Strength of Materials by R.K.Bansal_text

„

analysis of perfect frames 487

486 STRENGTH OF MATERIALS

Now consider the joint C. 2kN\ D

Joint C V Fvo/'
From Fig. 11.24 (6), we have \c

fcd ~ Fac = 4.216

(Compressive) Yfoe
\
and Fce = 2 kN (Compressive)
Now consider joint E. E
60"\/ 30^?^
<30* 60”\a60° Fig. 11.24 (fc)
F 4 D
iA Glx e
T Joint E [See Fig. 11.24 (c)] \/
I w 1 kN Resolving forces vertically, we get
F1 + 2 x sin 60° = ed x sin 60°

/or
Fig. 11.24 fed = 2 + = 3155 \2kN

Now taking moments about A, we get (Tensile) \ / ED
x 12 = 2 x AC + 1 x AD + 1 x AE
Resolving forces horizontally, we get °XX60 60 ° F
= 2 x 3.464 + 1 x 6.928 + 1 x4= 17.856
5.15-2 x cos 60° -Fed cos 60° -Fef = 0 < i _v_j >
E..lZfUl.49 kN Aa 5.15 kN E ^FfEeeF
or 5.15 - 2 x — - 3.15 x — — Fef = 0
"11 r 1 kN

Total vertical components of inclined loads Fef = 5.15 - 1 - 1.57 = 2.58 kN Fig. 11.24 (c)
= (1 + 2 + 1) x sin 60°
(Tensile)
= 4 x 0.866 = 3.464 kN
FFAt the joint G, two forces, i.e., bg and Da are in the same straight line and hence the

Total horizontal components of inclined loads Fthird force, i.e., gf should be zero.

= (1 + 2 + 1) cos 60° = 4 x 0.5 = 2 kN Fgf ~ 0

RNow a = Vertical components of inclined loads + 1.0 - Ii B Now consider the joint F. q.

- 4.464 -1.49 = 2.974 kN(t) Joint F [See Fig. 11.24 (d)3 \

and Ha = Sum of all horizontal components = 2 kN Resolving forces vertically, we get \
eQ°(\y^
Now the forces in the members can be calculated. Fdf x sin 60° = 0
F• DF - 0 ,„
Consider the equilibrium of joint A. E B
2.58 kN F 2.58 kN
Joint A 1 kN\ Resolving horizontally, we get
Fig. 11.24 (d)

Let Fae = Force in member AE \ Ffb~ Fef~ 2-58 kN
and Fac - Force in member AC Ffb = 2.58 kN (Compressive)
60^
Their directions are assumed as shown in Fig. 11.24 (a). Now consider the joint B.

Resolving the forces vertically, we get H 2 kktN A ^AE E Joint B

Fac x sin 30° + 1 x sin 60° = 2.974 k Resolving vertically, we get
For ac x 0.5 + 0.866 =: 2.974
2.974 kN Fbg x sin 30° = 1.49
Ra 1

fac ~ 2.974 - 0.866 F*„*„* - L4y

0.5 Fig. 11.24 (a) 0.5 = 2.98 kN (Compressive)
(Compressive)
= 4.216 kN (Compressive) Joint G Rn = 1.49 kN
Fig. 11.24 (e)
Resolving the forces horizontally, we get fod = fbc = 2-98 kN

F, = 2 + ac cos 30° - 1 x cos 60°
= 2 + 4.216 x 0.866 - 0.5 = 5.15 kN (Tensile)

488 STRENGTH OF MATERIALS analysis of perfect frames
The forces are shown in a tabular form as
Sol. First determine the reaction R B and R c.
The distance of line of action of 20 kN from point B is

AB x cos 60° or 2.5 x m1.25

Taking moments about point B, we get

Rc x 5 = 20 x 1.25
20 x 1.25 =5_ k. .M.

and dtB = 20 — 5 = 15 kN Fig. 11.26
Now draw a section line (1-1), cutting the membeis

AB and BC in which forces are to be determined. Now con-

sider the equilibrium of the left part of the truss. This part

is shown in Fig. 11.27.

Let the directions of FBA and FBC are assumed as

11.6. METHOD OF SECTIONS shown in Fig. 11.27.

When the forces in a few members of a truss are to be determined, then the method of Now taking the moments of all the forces acting on

section is mostly used. This method is very quick as it does not involve the solution of other the left part about point C, we get

joints of the truss. 15 x 5 + (Fba x AC)* = 0

In this method, a section line is passed through the members, in which forces are to be ( V The perpendicular distance between the line of Fig. 11.27
determined as shown in Fig. 11.25. The section line should be drawn in such a way that it does
not cut more than three members in which the forces are unknown. The part of the truss, on Faction of BA and point C is equal to AC) (v AC = BC x cos 30°)
any one side of the section line, is treated as a free body in equilibrium under the action of 75 + FBA x 5 x cos 30° = 0
external forces on that part and forces in the members cut by the section line. The unknown
forces in the members are then determined by using equations of equilibrium as p ~.Z5 =-17.32 kN

or 5 x cos 30°

FThe negative sign shows that BA is acting in the opposite direction towards point

2Ft = 0, ZF = 0 and 0. B) Hence force will be a compressive force.
y
p = 17.32 kN (Compressive). Ans.
moments of about point A, we get
all ^ forces acting on the left part
Again taking the
the

15 x Perpendicular distance between the line of action of FBC and point A

15 kN and point C = FBC x Perpendicular distance between

F15 x 2.5 x cos 60° = bc x 2.5 x sin 60°

15 x 2.5 x cos 60° 15 x 0.5
0.866
^BC ~ 2.5 x sin 60°

= 8.66 kN (Tensile). Ans.

(a) Given Truss (6) Left Part (c) Right Part These forces are same as obtained in Problem 11.1.

Fig. 11.25 Problem 11.12. A truss of span 5 mis loaded as shown in Fig. 11.28. Find the reactions

and forces in the members marked 4, 5 and 7 using method of section.

If the magnitude of the forces, in the members cut by a section line, is positive then the FBA into vertical
assumed direction is correct. If magnitude of a force is negative, then reverse the direction of
is 3(JO)
that force.
6QO =
taken by resolving the: force

5 - FBA componeni.
x5x
“a, aWC ^whereasand

^the
Problem 11.11. Find the forces in the members AB and AC of the truss shown in Fig. *Th Fmoment of the force BA about point C, is also

11.26 using method of section. (U.P. Tech. University, 2002-2003) horiamtS components at poinfs The moment of the
moment of vertical component will be CF* x on 60 ) x

(

—

ANALYSIS OF PERFECT FRAMES 491

STRENGTH OF MATERIALS

R RSol. Let us first determine the reactions A and B . Now takin g the moments of all the forces about point B acting on the right part, we get

ABDTriangle is a right-angled triangle having angle 12 x BE cos 30° + F x BE = 0
5

ADB = 90° 12 x cos 30° + F = 0
s
or F. = - 12 x cos 30° = - 10.392 kN
AD-AB cos 60° = 5 x 0.5 = 2.5 m 5
,.

The distance of line of action the vertical load 10 kN - ve sign indicates that F is compressive.
s
ADfrom point A will be
cos 60° or 2.5 x 0.5 = 1.25 m. F = 10.392 kN (Compressive). Ans.
s
Now taking the moments about point C of all the forces acting on the right parts, we get
From triangle ACD, we have

AC = AD = 2.5 m ^F^CE12 x (2.5 - BE cos 30°) + Rb *BC

mBC — 5 — 2.5 = 2.5 M

In right-angled triangle CEB, we have 60° 60°/ 12 x 2.5 - 2.5 x x = F x 2.5 x sin 30° + 10 x 2.5
7
,

—BE = BC cos 30° = 2,5 x C &@ £t

:m or 12 x (2.5 - 1.875) = F X 1.25 + 25 or 7.5 = 1.25F, + 25
7

The distance of line of action of vertical load 12 kN 7.5-25 = ~ 1, ,4.k.N.

or

Fig. 11.28

from point B will be BE cos 30° or BE x — Negative sign shows that F is compressive.
7
2
F = 14 kN (Compressive). Ans.
7

2 .5 x x - L875 m These forces are same as obtained in Problem 11.3.
=
Problem 11.13. A truss of span 9 mis loaded as shown in Fig. 11.30. Find the reactions
(
and forces in the members marked 1, 2 and 3.
AThe distance of the line of action of the load of 12 kN from point will be
RRSol. Let us first calculate the reactions A and B .
m(5 - 1.875) = 3.125
Taking moments about A, we get
Now taking the moments about A, we get
Rb x 9 = 9 x 3 + 12 x 6 = 27 + 72 = 99
Rb x 5 = 10 x 1.25 + 12 x 3.125 = 50
YFB = 99 = nkN
Rb ~ If = 10 kN and RA = (10 + 12) - 10 = 12 kN
and Ra = (9 + 12) - 11 = 10 kN
Now draw a section line (1- 1 ), cutting the members 4,
\ xo p ©C D E F
5 and 7 in which forces are to be determined. Consider the
equilibrium of the right part of the truss (because it is smaller
than the left part).

This part is shown in Fig. 11.29. Let F., F, andF, are
the forces in members 4, 5 and 7. Let their directions are
assumed as shown in Fig. 11.29.

Now taking the moments of all the forces acting on

Ethe right part about point , we get

Rb X BE cos 30° = F x (BE x sin 30°)
4

J3) V3 R
: F. x 2.5 x
10 x 2.5 x 4 x 0.5

2 Fig. 11.29

10 x = F x 0.5 Fig. 11.30
4
Now draw a section line (1-1), cutting the members 1, 2 and 3 in which forces are to be
— —F =10x
4 determined. Consider the equilibrium of the left part of the truss (because it is smaller than
x - 17.32 kN (Tensile).
2 0.5

"

STRENGTH OF MATERIALS analysis of perfect frames

the right part). This part is shown in Fig. 11.30 (a). Let Let us now find the forces in the members 1, 2 and y, h±

FFvF 3 by the method of section. Take a section Y-Y passing ! o.sm
2 and a are the forces members 1, 2 and 3 respec-
through the members 1, 2 and 3. Now consider the equi- 1J
tively. Let their directions are assumed as shown in Fig. “
|
11.30 (a). librium of left portion shown in Fig. 11.31 (a). /X
j 4gm
!
;j
Taking moments of all the forces acting on the left Let Fv F and F are the forces in the members 1,2 / F\;
part about point D, we get 2 3 !
/
4 and 3 respectively. Their assumed directions are also \2
/
10 x 3 = F x 4 shown in Fig. 11.31 (a).
3

F 10x3 Taking moments of all forces (here RA, Fv F and / \
2
A/ — —— [
• 4 F3) about point D, we get j
. j
1 (pA c i
3

= 7.5 kN (Tensile). Ans. J?A x 4 = Fj x 4.5 20 kN y' '

Now taking the moments of all the forces acting on Ra x 4 20 x 4 4 4 m *+« 4 m

the left part about point G, we get 4.5 4.5 Ha

10x3 + F x4 = 0 Fig. 11.30 (a) = 17.78 kN (Tensile). Ans. Fig. 11.31(a)
1

Fj = = - 7.5 kN Now taking the moments about C, we get

4 Ra x 8 = F x Perpendicular distance between F and point C ...(i)
3 3
FNegative sign shows that force is compressive.
l FTo find the perpendicular distance between the line of action of and point C, first find
Fj = 7.5 kN (Compressive). Ans.
s

Now taking the moments about the point C, we get angle CDH

F x3-9x3 + F x4 = 0 DE 4.5
2 a tan 6 ~ EC ~ 4.0

For 2 x 3 - 27 + 7.5 x 4 = 0
27 - 7.5 x 4 0 = tan-1 : 48.37° and tan a -

—. —a = tan-1
23 3 =- - 1.0 kN

= 7.125°

FNegative sign shows that force 2 is compressive. 4.0

F = 1.0 kN (Compressive). Ans. LCDH = 0 + a = 48.37 + 7.125 = 55.495
2
From triangle DEC, we know that
Problem 11.14. For the pin-joined, truss shown in Fig. 11.31, find the forces in the

members marked 1, 2 and 3 with the single load of 80 kN as shown. mCD 42 / \e7 T
= 2 + = 6.02 \

^4.5

R RSol. First calculate reactions A and B . Now from C, draw a perpendicular CL on the line of /\ '

Faction of as shown in Fig. 11.31 (5). / \1

3 /\

From right-angled triangle CDL,

CL / \\

sin (a + 0) =

CL = CD sin (a + 0) A Ec

© \l/ = 6.02 x sin (55.495) Fig 31 (j>)

I m= 4.96

EC FSubstituting the value of OL (i.e. , perpendicular distance between and C) in equation
+ 80 kN
3
H4 m — —— 4 m —H4 4 m —H-4— 4 m
(i), we get

fl.x8 = F,x 4.96

Fig. 11.31 _ Ra x8 _ = 32.26 kN (Compressive). Ans.

3 4.96 4.96

Taking moments about A, R„ x 16 = 80 x 12 F R F F FJTo find the force vertically. Hence, we get
v resolve the forces (i.e., A, 3, and

2

80 x 12 RA - F sin a + F sin 0 = 0
3 2
Jg-
~RB = = 60 kN

For x sin (48.37) = 0
Ra = Total vertical load - RB = 80 - 60 = 20 kN 20 - 32.26 x sin (7.125) +
2

—

STRENGTH OF MATERIALS ANALYSIS OF PERFECT FRAMES

or 20 - 4 +F x 0.7474 = 0 The section line (2-2), cutting the four members,
2 is shown in Fig. 11.32 (6). Let the forces in the members

or =F.-1, 16 =21.4 kN (Compressive). Ans. F FFare CA, co, FnG and BD . Let their directions are as- 2m
0.7474
sumed as shown in Fig. 11.32 (6). Consider the equilib- 2m
11.6.1. Method of Section, Cutting more than Three Members. In method of sec- rium of part above the section line (2-2). Taking the
moments of all the forces (acting on the upper part) about I
tion, in general a section should cut only three members, since only three unknowns can be point C, we get
±.
determined from three equations of equilibrium. However, there are special cases where we 1
n
Amay cut more than three members. It is illustrated in the following example. section line can -
C;
cut four members if the axes of the three of them intersect in one point, thus making it possible 20 x 2 + 20 x 4 + Fbd x 4 = 0

to determine the axial force in the fourth member by taking moments about the point of inter- or Fbd = - 40-- 80

section of the axes of the three members. ——- 120

Vroblemll.l5.FortheframeshowninFig. 11.32 q = = - 30 kN
T~ * 20 kN
find, the forces in the members BD, BG, GA, AC and AB /' 90” 'x 4
/2 m
of the bottom bay only. State their nature. H \ * - ve sign means the force FBD is compressive. 4m
/, , \F ^bd ~ 30 kN (Compressive). Ans.
Sol. Let us first find the reactions. The frame car- /\ Fig. 11.32 (6)
"E | Now taking moments of all forces (acting on the
ries horizontal loads. As the frame is supported on roll- f
upper part) about point D, we get
ers at B, hence the reaction R P will be vertical. /2 m / \90° \
FCA x4 = 20x2 + 20x4 = 120
At the point A, the frame is hinged and hence the / / \ \\/~ ,
r c Q 20 kN Fca = = 30 kN (Tensile). Ans.
Asupport reactions at will consist of a horizontal reac- D.rp*
/C\
H Rtion a and a vertical reaction A . // /2ir>./
\ /\ Now consider joint B. G\ D
Taking moments of all forces about A, we get [<£ Joint B
/\ (D \Tbg
x 4 = 20 x 2 + 20 x 4 + 20 x 6 = 240 Resolving forces vertically, we get

Fb = = 60 kN (f) wFro 45 + F = Rg

Now R Ra = Total vertical loads ~ g h* bX ©R/olle-rA^;B or Fbq x cos 45° + 30 = 60

= 0 - 60 = - 60 kN , _ 60 - 30
01 Ba cos 45°
- ve sign means, RA is acting downwards. R* R 30 45°X~
1A/2 A fba
Ra = 60 kN (J.) V
Rb = 60 kN
and Ha = Sum of all horizontal loads Fig. 11.32 Fig. 11.32 (c)

= (20 + 20 + 20) = 60 kN («-) = 30 x (Compressive)

Now draw a section line (1-1), cutting the members —H 2 m H Resolving horizontally, we get = 30 kN (Tensile)
BD, BG and BA in which forces are to be determined. Con-
CG D Fba = Fbg sin 45°

sider the equilibrium of the right part. This part is shown T .-CX = 30 x J2 x
;!
Fin Fig. 11.32 (a). Let BD ,FB0 and FBA are the forces in the
HGmembersBO, HAand respectively. Let their directions I / 90° \/
!

/1 ! \/uF Now consider joint A.
Joint A
are assumed as shown in Fig. 11.32 (o). X1 >/ fB bd C
30 kN
k /X? G

In this particular case, all the three forces are meet- A “ p<4 -A /
,/p
ing at one point B. Hence by cutting these members by Fag/
section line (1-1), we may not get the results. w' 60 kN MA FR FAt the joint ; A, tiA, AC and are known in mag-

R0 j

nitude and direction.

Let us draw a section line (2-2), cutting four mem- ... Resolving horizontally, we get Ha = 60 yy 5°

bers AC, CG, GD and BD in which forces are to be deter- Fig. 11.32 (a)

60 = 30 + Fag x cos 45° * A 30 kN*
60-30 30
mined. The axes of three members, i.e., AC, CG and GD are intersecting at point C. And hence B

taking moments about point C, we can find force in member BD.

Similarly the axes of BD, GD and CG are meeting at point D. And hence taking mo- 1 cos 45° ( 1 ) Ra = 60

ments about point D, we can find the force in member AC. UJ Fig. 11.32 (d)

= 30 x V2 (Tensile)

analysis of perfect frames

- 1.49 X 4

= - 2.98 kN

4 x sin 30°

- ve sign shows that the force FDG is compressive. 'C'-. / F0G
FnG = 2.98 kN (Compressive). Ans.

Now taking the moments about point D, we get / \/

Rb x BD cos 30° = Ffe x BD x sin 30° ~7* p 4n
FRb x cos 30° = fe x sin 30°
^ FE
1.49 x cos 30° = 1.49 x 0.866 ^

sin30° 0^5 Fig. 11.33 (a«)

= 2.58 kN (Tensile). Ans.

Now taking the moments of all forces acting on the right part about B, we get

Ffj) x 1 distance between Ffd and B = 0

Fpjj = 0. Ans.

(v F1 distance between fd and B is not zero)

11.7. GRAPHICAL METHOD

The force in a perfect frame can also be determined by a graphical method. The analyti-
cal methods (such as method of joints and method of sections) give absolutely correct results,
but sometimes it is not possible to get the results from analytical methods. Then a graphical
method can be used conveniently to get the results. The graphical
method also provides reasonable accurate results.

The naming of the various members of a frame are done ac-

cording to Bow’s notations. According to this notation of force is des-
ignated by two capital letters which are written on either side of the

A Aline of action of the force. force with letters and B on either side

of the line of action is shown in Fig. 11.34. This force will be called
AB.

The following steps are necessary for obtaining a graphical solution of a frame.

'(i) Making a space diagram

(ii) Constructing a vector diagram

(.Hi) Preparing a force table.

1. Making a space diagram. The given truss or frame is drawn accurately according
to some linear scale. The loads and support reactions in magnitude and directions are also
shown on the frame. Then the various members of the frame are named according to Bow’s

notation. Fig. 11.35 (a) shows a given truss and the forces in the members AB, BC and AC are
to be determined. Fig. 11.35 b( ) shows the space diagram to same linear scale. The member AB

is named as PS and so on.

2. Constructing a vector diagram. Fig. 11.35 (c) shows a vector diagram, which is
drawn as given below :

(i) Take any point p and draw pq parallel to PQ vertically downwards. Cutpo = 4 kN to

same scale.

(ii) Now from q draw qr parallel to QR vertically upwards and cut qr = 2 kN to the same

scale.

498 STRENGTH OF MATERIALS ANALYSIS OF PERFECT FRAMES 499

(a) Given Diagram (b ) Space Diagram (c) Vector Diagram

Fig. 11.35 ”

(Hi) From r draw rp parallel to RP vertically upwards and cut rp = 2 kN to the same scale. Fig. 11.36
(iv) Now from p, draw a line ps parallel to PS and from r, draw a line rs parallel to RS,
Now taking moments about B, we get
meeting the first line at s. This is vector diagram for joint (A). Similarly the vector diagrams Rc x 5 = 20 x 1.25 = 25
for joint (B) and (C) can be drawn.
—25
3. Preparing a force table. The magnitude of a force in a member is known by the
length of the vector diagram for the corresponding member, i.e., the length ps of the vector Rc = o = 5 kN and RB = 20 - 5 = 15 kN
diagram will give the magnitude of force in the member PS of the frame.
Now draw the space diagram for the truss alongvvith load of 20 kN and the reactions RB
Nature of the force (i.e., tensile or compressive) is determined according to the following and Rc equal to 15 kN and 5 kN respectively as shown in~Fig. 11.36 (6). Name the members
procedure : A8, AC and BC according to Bow’s notations as PR, QR and RS respectively. Now construct

(i) In the space diagram, consider any joint. Move round that joint in a clockwise direc- the vector diagram as shown in Fig. 11.36 (c) and as explained below :
tion. Note the order of two capital letters by which the members are named. For example, the
(£) Take any point p and draw a vertical line pq downward equal to 20 kN to some
members at the joint (A) in space diagram Fig. 11.35 b( ) are named as PS, SR and RP. suitable scale. From q draw a vertical line qs upward equal to 5 kN to the same scale to repre-
(ii) Now consider the vector diagram. Move on the vector diagram in the order of the
Rsent the reaction at C. Then sp will represent the reaction B to the scale.
letters (i.e., ps, sr and rp). (ii) Now draw the vector diagram for the joint (B). From p, draw a line pr parallel to PR

(Hi) Now mark the arrows on the members of the space diagram of that joint (here and from s draw a line sr parallel to SB, meeting the first line at r. Now prs is the vector
diagram for the joint (B). Now mark the arrows on the joint B. The arrow in member PR will be
joint A). towards joint B, whereas the arrow in the member RS will be away from the joint B as shown

(Hi) Similarly, all the joints can be considered and arrows can be marked. in Fig. 11.36 (6).
(o) If the arrow is pointing towards the joint, then the force in the member will be
compressive whereas if the arrow is away from the joint, then the force in the member will be (Hi) Similarly draw the vector diagrams for joint A and C. Mark the arrows on these

tensile. joints in space diagram.

Problem 11.17. Find the forces in the members AB, AC and BC of the truss shown in Now measure the various sides of the vector diagram. The forces are obtained by multi-

Fig. 11.36. plying the scale factor. The forces in the members are given in a tabular form as :

RSol. First determine the reactions Rg and c . m NProblem 11.18. A truss ofspan 7.5 carries a point load of 1000 at joint D as shown

mFrom Fig. 11.36 (a), AB = BC x cos 60° = 5 x — = 2.5 in Fig. 11.37. Find the reactions and forces in the member of the truss.

Distance of line of action of 20 kN from point B

m= AB cos 60° = 2.5 x | = 1.25

:

500 STRENGTH OF MATERIALS - analysis of perfect frames 501

RSol. First determine the reactions A and RB Problem 11.19. Determine the forces in all the members of a cantilever truss showrl in

Fig. 11-38. \

Sol. In this case the vector diagram can be drawn without knowing the reactions. First

Nof all draw the space diagram for the truss along with loads of 1000 of joints B and C. Name
VT
the members AB, BC, CD, DB, AD and BD according to Bow’s notation as PT, QS, SR, RV, r

:

and ST respectively. Now construct the vector diagram as shown in Fig. 11.38 (e) and as ex-

plained below

(a) Given Diagram (6) Space Diagram (c) Vector Diagram (i) The vector diagram will be started from joint C where forces in two members are

Fig. 11.37 Nunknown. Take any point q and draw a vertical line qr downward equal to load 1000 to some

Taking moments about A, we get suitable scale. From r, draw a line rs parallel to RS and from q draw a line qs parallel to QS,
meeting the first line at s. Now qrs is the vector diagram for the joint C. Now mark the arrows
Rb x 7.5 = 5 x 1000
on the joint C. The arrow in the member RS will be towards the joint C, whereas the arrow in

the member SQ will be away from the joint C as shown in Fig. 11.38 b( ).

r. Rar = -^25 _ 667 N and RA = 1000 - 667 = 333 N.

7.5

NNow draw the space diagram for the truss alongwith load of 1000 and reactions Ra
N Nand Rb equal to 333 and 667 respectively as shown in Fig. 11.37 (5). Name the members
AC, CB, AD, CD and DB according to Bow’s notations as PR, PQ, RT, QR and QS respectively.

Now construct the vector diagram as shown in Fig. 11,37 (c) and as explained below :

(i) Take any point s and draw a vertical line st downward equal to load 1000 N to some ,
Nsuitable scale. From t draw a vertical line tp upward equal to 333 to the same scale to
Rrepresent the reaction at A. The ps will represent the reaction B to the scale. j (a) Given Figure (b) Space Diagram (c) Vector Diagram
j
(ii) Now draw the vector diagram for the joint A. From p, draw a line pr parallel to PR Fig. 11.38
and from t draw a line tr parallel to RT, meeting the first line at r. Now prt is the vector
diagram for the joint A. Now mark the arrows on the joint A. The arrow in the member PR will D(ii) Now draw the vector diagram for the joints B and similarly.
be towards the joint A, whereas the arrow in the member RT will be away from the joint A as

shown in Fig. 11.37 (6).

(Hi) Similarly draw the vector diagrams for the joint C, B and D. Mark the arrows on Mark the arrows on these joints as shown in Fig. 11.38 (6).
Now measure the various sides of the vector diagram. The forces in the members are
these joints as shown in Fig. 11.37 (b).

Now measure the various sides of the vector diagrams. The forces in the members are given in a tabular form as :

obtained by multiplying the scale factor to the corresponding sides of the vector diagram. The Member

forces in members are given in a tabular form as : Force in member Nature offorce

According to given truss According to 1333 N Tensile
Bow’s notation 1333 N Tensile
AB 1666 N Compressive
BC PT 2500 N Compressive
CD QS 833 N Tensile
DE SR 1000 N Compressive
AD RV
BD VT
ST

From the vector diagram, the reactions 1?, and R at A and E can be determined in
2

magnitude and directions.

Reaction R = rv = 2500 N. This will be towards point E.
2

502 STRENGTH OF MATERIALS ANALYSIS OF PERFECT FRAMES 503
E
AReaction
R = up = 2000 N. This will be away from the point as shown in Fig. 11.38 (6). M M;X +Fcd x Fig. 11.39 (c)
1
(v Fbc = 200 kN)
The reaction R is parallel to vp.
t Fec - 0

_Problem 11.20. Determine the support reac- £ Fig. 11.39 (e)

tions and nature and magnitude of forces in the mem- \ FCS + FCD ~ T00 x ^
bers of truss shown in Fig. 11.39.
2m

,

±(U.P. Tech. University, 2001-2002) - 100 x 720

ASol. Let us start from joint where forces in I Resolving forces vertically, we get

two members are unknown. 2m fFqe s ' n 0 - cd sin 0 - Fbc = 0
F F- cd ) sin 0 = bc = 200
i

Joint A m—d B
K4
In triangle ABC, +H 4 m H

—tan 0 = BC = —2 Fig. 11.39 ~^QtcE ~ tcD 200 =_ 200
CA ' 200 kN
4 (_2_\

„ AC 4 4 lV20j

= 100 X

Adding equation (i) and (ii),

and sin 0 = —p== 2Fce = 200 x J20
720 F Jor ce = 100 x 2O (Tensile)

Refer to Fig. 11.39 (a). The forces are shown at joint A. Resolv- Fig. 11.39 (a) Substituting this value in equation (i), we get
ing forces vertically, we get
Fcd = 100 x 720 - 100 x 720 = 0
sin 0 = 200

200 200 x V20 Support Reactions
= 447.2 kN. Ans.
-“sinO 2/720 DTo find the support reactions, consider joint and E.
2

Resolving forces horizontally, we get Joint D

Fac - Fab cos 0 The force FBbDd = 400 kN

= (100 x 720 ) x -4= - 400 kN (Tensile). whereas Fpc ~ 0 - Hence at joint D, there will be only horizontal

720 Rreaction DH , which will balance force Fbd-
F-dh ~ Fbd = 400 kN.
Joint B

Refer to Fig. 11.39(6) Joint E

AABC = 90-6 FAt joint E, the force EC = 100 x 720 kN- To balance this

Resolving forces vertically, force, there will be horizontal reaction and vertical reaction at E.

FBC = ^AB C0S _ 0) Let Rev = Vertical component of reaction at E
REfl = Horizontal component of reaction at E
= FAB Sin 0
Resolving forces horizontally, we get
== x x -
(100 720 )
7

mR = Fec cos 0 = (100 x J20 ) x 4

= 400 kN. Ans.

Fab = 100 x V20 and sin 0 = - Fig. 11.39 (6)

Resolving forces vertically, we get

PBC =200kN (Tensile) 2

Resolving forces horizontally, we get Rev = Fec sin 0 = (100 x 720 ) x = 200 kN. Ans.

F^Fbd = F42 sin (90 - 0) = cos 0 Now the nature and magnitude of forces in the members are:
AB -* 447.2 kN (Compressive)
= (100 x J20 ) x -~L = 400 kN (Comp.) BC -* 200 kN (Tensile)
AC -» 400 kN (Tensile)
Refer to Fig. 11.39(c) BD -» 400 kN (Compressive)
CD-* 0
Resolving forces horizontally. CE 447.2 kN (Tensile).

F Ffcf. cos 0 + cd cos 0 = AC

: )

ANALYSIS OF PERFECT FRAMES 505

1. The relation between number of joints {/) and number of members (n) in a perfect frame is given (iv the truss is supported on rollers at one end and hinged at other end and carries inclined
by n = 2/ - 3.
loads.
2. Deficient frame is a frame in which number of members are less than (2/ - 3) whereas a redun-
dant frame is a frame in which number of members are more than (2'j - 3). 5. (a) What is the advantage of method of section over method of joints ? How will you use method

3. The reaction on a roller support is at right angles to the roller base : of section in finding forces in the members of a truss ?

4. The forces in the members of a frame are determined by . .

(i) Method of joints (ii) Method of sections, and (6) Explain with simple sketches the terms (i) method of sections and (ii) method of joints, as^

(iii) Graphical method. applied to trusses.

5. The force in a member will be compressive if the member pushes the joint to which it .is con- 6. How will you find the forces in the members of a joint by graphical method ? What are the
nected whereas the force in the member will be tensile if the member pulls the joint to which it
advantages or disadvantages of graphical method over method of joints and method pf section ?
is connected.
7. What is the procedure of drawing a vector diagram for a frame ? How, will you find out (i) magni-
tude of a force, and (ii) nature of a force from the vector diagram ?

8. How will you find the reactions of a cantilever by graphical method ?

9. What are the assumptions made in the analysis of a simple truss.
10 Explain what you understand by perfect frame, deficient frame and redundant frame.

6. While determining forces in a member by method of joints, the joint should be selected in such a B. Numerical Problems /
way that at any time there are only two members, in which the forces are unknown.
1. Find the forces in the members AB, AC and BC of the truss shown in Fig. 11.40.
7. If three forces act at a joint and two of them are along the same straight line then third force [Ans. AB = 4.33 kN (Comp.)
would be zero.
AC = 2.5 kN (Comp.)
8. If a truss (or frame) carries horizontal loads, then the support reaction at the hinged end will BC = 2.165 kN (Tens.)]
consists of (i) horizontal reaction and (ii) vertical reaction.

9. If a truss carries inclined loads, then the support reaction at the hinged end will consists of
(i) horizontal reaction and (ii) vertical reaction. They will be given as :
Horizontal reaction = Horizontal components of inclined loads

Vertical reaction = Total vertical components of inclined loads - Roller support reaction.

10. Method of section is mostly used, when the forces in a few members of a truss are to be deter- Fig. 11.40

mined.

11. The following steps are necessary for obtaining a graphical solution of a frame : A m N D2. truss of span 7.5 carries a point load of 500 at joint as shown in Fig. 11.41. Find the
(i) Making a space diagram,

(ii) Constructing a vector diagram, and reactions and forces in the members of the truss.

(iii) Preparing a force table. N[Ans. Ra = 166.5
Rg = 333.5 N
12. The various members of a frame are named according to Bow's notation.
F = 333 N (Comp.)
1

=NF
i
288.5 (Tens.)

EXERCISE 11 NF3 = 577.5 (Tens.)

F = 667 N (Comp.)
4

A. Theoretical Questions F N& = 577.5 (Tens.)]

1. Define and explain the terms : Perfect frame, imperfect frame, deficient frame and a redundant Fig. 11.41

frame. ( U.P. Tech. University, 2002-2003)

2. (a) What is a frame ? State the difference between a perfect frame and an imperfect frame. mA truss of span 7.5 is loaded as shown in Fig. 11.42. Find the reactions and forces in the

(6) What are the assumptions made in finding out the forces in a frame ? members of the truss.

3. What are the different methods of analysing (or finding out the forces) a perfect frame ? Which [Ans. AD = 3.464 kN (Comp.)
one is used where and why ?
AC = 1.732 kN (Tens.)

4. How will you find the forces in the members of a truss by method of joints when CD = 2.598 kN (Tens.)

(i) the truss is supported on rollers at one end and hinged at other end and carries vertical loads. CE = 2.598 kN (Comp.)

(ii) the truss is acting as a cantilever and carries vertical loads. DB = 3.50 kN (Comp.)

(iii) the truss is supported on rollers at one end and hinged at other end and carries horizontal BE = 5 kN (Comp.)
and vertical loads.
BC = 4.33 kN (Tens.)]

Fig. 11.42

ANALYSIS OF PERFECT FRAMES 507

506 STRENGTH OF MATERIALS Joint F

A4. truss is shown in Fig. 11.43. Find the forces in 10 kN 15 kN 20 kN C& 20 = 20 kN (Comp.)

all the members of the truss and indicate it is in 20 =0

tension or compression. 1
(U.P. Tech. University 2000-2001)
E

Joint E ZV = 0, 32.5 - 20 - Fce sin 60° = 0

12.5

sin 60° = Fce = 14.43 kN
ztr=o, Fed - Fce cos 60° = 7.215 kN]

Fig. 11.43

3m[Hint. In the problem, length of members are not given. Assume AD = DE = and UDAC = 5. Determine the forces in the various members of the truss shown in Fig. 11.44.

LDEC = 60 as from figure it appears that AD = DE and IDAC = LDEC [Ans. AS = 1200 N (Comp.)
BC = 800 N (Comp.)
M Ra = 0, 10 x 3 + 15 x 3 + 20 x 6 - 6 x E = 0, 400 N

30 + 45 + 120 CD = 800 N (Comp.)

: 32.5 kN

DE = 1200 N (Comp.)

and R. - 10 + 15 + 20 + 10 - R„ = 55 - 32.5 = 22.5 EF = 600 N (Tens.)

Start from joint B where forces in two members are unknown AF = 600 N (Tens.)

Joint B = 10 kN (Comp.) BF ~ DF - 400 N (Comp.)

=0 NFC = 400 (Tens.)]

Fig. 11.44

A6. plane truss is loaded and supported as shown in Fig. 11.45. Determine the nature and magni-

tude of forces in the members 1, 2 and 3.

[Ans. Fj = 833.34 N (Comp.)

Joint A TV = 0 NF1 = 1000 (Tens.)
Joint D
12.5 F = 666.66 (Tens.)]
3
: 14.43 kN

sin 60°

m=0 F F’ AD= .\c cos 60 ° = 7.215

10 kN

508 STRENGTH OF MATERIALS ANALYSIS OF PERFECT FRAMES 509
Joint C
SF,. = 0, Fbc cos 0 = Fcd cos 0
C F„= 17891
Fcd =Fbc = 17891 N (Comp.)

lF = 0, Fca - Fbc sin 6 + Fcd sin 0 = 0
y

fca “ 0 ( v fbc ~ Fa?

Joint A

Fad ~_ 12000 __
sin 45°

m >? 'B ZF, = 0. F, n = cos a - fab + fba = °
F„= 16002
+ 16002 = 0
;
16002 = 0
C”

N= 16002 + 11999 = 28001 (Tens.)]

10. Determine the forces in the truss shown in Fig. 11.49 which carries a horizontal load of 16 kN

and a vertical load of 24 kN. [Ans. AC - 24 kN (Tens.)

AD = 10 kN (Comp.)

CD = 24 kN (Tens.)

CB = 24 kN (Tens.)

BD = 30 kN (Comp.)]

Fig. 11.49

11. Find the forces in the member AB and AC of the truss shown in Fig. 11.40 of question 1, using

method of sections. [Ans. AB = 4.33 kN (Comp.)

AC = 2.5 kN (Comp.)]

12. Find the forces in the members marked 1, 3, 5 of truss shown in Fig. 11.41 of question 2, using

method of sections. N[Ans. F = 333 (Comp.)
y

NF3 = 577.5 (Tens.)

NF5 = 577.5 (Tens.)]

13. Find the forces in the members DE, CE and CB of the truss, shown in Fig. 11.42 of question 3,

using method of sections. [Ans. DE = 3.5 kN (Comp.)

CE = 2.598 kN (Comp.)

BC = 4.33 kN (Tens.)]

14. Using method of section, determine the forces in the members CD, FD and FE of the truss shown

in Fig. 11.43 of question 5. (Ans. CD - 800 N (Comp.)

FD = 400 N (Comp.)

NFE = 600 (Tens.)]

E)

510 STRENGTH OF MATERIALS

12

Deflection of Beams

12.1. INTRODUCTION

If a beam carries uniformly distributed load or a point load, the beam is deflected from

its original position. In this chapter, we shall study the amount by which a beam is deflected

from its position. Due to the loads acting on the beam, it will A B

be subjected to bending moment. The radius of curvature of I

c

— —the deflected beam is given by the equation - = . The ra-

1R (a) Beam position before loading

Rdius of curvature will be constant if = = constant.

The term (/ x E)/M will be constant, if the beam is subjected (b) Beam position afler loading
to a constant bending moment M. This means that a beam for
which, when loaded, the value of ( x I)/M is constant, will Fig. 12.1
bend in a circular arc.

Fig. 12.1 (a) shows the beam position before any load
is applied on the beam whereas Fig. 12.1 ( b shows the beam

position after loading.

12.2. DEFLECTION AND SLOPE OF A REAM SUBJECTED TO UNIFORM BENDING
MOMENT

A beam AS of length L is subjected to a uniform bend- __ _ D. _
x .- " "
Ming moment as shown in Fig. 12.1 (c). As the beam is sub- x
; *\
jected to a constant bending moment, hence it will bend into
a circular arc. The initial position of the beam is shown by ^
ACB, whereas the deflected position is shown by AC'B. !

Let R = Radius of curvature of the deflected beam, /
\
y = Deflection of the beam at the centre (i.e., dis-
[- \
tance CC'), j

£, = roungs modulus tor the beam material, and I! \

d = Slope of the beam at the end A (i.e., the angle 'O I

made by the tangent atA with the beam AS). I I

For a practical beam the deflection y is a Xt Xr /
v
small quantity. i r <H X /

\x 1 /
X(9O_0)
i
a \Jq r 7b

t

\ ~“ ” l *
2
_L c,

r2 .1.

Fig. 12.1 (c)

511

STRENGTH OF MATERIALS DEFLECTION OF BEAMS 513

Hence tan 0 =.6 where 9 is in radians. Hence 0 becomes the slope as slope is 12.3. RELATION BETWEEN SLOPE, DEFLECTION AND RADIUS OF CURVATURE

— = tan 0 = 0. Let the curve AS represents the deflection of a beam as shown in Fig. 12,2 (a). Consider
a small portion PQ of this beam. Let the tangents at P and Q make angle V and ip + dy with
Now AC ~ BC = — x-axis. Normal at P and Q will meet at C such that

PC = QC = R

Also from the geometry of a circle, we know that (v DC - DC' - CC‘ =2R-y)

AC x CB = DC x CC

-L x-L =(2R-y) xy

—or = 2Ry-y2

4

For a practical beam, the deflection y is a small quantity. Hence the square of a small
quantity will be negligible. Hence neglecting y2 in the above equation, we get

~I 2
= 2Ry

4

But from bending equation, we have Fig. 12.2

ME The point C is known as centre of curvature of the curve PQ.
= Let the length of PQ is equal to ds.

IR From Fig. 12.2 ( b ), we see that

M Angle PCQ = chp
PQ = ds = if.ahp
Substituting the value of R in equation (i), we get
—R= ds
—M8« x EI dip

The equation (12.1) gives the central deflection of a beam which bends in a circular arc. But if x and y be the co-ordinates of P, then .-(»)
...(ii)
Value of Slope (0) tan ij; = ~dy—
...(Hi)
From triangle AOB, we know that ax

AC (2) L —dy
Sln 6 “ AO R 2R
sm =
Since the angle 0 is very small, hence sin 0 = 0 (in radians)
ds
L
—dx
—M2 x
cos ip =
M Ly. as

AEquation (12.2) gives the slope of the deflected beam at or at B. Now equation (i) can be written as

sec

or =

(S)

1

514 STRENGTH OF' MATERIALS DEFLECTION of beams 515

Differentiating equation («) w.r.t. x, we get Differentiating equation (12.4) w.r.t. x, we get

sec2 i)i . d2y <*=EI^
dx
dx 2 dx dx 4

{<££ But ~dF” - w the rate of loading

dty d7x

dx 2 w = El^y ...(12.5)

sec ap dx 4

—Substituting this value of — in equation (in), we get Hence, the relation between curvature, slope, deflection etc. at a section is given by :
dx

2 3 Deflection =y

sec ip sec ip . sec ip sec \p

f d2y 1 d2y " (Wl S1°p e dy
dx 2
2 =i
\ /J
U-dx 2 —= El d2y
=-
Bending moment
sec 2 ip dx
Shearing force
l) —= El d 3y=-

Taking the reciprocal to both sides, we get dx

d2y d2y —d*y

_1 _ dx 2 _ dx 2 The rate of loading = El -r.
dx
R3 2 3/2
sec rp Units. In the above equations, E is taken in N/mm2
(sec \Jj)

d*y mm/ is taken in 4 y is taken in mm,

, x is taken in m.

dx^ M Nmis taken in and

(1+ tan 2 3/2 12.3.1. Methods of Determining Slope and Deflection at a Section in a Loaded
*p) Beam. The followings are the important methods for finding the slope and deflection at a
section in a loaded beam :
For a practical beam, the slope tan ip at any point is a small quantity. Hence tan2 ip can

be neglected.

...av) (t) Double integration method

R dx 2 ,, (ii) Moment area method, and

From the bending equation, we have ...(a) (in) Macaulay’s method

ME -(12.31 Incase of double integration method, the equation used is
=

1R M —4 M, = E„lr —da2y- or d 2y =
M1
——o0r
=
—First integration of the above equation gives the value of
R El

Equating equations (iv) and (u), we get or slope. The second inte-

M d2y gration gives the value of y or deflection.
The first two methods are used for a single load whereas the third method is sued for
EI~ dx 2
several loads.
M^EI^4
dx 1 12.4. DEFLECTION OF A SIMPLY SUPPORTED BEAM CARRYING A POINT LOAD
AT THE CENTRE
Differentiating the above equation w.r.t. x, we get
A simply supported beam AB of length L and carrying a point load IV at the centre is
dM_ = EI ^LZ
shown in Fig. 12.3.
dx dx 3
RAs the load is symmetrically applied the reactions A and f?s will be equal. Also the
~~But = F shear force (See page 288)
maximum deflection will be at the centre.

...(12.4)

—

STRENGTH OF MATERSALS

—w= .

x0

4

Ais the slope at and is represented by 6 A

®xfiA = -

6a 16El

The slope at point B will be equal to 0A , since the load is symmetrically applied.

UL0B 0A - _ 16El ...(12.6)

Equation (12.6) gives the slope in radians.

Deflection at any point
Deflection at any point is obtained by integrating the slope equation (iii). Hence inte-

grating equation (iii), we get

— — —Hxy W=
x3 WL2 x + nC2 -A,iv))

.

43 16

where C is another constant of integration. At A, x = 0 and the deflection (y) is zero.
2

Hence substituting these values in equation (if), we get

E/x0 = 0- 0 + C
2

Substituting the value of C in equation (to), we get
2

——El x y = VV—* 3 WL2 .x (u ;
12 16

The above equation is known as the deflection equation. We can find the deflection at
any point on the beam by substituting the values of x. The deflection is maximum at centre

L L
~point C, where x = represents the deflection at C. Then .. x — and, y — yc
• Let y- ^
substituting

in equation (o), we get

^Elxy 16~ X
12 2

WL3 Wl _ WL3 - 3WL3
'}_
~96
32 ~ 96

(Negative sign shows that deflection is downwards)

WL3

Downward deflection, y =

2

51 8 STRENGTH OF MATERIALS

A mProblem 12.1. beam 6 long, simply supported at its ends, is carrying a point load of
mm50 kN at its centre. The moment of inertia of the beam (i.e. I) is given as equal to 78 x 106
4

.

E mmIf for the material of the beam — 2.1 x 10s N/ 2 calculate : (i) deflection at the centre of the

,

beam and (ii) slope at the supports.

Sol. Given :

Length, L = 6 m = 6 x 1000 = 6000 mm

W NPoint load, = 50 kN = 50,000

mm/ = 78 x 10s 4

Value of E = 2.1 x 105 N/mm2

Let yc = Deflection at the centre and

0A = Slope at the support.

(i ) Using equation (12.7) for the deflection at the centre, we get

WL3

y<: ~ 48El

50000 x 6000 3
~ 48 x 2.1 x 10 5 x 78 x 10 s

= 13.736 mm. Ans.

(ii) Using equation (12.6) for the slope at the supports, we get

eB _ eA WI}
16 El

WL ((Numerically)

"= 16El

50000 x6000 s - radians
e
16 x 2.1 x 10 5 x 78 x
10

= 0.06868 radians

~= 0.06868 x degree ('' 1 radian = : degree

= 3.935°. Ans.

WProblem 12.2. A beam 4 metre long, simply supported at its ends, carries a point load

at its centre. If the slope at the ends of the beam is not to exceed 1°, find the deflection at the

centre of the beam.

Sol. Given :

Length, L = 4 m = 4000 mm

WPoint load at centre =

Slope at the ends, 0A = = 1° = 0.01745 radians
’
a 180

Let y = Deflection at the centre

Using equation (12.6), for the slope at the supports, we get

_ WL2 (Numerically)

A 16EI

or 0.01745 =

-f aa

520 STRENGTH OF MATERIALS deflection of beams 521

W ^Cci -- EFlIXx R6C a6- -

CSubstituting the value of in equations (i) and (iii), we get
x

—El W.b . X2, + El X 0„„c W.b. 2 ...(iv)
dx 2L
:

2L

Ely W.b x, f_ T W.b. 2
6L
} ...(»)

2L J*

Fig. 12.4 The equation (iv) gives the slope whereas equation (v) gives the deflection at any point

X(a) Now consider a section at a distance x from A in length AC. The bending moment in section AC. But the value of 0 C is unknown.

at this section is given by. X(b) Now consider a section at a distance x from A in length CB as shown in Fig. 12.5.

Mx = Ra xx Here x varies from a to L. The B.M. at this section is given by,
M Rx = a.x - W(x - a)
W —=
(Plus sign due to sagging) — .b . x - W(x - a)
-y-
L

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

M -El -d2~y
dx 1

Equating the two values of B.M., we get

„EI d 2 Wxb *X

y L

~dx 2 ~

Integrating the above equation, we get

—— —WEl ~dy~ =
x b x2 - ... Fig. 12.5
x + C,
dx L 2 1 ...U)

where Cj is the constant of integration. But B.M. at this section is also given by equation (12.3) as

Integrating the equation (i), we get M —- Eld2y

W — ——Ely = — dx 2
b x3 C+ C, .x +
192 ...(H) Equating the two values of B.M., we get
.

2L 3

where C is another constant of integration. The values of C, and C are obtained from bound- EI^1.= W-~ .x-W(x-a)
2 2
dx 2 L
ary conditions. Integrating the above equation, we get

(i) At A, * = 0 and deflection y = 0

Substituting these values in equation (ii), we get dyjvj>^_mx-a)2

0 = 0 + 0 + C, E1 dx L 2 2 3

C = 0 where C is the constant of integration.
2 3

Substituting the value of C in equation (ii), we get Integrating the equation (vi) again, we get
2

El.y - W.b . x3 + C, x. E„lr.Jy ~ W.b x3 W (x-a)3 „ C„ ,
3 ^,
...(iii) 2L . . + CrX + ...(vu)

23

—(ii) At C,x = a and slope where C is another constant of integration. The values of C and C are obtained from bound-
4 3 4
= 6 C . (Note that value of 0C is unknown).
ary conditions.

The value of C is obtained by substituting these values in equation (i). Hence, we get (i) At B, x = L and y = 0. Substituting these values in equation (vii), we get
1
„ W. b a2 „ W.b If W_ (L-af + C x L + C
34
2' 3

a) a

.

deflection of beams 523

The deflection (i.e., y at any point in CB is obtained by substituting the values of C and
3

C. in equation (uii). Hence, we get from equation (vii).

w-—(x„, -a)8 + EI.Bc W.b.a 2

-.x 3 2L

W ab

+ (a-b) - El . L . Qc

The deflection at the point C is obtained by substituting x = a in the above equation. Let

yc = the deflection at C. Hence, we get

E„Lryc = W.b.a 3 W, , 3 + f p/fl iy - 6 - q2
al Q.
~6L
2L )

Wab

+ (a - 6) - EI.L.QC

"

1 W.b.a 3 - 0 + El . a W.b.a* W.a.b

9c - (a - b) - EI.L.

El 6L

l|'W.b.a 3

El 6L

|

The deflection at the point C can also be obtained by substituting x = a in equation (v).

Hence, we get

W.b.a3 W.b.a 2
+ EI.Q C - 2L
£/.y c =

’

1 W.b.a 3 W.b.a c

+ EI.Qr-a
El 6L

Equating the two values of yc given by equations (A) and (B), we get

W1 .6. 3 W.b.a3 '] W1 f 6. . 3 W.b.a 3 W.a.b
2L
+ EI.Br.-a - 3
El 6L 2L El 6L
j [

+ El .a .Bq ~ EI.L. 0^]

W .a. b

or 0 “ 3 - (a - b) - EI.L.BC

——W.a.b
or EI.L. () 0r = - {a - b)
3

W.a.b •• (12 ' 8)

or °c ~ 3EI.L

_

The above equation gives the value of 0 C (.i.e., slope at point C). Substituting this value
of 9C in equation (iu), we get the slope at any point in AC. Hence, we get from equation (iv),

Wr, T dy .b ? W.a.b W.b.a 2
,

dx 2 L 3 El . L 2L

W.b , W.a.b. ,, W.b.a 2
2L 3L 2L

- a aa a—

524 STRENGTH OF MATERIALS deflection of beams

[3x2 + 2a{a — 6) - 3a2] [3x 2 - ( 2 + 2ab)] = 0
3x2 - (a2 + 2a6) = 0
= 2 - 2 ab - a2] ...(C) f WJ
[3s: a 1 + 2ab cannot be zero
6L
1* bEH

As the length AC is more than length CB, hence maximum slope will be at the support

A — AA, where x = 0. Let the slope at is represented by 6 .. Hence
at will be equal to 8,.

Substituting x = 0 in equation (C), we get a 1 + 2ab

^f _ YWL.db „ x 0 - 2a b - a 2 Substituting this value of x in equation (D), we get maximum deflection.

6L [3 ]

El.b. = - a2) W.b fa2 +2ab') a 2 + 2a6 (a" + 2ab)
^
-6rLe- (v - 2ab bEIL

W- ba.’. ' ( 3 j

or 6, = 6EL L (a + 2b) ...(12.9) mW.b f (a 2 + 2ab) (a2 + 2abf'2

[Negative sign with the slope means that tangent at the point A makes an angle in the bEIL 3 x V3 V3

[ ~

anti-clockwise or negative direction].

2 +2a6) 3/2 1 1

Value of Maximum Deflection .(a V3_
^bEIL
Since ‘a’ is more than ‘b’ hence maximum deflectipn will be in length AC. The deflection 3 x -J3
= Af_Ai (a 2 + 2a6) 3/2
ACat any point in length is given by equation (u) as

W.b. 2 bEIL 3j3

6L Elr, r . „0 C x = - ™- b - (a 2 + 2ab)312

{ 2L 9 WELL

-wXW.b X3 + W.b. 2 Negative sign means the deflection is in downward direction.
£„/T W.a.b,“- lx
{ 6) mymax =

-JL 2L J^g
Downward, f(a 2 + 2ab6)3/a ...( 12 . 10 )

W.a.b L

(a - b) from Eq..( 12 .8 ) Deflection under the point load

W.b W.a.b, W.b. 2 Let yc = Deflection under the point load
2L
= 3 + [-3L- (a - 6) X ACThe deflection at any point in length is given by equation (D), as

J Wh

W.b 3 3a 2 yc= bEIL k3 -*(a2 + 2“6 «
L= [.v - -
+ 2a (a b)x . x] The deflection under the point load will be obtained by substituting x = a in the above
6

Wb equation.

- [x3 + 2a2x - 2afa - 3a2x] y° = Wb [[“a 3 ~- °a(( “a 22 + 2“ab6))]

bL

— —W.b W.b -

= bLr [x 3 - a2x - 2abx] - bLr [x 3 - x(a2 + 2a6)] bEIL

Wb = ^EiZ l[a33 - a33 - 2aa22bb]
[x 3 -x(a 2 + 2ab)]
bEIL W.b ,~ 2„ a,bl)x = Wa 2b 2
6EIL*
—The deflection will be maximum if = 0 ~lML= C
ax
Negative sign means the deflection is downward.

dy W.b Downward,, —Wa 2b 2 ...( 12 . 11 )

[3x2 - (a2 + 2ab)] yc =
dx bEIL

—For maximum deflection, =0 Note. The above method for finding the slope and deflection is very laborious. There is a simple
method of finding the slope and deflection at any point in a beam. This method is known as Macaulay’s
dx
method which will be discussed later on.

526 STRENGTH OF MATERIALS

Problem 12.4. Determine : (i) slope at the left support, Hi) deflection under the load and

(Hi) maximum deflection of a simply supported beam of length 5 m, which is carrying a point
m mmload of 5 kN at a distance of3 from the left end. Take E = 2 x 105 N/mm2 and I = 1 x 10s 4

.

Sol. Given : L = 5 m = 5000 mm
Length, W = 5 kN = 5000 N
Point load,

Distance between point load and left end,

a = 3 m = 3000 mm
b = L- a = 5 - 3 = 2 m = 2000 mm

Value of E = 2 x 10s N/mm2

mm/ = 1 x 108 4

Let 0A = Slope at the left support,
yc = Deflection under the load, and

ymax = Maximum deflection.
(£) Using equation (12.9), we get

IV. a. b ,
0a = --6ELL +
{(ao 2b)

5000 x 300 0 x 2000
= - _ _ ^x (3000Q + 2 x 2000) (ra dians)
10 s 8
6 x 2 x x x 5000
10

= - 0.00035 radians. Ans.

ANegative sign means that the angle made by tangent at is anti-clockwise.

(it) The deflection under the load is given by equation (12.11), as

Wa 2 .b z

5000 x 3000 2 x 2000 2

3 x 2 x 10 5 x 10 s x 5000 = 0.6 mm. Ans.

(Hi) The maximum deflection is given by equation (12.10), as

W 0b (a2 + 2ab)312

yymax = ff-

973 ELL

5000 x 2000
(3000 2 + 2 x 3000 x 2000)372

9 x 73 x 2 x 10 5 x 10 8 x 5000

W_ (9000000 + 12GOOOOO)3/2
9 x 73 x 10 10

= 0.6173 mm. Ans.

12.6. DEFLECTION OF A SIMPLY SUPPORTED BEAM WITH A UNIFORMLY DIS-

TRIBUTED LOAD

A simply supported beam AB of length L and carrying a uniformly distributed load of w
Aper unit length over the entire length is shown in Fig. 12.6. The reactions at and/i will be

wxL

— —equal. Also the maximum deflection will be at the centre. Each vertical reaction - - .

Zi

: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.

deflection of cantilevers 575

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

Nmm= 7 x 1013 3

Substituting this value in equation (i), we get

- = 7 mm. Ans.

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

kNm

B.M. Diagram

dx

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

rectangle. 0 B = 0c = 2 El

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

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

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

6EI SEI

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

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

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

)

L = Length of beam,

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

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

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

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

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g Ej

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m

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

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

conjugate beam method determine :

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

Sol. Given :

Length, L=4m 1~

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

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

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

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

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

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

W= = = Q -000225 rad- Ans-

27io“8 7 io :*

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

0 C = Shear force at C* for conjugate beam

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

EI 2 EI El

= = °-°0015rad- AnS '
2 x 10 s^x 10'4

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

(ii) Deflection under each load

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

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

Now according to conjugate beam method,

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

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

2 x 10 s x 10' 4

; 0.2 mm. Ans.

Also yD - 0.2 mm.