Strength of Materials by R.K.Bansal_text

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

:3

528 STRENGTH OF MATERIALS deflection of beams

ELy= VLk x^HL. Negative sign indicates that deflection is downwards.
12 24
(v C, = 0) .-. Downward deflection,
w.L 3 if 4
u)LJ ——yc = 384 . El ...(12.14J
12 24
24 mm mmProblem 12.5. A beam of uniform rectangular section 200
wide and 300 deep is

WeThe equation (hi) is known as slope equation. can find the slope (i.e., the value of -^-1 simply supported at its ends. It carries a uniformly distributed load of 9 kN/m run over the

l dx) N/mmentire span of 5 m. If the value ofE for the beam material is 1 x 104 2 find :

,

at any point on the beam by substituting the different values of x in this equation. (i) the slope at the supports and (U) maximum deflection.

WeThe equation (iv) is known as deflection equation. can find the deflection (i.e., the Sol. Given
Width,
value of y) at any point on the beam by substituting the different values of x in this equation. Depth, b = 200 mm
mm: 300
Slope at the supports

300
—— mm.

/
M O_ = for = 200 x = .. x i1n088 4
I
Let 0A = Slope at support A. This is equal to f ~r~ ’ 4.5

\dxjj atA U.d.L, 12 12
Span,
w = 9 kN/m = 9000 N/m
Total load,
and 0 g = Slop at support B = f^-1 Value of L = 5 m = 5000 mm

W = w . L* = 9000 x 5 = 45000 N

—AtA, x - 0 and ax = 0.A. E = 1 x 104 N/mm2

1 Substituting these values in equation (hi), we get Let 0A = Slope at the support

—„ = wL w x 0_ wl} and yc = Maximum deflection.
24 (i) Using equation (12.12), we get
EI.B,A 4 6
x 0

w WL(v(... jy . £, -= = Total load) A-__~ W.L2

== 24 El
24 24
45000 x5000 2 — rad,i, ans
0A 24El ......( (1122..1122)) 24 x lx 10 4 x 4.5 x 10 s

= 0.0104 radians. Ans.

A(Negative sign means that tangent at makes an angle with AB in the anti-clockwise (ii) Using equation (12.14), we get

direction) JL Wy° __ 384 '

WL2 L-
By symmetry, -——6B = El
Maximum Deflection r ...(12,.13)
l
24 hj 5 45000 x 5 OOP 3

~ X lx 10 4 x4.5x 10 s

384

—The maximum deflection is at the centre of the beam i.e., at point C, where x = - Letyc - 16.27 mm. Ans.

2) mProblem 12.6. A beam of length 5 and of uniform rectangular section is simply
—= deflection at C which is also maximum deflection. Substituting y = yc and x =
in the supported at its ends. It carries a uniformly distributed load of 9 kN/m run over the entire
length. Calculate the width and depth of the beam ifpermissible bending stress is / N/mm and
“. equation (iv), we get ^ —*T0rc -
central deflection is not to exceed 1 cm.

Take E for beam material = 1 x 104 N/mm2.

-f-T “ *f-T -f-1 Sol. Given :
1,2 J Length,
12 V2 J 24 v2J 24 L = 5 m = 5000 mm
U.d.l.,
wL‘ _ _ 5iv.L4 w = 9 kN/m
48 ” 384
-_
96 384"

yr = 5 wL‘* 5 -W.—L-? W( v w.L = = Total, , *Here L should be taken in metre. Hence for calculating total load, L must be in metre and in
. other calculations L is taken in mm.
. li oad)

530 STRENGTH OF MATERIALS' deflection of beams 531

Total load, W = w.L - 9 x 5 - 45 kN = 45000 N A mProblem 12.7. beam of length 5 and of uniform rectangular section is supported at
Bending stress,
Central deflection, f- 7 N/mm2 its ends and carries uniformly distributed load over the entire length. Calculate the depth of the
Value of
Let mmyc = 1 cm = 10 section if the maximum permissible bending stress is 8 N/mm 2 and central deflection is not to
B=lx 104 N/mm2
mmb = Width of beam is exceed 10 mm.
d = Depth of beam in mm
Take the value of E = 1.2 x 104 N/mm2 .
r bd 3
Sol. Given : L = 5m = 5000 mm
Length,

Using equation (12.14), we get Bending stress, f= 8 N/mm2
Central deflection,
y° ~ 5 W.L5 Value of yc = 10 mm
Let
‘ El £ = 1.2 x 104 N/mm2

384 W = Total load

5 45000 x 5000 3 and d = Depth of beam

n —i111 —, Av The maximum bending moment for a simply supported beam carrying a uniformly dis-

384 lxl0 4 xfM!l tributed load is given by,

t 12 J M„„ = w.L2 = W.L W,(v = w.L,), - ...
88 ...(t)
5 45000 x 5000 J x 12 --
Now using the bending equation,
0r bd , = 384 X IxlO 4 xlO ‘
M f_
mm= 878.906 x 107 4 ...(i)
i~y
The maximum bending moment for a simply supported beam carrying a uniformly dis-

tributed load is given by, f*I 8x/

_ w.L2 __ W.L W(v = w.L = Total load) i
“ y (d/2)
88
M
45000x5 „Nm “ 45000 x 5 1000 Nmm d
Equating the two values of B.M., we get
~ 8 x

= 28125000 Nmm W.L 16/

Now using the bending equation as 8d

M f_ VV — 16 x 81 = 1287

i ~ Lxd Lxd
y

28125000 7 Now using equation (12.14), we get
=
Here y 5 Wl?
y° = 384 X El
(£] (f) mm Wyc = 10
5 1287 L3 and -
28125000 x 12 14
bd 3 ~ d 10 x x

384 L x d El

mm28125000 x 12 3 ~_ _5_ X 128 x 1}
= 24107142.85
384 d x E
Dividing equation (i) by equation (it), we get
5 128 xL2 5 128 x 5000 2
838.906 x 10 7
a= = 364.58 mm. Ans. £d = X “ 384 X 10 x 12 x 10 4

384 10 x

24107142.85 mm= 347.2 = 34.72 cm. Ans.

Substituting this value of ‘d’ in equation (it), we get

b x (364.58)2 = 24107142.85 12.7. MACAULAY’S METHOD

24107142.85 = 181.36 mm. Ans. The procedure of Ending slope and deflection for a simply supported beam with an
eccentric point load as mentioned in Art. 12.5, is a very laborious. There is a convenient method
364.58 2 for determining the deflections of the beam subjected to point loads.

x x

STRENGTH OF MATERIALS DEFLECTION OF BEAMS 533

This method was devised by Mr. M.H. Macaulay and is known as Macaulay’s method. where C, is a constant of integration. This constant of integration should be written after the
This method mainly consists in the special manner in which the bending moment at any sec- first term. Also the brackets are to be integrated as a whole. Hence the integration of (x - a) will

tion is expressed and in the manner in which the integrations are carried out. (x-a 2 and, not x2 ax.

12.7.1. Deflection of a Simply Supported Beam with an Eccentric Point Load. A be —)

Wsimply supported beam AB of length L and carrying a point load at a distance a from left 22

support and at a distance ‘b’ from right support is shown in Fig. 12.7 . The reactions at A and B Integrating equation (iv) once again, we get

x3 W(x-a) 3J
are given by, — —W.b ^r-TW.b + C„iX+c„ \; ...

Ra = Ely= 2 ~ Xv)

and, R„b = W.a i

where C2 is another constant of integration. This constant is written after C x. The _
t
integration

of ( - af will be • This tyPe of integration is justified as the constant of integrations

C Cand are valid for all values of x.
t2

The values of C, and C are obtained from boundary conditions. The two boundary con-
2

ditions are :

R = w.a I (i) Atx = 0, y = 0 and (ii) At* = L,y = 0
d

(t) At A, x = 0 andy = 0. Substituting these values in equation ( 0 ) upto dotted line only,

Fig. 12.7 we get

The bending moment at any section between A and C at a distance x from A is given by, 0 = 0 + 0 + Co

M =R,xx-—rr~xx (ii) At B, x - L and y = 0. Substituting these values in equation (v), we get
x Lt

The above equation of B.M. holds good for the values ofx between 0 and ‘a’. The B.M. at L L WW — aY—W.b_ 3

+0
+ x.O Is r yy (L -
. 0r, -
Aany section between C and B at a distance x from is given by, — ,-y

O-i
1
2L 3 2
M WRx = a. -
x (x - a) (: C = 0. Here complete Eq. (o) is to be taken)
2

m—= W.—b- . x - W(x - a)^ ~ W.b.L2 + Cn, xLT Wb3 L-a =(-.- b)
6 1 23
Is

The above equation of B.M. holds good for all values of x between x = a and x = b.

The B.M. for all sections of the beam can be expressed in a single equation written as 10 66

M W.b W (x - a) C= <L2 - 62)
=X
x > -liL

Stop at the dotted line for any point in section AC. But for any point in section CB, add Substituting the value of Cj in equation (iv), we get
the expression beyond the dotted line also.
—El dy = W.b x2 + \-WA(L2 -, W(x - af
The B.M. at any section is also given by equation (12.3) as dx 2
L L6 { 2

M —= El - W.b . x 2 W.b (L2 ~ 2 W (x - a) 2
2L
dx 5 6 2

QL ' )> \

Hence equating (i) and (it), we get The equation (vii) gives the slope at any point in the beam. Slope is maximum at A or B.
ATo find the slope at A, substitute * = 0 in the above equation upto dotted line as point lies in
EI^-^.x W{x - a)
AC.

Integrating the above equation, we get W.b X . Wb . - ... —fv at A = 0 A 1
dx L 2 ^^LELQ °-6L bZ>
(L2 l dx

)

W(x - af ...( iv ) _^{L= 2 - 62)

2

(L2 - b 2) (as given before)

) r

534 STRENGTH OF MATERIALS deflection of beams 535

Substituting the values of C and C in equation (l>), we get Sol. Given :
1 2
mmI = 85 x 10s 4 E = 2 x 105 N/mm2
KA. UJ--^—({LEIJyy= ^Ll. x2 ;
G6L
2 --bb2 ))\]Xx + 0 : --^ca-a 3 -...(»«) RFirst calculate the reactions RA and B.

L 6L J :® Taking moments about A, we get

The equation (viii) gives the deflection at any point in the beam. To find the deflection yc Rb x 6 = 48 x 1 + 40 x 3 = 168
under the load, substitute at = a in equation {viii) and consider the equation upto dotted line (as
168
point C lies in AC). Hence, we get Rar = kN
= —t~ 28

6

R , = Total load -fi B = (48 + 40) - 28 = 60 kN

= _^^(L2_ a2_ 62)
6L

^ —= - [(a + b)2 - a2 - b 2 L('•' = a + b)
6 Lv
]

—_ _ 6 L ^ [a 2 + b 2 + 2ab - a2 - 6 2 ]

____= Wa 2 2
W.a.b „r „ = - Fig. 12.8
..b
[22ab6]

X DBConsider the section in the last part of the beam (i.e ., in length ) at a distance x

v =_ •J’ . ...(same as before) from the left support A. The B.M. at this section is given by,

3 EIL

Note. While using Macaulay’s Method, the section x is to be taken in the last portion of the i2-j-ij\A.,X 480k - 1) - 40(k - 3)

mProblem 12.8. A beam of length 6 is simply supported at its ends and carries a point = 60k - 48(k - 1) -440(k - 3)

mload of 40 kN at a distance of 4 from the left support. Find the deflection under the load and ;:

maximum deflection. Also calculate the point at which maximum deflection takes place. Given Integrating the above equation, we get

mmM.O.I. of beam = 7.33 x 107 4 and E = 2 x 10s N/mm2 EI *lA2*1 + c :
.

Sol. Given : dx 2 1 2

I

Length, L = 6 m = 6000 mm = 30k 2 + c - 24(k - l)2 • 20(k - 3)2

Point load, W = 40 kN = 40,000 N ::

m mmDistance of point load from left support, a = 4 = 4000 Integrating the above equation again, we get

6 = i-s = 6-4 = 2m = 2000 mm Ely = 30k3 + Cn X + Cn - 24(k - 1)' - 20(k - 3
L2 3
Let yc = Deflection under the load = 10k 3 + CjX + C 3 ...(ii)
2: -• 8(k - l) 3
y = Maximum deflection : 20

-y0c-3)3

Using equation yc = W. a2 2 To find the values of and C use two boundary conditions. The boundary conditions
2,
.b -

^

40000 x 4000 2 x 2000 2 (e) at k = 0, y = 0, and (ii) at x - 6 m, y - 0.
= ~ 3 x 2 x 10s x 7.33 x 10 7 x 6000
(i) Substituting the first boundary condition i.e., at x = 0, y = 0 in equation (ii) and
= - 9.7 mm. Ans. considering the equation upto first dotted line (asK = 0 lies in the first part of the beam), we get

A mProblem 12.9. beam of length 6 is simply supported at its ends and carries two 0=0+0+C C =0
mpoint loads of 48 kN and 40 kN at a distance oflm and 3 respectively from the left support. 2 2

Find : (ii) Substituting the second boundary condition i.e., atK = 6 m, y = 0 in equation (ii) and

considering the complete equation (as x = 6 lies in the last part of the beam), we get

(i) deflection under each load, 0=10x63 + Cj:x6 + 0- 8(6 - l)3 - 20 (6 - 3)3

[ii) maximum deflection, and <5 (v C, = 0)

“

(iii) the point at which maximum deflection occurs. 0 = 2160 + 6^-8 x 53 - 90

mmGiven E -2 x 10s N/mm2 and I = 85 x 106 4 yor x 3s

.

= 2160 + 6C - 1000 - 180 = 980 + 6Cj
1

536 STRENGTH OF MATERIALS DEFLECTION OF BEAMS 537

- - 163.33 ...(i)

Now substituting the values of C and C in equation (ii), we get
l 2

Ely = 10x3 - 163.33.t - 8(* - l)3 —20 .. .(iii )

- (x - 3) 3
oj j
*

( i ) (a) Deflection under first load i.e., at point C. This is obtained by substituting x - 1 in

equation (iii) upto the first dotted line (as the point C lies in the first part of the beam). Hence,

we get

El. yc = 10 x l3 - 163.33 x 1

= 10 - 163.33 = - 153.33 kNm3

Nm= - 153.33 x 10 3 3

Nmm= - 153.33 x 103 x 109 3

Nmm= - 153.33 x 10 1Z 3

- 153.33 xlO 12 - 153.33 x 12

10

c El 2 x 10 s x 85 x 10 6

= - 9.019 mm. Ans.

(Negative sign shows that deflection is downwards).

3m( b ) Deflection under second load i.e. at point D. This is obtained by substituting i =

Din equation (iii) upto the second dotted line (as the point lies in the second part of the beam).

Hence, we get

EI.yD = 10 x 3s - 163.33 x 3 - 8(3 - l)3

= 270 - 489.99 - 64 = - 283.99 kNm3

Nmm= - 283.99 x 1012 3

yDn = - 283.99 xlO 12 = — 16.7 mm. .

2 x 10 5 x 85 x 10 e Ans.

(it) Maximum Deflection. The deflection is likely to be maximum at a section between C

and D. For maximum deflection, —dy should be zero. Hence equate the equation (i) equal to

dx

zero upto the second dotted line.

3 Ox2 + Cj - 24(x - l)2 = 0

or 3 Ox2 - 163.33 - 24(xz + 1 - 2x) = 0 ((v C,==-- 163.33)
or Gx 2 + 48x - 187.33 = 0 t

The above equation is a quadratic equation. Hence its solution is

J48 2 + 4 x 6 x 187.33
= 2.87 m.

(Neglecting - ve root)

mNow substituting x = 2.87 in equation (iii) upto the second dotted line, we get maxi-
mum deflection as

El^ymax = 10 x 2.873 - 163.33 x 2.87 - 8(2.87 - l) 3

= 236.39 - 468.75 - 52.31

Nmm= 284.67 kNm3 = - 284.67 x 10 12 3

- 284.67 x 10 12 = - 16.745 mm. Ans.

2 x 10 5 x 85 x 10 s

538 STRENGTH OF MATERIALS deflection of beams 539

Integrating again, we get Let * = 1, then R.H.S. of equation (iv)

Ely = 50 = 50 - 583.33 - 6.667 x 0 = - 533.33

20 (n - 4 20 (* - 4 Let* = 2, then R.H.S. = 50 x 4 - 583.33 - 6.667 x 1 = - 390.00
l)
4" C,X 4 Cn 5}

34 Let * = 3, then R.H.S. = 50 x 9 - 583.33 - 6.667 x 8 = - 136.69

+ C,x + c„ -ffe-D* 4 - (*- 5) 4 Let * = 4, then R.H.S. = 50 x 16 - 583.33 - 6.667 x 27 = + 36.58

o In equation (iu), when * = 3 then R.H.S. is negative but when * = 4 then R.H.S. is

where C and C2 are constants of integration. Their values are obtained from boundary condi- positive. Hence exact value of * lies between 3 and 4.
1

tions which are : Let * = 3.82, then R.H.S. = 50 x 3.82 - 583.33 - 6.667 (3.82 - l)3

(i) at x = 0,y = 0 and (ii) at * = 8 m, y = 0 = 729.63 - 583.33 - 149.51 = - 3.22

(i) Substituting x = 0 and y = 0 in equation (ii) upto first dotted line (as x = 0 lies in the Let * = 3.83, then R.H.S. = 50 x 3.83 2 - 583.33 = 6.667 (3.83 - l) 3

first part AC of the beam), we get = 733.445 - 583.33 - 151.1 = - 0.99

O^O + CjXO + Cj C = 0••• The R.H.S. is approximately zero in comparison to the three terms (i.e., 733.445, 583.33
2 and 151.1).

(ii) Substituting x = 8 and y = 0 in complete equation (ii) (as point x = 8 lies in the last .-. Value of * = 3.83. Ans.

part DB of the beam), we get

50 5 5 mHence maximum deflection will be at a distance of 3.83 from support A.
y C0 = <v C2 = 0)
* 83 + * 8 + 0 ~ (8 ~ 1)4 + (S ~ 5)4
i3 3
(c) Maximum deflection
= 8533.33 + 8C - 4001.66 + 135
: mSubstituting * = 3.83 in equation (Hi) upto second dotted line, we get the maximum

8Cj = - 4666.67 CDdeflection [the point x = 3.83 lies in the second part i.e.,

of the beam.]

„ - 4666.67 50 x 3-83 3 - 583.33 x 3.83 - -5
= - 583.33
VEI-ymax = o o (3.83 - l) 4

Substituting the value of C and C in equation (ii), we get = 936.36 - 2234.15 - 106.9 = - 1404.69 kNm3
12

- 583.33* |(*- 1)4 ; +|a*- 4)4 Nmm= - 1404.69 x 10 L2 3

(a) Deflection at the centre

mBy substituting* = 4 in equation (Hi) upto second dotted line, we get the deflection at

the centre. (The point x - 4 lies in the second part (i.e., CD) of the beam].

—Ely - 50 x 4s„ - 583.33 x 4 - -5 (4 - l)4

= 1066.66 - 2333.32 - 135 = - 1401.66 kNm 3

Nm= - 1401.66 x 1000 3

Nmm= - 1401.66 x 1000 x 109 3

Nmm= - 1401.66 x 1012 3

- 1401.66 x 10 12 - 140L66 x IQ 12
~
y~ El
2x 10 5 x 4.5 x10 s

mm= — 16.29 downward. Ans.

(b) Position of maximum deflection

The maximum deflection is likely to lie between C and D. For maximum deflection the

—dy

slope should be zero. Hence equating the slope given by equation (i) upto second dotted line
dx

to zero, we get

—0=100i-+C - o (*-l )3
21

0 = 50*2 - 583.33 - 6.667(x - l)3

The above equation is solved by trial and error method.

C) 5

STRENGTH OF MATERIALS DEFLECTION OF BEAMS

541

e„b = 10 X 9 = 1® kN A(a) Slope over the support
c
B bStlt,U ng "°' n equation u P t0
beam). dotted we
^ Apointt
Ra = Total load - RB = 10 - 15 = - 5 kN ABx =i'0a^lies in t!!he first part line, get the slope at support (the
Hence the reaction RA will be in the downward direction. Hence Fig. 12,11 will be modi-
of the
xied as shown in Fig. 12.12. Now write down an expression for the B.M. in the last section of
5 dy atA = 0 A 'l
the beam.
Elba = ~2 * 0 + 30 = 30 kNm2 _ 30 x 1000 Nm2 ( j

l dx

NmmNmm= 30 x 1000 x 106
2 = 30 x 10® 2

10 kN 30xl0 9

A B 1T 0A _ Ex I 30 x IQ 9
Ir b = 15kN
r~~~ r. "

i 2^ 10 5 x 5 x 10 s

r = 0.0003 radians. Ans.

1 Ra = 5 kN 1

(5) Slope at the support B

m(the
By substituting x: = 6 in equation (Hi) upto dotted line, we get the sloPpe at support aB
pomt x - 6 lies in the
Fig. 12.12 first part AB of the beam).

The B.M. at any section at a distance x from the support A is given by, E1.QB = - — x 62 + 30 = - 90 + 30 — at B = 0B

,2 = - 60 kNm2 = - 60 x 10® Nmm2 j
“
EV y R- -_ o „x x~ R+ r x (x - 6c\)
-
A

= - 5x ; + 15(x - 6) (v 5) -60x10® - 60 x IQ 9
2 x 10 s x 5 x 10 8
0B _ ExI

Integrating the above equation, we get = — 0.0006 radians. Ans.

dy - bx 2 . 15 (jc -6) 2 C(c) Slope at the right end i.e., at

dx 2 1 m “ 9m mequation
Integrating again, we get
be ng * = 9 equation > we Set the slope at C. In this case, complete
taiken as pomt r = lies
is to in the last part of the beam.

EI-y = - +c,* + c2 15 (x - 3 ~-|£/.0c =

6)

x 9 2 + 30 + (9-6) 2 at C = 0 c)

= --*3 +C i + Cj + ~(x - 6) 3 — — 202.5 + 30 + 67.5 ~ — 105 kNm2
1
Nmm= - 105 x 10®
2

vhere C and C are constant of integration. Their values are obtained from boundary condi-
1 2
-105x10®
tions which are : -105x10®
0 C ExI 2 x 10 s x x 10 8
(i) at x = 0, y = 0 and (ii ) at x = 6 m, y - 0.
” — 0.00105 radians. Ans.
(0 Substituting x = 0 and y = 0 in equation (ii) upto dotted line (as x = 0 lies in the first

part AB of the beam), we get (d) Maximum upward deflection between the supports

O^ + CjxO + C^ = 0•••

m(ii) Substituting x = 6 and y = 0 in equation (ii) upto dotted line (as x = 6 lies in the ^For maximum deflection between the supports, should be zero. Hence equating the

Vst part AB of the beam), we get slope given by the equation (iii) to be zero upto dotted line, we get

—0= _ g x 63 + Cjx6 + 0 (y C = 0) 0=--g *2 +30=-5*2 + 60
2

= - 5 x 36 + 6
:

C„ = + 5 x 36 30 5x 2 = 60 m‘ Vl2 = 3.464

‘6 mW SUbStitUting * = 3 464

Substituting the values of C and C in equations and (ii), we get deflectfrn a S wequafcion
1 2 (i (iv) u P to dotted line e get the maximum
-
~^|rr dy -: ^ X.92 +. o30n : +.
(x - 6)2

5 + —5 (x - 6)3 EIymax = - - x 3.464 3 + 30 x 3.464

Ely = - y x 3 + 30x : ...(iv)

6 2:

} i

STRENGTH OF MATERIALS deflection of beams 543
542

= - 34.638 + 103.92 = 69.282 kNm3 RNegative sign shows that A will be acting downwards. In order to obtain general ex-

Nmm mm= 69.282 x 1000 x 109 pression for the bending moment at a distance x from the left end A, which will apply for all
3 = 69.282 x 10 12 3
values of x, it is necessary to extend the uniformly distributed load upto point C, compensating
12
with an equal upward load of 4 kN/m over the span BC as shown in Fig. 12.14. Now Macaulay’s
69.282 x 10
method can be applied.
" 6 x 5 x 10 8

2 x 10

mm= 0.6928 (upward). Ans.

C(e) Deflection at the right end i.e., at point 4 kN/m

mBy substituting * = 9 in equation ( iv ), we get the deflection at point C. Here complete , B V V.VSVt„,
mequation is to be taken as point x = 9 lies in the last part of the beam. 1AAAAy'WV. '

4 kN/m^
m* 3 -

EIyc = - | x 9 3 + 30 x 9 + | (9 - 6}3

= - 607.5 + 270 + 67.5 Fig. 12.14

Nmm= - 270 kNm3 = - 270 x 10 12 3 The B.M. at any section at a distance x from the support A is given by,

- 270 x 10 12 R. x x - 4(x - 3) + RJx - 6) + 40c - 6)
yc = 2 x 10® x 5 x 10 s

mm= - 2.7 (downwards). Ans. = - 3x i - 2(x — 3)2 + 27(x - 6) | + 2(x - 6)2
:

mProblem 12.12. A beam ABC of length 9 has one support of the left end and the other Integrating the above equation, we get
msupport at a distance of 6 from the left end. The beam carries a point load of 1 kN at right end
mand also carries a uniformly distributed load of 4 kN/m over a length of 3 as shown in —3x 2 + 277((xx--66)) z + 2((xr --66)r3
~„ rdy— „: ” 2(x-—3) 3 :
+ Ci *
j&i .. ,(i)
2 1;
Fig. 12.13. Determine the slope and deflection at point C. dx 3 23

mmTake E = 2 x 105 N/mm2 and I = 5 x 10s 4 Integrating again, we get

. _3^ _2(^ 2273(^23£E/l.yy = -
23
Q v8 + Cc.x + Cc„ 27 (x - 3 »2 Or - 4
12 6)
Sol. Given : +; 6) +:
Point load,
W = 12 kN j 34 : 34
:

w - 4 kN/m C ^ ® Cor
U.d.l., El.y = - + C.jXx + C„ • - + ^-0(rx--66) 3 —+ (x - 6)4 ..(ii)
*
22 : Ua ; 6b

Value of E - 2 x 10s N/mm2 6
Value of
mm/ = 5 x 108 4 where C and C are constant of integration. Their values are obtained from boundary condi-
t 2

R RFirst calculate the reactions A and B . tions which are :

(i) at x - 0, y = 0 and (ii) at x = 6 m, y = 0.

(i) Substituting the x = 0 and y = 0 in equation (ii) upto first dotted line (as x = 0 lies in

ADthe first part of the beam), we get

0—0+C x0+C C = 0
2 2

(ii) Substituting x = 6 and y = 0 in equation (ii) upto second dotted line (as x = 6 lies in

the second part DB of the beam), we get

0-^0=
Fig. 12.13 -«! + Cl1 r6 +

Taking moments about A, we get 2 6

Ds x6 = 4x3x^3 + |j + 12x9 = - 108 + 6Cj - 13.5 = - 121.5 + 6C,

= 54 + 108 = 162 —or „ = 12r1—5 = 20.25

~ =27 kN(T) C,
6
b
= Total load - RB = 24 - 27 = - 3 kN (1
Substituting the values of C and C in equations (i) and (ii), we get
L 2

El dy = - -3 x2 2 (x - 3 —27 —2
3)
-dxf 2 + 20.25 + (x- 0)2 + -(x 6)3 ..(iii)
\

3

—Ely = - + 20.2 x x ; - ~(x - 3)4 3 +^(x-6)4
+f(*-6) i
\

STRENGTH OF MATERIALS deflection of beams 545

() Slope at the point C Negative sign shows that RA is acting downwards as shown in Fig. 12.16.

mBy substituting x = 9 in equation (Hi), we get the slope at C. Here complete equation
9mis to be taken as point j =
lies in the last part of the beam.

x9a + 20.25- | (9 - 3) 3 + - 2 (9 - 6)3
-| yEl. 0C = 6)
|(9 +

|(, C-« c Fig. 12.16

)

= - 121.5 + 20.25 - 144 + 121.5 + 18 = - 105.75 kNm2 The B.M. at any section at a distance x from A, is given by

Nmm Nmm= - 105.75 x 103 x 106 2
2 = - 105.75 x 10 9
,2

105.75 x 10 9 = _ q_0o|0575 radians. Ans. El = - 50* ^ + 300
dx 2
c 2 x 10 5 x 5 x 10® '

" = - 50tc + 300{x - 4)°

:

() Deflection at the point C Integrating the above equation, we get
By substituting r = 9min complete equation (if), we get the deflection at C.
dy ,-50xf
+ 3000c - 4)
+ 20.25 x 9 - | (9 - 3)4 + | (9 - 6)3 + | (9 - 6)4
yEl x yc = - Integrating again, we get

= _ 364.5 + 182.25 - 216 + 121.5 + 13.5 — —Ely = - 300 (x - 4)

Nmm= - 263.25 kNm3 = - 263.25 x 10 12 3 x + C.x + C,2
1
23

263.25 x 10' 2 _ _ 2.6325 nun. Ans. -y= xs + C a + C + 150{x - 4)2
c 2 x 105 x 5 x 10 s
1 2;

A mProblem 12.13. A horizontal beam AB is simply supported at and B, 6 apart. The where Cj and C2 are constants of integration. Their values are obtained from boundary condi-
mkNmbeam is subjected to a clockwise couple of 300
at a distance of 4 from the left end as tions which are :

mmshown in Fig. 12.15. IfE = 2 x 10s N/mm2 and I = 2 x 10s 4 determine : (£) atx = 0, y = 0 and m(ii) at * = 6 andy = 0.

,

(i) deflection at the point where couple is acting and (i) Substituting x = 0 and y = 0 in equation (ii) upto dotted line, we get

C C0 = 0 +
(ii) the maximum deflection. Cx 0 + .-. 2=0
l2

m(ii) Substituting x = 6 and y = 0 in complete equation (ii), we get

——0 = 25 x63 + C x6 + 0 + 150(6 - 4)2
1

_= - 1800 + 6C + 600
t

„ 1800 - 600 =200
g

Fig. 12.15

Substituting the values of C and C in equation (ii), we get
x2

Sol. Given : L=6m —Ely =- 25 x3„ + 200* + 150(x - 4)2 (v C„ = 0> ...(iii)
Length,
Couple - 300 kNm C(i) Deflection at

E = 2 x 105 N/mm2 y c(i.e., )

Value of mm1 = 2 x 108 4 By substituting x = 4 in equation (iii) upto dotted line, we get the deflection at C.
Value of

First calculate the reactions RA and RB . EIy, = - 43 + 200 x 4

:

Taking moments about A, we get = - 533.33 + 800 = + 266.67 kNm3

Rb x 6 = 300 Nmm= 266.67 x 10 12 3

KB = ^.=50kN<T) mm266.67 x 10 12 upwards. Ans.
and Ra = Total load - RB = 0 - 50 kN ( v There is no load on beam) = 6.66
2 x 10 5 x 2 x 10 8
= -50 kN

y
.

546 STRENGTH OF MATERIALS DEFLECTION OF BEAMS 547

(;ii ) Maximum deflection between the tangents at P and Q, . Hence the angle between the lines CP and DQ, will be
First find the point where maximum deflection takes place. The maximum deflection is 1 1

equal to dQ.

—likely to occur in the larger segment AC of the beam. For maximum deflection dy P^For the deflected part of the beam, we have

should be

zero. Hence equating the slope given by equation <i) upto dotted line to zero, we get PjQ = R.d0
t

But PjQj ~ dx

~Y*2 + 200 = 0 (V C = 200) dx = R.dQ
l
or - 25x2 + 200 = 0

1200 O2 nr MEBut for a loaded beam, we have

V2
mor -
x= x

V 25 _ El

Now substituting x = 2 x J2 in equation (Hi) upto dotted line, we get the maximum T= or
r

deflection. Substituting the values of R in equation (i), we get

EI*ymax ^- 2 x V2)3 + 200(2 x V2) Mdx dx

= - 188.56 + 565.68

Nmm= 377.12 kNm3 = 377.12 x 10 12 3 Since the slope at point A is assumed zero, hence total slope atB is obtained by integrat-

mm377.12 x 10 12 = 9.428 ing the above equation between the limits 0 and L,

2 x 10 5 x 2 x 10 a upwards. Ans.

12.8. MOMENT AREA METHOD Jo El El Jo

Fig. 12.17 shows a beam AS carry- (L

ing some type of loading, and hence sub- But M.dx represents the area of B. M. diagram of length dx. Hence I M.dx represents
jected to bending moment as shown in
Fig. 12.17 (a). Let the beam bent into the area of B. M. diagram between A and B.
AQjPj-B as shown in Fig. 12.17 (6).
But 9 = [Area of B. M. diagram between A and B]
Due to the load acting on the beam. Slope at B, jgy

Let A be a point of zero slope and zero E H— 0 = slope at B = 6
fi
deflection. J,_ B.M. Diagram ...(12.15)
——eB =_ Area of B. M. diagram between A and B
dy
Consider an element PQ of small AIf the slope at is not zero then, we have

length dr at a distance x from B. The A“Total change of slope between B and is equal to the erea ofB. M. diagram between B
Aand divided by the flexural rigidity El"
tcorresponding points on the deflected

beam are PjQj as shown in Fig. 12.17(6). ‘ de \\ or efi“0A= —Area of B. M. between A and B
\\
| ...(12.16)

Let R = Radius of curvature of de- y

P Qfleeted part Yl
1l
I Now the deflection, due to bending of the portion PjQ, is given by

d0 = Angle subtended by the -St-

are Pl Q1 at the centre O "\\ dy = x.dQ

M = Bending moment between \\ Substituting the value of de from equation (ii), we get

P and Q VV—-oe M.dx

PfC = Tangent at point P d,y=x '~m~ ••(»»)
1
ASince deflection at is assumed to be zero, hence the total deflection at B is obtained by
DQX = Tangent at point Q,
integrating the above equation between the limits zero and L.
The tangent at P and Q are cut-
xx \ — —'L xM.dx = 1 r L xM, ., d,x
o -
ting the vertical line through B at points El Jo
Fig. 12.17 Jo El;
C and D. The angle between the normals

at P, and Qj will be equal to the angle But x x M.dx represents the moment of area of the B.M. diagram of length dv about

point B.

t

STRENGTH OF MATERIALS deflection of beams 549

Hence f xM.dx represents the moment of area of the B.M. diagram between B and A 12.10. SLOPE AND DEFLECTION OF A SIMPLY SUPPORTED BEAM CARRYING A
POINT LOAD AT THE CENTRE BY MOHR’S THEOREM
Jo
WFig. 12.19 (a) shows a simply supported AB of length L and carrying a point load at
about B. This is equal to the total area of B.M. diagram between B and A multiplied by the
the centre of the beam i.e., at point C. The B.M. diagram is shown in Fig. 12.19 (6). This is a
distance of the C.G. of the B.M. diagram area from B.
case of symmetrical loading, hence slope is zero at the centre i.e., at point C.
yv — 1 x A x _ = Ax ...(12.17)
x El But the deflection is maximum at the centre.
El

where A = Area of B.M. diagram between A and B

Ax = Distance of G.G. of the area from B.

12.9. MOHR’S THEOREMS

The results given by equation (12.15) for slope and (12.17) for deflection are known as

Mohr’s theorems. They are state as :

I. The change of slope between any two points is equal to the net area of the BM.
diagram between these points divided by EL

II. The total deflection between any two points is equal to the moment of the area of

B.M. diagram between the two points about the last point (i.e., B) divided by El.

The Mohr’s theorems is conveniently used for following cases :

1. Problems on cantilevers (zero slope at fixed end).
2. Simply supported beams carrying symmetrical loading (zero slope at the centre).

3. Beams fixed at both ends (zero slope at each end).

The B.M. diagram is a parabola for uni- D —— Fig. 12.19
formly distributed loads. The following prop- i
AArea

erties of area and centroids or parabola are ->• q —Now using Mohr’s theorem for slope, we get
2 A, = Area of B.M. dia=g-r=a=m between Aand C
given as :

Let BC = d XWVW Slope at

CjL

But area of B.M. diagram between A and C

In Pig. 12.18, ABC is a parabola and = Area of triangle ACD

ABCD is a surrounding rectangle. A\Y\\V^^ 1L WL WL2

Let A, = Area of ABC _1 “2*2* _

from AD 4 16

x x = Distance of C.G. ofA
x

A = Area of ACD A.-. Slope at or 6 = -----
2 l,
CjI
x , = Distance of C.G. ofA2 from AD Fig. 12.18
Now using Mohr’s theorem for deflection, we get from equation (12.17) as
Gj = C.G. of area A
t Ax
y~
G, = C.G. of area A .
2 ~EI

A = Area of parabola ABC where A = Area of B.M. Diagram between A and C
t

WL2

_

A = Area ACD = Area ABCD - Area ABC 16
2
x - Distance of C.G. of area A from A

= b x d — — bd -—bd _~ 2 X L_L

33 ~
32 3

fi= ! 6 WL2 L
X
= 3 WL*

y‘
El 48£7

.. :

550 STRENGTH OF MATERIALS deflection of beams 551

12.11. SLOPE AND DEFLECTION OF A SIMPLY SUPPORTED BEAM CARRYING A HIGHLIGHTS
UNIFORMLY DISTRIBUTED LOAD BY MOHR’S THEOREM
1. The relation between curvature, slope, deflection etc. at a section is given by :
Fig. 12.20 (a) shows a simply supported beam AB of length L and carrying a uniformly Deflection = y

distributed load of tu/unit length over the entire span. The B.M. diagram is shown in Fig. 12.20 Slope =
b( ). This is a case of symmetrical loading, hence slope is zero at the centre i.e., at point C. ax

and ^B.M. = El
dx*

S.F. = El ^d-34v

As deflection is very small, hence slope is also given by = tan 0 = 0.

2'. Slope at the supports of a simply supported beam carrying a point load at the centre is given by

WL6, 2
.„
A “ B ~ 16El

where W = Point load at the centre, L = Length of beam

E = Young’s modulus, I = M.O.I.

3. The deflection at the centre of a simply supported beam carrying a point Load at the centre is

-™.WL3

givenby yc =

4. The slope and deflection of a simply supported beam, carrying a uniformly distributed load of

w/unit length over the entire span, are given by,

—„

A
= _ = WL2 and, y„c = 5 WL2
AnB 384
24 El •

El

5. Macaulay’s method is used in finding slopes and deflections at any point of a beam. In this
method :

(i) Brackets are to be integrated as a whole.

(ii) Constants of integrations are written after the first term.

(fit) The section, for which B.M. equation is to be written, should be taken in the last part of the

—6. For maximum deflection, the slope is zero.

dx

7. The slope at point B if slope oi A is zero by moment-area method is given by,

——o Bn = Area of B. M. diagram betw—e—en A and B
El
8. The deflection by moment area method is given by

where A ~ Area of B.M. diagram between A and B
x = Distance of C.G. of area from B

STRENGTH OF MATERIALS DEFLECTION OF BEAMS 553

EXERCISE 12 mA beam of length 10 is simply supported at its ends and carries two point loads of 100 kN and
m m60 kN at a distance of 2 and 5 respectively from the left support. Calculate the deflections

under each load. Find also the maximum deflection.

(A) Theoretical Questions mmTake / = 18 x 10 8 4 and E = 2 x 10 5 N/mm2.

1. Derive an expression for the slope and deflection of a beam subjected to uniform bending mo- mm mm[Ans. (i) - 4.35 (lit) ymw. = - 6.78 mm]
(ti>- 6.76
ment.
mA beam of length 20 is simply supported at its ends and carries two point loads of 4 kN and

M —2. Prove that the relation that = EI m m10 kN at a distance of 8 and 12 from left end respectively. Calculate : (i) deflection under
dx L
each load (it) maximum deflection.

mmTake E = 2 x 106 N/mm 2 and I = lx 109 4

.

Mwhere = Bending moment, E = Young’s modulus, l = M.O.I. mm[Ans. (i) 10.3 and 10.6 downwards, (ii) 11 mm]

3. Find an expression for the slope at the supports of a simply supported beam, carrying a point A mbeam of length 6 is simply supported at its ends. It carries a uniformly distributed load of

load at the centre. 10 kN/m as shown in Fig. 12.21. Determine the deflection of the beam at its mid-point and also
the position and the maximum deflection.
4. Prove that the deflection at the centre of a simply supported beam, carrying a point load at the

WL3 Take El = 4.5 x 10 s N/mm2 . [Ans. - 2.578 mm, x = 2.9 m, ymax = — 2.582 mm]

• 10.

centre, is bgiven byy y/cr = 48E7

Wwhere = Point load, L = Length of beam.

5. Find an expression for the slope and deflection of a simply supported beam, carrying a point load

W at a distance ‘a’ from left support and at a distance ‘b’ from right support where a > b.

6. Prove that the slope and deflection of a simply supported beam of length L and carrying a uni-

formly distributed load of w per unit length over the entire length are given by

Fig. 12.21

Slope at supports = , and Deflection at centre = A beam ABC of length 12 metre has one support at the left end and other support at a distance
mof 8 from the left end. The beam carries a point load of 12 kN at the right end as shown in
24 El 384 El

Wwhere = Total load = w x L. Fig. 12.22. Find the slopes over each support and at the right end. Find also the deflection at the

What is a Macaulay’s method ? Where is it used ? Find an expression for deflection at any section right end.
of a simply supported beam with an eccentric point load, using Macaulay’s method.
mm11. Take E - 2 x 10 s N/mm2 and / = 5 x 10 8 4

.

What is moment-area method ? Where is it conveniently used ? Find the slope and deflection of [Ans. 0A - 6.00364, 0S = - 0.00128, 6C = - 0.00224, yc = - 7.68 mm]
a simply supported beam carrying a (i) point load at the centre and (ii) uniformly distributed

load over the entire length using moment-area method.

(B) Numerical Problems

A m1. wooden beam 4 long, simply supported at its ends, is carrying a point load of 7.25 kN at
mm mmits centre. The cross-section of the beam is 140
wide and 240 Edeep. If for the beam =

6 x 103 N/mm 2 find the deflection at the centre. [Ans. 10 mm]
,
12.
A m W2. beam 5 long, simply supported at its ends, carries a point load at its centre. If the slope at Fig. 12.22

the ends of the beam is not to exceed 1°, find the deflection at the centre of the beam. An overhanging beam ABC is loaded as shown in Fig. 12.23. Determine the deflection of the

[Ans. 29.08 mm] beam at point C.

3. Determine : (i) slope at the left support, (ft) deflection under the load and (iff) maximum deflec- mmTake E = 2 x 10 5 N/mm 2 and / = 5 x 10 s 4 [Ans. yc = - 4.16 mm]
tion of a simply supported beam of length 10 m, which is carrying a point load of 10 kN at a
.

mdistance 6 from the left end.

mmTake E = 2 x 10 s N/mm 2 and I - 1 x 10 s 4 mm[Ans. 0.00028 rad., 0.96 and 0.985 mm]

.

mm mm4. A beam of uniform rectangular section 100 deep is simply supported at its
wide and 240

ends. It carries a uniformly distributed load of 9.125 kN/m run over the entire span of 4 m. Find

Ethe deflection at the centre if = 1.1 x 104 N/mm2 . [Ans. 6.01 mm]

A m5. beam of length 4.8 and of uniform rectangular section is simply supported at its ends. It

carries a uniformly distributed load of 9.375 kN/m run over the entire length. Calculate the Fig. 12.23
width and depth of the beam if permissible bending stress is 7 N/mm2 and maximum deflection
mA beam of span 8 MN-mand of uniform flexural rigidity El - 40 2 is simply supported at its
is not to exceed 0.95 cm. ,

Take E beam ~ x 104 N/mm2 mm[Ans. b = 240 and d = 336.S mm] ends. It carries a uniformly distributed load of 15 kN/m run over the entire span. It is also
.
for material 1.05 mkNmsubjected to a clockwise moment of 160
at a distance of 3 from the left support. Calculate

6. Solve prohlem 3, using Macaulay’s method. the slope of the beam at the point of application of the moment. [Ans. 0.0061 rad.]

13

Deflection of Cantilevers

13.1. INTRODUCTION

.h,„ <£Tu2x r “ ^b ';? In ™c ut

^finding deflections and slope of the simply supported blams been f° r

ssssr ™ ™*“ A POmr WAnAT ™.

p.i„« fe l ih * “po, B md “-** *

», ^load i, applied whereas «r b‘ f0r 'i

show8

Sid‘' * Fig. 13.1 «nd A. The B.M. at thie section is

given by“ * at * di“““ > f”» «»

^“byBut B.M. a. any section ta’.to equation 02.3) as

M = EI^L

y, dx 2
Equating the two values of B.M., we get

EI -jjr=-W(L-x) = -WL+ Wjc

Integrating the above equation, we get

^fEI fx =~WLx + + Cl

554

DEFLECTION OF CANTILEVERS

mProblem 13.1. A cantilever of length 3 is carrying a point load of 25 kN at the free
mm N/mmend. If the moment of inertia of the beam = 10s
4 and value ofE = 2.1 x 10s 2 find

,

(i) slope of the cantilever at the free end and (ii) deflection at the free end.

Sol. Given : l = 3m = 3000 mm
Length,
Point load, W = 25 kN = 25000 N

mm/ = 108 4

Value ofE = 2.1 x 105 N/mm2

(i) Slope at the free end is given by equation (13.1 A).

25000 x 3000 2
= 0.005357 rad. Ans.

2 x 2.1 x 10 5 x 10 8

(ii) Deflection at the free end is given by equation (13.2 A),

yJ„B = WL3 = 25000 x 3000 3 = 10.71 mm.
3x2.1x
3 El r 10sa Ans.
10 5
x

mProblem A13.2. cantilever of length 3 is carrying a point load of 50 kN at a distance

m mm N/mmof 2 from the fixed end. Ifl = 10s
4 and E =2 x 10s 2 find (i) slope at the free end

,

and (ii) deflection at the free end.

Sol. Given : L = 3 m = 3000 mm
Length, W = 50 kN = 50000 N
Point load,

Distance between the load and the fixed end,

a = 2 m = 2000 mm

mm/ = 108 4

Value of E = 2 x 105 N/mm2

(i) Slope at the free end is given by equation (13.3) as

50000 x 2000 Ans.
= 0.005 rad.
B
2EI\ 2 x 2 x 10 5 x 10 s

(ii) Deflection at the free end is given by equation (13.4) as

yB ~ Wa 3 Wa 2 (L ~ a)

3 El +

2 El

50000 x 2000 3 50000 x 2000 2
_ (3000 - 2000)

3 x 2 x 10® x 10 s 2 x 2 x 10 s x 10 8

= 6.67 + 5.0 = 11.67 mm. Ans.

13.4. DEFLECTION OF A CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD

A cantilever AB of length L fixed at the point A and free at the point B and carrying a
uniformly distributed load of w per unit length over the whole length, is shown in Fig. 13.3.

Consider a section X, at a distance x from the fixed end A. The B.M. at this section is

given by,

Mx = -w (L-x ) . X> (Minus sign due to hogging)

' 559

deflection of cantilevers

C CSubstituting the values of and in equation (i) and (ii), we get
l2

~and
EIy=-^-(L-x) + ...(iv)

24 6 24

The equation (Hi) is known as slope equation and equation (iv) as deflection equation.

From these equations the slope and deflection can be obtained at any sections. To find the

slope and deflection at point B, the value of x = L is substituted in these equations.

Let B0e = Slope at the free end i.e., at B

| j

yB = Deflection at the free end B.

From equation (Hi), we get slope at B as

w wL3 s
,t t
tvL

9B = -^6E7l =-^6Erl W=(v Total load = w.L) ...(13.5)
'

From equation (iv), we get the deflection at B as
—-—xLEI.yB =--(L-L)4
+

____wL4 wL4 ___3 wL4
r+lr=== wZ/4 =

yB = ~ wL4 = " WL3 ^=

8EI ~8EI

.-. Downward deflection at B,

wL4 WL3

yB ~ 8El “ 8El ...(13.6)

A mProblem 13.3. cantilever of length 2.5 carries a uniformly distributed load of

16.4 kN per metre length over the entire length. If the moment of inertia of the beam = 7.95
mm N mmx 107
4 and value ofE = 2 x 10s / 2 determine the deflection at the free end.

,

Sol. Given : L - 2.5 mm = 2500 mm
Length,

U.d.l., w = 16.4 kN/m

.-. Total load, W=u>xL = 16.4 x 2.5 = 41 kN = 41000 N
Value of
mml- 7.95 x 107 4

Value of E = 2 x 10s N/mm2

Let y g = Deflection at the free end.
Using equation (13.6), we get

41000 x 2500 3

8 LI 8x2 x 105 x7.95x 10
= 5.036 mm. Ans.

mAProblem 13.4. cantilever of length 3 carries a uniformly distributed load over the

entire length. If the deflection at the free end is 40 mm, find the slope at the free end.

I

STRENGTH OF MATERIALS 561
560

Sol. Given : 5 x 8 x 2 x 10 s x 8 x 10 7 16384 N/m
2.5 x 2500 3
Length, L = 3 m = 3000 m
16.384 kN/m. Ans.
mmDeflection at free end, yB = 40

Let - slope at the free end 13.5. DEFLECTION OF A CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD
FOR A DISTANCE ‘a’ FROM THE FIXED END
Using equation (13.6), we get

\ WL3 A cantilever AB of length L fixed at the point A and free at the point B and carrying

yB = SEI w/ma uniformly distributed load of length for a distance ‘a' from the fixed end, is shown

WL2 x L WL2 x 3000 in Fig. 13.4.
40 = war ~
&ei The beam will bend only between A and C, but from C to B it will remain straight since
BB.M. between C and is zero. The deflected shape of the cantilever is shown by AC'S' in which
WL2 40x8

El 3000 portion C'B 1 is straight.
Slope at the free end is given by equation (13.5),

WL2 Wl}_ =“_ ~ 40 x 8 X 1 0C = Slope at C, i.e., a* *
El 3000 6
~ _ *x 6611

v From equation (i). yc = Deflection at point C, and
yB = Deflection at point B.

= 0.01777 rad. Ams. L-a

mm mm mAProblem 13.4 (A). cantilever 120 w/m Length
wide and 200 deep is 2.5 long. What is

the uniformly distributed load which the beam can carry in order to produce a deflection of
(Annamalai University, 1991)
mm5 at the free end ? Take E = 200 GN/m2 .
iT
Sol. Given :

Width, b - 120 mm J*==
-R'V
Depth, d = 200 mm
Length, m mmL = 2.5 = 2.5 x 1000 = 2500 Fig. 13.4

mmDeflection at free end, yB = 5 ACThe portion of the cantilever may be taken as similar to a cantilever in Art. 13.4.
Value of E- 200 GN/m2 = 200 x 109 N/m2
G(" = GiSa = 10 )

_ 200 x 10 9 N °c ~ 6EI L-a[In equation (13.5) put ]

m mm)j ... ! 2 = (1000 2

]

mm2 2

(1000) y =c on ‘ [In equation (13.6) put L = a]
Jill
= 2 x 105 N/mm2

mmbd3 120 x 200 3 C BSince the portion C'B r of the cantilever is straight, therefore slope at = slope at

I- ~ 4,
Moment of inertia, =- 8 x 10 7
12 r, ^ WCV
Let m Nw = uniformly distributed load per length in ...(13.7)
c s ~ 6EI ...(13.8)
W = Total load
Now from Fig. 13.4, we have
= WX L (Here L is in metre)
yB = yc + ®c (L - a ')
= w x 2.5 = 2.5 x w N
wa 4 tv. a 3
Using equation (13.6), we get
+

8EI 6El

_ ml 13.6. DEFLECTION OF A CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD
FOR A DISTANCE ‘a’ FROM THE FREE END
yB - 8EI
AA ABcantilever and free at the point B and carrying
2.5 w x 2500 3 of length L fixed at the point

~ 8 x 2 x 10 5 x 8 x 107 a uniformly distributed load of w/ra length for a distance ‘a’ from the free end is shown in

Fig. 13.5 (a).

m

562 STRENGTH OF MATERIALS deflection of cantilevers 563

The slope and deflection at the point B is determined by considering : The upward deflection of point B due to upward uniformly distributed load acting on
the portion AC = upward deflection of C + slope at C x CB
(i) the whole cantilever AB loaded with a uniformly distributed load ofw per unit length
w(L-a)4 w.(L-a)3 (v CB = a)
as shown in Fig. 13.5 (6). =+
8El 6Yl X“
(ii) a part of cantilever from A to C of length (L — a) loaded with an upward uniformly
distributed load of w per unit length as shown in Fig. 13.5 (c). Net downward deflection of the free end B is given by

w(L - a4 w{L - ay
)

8 EI~ ...(13.10)

Problem 13.5. Determine the slope and deflection of the free end of a cantilever oflength

m m m3 which is carrying a uniformly distributed load of 10 kN! over a length of 2 from the

fixed end.

mmTake 1 - 10s 4 and E = 2 x 10s N/mm2
.

Sol. Given : L = 3m = 3000 m
Length,

U.d.l., w — 10 kN/m = 10000 N/m = 10000 N/mm = 10 N/mm

mLength of u.d.l. from fixed end, a = 2 = 2000 mm.

Value of mmI - 10s 4

Value of E = 2 x 10® N/mm2

Let 0B = Slope of the free end and
yB = Deflection at the free end.

Ci ) Using equation (13.7), we have

Fig. 13.5 wa 3 10 x 2000 3 0.00066. Ans.
6b= 6 6 x 2 x 10“ x 108
Then slope at B = Slope due to downward uniform load over the whole length
- slope due to upward uniform load from A to C (ii) Using equation (13.8), we get

and Bdeflection at - Deflection due to downward uniform load over the whole length
- deflection due to upward uniform load from A to C.
yB + (i Q)
(a) Now slope at B due to downward uniformly distributed load over the whole length
8EI 6El

10 x 2000 4 10 x 2000 3
5- +

8 x 2 x 10° x 10 8
5——= ;v x (3000 - 2000)
5
6 x 2 x x 10®
10

= 1 + 0.67 = 1.67 mm. Ans.

(6) Slope at B or at C due to upward uniformly distributed load over the length (L — a) mProblem 13.6. A cantilever of length 3 carries a uniformly distributed load of

w(L - 3 m mm N/mm10 kN/m over a length of 2 from the free end. If I = 10s 4 and E = 2 x 10s 2 ; find :

= a) (i) slope at the free end, and (ii) deflection at the free end.

6 El

Hence net slope at B is given by, Sol. Given : L = 3 m = 3000 mm
Length,
—e„n = 6EI
...(13.9) w = 10 kN/m = 10000 N/m = 10000 N/mm = 10 N/mm

6El U.d.l.,

The downward deflection of point B due to downward distributed load over the whole m mmLength of u.d.l. from free end, a = 2 = 2000

length AB

Value of mm/ = 10s 4

Value of E - 2 x 105 N/mm2

STRENGTH OF MATERIALS deflection of cantilevers 565

at B and Wa3 Wa? (' L-a)'
+|
Let Qb = Slope at the free end i.e., 3,2 - 3 El
2 El

y B = Deflection at the free end. _ 4000 x 2000 3 4000 x 2000 2 {(3000 _- 20Q00)
(i) Using equation (13.9), we get 3 x 2 x 10 s x 10® + __
10 5 10j®
2x 2 x x

wl? w(L - 3 mm= 0.54 + 0.40 = 0.94
a)

0a = 6EI 6 El .'. Total deflection at the free end

10 x 3000 3 10(3000 - 2000) 3 = y t + y2
_ 6 x 2 x 10 s x 10 8
= 0.9 + 0.94 = 1.84 mm. Ans.
6 x 2 x 10 s x 10s
A mProblem 13.8.
= 0.00225 - 0.000083 = 0.002167 rad. Ans. cantilever of length 2 carries a uniformly distributed load of
from the fixed end and a point load of 1 kN at the free
m2.5 kN/m run end if the section is rectangular 12 cm wide and 24 cm

end Find the
(ii) Using equation (13.10), we get for a length of 1.25
deflection at the free

wL* w(L - a)4 w(L - dr deep and E = 1 x 104 N/mm*. (Annamalai University, 1990)
8 El
y* ~ 8EI Sol. Given : L = 2 m = 2000 mm
Length,
10 x 3000* 10(3000 - 2000)? + 10(3000 - 2000) 3 v* w = 2.5 kN/m = 2.5 x 1000 N/m
6 x 2 x iO 5 x i° 8 U.d.l.,
~8x2xl0s xl08 L 8 x2xl05 xl0S
2.5 x 1000 N/mm = 2.5 N/mm
= 5.0625 - [0.0625 + 0.1667] = 4.8333 mm. Ans.

A mProblem 13.7. cantilever of length 3 carries two point loads of2 kN at the free end WPoint load at free end, = 1 kN = 1000 N
mand 4 kN at a distance of 1 from the free end. Find the deflection at the free end.
Distance AC, mmma = 1.25 = 1250

Take E = 2 x 10s N/mm2 and I = 10s mm4 Width, b = 12 mm
.

Sol. Given : Depth, d = 24 mm
Length,
L = 3 m = 3000 mm bd* 12 x 24 3

Load at free end, Wj = 2 kN = 2000 N Value of 1=

IT" 12 .
mm mm= 13824 cm4 = 13824 x 104
mLoad at a distance one from free end, . 4

4 = 1.3824 x 104

Distance AC, W = 4 kN = 4000 N Value of E = 1 x 10s N/mm2
2 Let
a = 2m- 2000 mm y = Deflection at the free end due to point load 1 kN alone

Value of E = 2 x 106 N/mm2 y2 - Deflection at the free end due to u.d.l. on length AC.

Value of mmI = 108 4

Let y L = Deflection at the free end due to load 2 kN alone
y2 = Deflection at the free end due to load 4 kN alone.

Fig. 13.7

(j) Now the downward deflection at the free end due to point load of 1 kN (or 1000 N) at

Fig. 13.6 — —the free end is given by equation (13.2 A) as = 1929mm
loooxj^ooo 3
Downward deflection due to load 2 kN alone at the free end is given by equation (13.2 A) " 3EI " 3 x 10 4 x 13824 x 10®

v, = WL? = 2000 x 3000 3 = 0.9 mm. (ii) The downward deflection at the free end due to uniformly distributed load of

3 El 3x2x ; x r m2.5 N/mm on a length of 1.25 (or 1250 mm) is given by equation (13.8) as
a 10®

10 wa 4 w.a 3

Downward deflection at the free end due to load 4 kN (i.e,, 4000 N) alone at a distance

m2 from fixed end is given by (13.4) as

^

566 STRENGTH OF MATERIALS DEFLECTION OF CANTILEVERS

2.5 x 1250 4 2.5 x 1250 3 (ii) Deflection at the free end

= -. s- + 1034 10®g (2000 - 1250) NLet y x = Deflection at the free end due to point load of 1000

8 x 10 4 x 13824 x 10 s 6 x x 13824 x y2 = Deflection at the free end due to u.d.l. on length BC.

= 0.5519 + 0.4415 = 0.9934 NThe deflection at the free end due to point load of 1000 is given by equation (13.2 A) as

/. Total deflection at the free end due to point load and u.d.l.

= y1 + y2 = 1.929 + 0.9934 = 2.9224 mm. Ans. WL3

A mProblem 13.9. cantilever of length 2 carries a uniformly distributed load 2 kN/m yi~ 3El (v Here y x = yB)

mover a length of 1 from the free end, and a point load of 1 kN at the free end. Find the slope 1000 x 2000 3

mmand deflection at the free end ifE = 2.1 x 10s N/mm2 and I = 6.667 x 107 4 = = =- = 0.1904 mm.

.

Sol. Given : (See Fig. 13.8) 3 x 2.1x 10 5 x 6.667 xlO 7

Length, L = 2 m = 2000 mm mThe deflection at the free end due to u.d.l. of 2 N/mm over a length of 1 from the free

w = 2 kN/m = —— — N/mm = 2 N/mm end is given by equation (13.10) as

U.d.l. wL4 4 w(L - a)3
a)
a = 1 m = 1000 mm y2 ~ 8 El w{L -
W = 1 kN = 1000 N
Length BC, 8EI + 6El * “
Point load,
Value of £ = 2.1 x 105 N/mm2 2 x 2000 4 2(2000 - 1000) 4
Value of [s x 2.1 x 10 5 x 6.667 x 10
mm/ = 6.667 x 107 4 .

. 8 x 2.1 x 10 s x 6.667 x 10 7

2(2000 - 1000) 3 x 1000 1
+

6 x 2.1 x 105 x 6.667 x 10 7 J

mm= 0.2857 - [0.01785 + 0.0238] = 0.244

Total deflection at the free end

=y +y2 = 0.1904 + 0.244 = 0.4344 mm. Ans.
1

Fig. 13.8

(t) Slope at the free end 13.7. DEFLECTION OF A CANTILEVER WITH A GRADUALLY VARYING LOAD

NLet = Slope at the free end due to point load of 1 kN i.e., 1000 A cantilever AB of length L fixed at the point A and free at the point B and carrying a
wBgradually varying load from 0 at to per unit run at the fixed end A, is shown in Fig. 13.9.
0 = Slope at the free end due to u.d.l. on length BC.

2

NThe slope at the free end due to a point load of 1000 at B is given by equation (13.1 A)

01 2 El (v 0 = 0 here)
fi t

— —= 2 x 2.1 x 10 s x 6.667 x 10j7 = 0.0001428 rad.

mThe slope at the free end due to u.d.l. of 2 kN/m over a length of 1 from the free end is

given by equation (13.9) as

wL3 w(L - 3 Fig. 13.9
a)
XConsider a section at a distance x from the fixed end A.
22 = ~66EElI m6 El v(v ' 0 =~ 0„ here)
fi 2

2 x 2000 3 2 x (2000 - 1000) 3 X — XCThe load at will be (JL- x) per unit run. Hence vertical height = j- (L-x).
6 x 2.1 x 10 s x 6.667 x 10 7 6 x 2.1 x 10 5 x 6.667 x 10 7
Hence the B.M. at this section is given by
= 0.0001904 - 0.000238 = 0.0001666 rad.
BXMx = - (Load on length Bx) x (Distance of C.G. of the load on
.'. Total slope at the free end from section X)

= + + = = - (Area of ABAC) x (Distance of C.G. of area BXC from X)

©! 0 = 0.0001428 0.0001666 0.0003094 rad. Ans. (Minus sign is due to hogging)

2

: )4

STRENGTH OF MATERIALS DEFLECTION of cantilevers 569

..(“ETOUirftaathffi] (a) Substituting x — L and = 0B in equation (in ), we get

EIQb =^(L-L)*-^=-^

But B.M. at any section is also given by equation (12.3) as

M -El d 2y 6b = _ 24E/ radianS -
-d-xI1t
(b) Substituting x = L and y =yB in equation (iv), we get

Equating the two values of B.M., we get w wL
^~ WEj?lty
wIj3
n L+'*~ r

iL
120 L L) s ~

EI df"~6L (L ~ X)3 24

Integrating the above equation, we get wL4 wL4 5wL4 + wL4

^24^ 4 120

"l20

dx 6L 4 =- ujXJ*

Integrating again, we get yB 30jHtl (Minus sign means downward deflection)

(L-x)4 + C :. Downward deflection of B is given by
l

wL4

yB = 3015/ ...(13.12)

- c '’* C ®- < mProblem 13.10. A cantilever of length 4 carries a uniformly varying load of zero
>
-T5T (t rf * intensity at the free end, and 50 kN/m at the fixed end.

where C, and C are constant of integrations. Their values are . from boundary condi- mmIfE = 2.0 x 105 N/mm2 and I = 10s 4 find the slope and deflection at the free end.

obtained ,

2

tions, which are : Sol. Given :

(i) atx = 0, y = 0 and (ii) at * = 0, —ay = 0. Length, L = 4m = 4000 mm

(/) By substituting x = 0 and y = 0 in equation (ii), we get Load at fixed end, w = 50 kN/m = 50 x 1000 ; 50 N/mm

’ 1000

0= ^I5l^-°)5 + C X0 + C <* C2=lW- Value of E = 2 x 10s N/mm2
l 2 Value of

—dy mm/ = 108 4

(ii) By substituting x = 0 and = 0 in equation (i), we get Let 0S = Slope at the free end and
yB = Deflection at the free end.
o=^tt-oi-,c,
(i) Using equation (13.11), we get

^n1_- ^ w - w^3 u>Ls———„ = - = 50 x (4000) 3 ~ - „0.„0„0„6„6„7 rad. AAns.
24 L B :
24 24 El
24 x 2 x 10 5 x 10 s

Substituting the values of C and C in equations (i) and (ii), we get (ii) Using equation (13.12), we get
x2

wL4 50 x (4000 4
“—
30 El 30 x 2 x 10 5 x 108
dx 24L -BIO —yBd — — —= z2l1.o3o3 mm. A.ns.

24

w 34
i^Z/
120L toZ/ mProblem 13.11. A cantilever of length 2 carries a uniformly varying load of 25 kN/m
!>?" * +
and T_ _ 20 Wv

N/mm mmat the free end to 75 kN/m at the fixed end. IfE - lx 10s
2 and I - 108 4 determine the

,

The equation (iii) is known as slope equation and equation (iv) as deflection equation. slope and deflection of the cantilever at the free end.
The slope and deflection at the free end (i.e ., point B) can be obtained by substituting x - L in
Sol. Given
these equations. Length, I = 2m = 2000 mm

BLet 25x 1000
B0B = Slope at the free end i.e., at and _ : 25 kN/m = = 25 N/mm

Load at the free end

yB — Deflection at the free end B.

— \

deflection of cantilevers 571

570 STRENGTH OF MATERIALS

Load at fixed end = 75 kN/m = 75 N/mm Total deflection at the free end

Value of E = lx 105 N/mm'2 = + y2 = 5 + 2.67 = 7.67 mm. Ans.

Value of mmI ~ 10s 4 13.8. DEFLECTION AND SLOPE OF A CANTILEVER BY MOMENT AREA METHOD

.

The load acting on the cantilever is shown in Fig. 13.10. This load is equivalent to a The moment area method is discussed in Art. 12.8, where this method was applied to a

uniformly distributed load of 25 kN/m (or 25 N/mm) over the entire length and a triangular simply supported beam. Let us apply this method to a cantilever. According to this method the
change of slope between any two points is equal to the net area of the B.M. diagram between
load of zero intensity at free end and (75 - 25 = 50 kN/m or 50 N/mm) 50 N/mm at the fixed end. these two points divided by El. If one of the points is having zero slope, then we can obtain the

Z —— ———— I slope at the other point.

0^ L Qpi 25 kN/m A BSimilarly if the deflection at a point is zero, then the deflection at the point accord-
1r 1
t— BT ing to this method is given by M
r A'
:1 i;

Fig. 13.10 y ~ El

where A = Aea of B.M. diagram between A and B, and

Ax = Distance of C.G. of the area from B.

(/) Slope at the free end A W13.8.1. Cantilever Carrying a Point Load at the Free end. Fig. 13.11 (a) shows a

Let = Slope at free end due to u.d.l. of 25 N/mm cantilever of length L fixed at end and free at the end B. It carries a point load at B.

0 = Slope at free end due to triangular load of intensity 50 N/mm at fixed end.

2

The slope at the free end due to u.d.l. of 25 N/mm (i.e., w = 25 N/mm) is given by equa-

tion (13.5) as

9U. = CP7 (Here 0. = 0OB', and w = 25)
L
l

25 X 2000 J
= 0.0033 rad.

6 x 1 x 10s x 10 b

The slope at the free end due to triangular load of intensity of 50 N/mm (i.e. w =

50 N/mm) is given by equation (13.11) as

wL3 C

02 = 24El Fig. 13.11

= 50 x 2000 3 (Here w = 50 N/mm). AThe B.M. will be zero at B and will be W.L at A. The variation of B.M. between and B

24 x 5 xlO 8r is linear as shown in Fig. 13.11 b( ).

lx 10 5 At the fixed end A, the slope and deflection are zero.

= 0.00167 rad. UJ0„ = Slope at B iLe“ f dy" at B and

.•. Total slope at the free end yB - Deflection at B

= 6^02 = 0.0033 + 0.00167 = 0.00497. Ans. Then according to moment area method,

(ii) Deflection at the free end Aea of B. M. Diagram between A and B

Let yx = Deflection at the free end due to u.d.l. of 25 N/mm =
0fl El
y2 = Deflection at the free end due to triangular load.

Using equation (13.11), we get deflection at the free end due to u.d.l.

^ mmy. =
1
= 25 x 2000 -xABx AC
; 5- = 5
8 El 2 (Aea of triangle ABC)
8 x 1 x 10 5 x 10 8

Using equation (13.12), we get deflection at the free end to uniformly varying load of — xLxW.L

zero at the free end and 50 N/mm at the fixed end. 2

—— mmwL* El

y2 = 3QEI
= 50 x 2000 4 =7 2.67

=

30 x 1 x 10 5 x 10 rs

. C0 —

STRENGTH OF MATERIALS

where A = Area of B.M. diagram between A and B -

x = Distance of C.G. of area of B.M. diagram from B = -----

TW . I? 2L
X W.L3
2

B _

El 3 El '

13.8.2. Cantilever Carrying a Uniformly Distributed load. Pig. 13.12 (a) shows a

Acantilever of length L fixed at end and free at the end B. It carries a uniformly distributed

load of w/unit length over the entire length.

w/Unit Length The B.M. will be zero at B and C. But B.M. at A will be -. The variation of B.M.

Abetween C and will be parabolic as shown in Fig. 13.13 (6). At the fixed end the slope and

deflection are zero.

— — —AArea of B.M. diagram
A —1 w.a2 w.a3
= a. . - =
(b)
.
w.L '
2: 3 26

lI and the distance of the C.G. of B.M. diagram from B,

B.M. Diagram —3a

x = (L - a) +
4

Fig. 13.12 B0S = Slope at i.e., at B and

yB - Deflection at B. j
Then according to moment area method,
The B.M. will be zero at B and will be at A. The variation of B.M. between A and

B is parabolic as shown in Fig. 13.12 b{ ). At the fixed end A, the slope and deflection are zero.

L ^. . W ®B El GEI

=
JArea of B.M. diagram (ABC), A =
J' -

3 26 Ax w.a 3 \, T . 3a 1 w.a 3 . T . w.a 4

and the distance of the C.G. of the B.M. diagram from B, ,

3L JB El GEI l 4 J GEI 8El

A mProblem 13.12. cantilever of length 2 carries a point load of 20 kN at the free end

mm mmand another load of 20 kN at its centre. If E -10s NI 2 and I = 10s 4 for the cantilever

Let 08 = Slope at B, i.e., and at B then determine by moment area method, the slope and deflection of the cantilever at the free

end.

yB = Deflection at B. Sol. Given : m_ L = 2
Then according to moment area method. Length,
Load at free end,
~B = wL? Load at centre, W1 = 20 kN = 20000 N
Area of B. M. diagram _ Value of WW.,2 == 20 kN = 20000 N
“ Value of
gjgjf

Ax w . L? 3 L w L4 £/? =- 1 5 N/mm2

mmI =- 108 4

13.8.3. Cantilever Carrying a Uniformly Distributed Load upto a Length ‘a’ First draw the B.M. diagram,

from the Fixed end. Fig. 13.13 (a) shows a cantilever of length L fixed at end A and free at B.M. at B =0

the end B. It carries a uniformly distributed load of w/unit length over a length ‘a’ from the B.M. at C Nmm= - 20 x 1 = - 20 kNm = - 20 x 103 x 103

fixed end.