• 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