properties: D = 550 mm (over all depth)
t (thickness of the web) = 11.2 mm
T (thickness of the flange) = 19.3 mm
d1=h1=depth of the web = 467.5 mm
Ixx= 64893.6 cm4= 64893.6*106 mm4, Zxx = 2359.8 cm3
ryy= 3.73 cm =37.3 mm

' ( .
Calculate: ( 0 ) = = 41.74 < 85
.
1 /.
( )= = 1.723 < 2
.
Consider Table 6.1B

.∗
( )= = 168.90
233 .

= = 28.50
1 /.
25 30
D/T
4
566
160 93 87
170 89 83

(a) For ( ) = 168.90 and = 25
233 1
./
= 93- (93-89)* = 89.44

(b) For ( ) = 168.90 and = 30
233 1
./
= 87- (87-83)* = 83.44

For ( ) = 168.90 and = 28.5
233 1
.
= 89.44- (89.44-83.44)* = 85.24 is permissible bending stress in
compression for the chosen section
∴ = 85.24
89:;<= 9> 5;?@?=A<B; 9> =ℎ; ?;B=@9< (M.R)= * Z provided
=85.24 * 2359.8 * 103
= 201.15 *106 N-mm > 198.45*106
N-mm (i.e. the moment of the resistance of the section is greater than the
maximum bending at the section calculated; hence the section is safe against
bending).
(Do the check for shear and check for deflection. Since the section provided is
ISMB 550 which is a higher section than the section provided for laterally
supported i.e. ISMB 450, it safe in shear and deflection. But you do and verify)
3. Design a beam of 5 m effective span, carrying a uniform load of 20 kN/m, if

2
the compression flange is laterally unsupported. (Assume fy=250 N/mm ).
(Similarly do as above)
4. A hall of clear dimensions 15 m x6 m is to be covered by RCC slab flooring

12 cm thick resting over RS joists (Rolled steel beams) spaced at an interval
of 3 m centre to centre. Terrazo finishing 2 cm thick is to be provided over
the RCC slab. The live load on slab is 4 kN/m2. The joists are resting over
30 cm thick walls. Design the floor joists if the permissible stresses in
bending and shear are 165 MPa (0.66fy) and 100 MPa (0.4fy) respectively.
Assume the unit of concrete and finishing as 24 kN/m3.
Solution:
The compression flange of the beam is laterally restrained thought.
Note: Always provide the beams in shorter span direction, since the span will be
less and maximum BM will be less and the section requirement is also less.
Load on beam per meter length:
(i) Slab weight = 1*3*0.12 ( thickness of the slab)*24= 8.64 kN/m

(ii) Finishing = 1*3*0.02(Floor finish thickness)*24= 1.44 kN/m
(iii) Assume self-weight of the beam = 1.0 kN/m
(iv) Live load = 1*3*4= 12 kN/m
Therefore total udl on the beam = (8.64+1.44+12+1) = 23.08 kN/m
. .
Effective span, leff = 6.0 + + = 6.30 m

∗ . ∗.

Maximum BM = = = 114.51 kN-m = 114.51 * 106 N-mm
. ∗.

Maximum SF = = = 72.7 kN = 72.7*103 N

Design:
(. ∗
Zrequired = = = 694000 mm3 = 694.0 cm3

Consider ISLB 350 @ 0.495 kN/m
Zxx = 751.9 cm3, D = 350 mm, t= 7.4 mm, Ixx = 13158.3 cm4
Check for shear:

. ∗!
Average shear stress τav = = 28.06 N/mm2 < 100 N/mm2 ( Hence safe
∗ .(
against shear)

.∗!
Maximum allowable deflection, δmax, allowable = = = 19.39 mm

) . ∗! (.∗! ))

Maximum deflection calculated, δmax,cal = = ∗ *
( *+ ( ! .∗. ∗ .∗)
= 17.99 mm < 19.39 mm, hence the beam is safe against deflection.
5. A beam ISMB 600 @ 1.23 kN/m has effective span of 9 m. determine the
safe load which the beam can carry.
(i) If the compression flange is laterally restrained thought.
(ii) If the compression flange is laterally unrestrained.
Solution:
(i) Given section ISMB 600 @ 1.23 kN/m

leff= effective span = 9.0 m
Properties of the section: Zxx= 3060.4 cm3
Therefore moment of resistance of the section MR = Zxx* =3060.4 * 103 * 165 =

504966000 N-mm
= 504.966 kN-m
If w kN be the total udl on the beam including self-weight

∗%DD ∗/
Max. BM = = = 10.125*w kN-m
Equating MR to Max. BM
10.125*w = 504.966
w= 49.873 kN
There safe udl the beam can carry = 49.873 – 1.23 = 48.643 kN say 48.0 kN/m
(ii)Try unrestrained

Solutions-DOS-II Unit-I

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1.

Assume the self-weight of the beam = 1.0 kN/m

Total udl on the beam = 51.0 kN/m

Consider ISMB 550 @ 1.03 kN/m

Section Properties: D = 550 mm (over all depth)

t (thickness of the web) = 11.2 mm

Check for shear:

Check for deflection:

Check for deflection:

2. A steel beam in a building has a span of 6.0m. It is simply supported over

Total udl on the beam w= 40.0 kN/m

Design of the beam:

Consider ISMB450 @ 0.724 kN/m

Check for shear:

Check for deflection:

(ii) (Compression flange is laterally supported only at the supports i.e.

Let  = 110 N/mm2

Consider ISMB 550 @ 1.03 kN/m

t (thickness of the web) = 11.2 mm

T (thickness of the flange) = 19.3 mm

d1=h1=depth of the web = 467.5 mm

Ixx= 64893.6 cm4= 64893.6*106 mm4, Zxx = 2359.8 cm3

ryy= 3.73 cm =37.3 mm

Consider Table 6.1B

3. Design a beam of 5 m effective span, carrying a uniform load of 20 kN/m, if

4. A hall of clear dimensions 15 m x6 m is to be covered by RCC slab flooring

The compression flange of the beam is laterally restrained thought.

Load on beam per meter length:

(i) Slab weight = 1*3*0.12 ( thickness of the slab)*24= 8.64 kN/m

∗ . ∗.

 . ∗.

Consider ISLB 350 @ 0.495 kN/m

Check for shear:

Check for deflection:

 ) . ∗! (.∗! ))

(i) Given section ISMB 600 @ 1.23 kN/m

Properties of the section: Zxx= 3060.4 cm3

Therefore moment of resistance of the section MR = Zxx* =3060.4 * 103 * 165 =

If w kN be the total udl on the beam including self-weight

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Let = 110 N/mm2

(i) Slab weight = 130.12 ( thickness of the slab)*24= 8.64 kN/m

∗ . ∗.

. ∗.

) . ∗! (.∗! ))

Therefore moment of resistance of the section MR = Zxx =3060.4 103 * 165 =