• A rectangular beam 200 mm deep and 300
mm wide is simply supported over a span of 8
m. What uniformly distributed load per metre
the beam may carry, if the bending stress is
not to exceed 120N/mm2. Also find max.
Central concentrated load it can carry.?kN
1
?kN
8m
?
200
300
R
E
yI
M==
σZ
y
IM
maxmax
.max
σσ ==
;2
;12
2
.max
3
bdIz
dy
bdI
==
==Z=2x106
2
6.max
bd
y
Iz ==
For a S.S. Beam with u.d.l, max.
NmmwNmwwwL
BM 800088
8
8
22
==×
==
mkNmNw
w
wZBM
/30/000,30
80002000000120
8000.max
==
=×
== σ
3kNW
NmmWNmW
WWL
120
20002
4
8
42000000120
=
==
×==×
• A beam is of square section of side ‘a’.
bending stress is
• Find the moment of resistance when the
beam section is placed such that(i) two sides
are horizontal,(ii) one diagonal is vertical. Find
also the ratio of the moment of resistance of
the section in two positions
σ
the section in two positions
4
a
a
2/a
2/a
6
66
;2
;12
3
1
32
.max
.max
3
aM
abd
y
Iz
dy
bdI
σ=
===
==
26
2
12
)2
(2
2
;2
;12
2
3
3
.max
.max
3
a
a
aa
y
Iz
ay
bhI
=
×
==
=×=
5
61M σ=
26
23
2
aM σ=
22
1=
M
M
Composite beam(Flitched beam)
• Consists of two different materials---eg. Timber
beam reinforced with steel plates at top and bottom
or on sides
steel
timber
• 2 materials are connected rigidly, strain same at any
distance from neutral axis
steel
EE 2
21
1
σσ=
Total resisting moment=sum of resisting
moment caused by individual material
M=M1+M2 6
A composite beam consist of two timber joists 100 mm
wide and 300 mm deep with a steel plate 200 mm deep
and 15 mm thick placed symmetrically in between and
clamped to them. Calculate the total moment of
resistance of the section if the allowable stress in the
joist is 9 N/mm2 assume Es = 20Ewjoist is 9 N/mm assume E 20E
A.
100 10015
200
mm
300
mm
6 N/mm29 N/mm2
120 N/mm2
Stress distribution for steelStress distribution
for timber
In the stress diagram for timber the allowable stress is 9
N/mm2 (maximum in joist)
at the level of steel, stress in timber =
We know that
2/6100
150
9mmN=×
w
w
s
s
EE
σσ=
E
This is the maximum stress in steel
2/120206 mmN
E
E
w
sws =×==∴ σσ
wwssWS ZZMMM σσ +=+=
3522
106
20015
6mm
bdZ s =
×==
36
22
1036
300200
6mm
bdZ s ×=
×==
ZZMMM wwssWS +=+= σσ
9
Nmm
wwssWS
6651039103910120 ×=××+×=
BEAM OF UNIFORM STRENGTH
• The beam considered so far has uniform cross section throughout the span.
• Stresses in the extreme fibres of the section vary from section to section along the span.
• All extreme fibres are not loaded to their maximum capacity.
10
• A simply supported beam with symmetrical loading,
BMmax occurs at midspan.
• Maximum bending stress corresponding to maximum
BM
• But towards the support BM reduces to zero
• This implies that there is wastage of material in the
cross section of the beam.
11
�A beam is designed in such that every extreme fibrealong the span is loaded to its maximum permissible stress by varying the cross section, then it is known as beam of uniform strength
�For extreme fibre stress to be kept same at every section, bending moment at any section should be section, bending moment at any section should be proportional to section modulus.
�Since ( is kept constant )
�The cross section of the ideal beam can be obtained by the following ways
ZMZM ασ ;.= σ
12
beam of uniform strength
i. Keeping the width constant throughout and varying
the depth.
ii. Keeping the depth constant throughout the span and
varying the width
iii. By varying both width and depth
Eg. The leaf spring or laminated spring used as shock absorbers in heavy vehicles.
13
Q. A cantilever, 2.5 m long carries a UDL of 20 kN/m.
The breadth of the section remains constant and is
equal to 100mm. Determine the depth of the section at
midspan of the cantilever and at fixed end if stress
remains same throughout as 120 N/mm2
A.A.
Bending moment at x distance from free end,
Zwx
M .2
2
σ==
mmNmkNw /2010
1020/20
3
3
=×
==
14
20kN/m
2.5 m X
88.38 mm
176.77 mm
10012010120
202
2
22d
xbdx ××
=→×=
15At x =l/2; d = 88.38mm& At x=l =2500; d = 88.38mm
200
1.
12000
60
6
10012010
6120
2
20
2
2
xx
d
dx
bdx
==
××=→×=
Q.Determine the cross section of a rectangular beam of
uniform strength for a simply supported beam of 6m
span subjected to a central concentrated load of 20kN
by keeping a depth of 300mm throughout .take
permissible strength as 8N/mm2
Ans: Bending moment at a section x distance from left
support,support,
16
20 kN
6 mX
X
17
?
plan
20 kN
300 mm
elevation
18
20 kN
6 mX
X
19
250 mm
plan
20 kN
300 mm
elevation
20
A: Since, the section is
unsymmetrical, first the position
of neutral axis has to be found out.
260 mm
500 mm
N A
y
20
12
4608
21
180 mm
Portion y, mm Ay
Bottom flange 180 x 20 10 36000
Web 460 x 8 250 920000
Top flange 180 x 20 490 1764000
Top plate 260 x 12 506 1578720
22
2wl
23
8
2wl