Strength of Materials by R.K.Bansal_text

deflection of cantilevers 575

B.M. at A = -20xl-20x2 = ~60 kNm = - 60 x 10® x 103 Nmm 20 +30 + 100 =70k
T=
B.M. diagram is shown in Fig. 13.14 (6). T~ m mm(v 3 = 109
Nmm= 70 x 103 x 109 3
3 )

Nmm= 7 x 1013 3

Substituting this value in equation (i), we get

- = 7 mm. Ans.

( 6) 60 —1 The slope i.e., or 0 of a cantilever at the free end is given by,

kNm

B.M. Diagram

dx

Fig. 13.14 0B„ = -h- when the point load is at the free end
when the point load is at a distance of ‘o’ from the fixed end
To find the area of B.M. diagram, divide the Fig. 13.14 (£>) into two triangles and one 2EI when it carries a uniformly distributed load over the whole length.

rectangle. 0 B = 0c = 2 El

area A = 1 x CD x BC = — x 20 x 1 _ 11
x

Nmm= 10 kNm2 = 10 x 10s x 106 m mm( v 2 wa 3
2 2 - 106 _ 6 El when it carries a uniformly distributed load over a length
)
n ec ‘a’ from the fixed end.
Nmm= 1010 2 B

Similarly area A = CD x AC = 20 x 1 = 20 kNm2 w . I? _ w .(L - 3 when it carries a uniformly distributed load over a
2 a)
distance 1 a’ from the free end
=

6EI SEI

area A = ^2, x FD x EF = - x 1 x 40 = 20 kNm2 L3 when it carries a gradually varying load from zero at the free end to
3 w/m run at fixed end.
2 qB _ us .

.-. Total area of B.M. diagram, 24EI

A = Aj + A + A = 10 + 20 + 20 = 50 kNm2 where W = Point load,
2 3
m mmr 2 = 106 w = Uniformly distributed load,
Nmm= 50 x 103 x 10® 2 <(v 2

)

L = Length of beam,

Slope and deflection at the fixed end is zero. I = Moment of inertia, and

Let 0B = Slope at the free end B. E = Young’s modulus.
Then according to the moment area method,
2. The deflection i.e., y of a cantilever of length L, at the free end is given by,
— —Area of B. M. diagram
yB = 3 El when the point load is at the free end
eB=

—_ 50 x 103 x 106 - 0.005 rad.i.ans. Ans. Wa? Wa2 when the point load is at a distance of * a’ from the
10 s 85 yB= ~3EI
X + (L ~ a) fixed end
10 ~2EI '

Let yB - Deflection at the free end B. yvs — W when it carries a uniformly distributed load over the whole length
Then according to moment area method,
8 El
_ Ax
yB ~ In Ta = wa 4 w a3 Wf ~ when it carries a uniformly distributed load over a

Now let us find .r or Ax. + length ‘ a’ from the fixed end.
Then total moment of the bending moment diagram about B is given by
g Ej

A A A A. x = 3 x3
x+ 2 ie2 +

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m

f ,34 STRENGTH OF MATERIALS CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS

mProblem 14.2. A simply supported beam of length 4 carries a point load of 3 kN at a The loading on the conjugate beam is symmetrical
R,4* - R;i* — Half of total load on conjugate beam
m mmdistance of 1 from each end. If E = 2 x 10s N/mm2 and I = 10s 4 for the beam, then using

conjugate beam method determine :

( i ) slope at each end and under each load = ^ [Area of trapezoidal A*B*F*E*\
f(.E*F* + A* B*) x E*C*
{ji ) deflection under each load and at the centre.

Sol. Given :

Length, L=4m 1~

-1 (2 + 4) 31 4.5
X
Value of E = 2 x 10s N/mm2 = 2 x 10 5 x 10s N/m2
22 El\~ El
|_

= 2 x 10 5 x 10 3 kN/m2 = 2 x 10 s kN/m2 (i) Slope at each end and under each load

mm —I = 108
Value of 8 m4 = ICr4 4 ALet 0A = Slope at for the given beam i.e., at A

=4 1 (l . 0a = Slope at B for the given beam j
0 C = Slope at C for the given beam and
10
D0D = Slope at for the given beam
As the load on the beam is symmetrical as shown in Fig. 14.4 (a), the reactions R and
i Then according to conjugate beam method,
Rb will be equal to 3 kN.

Now B.M. at A and B are zero.

0A - Shear force at A* for conjugate beam = RA *

W= = = Q -000225 rad- Ans-

27io“8 7 io :*

Re = b* = = 0.000225 rad. Ans.
fi

0 C = Shear force at C* for conjugate beam

= Ra* - Total load A*C*D*
M_I A A=
“ xX i xX =

EI 2 EI El

= = °-°0015rad- AnS '
2 x 10 s^x 10'4

Similarly, 0^ = 0.00015 rad. Ans. (By symmeettrry)

(ii) Deflection under each load

Due to symmetry, the deflection under each load will be equal

Let yc - Deflection at C for the given beam and
yD = Deflection at D for the given beam.

Now according to conjugate beam method,

Fig. 14.4 yc - B.M. at C* for conjugate beam
= Ra * x 1.0 - (Load A*C*E *) x Distance of C.G. of A*C*E* from C*
B.M. atC = i?A xl = 3xl = 3 kNm
B.M. atZ)=ffs xl = 3xl = 3 kNm 4.5/1 3] 1
Now B.M. diagram can be drawn as shown in Fig. 14.4 (6).
Now by dividing the B.M. at any section by El, we can construct the conjugate beam as El l 2 El) 3

shown in Fig. 14.4 (c). The loading are shown on the conjugate beam. : 45 _ 05 = 40 4 x 1000
EI EI EI
Let Ra * = Reaction at A* for the conjugate beam and
Rb* = Reaction at B* for conjugate beam 4

2 x 10 s x 10' 4

; 0.2 mm. Ans.

Also yD - 0.2 mm.

STRENGTH OF MATERIALS

Deflection at the centre of the beam
= B.M. at the centre of the conjugate beam
! x 2.0 - Load A*C*E*
x Distance of C.G. of A*C'*E* from the centre of beam
- Load C*H*J*E*
x Distance of C.G. of C*H*J*E* from the centre of beam

__9 2 L5J. _
~ El El lEl El

6.5 x 1000

r—

2 x 10 4
— m mm_ 6.5
" 2 x 10 8 4~r

x 10O'

= 0.325 nun. Ans.

mProblem 14.3. A simply supported beam, AB of span 4 carries a point of 100 kN
mmat its centre C. The value fl for the left half is 1 x 10s
4 and for the right half portion I is

mm2 x 10s 4 Find the slopes at the two supports and deflection under the load.

.

Take E = 200 GN/m2.

Sol. Given :

Length, L- 4 m
Length AC = Length BC = 2 m

Point load, W = 100 kN

Moment of inertia for AC

I = lx 108 mm4 m m4 = 10~4 4

Moment of inertia for BC

mm= 2 x 10s 4

m= 2 x 10-4 4 = 2/ (v =

Value of E = 200 GN/m2 = 200 x 109 N/m2

= 200 x 10s kN/m2.

AThe reactions at Band will be equal, as point load is acting at the centre.

fl A = fl =^r=50kN
fi

Now B.M. at A and B are zero.

B.M. at C = Ra x 2 = 50 x 2 = 100 kNm

Now B.M. can be drawn as shown in Fig. 14.5 (b).

Now we can construct the conjugate beam by dividing B.M, at any section by the prod-
uct of E and M.O.I.

The conjugate beam is shown in Fig. 14.5 (c). The loading are shown on the conjugate
beam. The loading on the lengthA*C* will beA*C*D* whereas the loading on length B*C* will

be B*C*E*.

The ordinate C*D* B.M. at C 100

" fix M.O.I. for AC El

CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS 589

(i) Slopes at the supports at A for the given beam

A= Slope at i.e.,

0B = Slope at B i.e., at B for the given beam

The according to the conjugate beam method,

e = Shear force at A* for conjugate beam = RA*

10 6 4 = 0.004166 rad. Ans.

3 x 200 x x 10

RShear force at B* for conjugate beam = *
B

200 — 200 10' 4, —= 0.003333 rad. Ans.
3 El
3 x _: x

200 x 10 6

(ii) Deflection under the load

Let y c = Deflection at C for the given beam.

Then according to the conjugate beam method,

= B.M. at point C* of the conjugate beam
- (Load A*C*D*) x Distance of C.G.
- RA * x 2
Oyc
of A*C*D* from

—1 X 2o X 100 x f— x 2
El )'\3 )
2
u3EI

_ 500 _~ 200 ~_ 100
" 3EI
3 El El

— m100 To find reactions RA and RD, take moments about A.
= r T4 Rd x 30 = 150 x 10 + 300 x 20 = 7500
s
200 x x 10
10

- m = —L_ x 1000 = 5 mm. Ans. 250 kN
200
Problem 14.4. A beam 200 simply supported at its ends A and, D^over a span , Ra - Total load - R p

metres. It is made up ABCD is of = (150 + 300) - 250 = 200 kN.
of three portions AB, BC and CD each 10 min length. The moments
30 of

inertia of the section of these portion are I, 31 and 21 respectively, where I 2 x Now draw B.M. diagram

beam carries a point load of 150 kN at B and a point B, C and D. Take E 2 B.M. at A and D = 0

of the beam calculate the slopes and deflections at A, B.M. at B = Ra x 10 = 200 x 10 = 2000 kNm
B.M. at C = RD x 10 = 250 x 10 = 2500 kNm
Sol. Given ;
B.M. diagram is shown in Fig. 14.6 (b).
Length, L = 30 m

Length AB = Length BC = Length CD = 10 m Now construct the conjugate beam as shown in Fig. 14.6 (c) by dividing B.M. at any
section by their product ofE and/. For the portion AB corresponding conjugate beam is A*B*C*,
X^ 11 0A_z2 for the portion BC corresponding conjugate beam is B*C*H*K* and for the portion CD the
MI..On.Ir. o-frAiBD, mm/I —= O2 xv. 110ft1l0O 4= m m4 °2 x" ”' 4
: corresponding conjugate beam is C*D*F*. The loading are shown in Fig. 14.6 (c).
.

M.O.I. of BC, m3/ = 6 x 10‘2 4 The ordinates B =:7?- , —B*K* = -~
M.O.I. of CD, m21 = 4 x 10-2 4
Point load at El 3El
Point load at B = 150 kN
Value of C = 300 kN 2500 , cm* = 2500
E = 2 x 10 2 kN/mm2 = 2 x 102 x 10 6 kN/m2 = 2 x 10 s kN/m2 c*f* =

STRENGTH OF MATERIALS ) 591
590
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS

77500 77500 .

H JTT . y. ~ 2500 2000 500 ~ 27EI 27 x 2 x 10 8 x 2 x 10~2
3 El " 3El
3EI = 0.0007176 radians. Ans.

Let Ra * = Reaction at A* for conjugate beam (c) Slope at C for the given beam
= Reaction at D* for conjugate beam.
RV* RTo find . beam , ,

and take the momenls of a11 l0adS aCting °n the COnjUgate = S.F. at C* for conjugate beam

d*’ = Rd* - Load D*C*F*

W6 £T0t R* X 30 = (j x A*B* X B*E*) x(| x A*B*) + (B*C* x B*K f )

fx (10 + ) + (| x K*J* x H*J*) x (10 + 10 x §) 293750 1 2500

27El 2 2EI

+ (| x C*F* x C*D*) x (20 + 10 x |) 293750 6250 293750-27x6250 125000

=

27EI EI 27 EI 27EI

xIOxt 125000
27 x 2 x 10 B x 2 x 10• '“i = 0.001157 rad. Ans.

Dd(i ) Slope at for the given beam

200000 300000 125000 437500 = S.F. at D* for conjugate beam

= " + +i | ~ 293750
3El 3EI 9El D 21EI
3El ' =

600000 + 900000 + 125000 + 13 12500 _ 2937500
m= 9 293 5 °
9£7 = 27 x 2 x 10 J8 x 2 x 10'2 = 0.00272 rad. A^s.
'

2937500 293750 DC(ii) Deflection at A, B, and

REd* ~- 9EIl Xx 30 “ 27El A() Deflection at for the given beam
RA * = Total load on conjugate beam - RD *
= B.M. at A* for the conjugate beam

293750 = 0. Ans.

27El ( ) Deflection at B for the given beam

r^r' 293750 = B.M. at B* for the conjugate beam
27EJI = R„* x 10- Load A*B*E* x Distance of C.G. ofA*B*E* from B*
10000 + 20000 + 2500 +> 6250 j
—= 347500 x 10 - (—1 x 10 x —2000 j x 10
~3Ef 3EI EI j.

f 300 00 + 20000 + 2500 + 18750 ^ _ 293750 U27 EI EI J 3
~ 3 EI 277E£I
= [ J

3475000 100000

71250 293750 641250-293750 347500 27 EI 3EI
=
~lEl 27El ~ WEI 27EI _ 3475000 - 900000 2575000

D(i) Slopes at A, B, C and 27EI 27EI

According to conjugate beam method m2575000

() Slope at A for the given beam = 7: ^ 8 10T'?2 = 0.02384
2
= S.F. at A* for conjugate beam 27 x x 10 8 x 2 x

= 23.84 mm. Ans.

9a = R a* = 347500 ~ 347500 Cc(i Deflection at for the given beam
21EI 27 x 2 x 10 8 x 2 x 10'2
= B.M. at C* for the conjugate beam

= 0.003218 rad. Ans. “a ~x 10 - Load D*C*F* x Distance of C.G. of D*C*F* from C*

() Slope at B for the given beam —293750 x l1n0 1 x 1l0n x 2500 x 10

- S.F. at B* for conjugate beam 27EI 2 2 EI 3

= Ra* - Load A*B*E* 2937500 2937500 - 62500 x 9

347500 1 in 2000 27EI

27 EI 2 EI

347500 10000 _ 347500 - 270000
HOT ~
: WEI

27EI

)*

592 STRENGTH OF MATERIALS CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS 593

23 75000 2375000 m= 0.02199
21EI
27 x 2x 10 s x2 x 10~2

= 21.99 mm. Ans.

D( d Deflection of for the given beam = 0. Ans.

14.5. RELATION BETWEEN ACTUAL BEAM AND CONJUGATE BEAM

The relations between an actual beam and the corresponding conjugate beam for differ-

ent end conditions are given in Table 14.1.

TABLE 14.1

S.No. Actual beam Conjugate beam

1. Simply supported or roller supported end Simply supported end B.M. = 0 but S.F. exists Fig. 14.7
(Deflection = 0 but slope exists)
2. Free end (slope and deflection exist) Fixed end (S.F. and B,M. exist) mProblem A14.5. cantilever of length 3 carries a point load of 10 kN at a distance of
3. Fixed end (slope and deflection are zero) Free end (S.F. and B.M. are zero)
4. Slope at any section S.F. at the corresponding section m mm2 from the fixed end. If E = 2 x JO3 N/mm2 and I = 10s 4 find the slope and deflection at
5. Deflection at any section B.M. at the corresponding section ,
6. Given system of loading The loading diagram is M/EI diagram
7. B.M. diagram positive (sagging) M/EI load diagram is positive (i.e., loading the free end using conjugate beam method.
is downward)
8. B.M. diagram negative (hogging) M/EI load diagram is negative (i.e., loading is Sol. Given : B=3m
upward) Length,
Point load, W = 10 kN
14.6. DEFLECTION AND SLOPE OF A CANTILEVER WITH A POINT LOAD AT THE Distance
FREE END Value of AC = 2 m

Value of E = 2 x 10 s N/mm2

WABFig. 14.7 (a) shows a cantilever = 2 x 10 5 x 10s N/m2 = 2 x 10 8 kN/m 2

of length L and carrying a point load at the free mm/ = 10s 4

end B. The B.M. is zero at the free end B and B.M. at A is equal to W.L. The B.M. diagram is

shown in Fig. 14.7 b( ). The conjugate beam can be drawn by dividing the B.M. at any section by —^= 10s x m4 = HP4 m4

El. Fig. 14.7 (c) shows the conjugate beam A*B* (free atA* and fixed atB*). The loading on the 10 12

conjugate beam will be negative (i.e., upwards) as B.M. for cantilever is negative. The loading B.M. at B = 0

on conjugate beam is shown in Fig. 14.7 (c). B.M. at C = 0

Let B= Slope at i.e., at B for the given cantilever and B.M. at A = - 10 x 2 = - 20 kNm

j Now B.M. can be drawn as shown in Fig. 14.8 (6). Now construct conjugate beam A*B*

ByB = Deflection at for the given cantilever. (free atA* and fixed at B*) by dividing the B.M. at any section by El, as shown in Fig. 14.8 (c).
The loading on the conjugate beam will be negative (i.e., acting upwards) as B.M. is negative.
Then according to the conjugate beam method,

0B = S.F. at B* for the conjugate beam —Let
= Load B*A*C* ( dy')

II
B0 S = and
Slope at the free end for the given cantilever i.e., at

1 x A*B* x A*C* = - x L x W.L W.L2 yB - Deflection at B for the given cantilever.
Then according to the conjugate beam method,
2 2 El 2EI
0S = S.F. at B* for conjugate beam
and yB = B.M. at B* for the conjugate beam = Load A*C*D * = L x A*C* x A*D*
= Load B*A*C x Distance of C.G. of B*A*C* from B*

0

594 STRENGTH OF MATERIALS CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS

595

20
*eTm1 * 2„ 20

2

s = 0.001 rad. Ans.

2 x 10 x 10"

3m
Given beam

Conjugate beam

Fig. 14.8

yB = B.M. at B* for the conjugate beam
= (Load A*C*D*) x Distance of C.G. of A*C*D* from B*

—1 x 2 x 20 1+—x 2

2 El 3

=_ 20 * 7 ~ 20 10" 4 * 7
El 3 10 8 x 3
2x

= 0.00233 m = 2.33 mm. Ans.

mProblem 14.6. A cantilever beam AB of length 2 is carrying a point load 10 h,N at B.
mmThe moment of inertia for the right hall of the cantilever is 1 s
4 whereas that for the left

mmhalf is 2 x 10s E4 = 2 x 10s kN/m2 find the slope and deflection at the free end of the
. If ,

cantilever.

Sol. Given : l = 2m
Length,
Point load, W = 10 kN
Length,
AC = length BC = 1 m

M.O.I. of length BC, mm2 x 10s 4 = 2 x Hr

M.O.I. of length AC

Value of E = 2 x 108 kN/m2

B.M. atB = 0

= 0.000625 rad. Ans.

— —* . *

STRENGTH OF MATERIALS CONJUGATE BEAM METHOD, PROPPEO CANTILEVERS AND BEAMS

597

yB = B.M. at B* for the conjugate beam
= Load A*C*F*H x Distance of its C.G. from B*

+ Load H*E*F* x Distance of its C.G. from B*

+ Load A*C*D* x Distance of its C.G. from B*

x 1.5 + f—i x lx —g _Wl —,l + ~2 x 1 + -xli x 10 X (| X 1)
3 ) \2 El
V 2 El

7.5 25 10 45 + 25 + 20
El + 6El + 2El "
0EI

90 15 15
- 6 El " El
2 x 10 O8 x -4A

10

= 0.00075 m = 0.75 mm. Ans.

mAProblem 14.7. cantilever of length 3 carries a uniformly distributed load of
mm80 kN/m length over the entire length. IfE = 2 x 108 kN/m2 and I = 10s4 and
find the slope
,

deflection at the free end using conjugate beam method.

Sol. Given : L=3m
Length,

U.d.l., w = 80 kN/m
E = 2 x 108 kN/m2
Value of

m m4 = 10-4 4 C*

Value of mmI = 108 4=

Fig. 14.10

B.M. at and yB - B.M. at B* for conjugate beam
B.M. at
A = - (ui.L) ,L^=-80x3x-3=- 360 kNm = Load A*C*B* x Distance of its C.G. from B*

The variation of B.M. between A and B is parabolic as shown in Fig. 14.10 (bl = fl x 3xM) x 3£ = 3^3x3 = 810
Now construct conjugate beam A*B* (free at A * and fixed at B*) by dividing the B.M. at V3 El J 4 El 4 El
any section by EL The loading diagram will be as shown in Fig. 14.10 (c).
810
Let 0 B = Slope at B for the given cantilever and m=
yB = Deflection at B for the given cantilever 72, x 1770s x 7TT1 = 0.0405 = 40.5 mm. Ans.
10
Then according to conjugate beam method,
14.7. PROPPED CANTILEVERS AND BEAMS
0B = S.F. at B* for conjugate beam
= Load B*A*C* or Area of B*A*C When a cantilever or a beam carries some load, maximum deflection occurs at the free

= - of the rectangle containing parabola end in case of cantilever and at the middle point in case of simply supported beam. The deflection
can be reduced by providing vertical support at these points or at any suitable point. Propped
3 cantilevers means cantilevers supported on a vertical support at a suitable point. The vertical
support is known as prop. The props which does not yield under the loads is known as rigid.
= | X (A*B* X A*C*) The prop (or support) which is of the same height as the original position of the (unloaded)
cantilever or beam, does not allow any deflection at the point of support (or prop) when the
—= —1 x 3, x 360 cantilever or beam is loaded. The prop exerts an upward force on the cantilever or beam. As
the deflection at the point of prop is zero, hence the upward force of the prop is such a magnitude
3 El as to give an upward deflection at the point of prop equal to the deflection (at the point of prop)
due to the load on the beam when there is no prop.
360 360
~ Hence the reaction of the prop (or the upward force of the prop) is calculated by equat-
~W= s 10' 4 ing the downward deflection due to load at the point of prop to the upward deflection due to
2 x x
10 prop reaction.

= 0.008 rad. Ans.

3 CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS

STRENGTH OF MATERIALS Equating equations (i) and (»), we get
598
PL3 SWL3
14.8. LS.OF.AADNADTB.TMH.EDCIAEGNRTARMESAFNODRPARPORPOPPEPDEADTCTAHNTEIFLREEVEERECNADRRYING A POINT
A AFig. 14. 11 (a) shows a cantilever B of length L fixed at and supported on a prop at B 3El 48El
„ 5 ...
Wcarrying a point load at the centre.
..(14.1)
(i) S.F. Diagrams B=-P
(Minus sign due to right upwards)
S.F. at

The S.F. will remain constant between B and C and equal to (-)

W W WC„ =5
SQ.1F?. at, ... 11
+ =+

16 16

The S.F. will remain h 11W between C and A.

16

The S.F. diagram is shown in Fig. 14.11 b{ ).

(ii) B.M. Diagram

S.F. Diagram B.M. at B-0

SWL Y+? B.M. at 5W L 5WL
B.M. at
32 16 xX 2 "_ 32

8.M. Diagram —A. = 5W x Lr W.L

16 2

_ SWL - 8WL 3 WL

16 16

The B.M. diagram is shown in Fig. 14.11 (c). As the B.M. is changing sign between C and

Fig. 14.11 A, hence there will be a point of contrafluxure between C and A. To find its location, equate the

Let P = Reaction at the rigid prop. B. M. between A and C to zero.

WTo find the reaction P at the prop*, the downward deflection due to at the point of AThe B.M. at any section between C and at a distance * from B

prop should be equal to the upward deflection due to prop reaction at B. = ™« x -wL-k)

WNow we know that downward deflection at point B due to load 16 V 2 J

/T\ /J Equating the above B.M. to zero, we get

— —= LjlL + ) (See equation 13.4) —5W .x-„W7 (U-—L\ =0

3El 2El V2J

WL3 WL3 Sx —L =0 •

~+ n16 2
24El 16 El _ X _~ _ l
1Q 2
_ 2 WL3 + 3WZ,3 5 WL3

48 El “ 48El X ~_ 16L 8L
11x2 ” 11
The upward deflection at the point B due to prop reaction P alone

Hence the point of contraflexure will be at a distance 8L/11 from B or 3L/11 from A.

W*Never calculate P by equating the clockwise moment due to the load to the anticlockwise 14.9. S.F. AND B.M. DIAGRAM FOR A PROPPED CANTILEVER CARRYING A UNI-
FORMLY DISTRIBUTED LOAD AND PROPPED AT THE FREE END
moment due to P at the fixed end, as at the fixed end there exist a fixing moment.
AFig. 14.12 (a) shows a cantilever AB of length L fixed at and propped at B, carrying a

uniformly distributed load of ic/unit length over its entire length.

STRENGTH OF MATERIALS ’conjugate beam method, propped cantilevers and beams

S.F. Diagram — 'j 3wL
-J .8.

B.M. Diagram

Fig. 14.12

Let P = Reaction at the prop.
To find the reaction P at the prop, the downward deflection due to uniformly distributed
load at B should be equated to the upward deflection due to prop reaction at B.
We know that downward deflection at point B due to u.d.l.

= —— ...(i) {See equation 13.6)

8 El ...(ȣ)

The upward deflection at point B due to prop reaction P alone

™=
3El

Equating equations (?) and (ii), we get

PZ? w_jS_

=

3 El 8El

P-- W . L ...(14.2)

(i) S.F. diagram B =- P (Minus sign due to right upwards)

S.F. at

= - - wL

8

The S.F. at any section at a distance x from B is given by

F =-~w.L+w.x ...{Hi)
o

A AThe S.F. varies by a straight line law between and B. S.F. at is obtained by substi-

tuting x = L in the above equation.

L2.

602 STRENGTH OF MATERIALS CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS

(v) Maximum deflection

El d 2 y£ - —3 W.Lr.X wx 2 Maximum deflection takes place where
dx 2 8 2
is zero. Differentiating equation (vii) w.r.t.

Integrating the above equation, we get x, we get

dy 3w . L.x 2 -H * - 3wL 2 4iv.x 3 wl}

+ Cl El
23 1 dx

-O.w.LT.x 2 --W.*3. + C, 24 48

w L3 . o w .x 3 w 3

J•LV .L

Integrating again, We get 16 48

— —w3 x Putting, 0: we get
LT ,
.
+ CjX + C
2
16 3 6 4 — UnU _= 3 w.L.x2 wx 3 w .

w.L.x 3 i . x4 + CjX + C 16 6 48
1 2
16 24 0 = 9w .L.x2 - 8w x3 - wLs
. .

where C and C2 are constant of integration. At the fixed end the slope and deflection are zero. The above equation is solved by trial and error. Hence we get
y

At the end B, deflection is zero. Hence at B, x = 0 and y = 0. x = 0.422L

Substituting x = 0 and y - 0 in equation (vi ), we get ...(14.4)

0 = C0 Substituting this value in equation (vii), we get maximum deflection.

Substituting x = L and y = 0 in equation (oi), we get ~ ^x (.422L)3 - x (0.422L)
(0.422L)4 -
lb z4 48
0 = W L L- - _
. La + C. . L + 0 (v C9 = 0) = - 0.005415mL4

16 24 0.005415tc.L4

wl} w.L3 „ ymax _ EI

~

16 24 1 .'. Maximum downward deflection

w&£ ~_ wl} _ wl} —= 0.005415 w Lri,
1 *24 wl} 2wl} - 3 48"

=

Iff" 48 ...(14.5)

Substituting the values of C and C in equation (vi), we get mA kNProblem 14.8. cantilever of length 6 carries a point load of 48 at its centre. The
x2

——E_lxy = w.L.x3 cantilever is propped rigidly at the free end. Determine the reaction at the rigid prop.
w4 - wl} x.

x. •

16 24 48 Sol. Given :

The above equation gives the deflection at any section of the cantilever. Length, L=6m

—The deflection at the centre of the cantilever is obtained by substituting x = in equa- Point load, W = 48 kN

Let p = Reaction at the rigid prop

tion (vit). If yc is the deflection at the centre then, we have Using equation (14.1), we get

UJUJ—wL (( LY
El . = Xx w fLY P=-|xW
c 16
16 1 24
—5
wl} wL4 wl}
= x 48 = 15 kN. Ans.
16 x 8 " 24 x 16 " ~96~
lb

3wl} - wl} - AwLi 2 wl} mProblem 14.9. A cantilever of length 4 carries a uniformly distributed load of
24x16 24x16 rigidly at the free end. If the value
lkN/m run over the whole length. The cantilever propped
is

mmof E = 2 x 105 N/mm2 and 1 of tlu cantilever = 10s 4 then determine :

,

Ci ) Reaction at the rigid prop,

(Negative sign means that deflection is downwards) (ii) The deflection at the centre of the cantilever,

Downward deflection, (Hi) Magnitude and position of maximum deflection.

...(14.3) Sol. Given : L = 4m
Length,

STRENGTH OF MATERIALS P

CONJUGATE BEAM METHOD. PROPPED CANTILEVERS AND BEAMS

U.d.l. w = 1 kN/m run

Value of ~ E = 2 x 105 N/mm2 = 2x 105 x 10s N/m2

Value of = 2 x 10 1L N/m2

mm m mI = 108
4 = 108 x 10" 12 4 = 10'1 4

(i) Reaction at the rigid prop Fig. 14.13

Let P = Reaction at the rigid prop To find the reaction P at the prop, the downward deflection due to uniformly distributed
Using equation (14.2), we get load on the AB at point C should be equated to the upward deflection due to prop reaction at C.

P = —3 x w . L We known that downward deflection at point C due to u.d.l. on length AB is given by,

8 —wL, 4

= —g x 1 x 4 = 1.5 kN. Ans. ~r r- +

8 8El GEI

(ii) The deflection at the centre of the cantilever 1 x 4 4 lx4 3

Let yc - Deflection at the centre of cantilever 8El 6El
96 + 64 _ 160
Using equation (14.3), we get
3El 3El
wl 4 (v w = 1 kN = 1000 N)
y c = Eli1r9\2n lT'T The upward deflection at point C due to prop reaction P alone

1000 X 4 4 PI? Px 6 3 72P

192 x 2 x 10 11 x 10'4 ~ 3El ~ 3El ~ El
Since both the deflections given by equations (i) and (ii) should be equal.
256 m 2 1000
^ 4a min 160 72
:,
3 10 3El~ El
4 Ans.
160
384 x 10
P = -r~~- = 0.741 kN. Ans.
= 0.0667 mm.

(iii) Magnitude and position of maximum deflection 14.10. S.F. AND B.M. DIAGRAMS FOR A SIMPLY SUPPORTED BEAM WITH A UNI-
FORMLY DISTRIBUTED LOAD AND PROPPED AT THE CENTRE
The position of the maximum deflection is given by equation (14.4).
Fig. 14.14 (a) shows a simply supported beam AB of length L propped at its centre C and
x = 0.422 xL
carrying a uniformly distributed load of m/unit length over its entire span.
= 0.422 x 4 = 1.688 m.
Let P = Reaction of the prop at C
mHence maximum deflection will be at a distance 1.688 from the free end of the canti-
To find the reaction P at the prop, the downward deflection at C due to uniformly dis-
lever. tributed load should be equated to the upward deflection at C due to prop reaction.

Maximum deflection is given by equation (14.5) The downward deflection at the centre of a simply supported beam due to uniformly
ymax ~_ 0,00 5415m . L4
distributed load is given by,
0.005415 x 1000 x44 ^ (v ^ = l kN = 1000 N)

2 x 10 u x 10' 4

mm= 0.005415 x 1000 x 256 x 1000 y° - 5wL4 W'"
2 x 10 7 384 El

= 0.0693 mm. Ans. The upward deflection of the beam at C due to prop reaction P alone is given by,

Problem 14.10. A cantilever ABC is fixed at A and rigidly propped at C and is loaded as PL3 ~(H)
y°~
shown in Fig. 14.13. Find the reaction at C.
48E/
Sol. Given : Equating equations (i) and (ii), we get
Length,
L=6m PL3 ~ 5wL4
U.d.L,
w = 1 kN/m 48El 384El
Loaded length,
Let L =4m
l
P = Reaction at the prop C.

— s— STRENGTH OF MATERIALS F 607

606 CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS (v W=w.L)

5 wL 48BI — Cx = and hence S.F. at will be,
A
= — .w .L = —
„ 3W wL
88
16"
^w/Unit Length _3W W

16 2 5W

3W - 8W

t X G‘ J — —Hence for the span AC, the S.F. changes uniformly from + - at A to - — at C.
lb lb
IrA P rJ

X ~T 5W 3W
5 wL
Similarly for the span C/1 the S.F. will change uniformly from + ~~z at C to at LI.
16
, lb 16
s *sss *
5 wL ALet at a distance x from in the span AC, the S.F. is zero. Equating S.F. as zero in
S.F. Diagram
16 equation (i), we get

jl 0=3~W-wxx

22 = ~T16X~~ WX W(v = w . L)

/+ X?9wL 9wL 3L
512
/. + . *
y/ / / j 16

3L

Hence S.F. is zero at a distance — —3L 3L
from A. Also S.F. will be zero at a distance from
lb lb

B.M. Diagram B due to symmetry. Now the S.F. diagram can be drawn as shown in Fig. 14.14 b{ ).

Fig. 14.14 (ii) B.M. Diagram

Now reactions RA and RB can be calculated. Due to symmetry, the reactions RA and RB AB.M. at is zero and also at B is zero.

would be equal. X AB.M. at any section at a distance x from is given by,

But Ra + Rb + P = Total load on beam M R w —=
x
= w.L = W , .x- x. .
2

^=W Rb = Ra and P = ~_ 3wL- _ w.x 2 R„4a = 3W or 3w.L
le '
2~ 16 16
*
—R„a = r1(W.„--5—W \ =1-x3—W = 3W
8J 2 8 —The B.M. at C will be obtained by substituting x = in the above equation.

R„a- r„b~ 3W mc =^.

- 16 4-

16 2

( i ) S.F. Diagram WA, - R„a - 3 3w.L2 w . l} 3wL2 - 4wL2

S.F. at 8“"

X AThe S.F. at any section at a distance x from is given by, 32 32

v = 3W wx ...(14.6)
x 16

'

STRENGTH OF MATERIALS CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS 609

Now the B.M. will be maximum where S.F. is zero after changing its sign. But S.F. is 14.11. YIELDING OF A PROP

"zero after changing its sign at a distance x = 3L fr rom A. In case of a rigid prop the downward deflection due to load is equal to the upward
deflection due to prop reaction. But if the prop sinks down by some amount say 6, then down-
Hence by substituting x = yj-g- in equation (it), we get maximum B.M. ward deflection due to load is equal to the upward deflection due to prop reaction plus the
amount by which the prop sinks down.

LYwL3 If y l = Downward deflection of beam at the point of prop due to load,
|S.t-i 3 y2 = Upward deflection of the beam due to prop reaction, and
6 = Amount by which the prop sinks down
Max. B.M.

16 16 2 v 16 J

9ivL2 9wL2 18wL2 - 9wl? Then y=y2 + 6 ...(14.7)

256 2x256 2x256 AProblem 14.12. cantilever of length L carries a uniformly distributed load w per unit

9wL2 length over the whole length. The free end of the cantilever is supported on a prop. If the prop

512 sinks by 5, find the prop reaction.

To find the position of point of contraflexure, the B.M. must be equated to zero. Hence Sol. Given :

Msubstituting x = 0, in equation <ii), we get Length =L
=w
„ 3wL wo U.d.l.
0
= .x x. Sinking of prop =6

16 2

or t w(Cancelling . x to both sides) The downward deflection (y j ) of the free end of cantilever due to uniform!” distributed
=

16 2 W ^‘

3L „ 3L load is equal to .
*
0r * 16 8

“

Now the B.M. diagram can be drawn as shown in Fig. 14.14 (c). The upward deflection (y 2) of the free end due to prop reaction P will be equal to

A mProblem 14.11. uniform^girder of length 8 is subjected to a total load of 20 kN Now using equation (14.7), we get

uniformly distributed over the entir'e length. The girder is freely supported at its ends. Calcu- J'i=y2 + 6

late the B.M. and the deflection at the centre. wL? PL3

If a prop is introduced at the centre of the beam so as to nullify this deflection, find the + s,

net B.M. at the centre. 8El 3El
PL3 wL4
Sol. Given : L-8m
Length, ~
Total load, W = 20 kN
W 20 3El 8El

= 2.5 kN/m.

(i) The deflection at the centre of a simply supported beam carrying a uniformly distrib- Problem 14.13. A simply supported beam of span 10 rn carries a uniformly distributed

uted load is given by (without prop) ^ Nload of 1152 per unit length. The beam, is propped at the middle of the span. Find the amount,

_ 5wL4 _ 5 x 2.5 x 8 4 = 400 by which the prop should yield, in order to make all the three reactions equal.

y' 384 El 3EI mmTake E - 2 x 10s N/mm2 and I for beam = 10B 4

384 £/ .

where El - Stiffness of the girder. Sol. Given :
Span,
(it) The B.M. at the centre of a simply supported beam due to uniformly distributed load L = 10 m

only (t.e., without prop) is given by U.d.l., w = 1152 N/m

—M —= = — = 20 kNm. Ans. Value of E = 2 x 10s N/mm2 = 2 x 105 x 106 N/m2
= 2 x 10u N/m2
88 Value of
Total load on beam, mm m mI = 108
(Hi) Net B.M. at the centre when a prop is introduced at the centre 4 = 10 s x 10“12 = 104 f| 4

MLet c = Net B.M. at centre when a prop is provided. W - w . L - 1152 x 10 = 11520 N

Now using equation (14.6), we get R RIf all the three reactions (i.e., A , B and P) are equal, then each reaction will be one

M,,„c = wL2 2.5 x 8 2 — _ k. N-jm. A.ns. third of the total load on the beam.
32
= 5
32

STRENGTH OF MATERIALS ): 611
610
conjugate beam method, propped cantilevers and beams

M(ii) B.M. at fixed end, =

—(Hi) Point of contraflexure, x =

wL4

(to) Deflection at the centre, yc = TKnpj

Fig. 14.16 10. (.o)..Ma.x.imum ,„ - 0.005415wL4

deflection, y =
El
=-W
Ra Rb p 11520 where w = Uniformly distributed load,
3 = 3840 N.
= =
x - Distance from free end.

Let 5 = Amount by which the prop should yield if all the three reactions are equal. For a simply supported beam, carrying a uniformly distributed load over the entire span and

Now the downward deflection of the beam at the centre due to uniformly distributed propped at the centre, we have

load alone is given by, — W5

1152 x 104 (f) Prop reaction, P =

5 wL4 2 x 10 u x 10‘4

384 El
m3,1 ~
~_ 5 R(it) Support reactions, A = RB = 3W
384 *

m mm mm= -^4- = -^4 x 103
= 7-5 - M(Hi) B.M. at centre, = -
oA
3 3

10 10

The upward deflection due to prop reaction at the point of prop is given by, (in) Point of contraflexure, x - ~3~L~

PL3 3840 x 10 3 o
48El 48 x 2 x 10 n x 10'4
my* ~ ~_ (v P = 3840 N) Wwhere = Total load on beam

= w.L

= iP_ m= i°JLl0i mm = 4mm w = Uniformly distributed load on beam
10 4 10 4
x = Distance from the support.
No\V using equation (14.7), we get
EXERCISE 14
yx = y2 + 8

6 = y 1 - y2 = 7.5 - 4.0 = 3.5 mm. Ans,

(A) Theoretical Questions

HIGHLIGHTS 1. Define and explain the terms : Conjugate beam, conjugate beam method, flexural rigidity and
propped beam.
1. The conjugate beam method is used to find the slope and deflections of such beams whose flexural
2. What is the use of conjugate beam method over other methods ?
rigidity (i.e., El) is not uniform throughout of its length. 3. How will you use conjugate beam method for finding slope and deflection at any section of a

2. Conjugate beam is an imaginary beam of length equal to that of original beam but for which load given beam ?
diagram is M/El diagram. 4. Find the slope and deflection of a simply supported beam carrying a point load at the centre,

3. The load on conjugate beam at any point is equal to the B.M. at that point divided by El. using conjugate beam method.
4. The slope at any section of the given beam = S.F. at the corresponding section of the conjugate ,

beam. A5. cantilever carries a point load at the free end. Determine the deflection at the free end, using
5. The deflection at any point of the given beam = B.M. at the corresponding point of the conjugate
conjugate beam method.
beam. 6. What is the relation between an actual beam and the corresponding conjugate beam for different
6. Propped cantilevers means cantilevers supported on a vertical supported at a suitable point.
7. The rigid prop does not allow any deflection at the point of prop. end conditions ?
8. The reaction of the prop (or the upward force of the prop) is calculated by equating the downward
7. What do you mean by propped cantilevers and beams ? What is the use of propping the beam ?
deflection due to load at the point of prop to the upward deflection due to prop reaction. 8. How will you find the reaction at the prop ?
9. For a cantilever carrying a uniformly, distributed load over the entire span and propped rigidly
9. A cantilever of length L, carries a uniformly distributed load of w/m run over the entire length .
at the free end, we have
It is rigidly propped at the free end. Prove that
~(i) Prop reaction, P = w.L
o —3
(i) Prop reaction = w . L and

o

(ii Deflection at the centre = WL4--

.

STRENGTH OF MATERIALS
612

10. A simply supported beam of length L, carries a uniformly distributed load of uitm run over the

entire span. The beam is rigidly propped at the centre. Determine :

(i) Prop reactions. 15

(ii) Support reactions, Fixed and Continuous Beams

(in) B.M. at the centre, and 15.1. INTRODUCTION
A beam whose both ends are Fixed is known as a fixed beam. Fixed beam is also called a
(to) Point of contraflexure, if any.
built-in or encaster beam. In case of a fixed beam both its ends are rigidly fixed and the slope
(B) Numerical Problems and deflection at the fixed ends are zero. But the fixed ends are subjected to end moments.
Hence end moments are not zero in case of a fixed beam.
A m1 beam 6 long, simply supported at its ends, is carrying a point load at 50 kN at its centre.
mm EThe moment of inertia of the beam is 76 x 10s = 2.1 x 10 5 N/mm2 determine the slope (a)
4 If ,
(i»
.
Deflection curve
at the supports and deflection at the centre of the beam using conjugate beam method. Fig. 15.1

[Ans. (t) 3.935 and 13.736 mm] In case of simply supported beam, the deflection is zero at the ends. But the slope is not
zero at the ends as shown in Fig. 15.1 (a).
m2. A simply supported beam of length 10 m, carries a point load of 10 kN at a distance 6 from the
mmleft support. If E - 2 x 10 5 N/mm3 and 1 = 1 x 10a 4 determine the slope at the left support In case of fixed beam, the deflection and slope are zero at the fixed ends as shown in
, Fig. 15. 1 (6). The slope will be zero at the ends if the deflection curve is horizontal at the ends.
To bring the slope back to zero (i.e., to make the deflection curve horizontal at the fixed ends),
and deflection under the point load using conjugate beam method.
M M M Mthe end moments A and g will be acting in which A will be acting anti-clockwise and B
[Ans. 6.00028 rad. and 0.96 mm]
will be acting clockwise as shown in Fig. 15.1 b( ).
m3. A beam of length 6 is simply supported at its ends and carries two point loads of 48 kN and
m m40 kN at a distance of 1 and 3 respectively from the left support. Find the deflection under A beam which is supported on more than two supports is known as continuous beam.
mmeach load. Take E - 2 x 10s N/mm2 and 1 = 85 x 10 6 beam method.
4 Use conjugate This chapter deals with the fixed beams and continuous beam. In case of fixed beams the B.M.
diagram, slope and deflection for various types of loading such as point loads, uniformly dis-
. tributed load and combination of point load and u.d.l., are discussed. In case of continuous
beam, Clapeyron’s equation of three moments and application of this equation to the continu-
mm[Ans. 9.019 and 16.7 mm] ous beam of simply supported ends and fixed ends are explained.

W4. A beam AB of span L is simply supported at A and B and carries a point load at the centre C 613

of the span. The moment of inertia of the beam section is / for the left half and 21 for the right

half. Calculate the slope at each end and deflection at the centre.

WL2 WI? WLS
„ 5„ ,
9s and yC 68 El
A, nSl 0/1 =
96El’
24El

m mmA5. cantilever of length 3 is carrying a point load of 25 kN at the free end. If I = 10s 4 and E

' ~ 2.1 x 105 N/mm 3 then determine : (i) slope of the cantilever at the free end and (it) deflection
,
at the free [Ans. 0.005357 rad. and 10.71 mm]
end using conjugate beam method.

m6. A cantilever of length 3 mis carrying a point load of 50 kN at a distance of 2 from the fixed
2 find (i) slope at the free end, and (ii) deflection at the
mm Eend. If I - 10s N/mm= 2 x 105 ,
4 and [Ans. 0.005 rad. and 11.67 mm]

free end using conjugate beam method.

mA7. cantilever of length 5 carries a point load of 24 kN at its centre. The cantilever is propped
10. rigidly at the free end. Determine the reaction at the rigid prop. [Ans. 7.5 kN]

A m8. cantilever of length 4 carries a uniformly distributed load of 2 kN/m run over the whole
mmElength. The cantilever is propped rigidly at the free end. If — 1 x 10 6 N/mm- and I = 10
,

then determine :

(i) reaction at the rigid prop

(ii) the deflection at the centre of the cantilever, and

(iii) magnitude and position of maximum deflection.

mm[Ans. (iii)x = 1. 688 m, ymwc = 0.0693 mm]
kN(i) 3 (ii) 0.0667

m9. A simply supported beam of length 8 carries a uniformly distributed load of 1 kN/m run over
the entire length. The beam is rigidly propped at the centre. Determine : (i) reaction at the prop

(ii) reactions at the supports (iii) net B.M. at the centre and (w) positions of points of contraflexures.

kNm mkN kN(Ans. (i) 5
(ii) 1.5 (iii) - 2.0 (iv) 3 from each support]

A mcantilever of length 10 carries a uniformly distributed load of 800 N/m length over the
whole length. The free end of the cantilever is supported on a prop. The prop sinks by 5 mm. If
[Ans. 2750 N]
mmE = 3 x 105 N/mm2 and I = 108 4 then find the prop reaction.
,

614 STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 1

15.2. BENDING MOMENT DIAGRAM FOR FIXED BEAMS 615

W WFig. 15.1 (e) shows a fixed beam AB of length L subjected to two loads and 2 at AB.M. at = 0, B.M. at B = 0.

distance of ~r from each ends. B.M. at C = 5 W x —L = 5WL
4
4 4 16
Let Ra ~ Reaction at A
—W —D„ x L = 1WL

B.M.
„ = 7

at Z)

44 16

Now B.M. diagram can be drawn as shown in Fig. 15.2 (6). In this case, B.M. at any

A point is a sagging (+ve) moment.

A(it) simply supported beam subjected to end moments only (without given loading) as

Fig. 15.1 (c) shown in Fig. 15.3.

RB = Reaction at B MLet a = Fixed end moment at A
Ma = Fixed end moment at A MB = Fixed end moment at B

R = Reaction* at each end due to these moments.

Mb = Fixed end moment at B WAs the vertical loads acting on the beam are not symmetrical (they are at distance
M MR RThe above four quantities i.e., A , B , A and
B are unknown. 2WLI4 from A and at a distance L/4 from B), the fixed end moments will be different.

M MThe values of RA, Rg , A and g are calculated by analysing the given beam in the M MSuppose B is more than , and reaction R at B is acting upwards. Then reaction R at

following two stages : A will be acting downwards as there is no other load on the beam. (ZFy = 0). Taking moments

A(i) simply supported beam subjected to given vertical loads as shown in Fig. 15.2. about A, we get clockwise moment at A = Anti-clockwise moment at A.

Consider the beam AB as simply supported. M MB = a + R.L

R ALet* = due to vertical loads -Ma
a Reaction at R„ Mg
L
Rb * = Reaction at B due to vertical loads. = ~ -(A)

Taking moments about A, we get

W 2W

T 7 / / S\ "T”

m(b) a J$ Me
jL

B.M. diagram considering beam as simply supported B.M. diagram due to end moments

Fig. 15.3

M MAs B has been assumed more than A , the R.H.S. of equation (A) will be positive. This

Rmeans the magnitude of reaction at B is positive. This also means that the direction of
i reaction R at B is according to our assumption. Hence the reaction R will be upwards at B and

Adownwards at as shown in Fig. 15.3 (a). The B.M. diagram for this condition is shown in

Fig. 15.3 (6). In this case, B.M. at any point is a hogging (-ve) moment.
Since the directions of the two bending moments given by Fig. 15.2 (b ) and Fig. 15.3 (b)

are opposite to each other, therefore their resultant effect may be obtained by drawing the two
j moments on the same side of the base AB, as shown in Fig. 15.4.

*The reaction at each end will be equal. There is no vertical load on the beam hence reaction at

A + reaction at B = 0. Or reaction at A = - reaction at B.

—

616 STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS

Let a = Area of B.M. diagram due to vertical loads
a ' = Area of B.M. diagram due to end moments.

AC Then Mf dx =a
D
Resultant B.M. diagram Jo

B cL

and I M'.dx - a'

Jo

Substituting these values in equation (ii), we get

Fig. 15.4 0 = a - a'

Now the final reactions RA and RB are given by or a=a ' ...(15.1)

ra =ra*~r The above equation shows that area of B.M. diagram due to vertical loads is equal to the
area of B.M. diagram due to end moments.

and Rg - Rb* + R Again consider the equation (i)

RIn the above two equations, * and RB * are already calculated. They are : RA * = 5W/4

A ,2
M M Mand
R* = 7W/4. But the value of R is in terms of B and A . It is given by R = (MB - A )/L. EI-X = M -M'

b
M MHence to find the value of R, we must calculate the value of B and A first.
dx 1 X* *

M MTo find the values of A and a Multiplying the above equation by x, we get
M ALet x = B.M. at any section at a distance x from due to vertical loads
M A'= —yd 2y
x
B.M. at any section at a distance x from due to end moments. EI.X. = X.M - X.M '

dx 1 xx

The resultant B.M. at any section at a distance x from A

= M - M' (M is +ve but AL. is ~ve) Integrating for the whole length of the beam i.e., from 0 to L, we get
v
'

'L y rL i*L
EI.x—-T
But B.M. at any section is also equal to El J0 dx .dx - x.Mx .dx- x.M ‘dx
Jo Jo

—rL d^v rt i“t'

EI\ x. 25-.c!x= x.M .dx-\ x.M'x .dx
*x
Jo clx Jo Jo

Integrating the above equation for the entire length, we get MIn the above equation, x.dx represents the area of B.M. diagram due to vertical loads

at a distance x from the end A. And the term (x.M .dx) represents the moment of area of B.M.

x

—dy rL

diagram about the end A. Hence j^x.Mx .dx represents the moment of the total area of B.M.

But represents the slope. And slope at the fixed ends i.e., at A and B are zero. The diagram due to vertical loads about A, and it is equal to total area of B.M. diagram due to
vertical loads multiplied by the distance of C.G. of area from A,
above equation can be written as

H-e at x = 0 rL

x.Mr dx = ax

Jo

where x = Distance of the C.G. of B.M. diagram due to vertical loads.

8 MII flo j
rL Similarly rL = x'
i x 'dx ! a'
i j^x.Mx ‘ .dx
\Q (

^MrL rL j

EI[_0-0] = x .dx- where x' = Distance of the C.G. of B.M. diagram due to end moments.
Substituting the above values in equation (iii), we get

Cmor -'* 2yy-., dx=a-x~a, x
0 = j^Mx . dx - ...(.ii) „.r L d
El I x.

Jo dx

& rL
M MNow f x . dx represents the area of B.M. diagram due to vertical loads and
dx' —El Tx ~dy L _ d ( dy d2y dy} dy d 2y
a ax - a x dx. I dx dx 2 dx j dx dx 1
x. 1

y
represents the area of B.M. diagram due to end moments. dx _ o

STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS
618
MLet a = Fixed end moment at A
mor El B -yB) - (0 x 0A -yA) ] = Mb = Fixed end moment at B
ASince slope and deflection at and B are zero, hence 0A, 0B , and yB are zero.
Ra = Reaction at A
0 - ax - x' Rb = Reaction at B.
a'
M MR RThe above four are unknown i.e., A , B , A and B are unknown.
or ox = ax ...(15*2)
({) B.M. Diagram
But from equation (15.1), we have
M MDue to symmetry, the end moments A and B will be equal. Hence the B.M. diagram
- _- x ...(15.3) M Mdue to end moments will be a rectangle as shown in Fig. 15.5 (b ) by AEFB. Here the magnitude
x
of a arid B are unknown. The bending moment diagram for a simply supported beam carrying
Hence the distance of C.G. of B.M. diagram due to vertical loads from A is equal to the a point load at the centre will be a triangle with the maximum B.M. at the centre equal to

W.L

-——-The B.M. diagram for this case is shown in Fig. 15.5 (6) by a triangle ADS in which

distance of C.G. of B.M. diagram due to end moments from A.

M MNow by using equations (15.1) and (15.3) the unknowns A and B can be calculated.

M MThlc also means that A and B can be calculated by Now according to equation (15.1), area of B.M. diagram due to vertical loads should be

(i) equating the area of B.M. diagram due to vertical loads to the area of B.M. diagram equal to the area of B.M. diagram due to end moments.

due to end moments. Equating the areas of the two bending moment diagrams, we get
(u) equating the distance of C.G. of B.M. diagram due to vertical loads to the distance of
Area of triangle ADC = Area of rectangle AEFB
C.G. of B.M. diagram due to end moments. The distance of C.G. must be taken from the same
or — x AB x CD - AB x AE
end in both cases.

15.3. SLOPE AND DEFLECTION FOR A FIXED BEAM CARRYING A POINT LOAD —W—or L-
AT THE CENTRE —1 x Lt x = LT x Mw-,A
4
WFig. 15.5 (a) shows a fixed beam AS of length L, carrying a point load at the centre C 2

of the beam.

...(15.4)

Now the B.M. diagram can be drawn as shown in Fig. 15.5 (6).

(«) S.F. Diagram

Equating the clockwise moments and anti-clockwise moments about A, we get

Rb xL + M =Mb+ W. |
a

MMBut a = b

W -.-. R„ x L =
.

A c B.M. diagram B Due to symmetry, —RA =

S.F. diagram Now the S.F, diagram can be drawn as shown in Fig. 15.5 (c).
Fig. 15.5
—There will be two points of contraflexure at a distance of from the ends.
4

FIXED AND CONTINUOUS BEAMS 621

WNote. The deflection at the centre of a simply supported beam carrying a point load at the

WL?

centre is • Hence the deflection of the simply supported beam is four times the deflection of the

fixed beam.

Or in other words, the deflection of a fixed beam is one fourth times the deflection of the simply
supported beam. Hence when fixed beams are used, the deflection will be less.

mProblem 15. 1. A fixed beam AB, 6 kNlong, is carrying a point load of 50 at its centre

mmThe moment of inertia of the beam is 78 x l(fi 4 and value of E for beam material
is 2.1 x 10s N/mm2. Determine :

(i) Fixed end moments at A and B, and

(ii) Deflection under the load.

Sol. Given :

Length, L = 6 m = 6000 mm

WPoint load, = 50 kN = 50000 N

M.O.I., mmI = 78 x 10s 4

Value of E = 2.1 x 105 N/mm2
Let
Ma = Fixed end moment at A,
MB = Fixed end moment at B,

y max ~ Deflection under the central point load.

Using equation (15.4), we get

m Ma = b -

= 37.5 kNm. Ans.
Using equation (15.5), we get

WL3

192 El

50000 x 6000 3
= 3.434 mm.
Ans.
192 x 2.1 x 10 5 x 78 x 10 8

Alternate Method

Fig. 15.5AC6) shows the simply supported beam, which is having Max. B.M. at the cen-

tre equal to RA * x 3 = 25 x 3 = 75 kNm. Fig. 15. 5A (e) shows the B.M. diagram for simply

supported beam.

M MFig. 15.5A(d> shows the fixed beam with end moments only. Due to symmetry end mo-

ments are equal. Hence A = g . Fig. 15. 5A (el shows the B.M. diagram due to end moments
only. This diagram is a rectangle.

STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 623

622

(i) B.M. Diagram

M MAs the load is not acting symmetrically, therefore A and g will be different. In this
M Mcase B will be more than A as the load is nearer to point B. The B.M. diagram due to end
Mmoments will be trapezium as shown in Fig. 15.6 ( b ) by AEFB. Here the length AE (i.e.,
MBFand A)
R ) are unknown.
(i.e.,

The B.M diagram for a simply supported beam carrying an eccentric point load will be

—Wtriangle with maximum B.M. under the point load equal to ab' '

The B.M. diagram for this

1j

—case is shown in Fig. 15.6 ( b ) by a triangle ADB in which CD = a-°\

B.M. diagram due to end moments only A CB

Fig. 15.5A Fig. 15.6

Equating the areas of two B.M. diagrams, we get Equating the areas of the two bending moment diagrams, we get

Area of B.M. diagram for simply supported beam moments. Area of trapezium AEFB = Area of triangle ADB
= Area of B.M. diagram ~(AE + BFYAB = - x AB x CD
due to end
22
i.e
|(MA+ Ma ).L = |xL-^A
or M Ma + b -

,5.4. SLOPE AND DEFLECTION FOR A FIXED BEAM CARRYING AN ECCENTRIC Now using equation (15.3),

POINT LOAD Aoi Distance of C.G. of B.M. diagram due to vertical loads from = Distance of C.G. of B.M.
WFig. 15.6 (a) at C a^a
and M, diagram due to end moments from A.
Vdiste.ce of from

Aalso reactions at
shows a fixed beam AB of length L, carrying a point load
. distance off fined end
A and at from B. The moment M,
Band i.e.
BA, and R„ an shown in the same figure.

M1 b

.

FIXED AND CONTINUOUS BEAMS

2M(M..\ + g) . L _ a + L
+M ~~3~3 (Ma
b)

l^i

L_ ( a + ) W.a.1 M M —W.a.b
L 'L a + b = L from equation (i)

—= +,
(a
Irf), . W.a.b

5

Subtracting equ : <,ton (i) from equation (ii), we get

M —W-,1/f a- b
r
= <(a + Lt)\ W.a.b

W.a.b (a+L

W ,a,b ( a + L - L\ W.a2 .b

MSubstituting the value of g in equation (i), we get

W.a2 .b W.a.b

W.a.b Wa 2b

~~WL''

W.a.b (L-a) = W.a.b.

-~

L2

W.a.b 2

M MNow a and B are known and hence bending moment diagram can be drawn. From
M Mequations (Hi) and (iv), it is clear that if a > b than B > A .

(ii ) S. F. Diagram

Equating the clockwise moments and anticlockwise about A,
——M M WRb xL + a = b + .a
~MR„b = C B A ) + W.a

Similarly U W(M_A_- B ) <• - 6

AL
M MBy substituting the values of A and B from equations (Hi) and (iv), in the above
Requations, we shall get RA and B. Now S.F. can be drawn as shown in Fig. 15.6 (c).

(Hi) Slope and Deflection

The B.M. at any section between AC at a distance x is given by

(v a + b = L) MR-- xx- a
a

3

STRENGTH OF MATERIALS
626

Substituting the value of RA in the above equation, we get

d*y [(Ma -Mb ) + W-61
EI

(Ma -Mb ) x + WJ>

LL

= YLA' X -\ma ±{Mb -ma )^
L
L

Substituting the values of and we get

d2y = W.b •*“ \ W.a.b2 (w . a 2 .b W .a. b2 )x

.0 t r2 + T 2 T.

W.b.x W.a.b 2 f W.a2 .b W.a.b2 } x_
L2 { L2 ~ L2 )L
~~~L
W.a.b 2 W.a.b £.
TW.b.x L2 L2
" } 'L
= W.a.b, t, s W.a.b 2

—IV. 6.*

=

i/ JU

—= w.b, r i - a2 + ab,)s x w.a.b 2

{h ~2

L/ A/

But L = a + b

L2 = (a + b)2 = a2 + b2 + 2ab.

Substituting the value of 2 in the above equation, we get
Z,

—= V" (a2 + b 2 + 2ab - a2 + ab)* - W.a.b 2
L2

W.b,L 2 W.a.b 2
(b
+ 3ab)*

— —W.b2 <,b, + 3„a), x W.a.b 2
5
“2

Integrating the above equation, we get W.a.b 2

^(birr^ = IV- 62 ^ +. 3o.ai).*-2

where C, is a constant of integration.

At x = 0, = 0. Hence C = 0.
1
EI^L =_WZ.Lb^2(j, ) + 3a).x i - W.a.b2
dx 2 Lr

~ L
:

628 STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS Q29

W.b2 ( 2aL r (b + 3a).-^-~SaL 2 45 X 2 3 X l2 16 X 45

31? [fe + 3aj b(i + 3a) 3 x 1 x 10 4 (1 + 3 x 2 3 x 10 4 x 49
2)

Wb2 ( 2aL ? m— — 0.00049 = — 0.49 m. Ans.

A(iv) The distance of maximum deflection from point is given by equation (15.7) as

61? \6 + 3aJ 2a.
X ~ (b + 3a)
4a 2 2

' (b + 3a) 2
Wb L2 . ah - - 2

61? 3 ' (6 + 3a) T m2x2x3 12

y =_ 2 Wa 3ub 2 = 1+3x2 = = 1,714 - Ans-

...(15.8)

3El (6 + 3a) 2 Alternate Method

A mProblem 15.2. A fixed beam B of length 3 carries a point load of 45 kN at a distance Fig. 15.7A (b) shows the simply supported beam with vertical load of 45 kN at a distance

m kNmof 2 from A. If the flexural rigidity (i.e., El) of the beam is 1 x 104 2 determine : m2 from A.
,

(i) Fixed end moments at A and B, RThe reactions Ajf? * and * due to vertical load will be :
B

( ii ) Deflection under the load, 3Rb * = 45 x 2 or RB * = 90/3 = 30 kN and RA * = 45-30 = 15 kN.

(Hi) Maximum deflection, arid Fig. 15.7A (c) shows the B.M. diagram with max. B.M.at C and equal to R, * x 2 = 15 x 2

(in ) Position of maximum deflection. = 30 kNm.

Sol. Given : Fig. 15. 7A (d) shows the fixed beam with end moments and reactions. As the vertical
Length, M M Mload is not acting symmetrically, therefore
L = 3m A and B will be different. In this case will be

fj
Mmore than A , as load is nearer to point B. The B.M. diagram is shown in Fig. 15.7A(e)
Point load, IV = 45 kN
M M(i) Fixed end moments at A and B. To find the value of A and B , equate the areas of
kNmFlexural rigidity, El = 1 x 10 4 2

two B.M. diagrams.

Distance of load from A, .'. Area of B.M. diagram due to vertical loads
= Area of B.M. diagram due to end moments
a=2m

Distance of load from 8,

b-1m Aj + A = A + A where Aj = 30, A,
2 3 4
M MLet a and B - Fixed end moments,
mb~MmA 3M A4~= ^
yc = Deflection under the load 3 - <~ a )x3
A’
ymax = Maximum deflection and 2
x = Distance of maximum deflection from A.
= 1.5 (Mn -MJ
M M M30 + 15 = 3 a + 1.5 b - 1.5
(.i ) The fixed end moments at A and B are given by M45 = 1.5 a + 1.5Mb a

M — ^a ~
W.a.b 2 = 45x2x1 kNm= 10 , T - A. "5

2 — M M45 a + Mg = 30
=
+ B or

a2 45 x 2 2 x1
,
M p—bW. b = ~ = 20 kNm. Ans. Now equating the distance of C.G. of B.M. diagram due to vertical load to the distance of

= C.G. of B.M. diagram due to end moments from the some end (i.e., from end A)

(ii) Deflection under load is given by equation (15.6) as

_ iW.fe 3 = _ 45x 23 xl3 m- 0.Q000(0444 Ajiq + A2 X2 _ Ag x A+ 4 x 4
c 3 Ell} 3 x lx 10 4 x 3 s
A, + A2 A A„
= — 0.444 mm. Ans. 3 +

-ve sign means the deflection is downwards. —4 ( -l M3Ma x | + 1.5 (MB ~ a ) x 2
2
(iii) Maximum deflection is given by equation (15.8) as (330 x + 15 x +

3

2 Was .b2 30 + 15

ynULX =~ _ 3EI (6 + 3a) 2 M Mor
* 40 + 35 _ 4.5 a + 3MB - 3MA L5 A + 3Mg

M M45 15 a + 1.5 B

)

STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 631
630

MM MM MMor ATaking the moments about for Fig. 15.7A(cO, we get clockwise moment at A = Anti-
75 —_ L5(Af A + 2 b ) °r 5 _ A +2 B
45 L5(Ma + b ) clockwise moments at A
a+ b
3 M -Mb a + R-x. 3
r _ Mb -Ma 20-10
Mor 5Ma + 5 b = 3i¥A + GMb —{H) ^10
MOr 2Ma = b
3 33
Solving equations (i) and (ii), we get
MkNmAj|/f = 10 ANow the total reaction at and B will be,
B = 20 kNm. Ans.
and ^BA4 = RAa*~R = 15 -

45 kN 3 = ~kN
3
— —and
RBr = RB* + R = 30 + = kN

(a) Aj- 33

-2m- C Now, consider the fixed beam as shown in Fig. 15.7B.

-H MThe B.M. at any section between AC at a distance x from A is given by RA x x - A

-3m-

45 kN

MEl d^y
2
Ra xx - a

etc

B.M. diagram for simply supported beam with vertical loads f= X *"l°

Fig. 15.7A Integrating, we get

Let us now find the reaction R due to end moments only. As the end moments are ~ — —EI = x - 10* + C 1L
Adifferent, hence there will be reaction at and B. Both the reactions will be equal and opposite dx 3 2

Min direction, as there is no vertical load, when we consider end moments only. As B is more, at x = 0, =0 C =0
1
the reaction R will be upwards at B and downwards at A as shown in Fig. 15.7A id). dx

—EI~ = x 2 - 10* ...( Hi
ax 6
...(iv)
Integrating again, we get

— —35 x3 10*

El x y = x + Co

63

at x = 0, y - 0, .-. C2 ~ 0

El x yJ = * 3 - 5x 2

18
(ii) Deflection under the load

From equation (iv), we have

~xy = 1
18'*z -55x'T 2
El

STRENGTH OF MATERIALS

mTo find the deflection under the load, substitute x - 2 in the above equation, For the sake of convenience, let us first calculate the fixed end moments due to loads at

i r 35 -3 _ 0 2 C and D and then add up the moments.

y El 18 J (t) Fixed end moments due to load at C.

— 1“ 35x_S m mFor the load at C, a = 2 and 6 - 4

=1 _ 20 (( v El = lx 104) WMa ~ 2
1 x 10 4 L 18
c .a.b
= - 0,000444 m = - 0.444 mm. Ans.
L2

{- ve sign means the deflection is downwards). 160x2x4^

(tit) Maximum deflection =
62

^B W kNm=
—Deflection (y) will be maximum when dy = 0. 2 = 160 x 2 2 x 4 = 71.11
“j :
c .a .b
-j
Ll
6

—Hence substituting the value of (it) Fixed end moments due to load at D.

= 0 in equation (fit), we get m mSimilarly for the load at D, a = 4 and b = 2

dx

—x35 - 10*2 , jn. ma =
L
0= 120 x 4 x 2a = 53.33 kNm

6

0 = 35x 2 - 60*

0 = x (35* - 60)

This means that either * = 0 or 35* - 60 = 0 for maximum deflection. W2 160x4 2 x2
But * cannot be zero, because when * = 0, y - 0. D .a .b
- = 106.66 kNm

35* - 60 = 0

m60

x ~ 35
12
“ 1, -.714,

7

mSubstituting * = 1.714 in equation (to), we get maximum deflection.

gd-714)3 -5(L714) 2

y max = J_r35(i.7i4) s -5(1714) 2
El L 18
j

= — —L [9.79 - 14.69]
10 4r
lx L

= 0.00049 m = 0.49 mm. Ans.

(iu) Position of maximum deflection

mThe maximum deflection will be at a distance of 1.714 x—(i.e., 1. 1 14 m) from end A.

Problem 15.3. A fixed beam AB of length 6 m carries point loads of 160 kN and 120 kN
m mat a distance of 2 and 4 from the left end A. Find the fixed end moments and the reactions

at the supports. Draw B.M. and S.F. diagrams.

Sol. Given : =6m Fig. 15.8
Length
Load at C, Wc = 160 kN Total fixing moment at A,
Load at D, WD = 120 kN
Distance M M Ma = Ai + Ai = 142.22 + 53.33
Distance AC = 2 m
= 195.55 kNm. Ans.
.AD = 4 m

)

STRENGTH OF MATERIALS fixed and continuous beams 635
634

and total fixing moment at B, Now the B.M. diagram due to vertical loads can be drawn as shown in Fig. 15.8A(c)

M M Mb = Bi + Bi = 71.11 + 106.66 Fig. 15.8A d( ) shows the fixed beam with end moments only. As the load 160 kN is

= 177.77 kNm. Ans. M Mnearer to end A, hence A will be more than R. The B.M. diagram due to end moments is

B.M. diagram due to vertical loads Let RA * and RB* are the reactions at A and shown in Fig. 15.8A(e).

Consider the beam AB as simply supported. M MTo find the values of A and B , equate the areas of two B.M. diagrams.

B due to simply supported beam. Taking moments about A, we get Area of B.M. diagram due to vertical loads

= Area of B.M. diagram due to end moments

Rb * x 6 = 160 x 2 + 120 x 4

—= 320 + 480 = 800
«**= = 133.33 kN

Ra * = Total load - RB*= (160 + 120) - 133.33
= 146.67 kN

B.M. at A = 0
B.M. at C = Ra* x 2 = 146.67 x 2 = 293.34 kNm
B.M. at D = Rb* x 2 = 133.33 x 2 = 266.66 kNm

B.M. atB = 0.

Now the B.M. diagram due to vertical loads can be drawn as shown in Fig. 15.8 b( ).

In the same figure the B.M. diagram due to fixed end moments is also shown.

S.F. Diagram

ALet Ra = Resultant reaction at due to fixed end moments and vertical loads

Rb - Resultant reaction at B.

Equating the clockwise moments and anti-clockwise moments about A, we get

M MRb x 6 + a = 160 x 2 + 120 x 4 + B

or Rb x 6 + 195.55 = 320 + 480 + 177.77
^RBr =
and 800 + 177.77 - 195.55 = 130 37
6

~ Total load — RB B.M. diagram for simply supported beam with vertical loads
= (160 + 120) - 130.37 = 149.63 kN Fig. 15.8A

AS.F. at = = 149.63 kN
S.F. at C = 149.63 - 160 = - 10.37 kN
S.F. at D = - 10.37 - 120 = - 130.37 kN

S.F. atB=- 130.37 kN

Now S.F. diagram can be drawn as shown in Fig. 15.8 (c).

Alternate Method

Fig. 15.8A ( b shows the simply supported beam with vertical loads.

Let RA* and RB * are the reactions at A and B due to vertical loads. Taking moments

about A, we get

Rb* x 6 = 160 x 2 + 120 x 4 = 320 + 480 = 800

RB* = = 133.33 kN

63

and Ra -’ Total load - Rn *

= (160 + 120) - 133.33 = 146.67 kN

B.M. at A = 0
B.M. at C = Ra* x 2 = 146.67 x 2 = 293.34 kNm
B.M. at D = Rs * x 2 = 133.33 x 2 = 266.66 kNm

2—

STRENGTH OF MATERIALS fixed and CONTINUOUS beams

a +a +a +a =a +a -W Combined B.M. Diagram
1 23 456
M Ma — 195.55 kNm and B = 177.77 kNm. Now the combined B.M. diagram can be drawn
wh, ere A,, = ACxCE 2x293.33 = 2„9„3„.3c3o
as shown in Fig. 15.8 6( ).
=
To draw the S.F. diagram, let us first find the values of resultant reactions due to verti-
12 2
cal Ioads and fixed end moments RA and Rr . Refer to Fig. 15.8A<a). Taking moments about A,
A = CD x DF = 2 x 266.67 = 533.34 we get clockwise moments at A = Anti-clockwise moments at A
2
M M160 x 2 + 120 x 4 + B = a + RB x 6
„A, Mor 320 + 480 + Mg — A + 6Rg
GFxGE 2x26.66 ..
= 66.66 800 + 177.77 = 195.55 + 6R„
==

32 2
~F” ~A = DB x DF = 2 x 266.67 = „2„6„6 '67
*
M m M MA =
5
6 x (MA
zA = jI =3 (MA - 3M 3M=fl)
b X6 = X- fl Rb = 800 + 177.77 - 195.55 = 130.37 kN
B, 6

Substituting these values in equation (i), we get Ra = Total load - RB = (160 + 120) - 130.37 = 149.63 kN

M M M293.33 + 533.34 + 66.66 + 266.67 = 6 B + 3 A - 3 B S.F. Diagram
M Mor 1119.98 = 3Mb + 3 A = 3 (MB + A)

M M+b + a =: = 373.33 ...(h) S.F. at A = Ra = 149.63 kN
S.F. at C = 149.63 - 160 = - 10.37 kN
3
M MTo get the other equation between A and B , equate the distance of C.G. of B.M.
diagram due to vertical ioads to the distance of C.G. of B.M. diagram due to end moments from, S.F. at D = - 10.37 - 120 = - 130.37 kN

S.F. at B = - 130.37 kN

end A. Now S.F. diagram can be drawn as shown in Fig. 15.8(c).

A^x^ + A^x^ + Ag^g + _ AgXg + A^ffg A mProblem 15.4. fixed beam of length 6 carries two point loads of 30 kN each at a

Aj + Ag + Ag + A4 As + A$ distance of2m from both ends. Determine the fixed end moments and draw the B.M. diagram.

Sol. Given :

Length, L=6m

293.33 x | + 533.34 x 3 + 26.66 x ^2 + | j + 266.66 x ^4 + | Point load at C, IFj = 30 kN

293.33 + 533.34 + 26.66 + 266.66 WPoint load at D, 2 = 30 kN

6Mb 3(Ma — Distance AC = 2 m

x 3 + - SMq) x o x 6 AD = 4 m

6M M M_____ Distance
B +3
a -3 b DThe fixing moment at A due to loads at C and is given by
M M39L1+ 1600 + 70.91 +1245.35
3(6M +2 A -2 B) Ma = Fixing moment due to load at C + Fixing moment due to load at D
fl

1119.98 M" 3 (MB + a ) VVA W2 2a2 .b22

—M M+M2.95 = 4 rr + 277^a- ~ L2 I?
ba
30 x 2 x 30 x 4 x 2* — —80 40
M M2.95 b + 2.95Ma = 4MB + 2 A + + = 40 kNm.
M M2.95 a - 2 a = 4MB - 2.95MB 62
Z O <5

6

Since the beam and loading is symmetrical, therefore fixing moments at A and B should

0.95MA = 1.05MB be equal.

MSubstituting this value of A in equation Hi), we get M MB = a = 40 kNm. Ans.
M Mr + 1.1 B = 373.33 .
To draw the B.M. diagram due to vertical loads, consider the beam AB as simply sup-
ported. The reactions at the simply supported beam will be equal to 30 kN each.

B.M. at A and'R = 0

_ 373.33 B.M. at C = 30 x 2 = 60 kNm
B
= 177 77 kNm- Ang. B.M. at D - 30 x 2 = 60 kNm.

2.1

MFrom equation (Hi), A = 1.1 x 177.77 = 195.55 kNm. Ans. Now the B.M. diagram due to vertical loads and due to end moments can be drawn as

shown in Fig. 15.9 (b).

}.

638 STRENGTH OF MATERIALS fixed and continuous beams 639

Pig. 15.9 Fig. 15.10

15.5. SLOPE AND DEFLECTION FOR A FIXED BEAM CARRYING A UNIFORMLY Hi) S.F. Diagram
DISTRIBUTED LOAD OVER THE ENTIRE LENGTH
Equating the clockwise moments and anti-clockwise moments about A, we get
Fig. 15.10 (a) shows a fixed beam of length L, carrying uniformly distributed load of
M ML
uVunit length over the entire length.
Rg x L + a -w.L.— + b
MLet a = Fixed end moment at A
Mb = Fixed end moment at B M MBut a = b
L
Ra = Reaction at A
Rb = Reaction at B. —RB x L = w.L.—
or w.L
(i) B.M. Diagram
RB =
M MSince the loading on the beam is symmetrical, hence A = B . The B.M. diagram due to
Due to symmetry,
end moments will be a rectangle as shown in Fig. 15.10 (6) by AEFB. The magnitude ofMA or
=-Ra = RS w.-L ...(15.10)
Mb is unknown.
2
The B.M. diagram for a simply supported beam carrying a uniformly distributed load
will be parabola whose central ordinate will be w.L2/8. The B.M. diagram for this case is shown Now the S.F. diagram can be drawn as shown in Fig. 15.10 (c).

—in Fig. 15.10 (6) by parabola ADB in which CD = w 1 (tii) Slope and deflection

8 The B.M. at any section at a distance x from A is given by,

Equating the areas of the two bending moment diagrams, we get El ^~r = RA, x x - M.A - w.x --

Area of rectangle AEFB = Area of parabola ADB dx 2 2

~ w.L X wL2 wx2
2~
~~2~

12

~~wL2.x wx2 wl}
~

2 12

Integrating the above equation, we get

EI dy _ w.L x 2
dx 2 2'

wL= r ,2 -

.

4

Cwhere is a constant of integration.
l

8

STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 641

As LI2 represents the centre of the beam. Hence the two points of contraflexures occur

at a distance of L!2^3 from the centre of the beam.

mAProblem 15.5. fixed beam, of length 5 carries a uniformly distributed load of

m m9 kNI and E 107 kN/m2
run over the entire span. If I = 4.5 x i(H 4 = 1 x , find the fixing

moments at the ends and the deflection at the centre.

Sol. Given : L=5m
Length,

U.d.l. w = 9 kN/m

Value of mI = 4.5 x 10"4 4
Value of E = 1 x 10 7 kN/m2 .

(i) The fixing moments at the ends is given by equation (15.9) as

M.A = M„B = ----- = ----- = 18.75 kNm. Ans.
12 12

(ii) The deflection at the centre is given by equation (15.11) as

y‘~~ wLa _~ 9 x 54
384 x 1 x 10 7 x 4.5 x 10'4
384 El

m= 0.003254 = - 3.254 mm. Ans.

Problem 15.6. Find the fixing moments and support reactions of a fixed beam AB of
mlength 6 m, carrying a uniformly distributed load of 4 kNI over the left half of the span.

Sol. Given :

mLength, L = 6

U.d.l., w = 4 kN/m

( i ) B.M. diagram due to end moments

MLet a - Fixing moment at A
M .= Fixing moment at B.

fl

M MThe value of A will be more than B as load due u.d.l. is nearer to point A.

The B.M. diagram due to end moments will be trapezium as shown in Fig. 15.11 (6) by

AEFB.

The area of B.M. diagram due to end moments is given by,

M Ma' = \ (Ma + b) x 6 = 3(Ma + b) ...(£)

(ii) B.M. diagram due to vertical loads

Now draw the B.M. diagram due to u.d.l. for a simply supported beam.

ALet Ra* - Reaction at for a simply supported beam

Rb * - Reaction at B for a simply supported beam.
Taking moments about A for a simply supported beam, we get

R b * x 6 = 4 x 3 x 1.5 = 18

rb* = 1 = 3 kN

~0~

and Ra * = Total load - RB*

=4x3-3=9kN

The B.M. at A and B are zero.

642 . STRENGTH OF MATERIALS

FIXED AND CONTINUOUS BEAMS 643

Equating the two areas given by equations (i) and (iii), we get ...(io)

M3<Ma + B) = 36.0

Ma + = 12.0
Now moment of B.M. diagram due to vertical loads about A is given by

— 3 + Area of triangle BCD
ax =
f x.Mx .dx

x Distance of C.G. of BCD from A

= f x(9x -2* 2 ).<ic + — x9x3xf3 + — X 3 J

2 3i, )

= (*3 (9x 2q - 2x 3 ) .dx + 54

Jjo

—= 9x3 3
—2x 4 1 r 1 1
L3 +54= 3 x 3 3 - — x 3 4 ] + 54
4 2J
Jo L

= (81 -40.5) + 54 = 94.5 (lj)

(c) B.M. diagram for vertical loads Moment of B.M. diagram due to end moments about A is given by [see Fig. 15.11 ( 6 )].

Fig. 15.11 a'x ' = Area ABFH x Distance of C.G. of ABFH from A

B.M. at C = Rb * x3=3x3=9 kNm. + Area HFE x Distance of its C.G. from A

The B.M. diagram from A to C will be parabolic and from C to A the B.M. diagram will m= < b x L) x M1 x L x (Ma - B) x ~ x L

follow a straight line law as shown in Fig. 15.11 (c).

The area of the B.M. due to vertical loads is given by m M= b x 6 x g + | x 6 x (Ma - B) x |

a = Area of parabola ACD + Area of triangle BCD = 18Mb + 6Ma - 6Mb

= Area of parabola ACD + y x 9 x 3 ...(it) M- GMa + 12Mb = 6(Ma + 2 B) ...(„•)

To find the area of the parabola ACD, consider a strip of length ‘dx’ at a distance* from M —94.5 = 6(Ma + 2Afg)

A in portion AC.

The B.M. at a distance * from A is given by 94.5

Mx = Ra* x * - 4 x * . f = 9r- 2x2 a+ = = 15.75

(v RA * = 9) Substracting equation (in) from (oii), we get

Area of B.M. diagram of length dx MB = 15.75 - 12.0 = 3.75 kNm. Ans.

M= x.dx = (9x - 2x2).dx Substituting this value in equation (iv), we get

Total area of parabola from A to C is obtained by integrating the above equation between Ma = 12 - 3.75 = 8.25 kNm. Ans.

the limits of 0 and 3.

Area of parabola ACD Support reactions

Let Ra = Resultant reaction at A

= f (9* - 2x 2 ).dx ifB = Resultant reaction at B.

Jo

— 2 MEquating the anti-clockwise moments and clockwise moments about A,
2
-= - .? - iiiil = 40.5 - 18 = 22.5 Rg X 6 + a = 4 X 3 X 1.5 + Mg

3 2 3 Rb x 6 + 8.25 = 18 + 3.75 = 21.75

Jo — — —n

Substituting this value in equation (it), we get _ 21.75-8.25 13-.50 = 2.25 kN. Ans.
=

a = 22.5 + | x 9 x 3 = 36.0 Ra = Total load - Rn

= 4 x 3 - 2.25 = 9.75 kN. Ans.

~ 645

6 44 STRENGTH OF MATERIALS Substituting this value in equation ( iv ) {Here complete equation is taken], we get

Second Method for Problem 15.6 M ¥ ¥ _» Ra X* 6°3 Ma*a2XxG6 2
Macaulay’s method can be used and directly the fixing moments and end reactions can 1 ,1
* 6 +. x (6 “ 3)4
be calculated. This method is used where the areas of B.M. diagrams cannot be determined
: 36.KA - 18Ma - 216 + 13.5
conveniently.

M202.50 = 36Ra - 18 A
M101.25 = 18UA - 9 a

—At a; = 6 m, = 0.

dx

Substituting these values in the complete equation (ii), we get

0=J?A x-|--MA x6-|x63 +|(6-3)3

Fig. 15.12 M= 182?a - a x 6 - 144 + 18
M126 = 18iJA - 6 a
For this method it is necessary that u.d.L. should be extended upto B and then compen-
sated for upward u.d.l. for length BC as shown in Fig. 15.12. Substracting equation (u) from equation (vi), we get

AThe B.M. at any section at a distance x from is given by M126 - 101.25 = 9 a - GMa

= R..x -M.-wxxx + w x (x - 3) x 24.75 = 3Ma

- 2 M —,, - 24g.7—5 = 8.25 kNm. Ans.
3) a
m= rax ~ a 4(*

Substituting this value in equation <ui), we get

M= R,.x - < - 2x 2 2 126 = 187?a - 6 x 8.25
- ———R„Aa =
+ 2(x 3) 126 + 6x8.25
= 9.75 kN.
Ans.
18
Integrating, we get
Now Rb = Total load - Ra

dy .xZ2--M 2x 3 \ 2(x - 3 = 4x 3- 9.75 = 2.25 kN. Ans.
3)
EI-t=R .x-==-
a a + MTo find the value of B , we must equate the clockwise moments and anti-clockwise

3:

when a: = 0, —r~ = 0. moments about B. Hence

ax Clockwise moments about B = Anti-clockwise moments about B.

Substituting this value in the above equation upto dotted line, we get M MRB + a x 6 = a + 4 x 3 x (4.5)

C, = 0. Mor b + 9.75 x 6 = 8.25 + 54 M(v RA = 9.75 and A = 8.25)
Mor b + 58.50 = 62.25
Therefore equation (i) becomes as

2(x - 3 Mb = 62.25 - 58.50 = 3.75 kNm. Ans.
3)

3 Problem A15.7. fixed beam of length 20 m, carries a uniformly distributed load of

Integrating again, we get m8 kN/m on the left hand half together with a 120 kN load at 15 from the left hand end.

—REl Jv = M —2 2 (s-3) 4 Find the end reactions and fixing moments and magnitude and the position of the maximum

-Aa .x

O2
a x3 - -2 x i + c2 mmdeflection. Take E = 2 x 10s kN/m3 and /'= 4 x 10s 4
.
3O 4A .
2o ' 3o
' 4 Sol. Given :
Length,
3

when x - 0, y = 0. L = 20 m
Substituting this value upto dotted line, we get
U.d.l., w = 8 kN/m
C = 0
2 Point load, W = 120 kN

Therefore equation (Hi) becomes as Value of E = 2 x 10s kN/m2
Value of
Ra.x3 M,.r mm m/ = 4 x 10s
+ -(x- 4 4 = 4 x 10-4 4
3)
AC = 10 m, AD = 15 m
6 Lengths,

when x = 6, =.y 0. Fig. 15.13 shows the loading on the fixed beam.

—x

646 STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 647

Integrating again, we get

M —— ——j

—Rh,ly — a .x - -

13 ~r~6
2 4x* + : — 60(x - 3 4(*-10) 4 "-{m)
:
„a .x — 15) +
3x42 34
Co2 3:!

when x = 0, y = 0. Substituting this value in the above equation upto Erst dotted line, we get

C = 0. Therefore equation (Hi) becomes as
2

—Ely = M2 - 20(re - 2 + ( 10)
"^ 6 a .x
2 15)

3i 3:

when x = 20, y = 0. Substituting these values in complete equation (iv), we get

—. R, x 20d M- a * 20 - - 20(20 - 15)3 + (20 - 10)4

0=

Z3 o

_ 20 ^A „^_20__125 + 1 10 4 (Dividing by 20 2 )
2 3 20 3 400
6 *

M~ 20 RA a 400 12.5 25
6 +

:

2 3 23

3M_ 20-ra - A - 800 - 37-5 + 50

m20Ra - 6
A = 800 + 37,5 - 50 = 787.5

At x = 20, —U,y = 0. Substituting these values in complete equation (ii), we get

M0 = x 202 - a x 20 - \ x 203 - 60(20 - 15)2 + - (20 - 10) 3

-M= 10Ra - 4 x 400 3 x 25 + (Dividing by 20)

a3
~= 10R, - M, -
- 75 +

=^10Ra -Ma 200 1400
+ 75-

M10i?A - a = 541.66 (Multiplying by 2 both sides) ...(vi)
M20iffJA4 - 2 a = 1083.32

Substracting equation (o) from equation (vi), we get

Ma = 1083.32 - 787.50 = 295.82 kNm. Ans.
MSubstituting this values of A in equation (vi), we get

20Ra - 2 x 295.82 = 1083.32

R„*~ 1083.32 + 2 x 295.82

20
= 83.748 kN. Ans.

Now Rb = Total load on beam - RA
= (10 x 8+ 120)- 83.748= 116.252 kN. Ans.

Equating the clockwise moment and anticlockwise moment about B, we get

M mb + ra x 20 = a + 120 x 5 + 8 x i0 x 15
Mor b + 83.748 x 20 = 295.82 + 600 + 1200
Mor b = 2095.82 - 83.748 x 20 = 420.86 kNm. Ans.

STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 649

Maximum, deflection and position of maximum deflection

Since the point load is more than the toal distributed Load and acts at an equal distance

from the nearest end, hence maximum deflection will be in the portion AD. For maximum

—deflection
should be zero. Substituting the value of -p- = 0 in equation (ii) [the term
dr dx

- 60(x - 15) 2 in equation (ii) should be ignored as this term is for the portion DB], we get

—B0 = x x,, —Mt.x —4 4
x3 + , - .

(x lOr

2 33

•XX- 295.82x - - 4(x - lOF

3

-, 41.874x 2 - 295.82x - -x 3 + [x 3 - 1000 - 3x x 10(x - 10)]

oo

: 41.874x2 - 295. 82x —4 x 3x x 10 x x + —4 x 3x x 10 x 10

= 41.874x2 - 295. 82x - - 40x 2 + 400x

= 1.874x 2 + 104.18* - 1333.33
This is a quadratic equation. Hence its solution is

- 104.18 ± 104.18 2 + 4 x 1.874 x 1333.33

*” 2x1.874

-104,81+144.387 (,„NTeg6l,ecting -ve root)
=

2x 1.874

= 10.727 m. Ans.

mHence maximum deflection occurs at a distance of 10.27 from A. Maximum deflection
mis obtained by substituting x = 10.727 in equation (iv) [neglecting the term - 20(x - IS) 3 ]

ElJymax = mRa^1_ 2 fa- 10)4

a-* _ 33

q2 295.82'x 10.727 2

83.748 x 10.727 3 4

62 10.727- +( -11 (10.?79277 _- l1n0i)44
= 17228.9 - 17019.8 - 4413.6 + 0.09
33

= - 4204.5

- 4204.5 - 4204.5

ymax - EI " 2 x 10® X 4 X 10-4

— 0.05255 m = 52.56 mm. Ans.

15.6. FIXED END MOMENTS OF FIXED BEAM DUE TO SINKING OF A SUPPORT

If the ends of a fixed beam are not at the same level, then the support which is at a lower

level is known as sinking support. Fig. 15.15 (a) shows a fixed beam A6 of length L whose ends
A and B are fixed at different levels. The end A is at a higher level than the end B. The beam

carries no load. Hence rate of loading on the beam is zero.

.

Let 5 = Difference of level between the ends

Ma = Fixing moment at the end A
Mb = Fixing moment at the end B

FIXED AND CONTINUOUS BEAMS

L* '

2 GEI6 6 E/8
-1FX X'

L L2

12£7S 6EI6
7 .X -
L? L2

At x = L, El represents B.M. at B i.e., El MB . Hence the above equation
becomes as

12E/3 r 6E/8
12E/5 6EI& GEIb

M MHence numerically A = B = . . This means that if the ends of a fixed beam are at

2

different levels (or one end sinks down by an amount 6 with respect to other end), the fixing

moment at each end is equal. At the higher end, this moment is a hogging moment and at the

lower end this moment is a sagging moment. The B.M. diagram is shown in Fig. 15.15 (f>).

Problem A15. 8. s fixed beam AB of length 3 m is having moment of inertia
mmI = 3 x 106
4 The support B sinks down by 3 mm. If E = 2 x 10s N/mrn2 find the
,
.

fixing moments.

Sol. Given :

Length, L = 3 m = 3000 mm

mmValue of / = 3 x 10® 4

Value of E = 2 x 105 N/mm2

The amount by which the support B sinks down,

8 = 3 mm.

The fixing moments at the ends is given by,

m 6E/6

a ~ Mb -
jj.

6 x 2 x 10 6 x 3 x 10 6 x 3

= 12 x 10s Nmm = 12 x 103 Nm = 12 kNm. Ans.

The fixing moment at A will be a hogging moment whereas at B it will be a sagging

moment.

15.7. ADVANTAGES OF FIXED BEAMS

The following are the advantages of a fixed beam over a simply supported beam :
(i) For the same loading, the maximum deflection of a fixed beam is less than that of a
simply supported beam.

(ii) For the same loading, the fixed beam is subjected to a lesser maximum bending

moment.

(Hi) The slope at both ends of a fixed beam is zero.
(iv ) The beam is more stable and stronger.

: [

STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 653

15.8. CONTINUOUS BEAMS x = Distance of C.G. of the B.M. diagram due to vertical loads on BC from B
l
Continuous beam is a beam which is supported on more than two supports. Fig. 15.16
shows such a beam, which is subjected to some external loading (here a uniformly distributed x2 = Distance of C.G. of the B.M. diagram due to vertical loads on CD from D.
load). The deflection curve for the beam is shown by dotted line. The deflection curve is having
convexity upwards over the intermediate supports, and concavity upwards over the mid of the The equation (15.12) is known as the equation ofthree moments or Clapeyron’s equation.
span. Hence there will be hogging moments (i.e., negative) over the intermediate supports and
sagging moments {i.e., positive) over the mid of the span. The end supports of a simply sup- 15.9.1. Derivation of Clapeyron’s Equation of three Moments. Fig. 15.17 shows
ported continuous beam will not be subjected to any bending moment. But the end support of
fixed continuous beam wall be subjected to fixing moments. If the moments over the interme- the length BCD (two consecutive spans) of a continuous beam which is shown in Fig. 15.16. Let

diate supports are known, then the B.M. diagram can be drawn. M M M Db , c and D are the support moments at B, C and respectively.

w/Unit length

- Deflection curve

Fig. 15.16

The Fig. 15.16 shows a simply supported continuous beam. In this figure the end sup-

A E M Mports at and will not be subjected to any bending moment. Hence in this case A = E = 0.

Fig. 15.16 (a) shows a continuous beam with fixed ends atA andB. Here the end supports

M Mat A and E will be subjected to fixing moments. Hence A and E will not be zero.

Fig. 15.16 (a) BCD

15.9. BENDING MOMENT DIAGRAM FOR CONTINUOUS BEAMS

In Art. 15.8 it is mentioned that if the moments over the intermediate supports of a Resultant B.M. Diagram

continuous beam are known, then the B.M. diagram can be drawn easily. The moments over Fig. 15.17
the intermediate supports are determined by using Clapeyron’s theorem of three moments

which states that Let Lj = Length of span BC

KBC and CD are any two consecutive span of a continuous beam subjected to an external Lrj - Length of span CD

M M Mloading, then the moments B , c and D at the supports B. C and D are given by, a = Area of B.M. diagram due to vertical loads on span BC
l
+ 2MC{L1 B2) Mq.L 60]*! 603X2 ...(15.12)
2
+ + — due on span CD
M Ma
2
= Area of B.M. diagram to vertical loads

= Length of span BC af = Area of B.M. diagram due to support moments B and c
M Ma.f - Area of B.M. diagram due to support moments
L., = Length of span CD . and D
a - Area of B.M. diagram due to vertical loads on span BC
;
x
x = Distance of C.G. of B.M. diagram due to vertical loads on BC
a = Area of B.M. diagram due to vertical loads on span CD l
2
x2 - Distance of C.G. of B.M. diagram due to vertical loads on CD

xf — Distance of C.G. of B.M. diagram due to support moments on BC

x 2 = Distance of C.G. of B.M. diagram due to support moments on CD.

w

.

654 STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 655

Fig- 15.17 (6) and (c) shows the B.M. diagrams due to vertical loads and due to supports ~-13Mb + 2MC - 2MB ] (Mb + 2Mc
Mb + Mq
moments respectively. \ Mb + Mq J 3

(i) Consider the span BC Substituting the values of a, and xf in equation (ii), we get

MLet = B.M. due to vertical loads at a distance x from B (sagging) MELL^c = a x * ^
x 1 - 1 (Mb C ).LX
Mf = B.M. due to support moments at a distance x from B (hogging) 1

.-. Net B.M. at a distance x from B is given by,

M Md 2 y Mg +

El —4r -
dx 2 -'

Multiplying by x to both sides, we get =ax M+ 2 c)
ll

—EI.x. d 2 y = x.M - x.M'x Mor
dx ,1, 6EI.SC = - L {(Mb + 2 c ) ...(Hi)

Integrating from zero to L v we get "i

(ii) Consider the span CD

f 'El.x.^—^.dx = f 'x.Mx .dx - f x.Mx ‘.dx DSimilarly considering the span CD and taking as origin and x positive to the left, it
Jo dx 2 Jo Jo
can be shown that

§a2 x 2
cw -L 2M6E7.Ct- —
aec)\ (Mr otm ^
= rn/t +. c)

- x — 2d
x
= ct]*! a
l

Jo C[In the above case the slope at (i.e., 6 C) will have opposite sign than that given by
Dequation (iii). The reason is that the direction of x from B for the span BC, and from for span
M( .dx = Area of B.M. diagram of length dx. And x.Mx.dx CD are in the opposite direction].

x Hence the above equation becomes as
- Moment of area of B.M. diagram of length dx about B.

Hence f 'x.Mx .dx = ax . And so on)
xx
Jo M- LJMn 4 2 r)
_ 6EIQ =
...(iv)

Substituting the limits in L.H.S. of equation (i), we have

Adding equation (iii) and (iv), we get

“LlHsj..-* 0 = _ Ly(MB 4 2MC) 4 -L2(Md + 2Mc)

= a ix l ^ m M M= “p- + 6a
X -L 2L -LJ1d - 2L
i b~ x c 2 c

EIl(L vQc - yc) - (Q - yB )] = ax - x' =9 C M M Mor
1l x dxkiC
Lv
b 4 L d42 c (L + L2) =
2 x
Ly
But deflection at B and C are zero. Hence yB = 0 and yc = 0. Hence above equation
M M ---or
becomes as / bL x 4 2 C(L X + l2 ) 4 = 1 1- 4

[EI.L V 9C = a x - a 'x ' ...iii) Lj L2
xx
x x

But a/ = Area of B.M. diagram due to supports moments 15.9.2, Application of Clapeyron’s equation of Three Moments to Continuous
and Beam with Simply Supported ends. The fixing moments on the ends of a simply supported
= Area of trapezium BCKJ
beam is zero. The continuous beam with simply supported ends may carry uniformly distrib-
M= 4(ilfB + c)xL1
uted load or point loads as given in the following problems:

*' = Distance of C.G. of area BCKJ from B Problem A15.9. continuous beam ABC covers two consecutive span AB and BC of lengths
l m4 and 6 m, carrying uniformly distributed loads of 6 kN/m and 10 kN/m respectively. If the
ends/A and C are simply supported, find the support moments at A, B and C. Draw also B.M.
M MB .LX .^- + | x (Mc - b ).L, x
and’S.F. diagrams.

M -MB .L1 + ±(MC B ).LX Sol. Given :
Length AB,
L, MZMBLX + 2Li(Mc - B ) Length BC, L =4m
x
B) x g
M M —.—L x L = 6m
c x + (Mc - 2Mg +Mc -Mb

M -Mb + (Mc b ).±

fixed and continuous beams 657

:-x4x

3

mXl
2
22

a = Area of B.M. diagram due to u.d.l. on BC
2

—= —2 x BC x w2 L0 2 = -2x6„>i 10 x 6 2 = ,18„0„

3 83 8

—h- = = —6 = Q3 m.

x,1 2
2

Substituting these values in equation (i), we get

M —3x32 + 180 x 3
20 Bb =
2

= 96 + 540 = 636

M —Bn = =: 31.8 kNm.
20

Now B.M. diagram due to supports moments is drawn as shown in Fig. 15.18 (b) in

M M Ma - 0, c = 0 and B = 31.8 kNm.

The B.M. diagram due to vertical loads (here u.d.l.) on span AB and span BC are also

shown by parabolas of altitudes
^w,L, = 6 x4 2 = 12 kNm and w2L22 __ 10 x 62 = 45 kNm

88 8

respectively in Fig. 15.18 (b).

S.F. Diagram.

RFirst calculate the reactions J?A , B and Rc at A, B and C respectively. For the span AB,

taking moments about B, we get

R. x4-6x4x — = MR M(The support B has moment B)

= -31.8

M( v b = 31.8. Negative sign is taken as the moment at B is hogging)

4Ra - 48 = - 31.8

R„.a = - 31.8 + 48 = 4.05 kN.
4

Similarly for the span BC, taking moments about B, we get

MRc x6-6xl0x — = b = - 31.8

6i?c - 180 = - 31.8
180-31.8
= 24.7 kN.

= Total load on ABC - (RA + Rc)
= (6x4+ 10x6)- (4.05 + 24.7) = 55.25 kN.

Now complete the S.F. diagram as shown in Fig. 15.18 (c).

M

658 STRENGTH OF MATERIALS FIXED AND CONTINUOUS BEAMS 659

mProblem 15.10. A continuous beam ABCD of length IS rests on four supports cover- For ABC, we get

ing 3 equal spans and carries a uniformly distributed load of 1.5 kN/m length. Calculate the M M2MaLx +
B(Ly + L2) + c .L = +
moments and reactions at the supports. Draw the S.F. and B.M. diagrams also. 2
^Li L.2
Sol. Given : M0x5 + 2MBfl(5 + 5) + c x 5 = X
+ 2
.

Length AB, L = 5 in 55
1

mLength BC,
L =5 M ~20 +<z 2 2 )
2 b+ 5 c= (a x +
]l
Length CD, L = 5 m o
3
Now a = Area of B.M. Diagram due to u.d.l. on AB when AB
-wU.d.l., 2 = u>3 = 1-5 kN/m. 1
w are simply supported, the support moments at
l
D ASince ends A and Dand will be zero. is considered as simply supported beam

From symmetry M Ma = 0 and D = 0 = —2 x AB x Altitude of parabola
MB = Mc
O

WiL L5x5 2
rl
To find the support moments at B and D, Clapeyron’s equation of three moments is — —= t2 x5x = —2 x5x = 15.625

applied for ABC and for BCD. d oo 8

L
i
1 .5 kN/m Y m- = 5 = „„
= 2 2,5
*2

Due to symmetry a = a = 15.625 and x2 - x = 2.5
x l
2

Substituting these values in equation (i), we get

M20MS + 5 c = -6 (15.625 x 2.5 + 15.625 x 2.5)

x 2 x 15.625 x 2.5 = 93.750

M20 b + 5M = 93.750 M M(v b - c due to symmetry)
fi

A B CD Mb = 93^50 _ 3 ?5 kNm

B.M. diagram due to vertical toads i

M =Mb c = 3.75 kNm. Ans.

Now the B.M. diagram due to supports moments is drawn as shown in Fig. 15.19 (c), in

which

M M M Ma = 0, d = 0, b = c = 3.75 kNm.

The B.M. diagram due to vertical loads (here u.d.l.) on span AB, BC and CD

wL^

(considering each span as simply supported) are shown by parabolas of altitudes =1 1

8

— kNmL5 x 5 2 each in Fig. 15.19 (6). Resultant B.M. diagram is shown in Fig. 15.19 d( ).
O0 = 4.6875

Support Reactions

AB C R DLet a, Rb , Rc and RD are the support reactions at A, B, C and respectively.

S.F. diagram Due to symmetry, RA = RD
Fig. 15.19
Rr = Rc

For the span AB, taking moments about B, we get

Mg = RA x 5 - 1.5 x 5 x —

- 3.75 =Ra x 5- 18.75 M(v n

5R. = 18.75 - 3.75 = 15

—Ra = = 3.0 kN. Ans.

-

STRENGTH OF MATERIALS fixed and continuous beams 661

Due to symmetry, RD = RA = 3.0 kN. Ans. M —M„or 1* 1 + gtz2*2 -d)
w Ra + RB + Rc + Rd = Total load on ABCD 0+2 (6 + 5) + r X 5 =

ra + r b + rb + ra : 1.5 X 15 65

R2(fi, + b) = 22.5 R RR R('.' r - B’ D ~ A^ or 22MS 5MC = _ + —6 _
a2x2
+ ax
11

ra +rb~ '

Rb = 11.25 - Ra = 11.25 - 3.00 = 8.25 (V «a = 3.0)

Rb = Rfi c = 8.25 kN. Ans. (a)

Now the S.F. diagram can be drawn as shown in Fig. 15.19 (e).

Problem 15.11. A continuous beam ABCD, simply supported atA,B,C and D is loaded

as shown in Fig. 15.20 (a). Find the moments over the beam and draw B.M. and S.F. diagrams.

Sol. Given :

Length AS, Lj = 6 m (6)

Length BC, L = 5 m
2

Length CD, L - 4 m
3

WPoint load in BD, = 9 kN

1

WPoint load in BC, 2 - 8 kN (c)

U.d.l. on CD, w = 3 kN/m. A, C D

(i) B.M. diagram due to vertical loads taking each span as simply supported B !

Consider beam AB as simply supported |

B.M. diagram due to support moments

|

MRB.M. aaft. mpoiinntt llnoaardi at E =ft - — x ax b 9x2x4== ,.. Here a„ = 2 m, b = 4 m)

V.

= 12 kNm w

Similarly B.M. at F, considering beam BC as simply supported AE BF C
Resultant B.M. diagram
= Wz.a.fc _ 8x2x3 |I
= = 3 and L - |
L5 Here 2 5)
2- .. a 2, 6
(
= 9.6 kNm

The B.M. at the centre of a simply supported beam CD, carrying u.d.l. (e)

= w x jal = = 6 kNm. Now S.F. diagram

88 Fig. 15.20

Now the B.M. diagram due to vertical loads taking each span as simply supported can

be drawn as shown in Fig. 15.20 b( ).

(U) B.M. diagram due to support moments Dand respectively. But a1x 1 = Moment of area of B.M. diagram due to .vertical load on AB
when AB is considered as simply supported beam about point A.
M Mthe D are the supports moments at A, B, C bending moment. Hence
Let M., M„, a c and
end supports of simply supported beam are not subjected to any

the support moments at A and D will be zero. = ~ x 2 x 12 x + -x4xl2x|2 + ix4|

M MTo 2 32 V3;

a=0 and D=0 equation of three moments in = 16 + 80 = 96
find the support moments
at B and C, Clapeyron’s

applied for ABC and for BCD. a2 :t 2 = Moment of area of B.M. diagram due to vertical load on BC
(a) For spans AB and BC from equation of three moments, we have when BC is considered as simply supported beam about point C

M M M LLA .L l + 2 b (Lj + 2) + c . 2 = = — x 3 x 9.6 x — x 3 + — x2x 9.6 x \3 + — x 2)

2 32 V3J

= 28.8 + 35.2 = 64.0

—-

662 STRENGTH OF MATERIALS fixed and continuous beams 663

Substituting these values in equation (i), we get or -4.48 = 47^-24 M(v c = -4.48)

M22MS + 5 c = 96 + -6 x 64 24 - 4.48
Rd = - = 4.88 kN. Ans.

= 172.8 Now taking moments about C for ABC, we get

(6) For spans SC and CD from equation of three moments, we have Mc = Ra x (6 + 5) - 9 (5 + 4) + Rb x 5 - 8 x 3
Mor
6aoX-> 6a3x* -4.48 = 4.86 x 11-9 x 9 + Bs x 5-24 (v c = - 4.48, RA = 4.86)
M 2M M —r —B ^2 +
L L„ . . , _ -
C (L + 3) + d. + 5RB = 81 + 24 - 4.86 x 11 - 4.48 = 47.06
3
2

C(5 + 4) + 0 = 6a 2 .v 2 6°3*3 47.06
R„a = -
2MM„ x 5 + = 9.41 kN. Ans.
5

M M5 ~a2x2 + -a3 * a Now Rc = Total load on ABCD - (RA + RB + RD)
= (9 + 8 + 4x3)- (4.86 + 9.41 + 4.88)
B + 18 --

c

where a2 x 2 = Moment of area of B.M. diagram due to vertical load on SC when = 9.85 kN. Ans.
SC is considered as simply supported beam, about point B
Now complete the S.F. diagram as shown in Fig. 15.20 (e).
= -^x2x9.6x|x2+^x3x9.6x^2-t--|x3j
15.9.3. Clapeyron’s Equation of Three Moments Applied to Continuous Beam
an d = 12.8 + 43.2 = 56.0 with Fixed end Supports. We have seen in Art. 15.9.2 that fixing moments on the ends of a

a 3 .r3 = Moment of area of B.M. diagram due to u.d.l. on CD, when CD is simply supported continuous beam are zero. But in case of a continuous beam fixed at its one
or both ends, there will be fixing moments at the ends, which are fixed. To analyse the continu-
Dconsidered as simply supported beam, about point ous beam which is fixed at the ends by the equation of three moments an imaginary support of
zero span is introduced. The fixing moment at this imaginary support is always equal to zero.

(—2 „ .. ,A —Ba—se
U- . x
Base Altitude
x x

J2

= -2 x4x6x-4 = 32

Substituting these values in equation (tit), we get

M5 b + 18MC = |x56+|x32 = 115.2 -(w) A(a) Continuous beam fixed at (6) Continuous beam with zero span

Solving equations (ii) and (it)), we get Fig. 15.21

M Mb = 6.84 kNm and c = 4.48 kNm. If the beam is fixed at the left end A, than an imaginary zero span is introduced to the

Now the B.M. diagram due to supports moments is drawn as shown in Fig. 15.20 (c), in Aleft of as shown in Fig. 15.21 b( ). But if the beam is fixed at the right end, then an imaginary

which zero span is introduced to the right end support. After this Clapeyron’s equation of three
moments is applied.
M M M Ma = 0, b = 6.84, c = 4.48 and D = 0.
Problem 15.12. A continuous beam ABC of uniform section, with span AB and BC as
The B.M. diagram due to supports moments will be negative. Resultant B.M. diagram is m A4 each, is fixed at and simply supported at B and C. The beam is carrying a uniformly

shown in Fig. 15.20 (d). distributed load of 6 kN/m run throughout its length. Find the support moments and the reac-
tions. Also draw the bending moment and S.F. diagrams.
(Hi) Support Reactions
Sol. Given :
DLet Ra, Rb , Rc and RD are the support reactions at A, B, C and respectively.
Length AB, L^ = 4 m
For the span AS, taking moments about B, we get

= RA x6-9x4 mLength SC,
Mr = - 6.84) L = 4
6.84 = 6Rb - 36 2

U.d.l., w = 6 kN/m.

R 36 ~ 6,84 (i) B.M. diagram due to u.d.l. taking each span as simply supported

a = R = 4.86 kN. Ans. Consider beam AB as simply supported. The B.M. at the centre of the span AB

For the span CD, taking moments about C, we get _ 6 x 42
~8
M Sd x4~3x4x — - W - Li = 12 kNm
c
= 8

STRENGTH OF MATERIALS

Similarly B.M. at the centre of span BC, considering beam BC as simply supported

The B.M. = w.L? = 6x4 2 = 12 ,kNm

Fig. 15.22 (c). 88

diagram due to u.d.l. taking each span as simply supported is drawn in

(ii) B.M. diagram due to support moments

AAAs beam is fixed at A, therefore introduce an imaginary zero span 1 Ato the left of as

Ashown in Fig. 15.22 (£>). The support moment at is zero.
t
MLet
g = Support moment at A and is zero
x
Ma = Support moment at A
Mb = Support moment at B
Mc = Support moment at C.
M M MThe extreme end C is simply supported hence c = 0. To find A and B theorem of

three moments is used.

Fig. 15.22

Applying the theorem of three moments for the spans AjA and AS, we have

¥c x0 + 2Ma{0 + Lx ) + L0 l
x

666 STRENGTH OF MATERIALS fixed and continuous beams 667

Now B.M. diagram due to support moments is drawn as shown in Fig. 10.22 (c) in which Mb = Support moment at B
Mc = Fixing moment at C.
M M Mb = 6.86, b = 10.28, and c = 0. The B.M. due to supports moments will be negative. M M MTo find a , b and c, theorem of three moments is used.

Resultant B.M. diagram is also shown in Fig. 15.22 (c). (a) Applying the theorem of three moments for the spans AjA and AB, we get

(iii) Support Reactions M M Mx 0 + 2 L= + ggigi
0
RRLet a , b and R c are the support reactions at A, B and C respectively. a (0 + Lj) + B. L0 L
t x
For the span BC, taking moments about B, we get

= Rc X 4 - I6x4x-4 Mb is negative) 12 kN

- 10.28 : 4Rc - 48

= 9.43 kN. Ans.

For the span AB, taking moments about B, we get

M Mb = a + Ra x4- 6 x 4 x

M Mor
- 10.28 = - 6.86 + ARa - 48 ( v Bb and A are negative);
—or
R„a = 48 + 6.86-10.28 = 11.14 kN. .

Ans.

and RB = Total load — iRA + Rq)

= 6 x 8- (11.14 + 9.43) = 27.43 kN. Ans.

Now complete the S.F. diagram as shown in Fig. 15.22 (d).

Problem 15.13. A continuous beam ABC of uniform section, with span AB and BC as
m A6 each, is fixed at and C and supported at B as shown in Fig. 15.23 (a). Find the support

moments and the reactions. Draw the S.F. and B.M. diagrams of the beam.

Sol. Given :

Length AB, L -6m
Length BC, x

L = 6 m
2

U.d.l. in AB, w = 2 kN/m

WPoint load in BC, = 12 kN.

(i) B.M. diagram due to vertical loads taking each span as simply supported

Consider beam AB as simply supported. The B.M. at the centre of AB

—= w L, 2 = 2 x 62 = _ ,k,NTm.

9

88

Consider beam BC as simply supported. The B.M. at the centre of BC

WxL mkNm= ^2 = 12 x 6 = 1i 8q i

44

The B.M. diagram due to vertical loads is drawn as shown in Fig. 15.23 (c).

(ii) B.M. diagram due to support moments

As beam is fixed at A and C, therefore introduce an imaginary zero span AA and CC to
t l

Athe left of and to the right of C respectively as shown in Fig. 15.23 b( ). The support moments

at Aj and C are zero.
1
MLet
0 = Support moment at A, and C and it is zero
j
Ma = Fixing moment at A

+ —

'

STRENGTH OF MATERIALS

M Mor a x 6 + 2MS(6 + 6) + c x6= £ + g Now Rb = Total load - (RA + Rc)

= (6 x 2 + 12) - (5.625 + 6.375) = 12 kN. Ans.

6M M 6Mor a x=C 1 l + cu2*2 - —(iii) The S.F, is shown in Fig. 15.23 d( ).
a + 24 b +

where a-pc x = —2 x6x9x —6 = 108 HIGHLIGHTS

o £

a2x2 = Moment of area of B.M. diagram due to point load on BC when it is 1. A beam whose both ends are fixed is known as fixed beam. And a beam which is supported on

considered as simply supported beam about C more than two supports is known as a continuous beam.
2. In case of a fixed beam :
= -^x6xl8x3 = 162
(t) a = a' (ii)ax-a'x' and x = x
Substituting these values in equation (iii), we get

M M6 a + 24Ma + 6 c = 108 + 162 = 270 Or
M M Mor a + 4 B + c = 45 (i) The area of B.M. diagram due to vertical loads is equal to the area of B.M. diagram due to end
...(iv)
moments.

(c) Now applying the theorem of three moments for the span BC and CCV we get (ii) Distance of C.G. of B.M. diagram due to vertical loads is equal to the distance of C.G. of B.M.
diagram due to end moments from the same point.
M M Mb .L2 + 2 c(L2 + 0) + x0=
3. The deflection at the centre of a fixed beam carrying a point load at the centre is given by
0

or Mb x 6 + 2MC(6 + 0) + 0 = +0 WI?
or
M6 b + 12MC = a2x2 —ip) where ~
where yc 192 El

W = Point load,

a2x2 = Moment of area of B.M. diagram due to point load on BC when it is considered L = Length of beam.

as simply supported beam about B 4. The deflection at the centre of a fixed beam carrying a point load at the centre is one-fourth of the

deflection of a simply supported beam.

= — x 6 x 18 x 3 = 162. 5. The deflection of a fixed beam with an eccentric load, under the point load is given by,

Substituting this value in equation (o), we get

MQMB + 12 c = 162 6. (a) For a fixed beam carrying uniformly distributed load over the whole length :
Mb + 2MC = 27
WEnd moments = x 1?
Solving equations (ii), (iv) and (vi), we get
12
Ma = 5.25 kNm, MB = 7.5 kNm
Mc = 9.75 kNm. —Max. deflection = ~~

Now B.M. diagram due to support moments is drawn as shown in Fig. 15.23 (c). The 384 El
(b) The deflection at the centre of a fixed beam carrying uniformly distributed load over the
B.M. due to support moments is negative.
whole span is one-fifth of the deflection of a simply supported beam.
(iii) Support reactions 7. The end moments of a fixed beam due to sinking of a support is given by

Let Ra, Rb and R c are the support reactions at A, B and C respectively.

For the span AB, taking moments about B, we get

M - Ra x6-6x2x3 + Ma where 6 = Sinking of one support with respect-to the other. At the higher end this moment is -ve
B whereas at the lower end it is positive.
M<v R and M. are negative)
- 7.5 = Ra x 6 - 36 - 5.25 8. Clapeyron’s theorem of three moments for a continuous beam ABC is given by.

36 + 5.25-7.5 =556.62255kkIN. Ans. 2M MMj.L.
i
A 6 +• L xLg(LI +1 2 > + c ~ 2 = ;

For the span BC, taking moments about B, we get Ll L2

M MB = RC x 6 - 12 x 3 + c where a = Area of B.M. diagram due to vertical loads on span AB
l
M- 7.5 = Rc x 6 - 36 - 9.75
a = Area of B.M. diagram due to vertical loads on span BC
2
(v Mo and c are negative)
^or Xi = Distance of C.G. of B.M. diagram due to vertical loads on AB from A

36 9.75 - 7. 5 _ fi o 75 Ans. BCx2 - Distance of C.G. of B.M. diagram due to vertical loads on from point C.

6

670 strength of materials FIXED AND CONTINUOUS BEAMS 671

9. To apply the theorem of three moments to a fixed continuous beam, an imaginary support of zero A m5. fixed beam AB of length 6 carries a uniformly distributed load of 3 kN/m over the left half of
span is introduced. mthe span together with a point load of 4 kN at a distance of 4.5 from the left end. Determine the

fixing end moments and the support reactions.

M M[Ans. a - 7.3 kNm, B = 6.2 kNm, RA = 7.93 kN, RB = 5.07 kN]

m mm6. A fixed beam AB of length 6 is having moment of intertia / = 5 x 106 4 The support B sinks

.

down by 6 mm. If E = 2 x 10 s N/mm2 find the fixing moments. M M Nm[Ans. A = B = 1000 ]

(A) Theoretical Questions m7. A continuous beam ABC of length 10 rests on three supports A, B and C at the same level in

1. What do you mean by a fixed beam and a continuous beam ? mwhich span AB = 6 and span BC = 4 m. In span AB there is a point load of 3 kN at a distance
2. Prove that for a fixed beam : ,

(i) Area of B.M. diagram due to vertical loads is equal to the area of B.M. diagram due to end mof 2 from the end A, whereas in the span BC, there is a uniformly distributed load of 1 kN/m
moments.
run over the whole length. Determine the support moments and support reactions. Draw S.F.
di) Distance of C.G. of B.M. diagram due to vertical loads is equal to the distance of C.G. of B.M. M M[Ans. (i) A = c = 0,
diagram due to end moment from the same point. and B.M. diagrams also. = 2.4 kNm,

3. Find an expression for the deflection for a fixed beam carrying a point load at the centre. Also (w) Ra = 1.6 kN, Rb = 4 kN, Rc = 1.4 kN]
obtain the value of maximum deflection.
m m8. A continuous beam consists of three successive span of 8 m, 10 and 6 and carries loads of
4. Prove that the deflection at the centre of a fixed beam is one-fourth the deflection of a simply
6 kN/m, 4 kN/m and 8 kN/m respectively on the spans. Determine the bending moments and
Wsupported beam of the same length, when they carry a point load at the centre. M M M M[Ans. (i) A = D = 0, c = 32.2 kNm, B = 40.16 kNm,
reactions at the supports.
5. Draw the S.F. and B.M. diagrams for a fixed beam, carrying an eccentric load.
6. Prove that the deflection at the centre of a fixed beam, carrying a uniformly distributed load is R R10. (,ii ) a = 18.98 kN, RB = 49.82 kN, Rc = 48.57 kN, d = 18.63 kN]
m m9. A continuous beam ABC consists of two consecutive spans AB and BC of length 8 and 6
given by
respectively. The beam carries a uniformly distributed load of 1 kN/m throughout its length. The
wL4
end A is fixed and the end C is simply supported. Find the support moments and the reactions.
y‘ ~ 384B/ M M M[Ans. (i) A = 5.75 kNm, B = 4.5 kNm, c = 0,
Also draw the S.F. and B.M. diagrams.
Determine the position of points of contraflexures also.
(ii) Ra - 4.15 kN, RB = 7.6 kN, Rc = 2.25 kN]
7. Derive an expression for the fixing moments, when one of the supports of a fixed beam sinks
down by 5 from its original position. mDraw the S.F. and B.M. diagram of a continuous beam ABC of length 10 Awhich is fixed at

8. What are advantages and disadvantages of a fixed beam over a simply supported beam ? and is supported on B and C. The beam carries a uniformly distributed load of 2 kN/m length
9. What is the Clapeyron’s theorem of three moments ? Derive an expression for Clapeyron’s theo-
mover the entire length. The spans AB and BC are equal to 5 each.
rem of three moments.
M U M[Ans. (i) a = 3.57 kNm, B = 5.357 kNm, c = 0,
10. How will you apply Clapeyron’s theorem of three moments to a
R R R(ii) a = 5.357 kN, a = 8.571 kN, c = 6.071 kN]
(i) continuous beam with simply supported ends

(ii) continuous beam with fixed end supports ?

(B) Numerical Problems

m1. A fixed beam AB, 5 long, carries a point load of 48 kN at its centre. The moment of inertia of
mmthe beam is 5 x; 1'674 and value of B for the beam material is 2 x 103 N/mm 2 . Determine :

(i) Fixed end moments at A and B, and

(ii) Deflection under the load. M M[Ana. (i) A - B = 30 kNm, (ii) 3.125 mm]

m m2. A fixed beam of length 5 carries a point load of 20 kN at a distance of 2 from A. Determine
the fixed end moments and deflection under the load, if the flexural rigidity of the beam is
M MkNm1 x 104
2 [Ans. A = 14.4 kNm, B = 9.6 kNm, yc - 1.15 mm]

.

A m m m3. fixed beam of length 6 carries point loads of 20 kN and 15 kN at distances 2 and 4 from
the left end A. Find the fixed end moments and the reactions at the supports. Draw B.M. and
M M RS.F. diagrams.
[Ans. A = 24.44 kNm, B ~ 22.22 kNm, RA — 18.70 kN, B = 16.30 kN]
m mA4. fixed beam of length 3 carries two point loads of 30 kN each at a distance of 1 from

both the ends. Determine the fixing moments and draw the B.M. diagram.

M M[Ans. A = b = 20 kNm]

TORSION OF SHAFTS AND SPRINGS 673

<|> = LDCD' also equal to shear strain
0 = A.DOD' and is also called angle of twist.

16

Torsion of Shafts and Springs

16.1. INTRODUCTION Fig. 16.2. Shaft fixed at AA and subjected to torque T at BB.

A shaft is said to be in torsion, when equal and opposite torques are applied at the two

ends of the shaft. The torque is equal to the product of the force applied (tangentially to the
ends of a shaft) and radius of the shaft. Due to the application of the torques at the two ends,
the shaft is subjected to a twisting moment. This causes the shear stresses and shear strains in

the material of the shaft.

16.2. DERIVATION OF SHEAR STRESS PRODUCED IN A CIRCULAR SHAFT Now distortion at the outer surface due to torque T
SUBJECTED TO TORSION
- DD'

When a circular shaft is subjected to torsion, shear stresses are set up in the material of Shear strain at outer surface

the shaft. To determine the magnitude of shear stress at any point on the shaft, consider a = Distortion per unit length

shaft fixed at one end AA and free at the end BB as shown in Fig. 16.1. Let CD is any line on Distortion at the outer surface _ DD'

the outer surface of the shaft. Now let the shaft is subjected to a torque T at the end BB as Length of shaft L

shown in Fig. 16.2. As a result of this torque T, the shaft at the end BB will rotate clockwise DD'

Dand every cross-section of the shaft will be subjected to shear stresses. The point will shift to CD* = tanifi

CD ODD' and hence line = ~(if <(> is very small then tan <ji <|>)
will be deflected to CD' as shown in Fig. 16.2 (a). The line will be <j)

shifted to OD' as shown in Fig. 16.2 (6). .-. Shear strain at outer surface,

DD' ~W

^~T (v OD = R = Radius of shaft)

Now from Fig. 16.2 b( ).

Arc DD' = OD x 6 = R0

Substituting the value of DD' in equation (i), we get

Shear strain at outer surface

<j>= -— —12x0 ••(«)

is

Now the modulus of rigidity C( ) of the material of the shaft is given as

AAFig. 16.1. Shaft fixed at one end before torque T is applied. £ _ Shear stress induced _ Shear stress at the outer surface

Let R = Radius of shaft Shear strain produced Shear strain at outer surface

_ —I— —(
L - Length of shaft i?0 ^
T = Torque applied at the end BB
From equation (ii), shear strain =

x = Shear stress induced at the surface of the shaft due to torque T

C - Modulus of rigidity of the material of the shaft

...(16.1)

672