Rigid Body:
A rigid body may be defined as a body in which the relative
positions of any two particles do not change under the action of
force.
OR
A rigid body may be defined as a body which can retain its size or
shape even if it is subjected to force.
Deformable
body:
A body which undergo certain deformation under the
action of force.
Force: _
Itis the action of one body on another tending to produce a change in its
state of rest or of motion. It is a vector quantity.
Characteristics:
LoA
F(magnitude)
g
Direction
PoA
Magnitude (F): It is the number (say 10 N)
Line of Action (LoA): It is the infinite line along which the force acts
Point of Application (PoA): It is the point where the force acts
Direction (0): It is the angle of inclination of LoA with the reference axis
Principle of Transmissibility:
The principle of transmissibility of a force states that its point of application
may be transmitted to anywhere along the line of action without changing the
conditions of equilibrium.
OR
The state of rest or of motion of a rigid body is unaltered if a force acting on
the body is replaced by another force of same magnitude and direction but
acting anywhere on the body along the line of action of the replaced force.
-
F
-
o
-
System of Forces:
When several forces of different magnitude and direction act upon a body they
constitute a system of forces.
When the line of action of all forces of a system lie in single plane the system
is said to be Coplanar whereas when the line of action of all forces of a system
do not lie in single plane the system is said to be Non-Coplanar.
When the line of action of all forces of a system passing through a common
single point is said to be Concurrent whereas when the line of action of all
forces of a system do not pass through a common single point is said to be
Non-Concurrent.
Examples:
y
y
Fy
Z
sr,
1
F,
x
Coplanar — Concurrent
x
Coplanar — Non Concurrent and
Parallel
Non- coplanar - parallel
Non- coplanar — Concurrent
y
F,
>*
Fy wt
<_
F,
Z
Non- coplanar — Non concurrent
Free Body Diagram (FBD):
It isa diagrammatic representation of the body which is isolated from all the contact
surfaces, and showing
only the internal (reaction) and external
VIIP TATA
TT TTT TT
3
Physical Diagram
th
Free Body Diagram
AAS
AAS SAY AVA
ON
ARN
acting on the body.
(applied) forces
COPLANAR
Resultant:
CONCURRENT
FORCE
SYSTEM
It is a single force which acting alone can produce the same effect
on the body as that due to the given system of forces. The method
of finding the resultant is called as Composition of forces.
Composition of system of forces:
It is the process of replacing a given system of forces by a single
force [Resultant]
such that this single force produces the same
effect on the body as that due to the given system of forces.
Resolution of force:
It is the process of replacing a given single force by two or more
forces [components]
such that they together produce the same
effect on the body as that due to the given single force.
F. =F sin 0
0
A
F =F cos 0
Wossseeweel
Resolving the Force in to Rectangular components:
Graphical Method to determine Resultant:
Parallelogram Law:
If two forces acting simultaneously on a body at a point are
represented in magnitude and direction by the two adjacent
sides of a parallelogram,
their resultant
is represented
in
magnitude and direction by the diagonal of the parallelogram
which passes through the point of intersection of the two
sides representing the forces.
Triangular Law:
If two forces acting simultaneously on a body at a point are
represented in magnitude and direction by the two sides of a
triangle taken
in order,
their
resultant
is represented
in
magnitude and direction by the closing side of the triangle
taken in opposite order.
Polygon Law:
If number of concurrent forces acting simultaneously on a
body are represented in magnitude and direction by the sides
of a polygon taken in order, their resultant is represented in
magnitude and direction by the closing side of the polygon
taken in opposite order.
Analytical Method to determine Resultant:
(a) When two concurrent forces acting:
Considering two forces F; and F2 acting on a body as shown. Let the angle
between them be 8. The diagonal AC of parallelogram ABCD represents the
resultant of F; and F.
R=AC
= (AE?
+ CE?)
AB=F,;
=
BE=F>2cos 0;
/(AB+ BE)? + CE?
CE = F2sin@
R=/(F, + F,cos0)2 + (F,sin@)?
R= Fi + F5+2F,F,cos0
.
ana=
CE
—
AE
Wheno =909
Whenod =0°
When
= 1809
F,sin@
= ——
F,+F,cosé
R=
JF?2+ Fe
R=Fit Fo
R=1
Fi - Fe!
(b) When number of concurrent forces acting:
F,
Foy
Fiy
A
|
Let Fi, F2,F3 and F4 be system of 4 forces acting as shown. Let Fix, Fax, F3x and
Fa, and Fiy , Fay, Fay and Fay are the components of Fi,
F2,F3 and Fain ‘x’ and ‘y’
direction respectively.
Algebraic sum of components of all the forces in ‘x’ direction
Fy = Faxt Fox+ Fax+ Fay
Algebraic sum of components of all the forces in ‘y’ direction
DFy = Fayt Fay+ Fay+ Fay
Since, XF, and =F, are the two forces perpendicular to each other and
are the rectangular components of the resultant in ‘x’ and ‘y’ direction
respectively.
1
I
DF,= R, = R sina
!
\a
5I
iF,= R, = R cosa
R=
{(2F,)?
+ (ZF)
a= tan-*|
Fy
F|
Equilibrium of a Body:
A body is said to be in equilibrium when it is at rest or continues to be in steady
linear motion. According to Newton’s law of motion it means that the resultant
of system of forces acting on a body in equilibrium is zero. In other words, the
net force acting on the body should be zero for equilibrium.
Mathematically,
XF =0
The above equation can be written as
ZF, = 0
ZF, =0
Where F, & Fy are the horizontal & vertical component of the system of forces
Equilibrant:
If another force acting on the body which is equal in magnitude but
opposite in direction to the force causing the motion then such a
force is known as Equilibrant. It is equal and opposite to resultant
force.
Lami’s Theorem:
If a body
is in equilibrium
under
the action
of three
concurrent forces then each force is proportional to the Sine
of the angle between the other two forces.
P
P
Q
S
Sina SinB Siny
Determine the resultant of system of forces acting on a body as shown in Fig.
200 N
400 N
3
4
\ a3”
a
3
30°
15ON
1
270 N
Solution:
Tan 0; = % 0; = 36.87°
> Fx
(——+)
= 400 sin 55° + 200 sin 36.87° — 150 cos 30° - 270 sin 18.26°
= 233.1N (J)
R=
Q2 = 18.26°
= 400 cos 55°- 200 cos 36.87° — 150 cos 60° + 270 cos 18.26°
=250.8N
> Fy
Tan 02 = 1/3
[(oF x )2 + ( (IF y)
_
-1
a=tan
[2%
=|
R=,
2
"A
=
342.44N
_=
42.9°
oR
AK
i CR
= SE
The resultant of the forces acting on a block on an inclined plane as shown in figure,
is parallel to the plane. Determine the force P and the Resultant.
20 kN
Solution:
= forces parallel to the incline plane = R
— 20 sin 30°—P sin 15° ------------- (1)
R = 30 cos 30°
= forces perpendicular to the incline plane
i.e.
=O
-20 cos 30°— 30 sin 30° + P cos 15° =0
P = 33.46 kN
Substituting for P in equation (1) we get,
R= 7.32 kN
(Ris parallel to the incline plane)
ed to start the wheel
Determine the amount and the direction of the smallest force P requir
e. What is the reaction at the block?
of weight W=2000N over the block as shown in figur
Shs
Solution:
P
Fo«u
+o be
Forw CP) and
th
leart,
th
ReactromCR)
Sine = Seo
betinee 2
ong
hes te be To
6: aes
,
rmine the
N, respectively as shown in Fig. 2. Dete
The smooth disks D and E have a weight of 200 N and 100
up
of disk E without ca using the disk D to move
center
the
to
applied
be
can
that
P
force
tal
largest horizon
the incline.
Solution:
QOOonN
Lpraemequeear
100N
fer
isk
©
R
Rpe
Sim
Roe
:.
200
Sin 126-87
131-6
Roe
=
2I4N
W53
J
Le
D
Equikbsivn
—b&
2
Disk
ors
4ve
:
oe
Rre
p-
Cosiiss
—
209-68N
Peo
FEELS
Two cylinders A & B rest in a horizontal channel as
shown in figure. The cylinder A has a weight of 1000
AAPLAASSA AEA
N and the radius of 90 mm. The cylinder A has a
weight of 400 N and the radius of 50 mm. Determine
the reaction at the contact surfaces.
Solution:
y = 90 tan 30° = 52 mm
JS
x =180 — 50-52 = 78mm
o”
SY
LLAS SSL
8 = cos! (78/90) = 56.14°
ae
ge?
~
R
A2
SALSA,
y
/
oy
ppl tSAL
R Al
Ae
SSS
x
180
fo
400 N
Cylinder B
D=Fy,=0;
Rag cos 56.14°-Rg=0
XFy=0;
Ras sin 56.14°- 400=0
Ras = 481.9 N
Re = 268.5 N
Cylinder A
ZF,=0;
Raz
XFy=0;
Raz sin 60°— Rag cos 56.14°=0
=310N
Rai—1000- Rag sin 56.14°+ Raz cos 60° = 0
Rai = 1248.2 N
The man attempts to pull the log at C by using the three ropes as shown in Fig. 2, Determines the direction 0
in which he should pull on his rope with a force P = 80 N, so that he exerts a maximum, force on the log.
What is the force on the log for this case? Also, determine the direction in which he should pull in order to
maximize the force in the rope attached to B. What is this maximum force? Take ol Se.
Solution:
FaB:
Peorme
Sim Co-40)
—CoLO Cos (g-4o'
CosC¢-4o')
Faa:
—
PCot
CO +o-F0)
Ce. Cg-4Fo )
Te
Ordy
to
Maxtentse
i-@
+R
@©+6-40
@=
FAB
=
l60N
Fore
= 180
IZo
CosCe+o-Fo)
=1
the rope which is knotted at point A as
Romeo tries to reach Juliet by climbing with constant velocity up
of 2 kN before it
shown in figure, Any of the three segments of the rope can sustain a maximum force
rope, and if so, can he along with
breaks. Determine if Romeo, who has a mass of 65 kg, can climb the
his Juliet, who has a mass of 60 kg, climb down with constant velocity?
If the tension developed in each of the four wires is not allowe
d to exceed 600 N, determine the
maximum
mass of the chandelier that can be Supported as shown
in Figure,
decorative ceiling-mounted light fixture]
[A chandelier is branched.
Solution:
FRD
@
,
TJetmt
D
a
+Ve
Shr-0
;
Es
>
Iep
COS30—
>
Tea
S$tm 30
tu
!@p
:
Cos4s
;
{zry-0
=O
—@)
:
:
+ Tap Stw4S
~ Mx4-81=0
L(2)
GQ)
Solutms
Ten
4
@,
7-18
™
rn (4-81)
lene
FRD
®
Formt
2-79
mM
B
+ve
TAB
A
;
oe
»
Tag
TAB?
Tec
Te
obfexve
‘
Sim4s-o
12-44mM
3 Taalel 304+ Tec + Tap los4s =0
Tee = 4455 m
foorm
“Kes
Tan
—Ptve
ZFRH=C
Ne
Sfm30'—
xeoitutts
GOON=
wT) =
(anes
48
that
12-4411)
-24
a
Tae
har
I
\-
ten
Coron?
COPLANAR
NON-CONCURRENT
FORCE
SYSTEM
Moment of a force:
The moment of a force about a point represents the tendency to rotate the
moment arm (or the body on which the force acts) about an axis perpendicular
to the plane which defines the force and its moment arm. It is a measure of
rotational effect of the force. It is determined as the product of magnitude of
the force and the perpendicular distance (moment arm) from the point to the
line of action of the force. The point about which the moment is considered is
called moment centre and the perpendicular distance from the point to the
line of action of the force is known as moment arm.
AN
Zz
y
GM,
Magnitude of Moment: Ma=F xd
Unit: kN-m or N-m
Sign Convention:
Anticlockwise
Clockwise
-ve
+ve
Theorem of Moments — Varignon’s Theorem:
Algebraic sum of the moments due to system of coplanar forces
about a moment centre in their plane is equal to the moment due
to their resultant force about the same
moment
A
centre.
X
Let ‘R’ be the resultant of ‘P’ & ‘Q’ acting at ‘A’. Let ‘B’ be any point in its plane.
Let d, di & d2 are be the perpendicular distances from point ‘B’ to line of
actions of forces R, P & Q respectively.
According to the statement of theorem,
Rd=Pdi+Qd2
Select x-axis to be the line normal to AB and y axis along the line AB. Let 0, 01
& 02 are be the angle made by the forces R, P & Q respectively with the x-axis.
Moment due to Resultant ‘R’ about point B =Rd
=RABcos 0
= AB (R cos@)
Moment due to Force ‘P’ about point B
=Pdi
=P AB cos 01
= AB (P cos@:)
= AB P, ---------------(2)
Moment due to Force ‘Q’ about point B
=Qd)
= Q AB cos 62
= AB(Q cos@2)
Adding (2) & (3) we get,
=Pdi+Qdp2
= AB (Px + Q,)
= AB (Rx)
From (1)
Hence,
=Rd
Rd=Pdi+Qd2
Couple:
A couple is a pair of equal and oppositely directed forces. In the Fig. the
perpendicular distance ‘d’ between the line of action of forces is called
the moment arm of the couple.
F <—_—_——————_
Properties:
(i)
A couple consists of a pair of equal and opposite parallel
forces separated by a definite distance.
(ii)
The translatory effect of a couple on a body is zero (i.e.
vector sum of these two forces is zero).
(iii)
The rotational effect (moment) of a couple about a point
is a constant
and
it is equal
to the
product
of the
magnitude of the force and the perpendicular distance
between the two forces. It can be rotate through
any
angle and shift to any position. It is independent of point
of application.
(iv)
A couple is represented by the magnitude of moment
it has only rotational effect.
as
Resolution of Force in to Force-Couple system:
A force acting at some point ‘A’ and can be moved to a point ‘B’ (away/across
from its line of action) provided a couple is added. The moment of a couple to
be added is equal to the moment of force about the new point ‘B in its original
position ‘A’.
Consider a force F acting at ‘A’ as shown in Fig.(a). According to the principle
of transmissibility the force F can be moved to any other point on its line of
action. But it can’t be moved to a point ‘B’ away from its line of action without
changing its effect. But we can apply two forces acting at ‘B’ in the opposite
direction and of same magnitude as shown in Fig. (b). Now, we could see from
the Fig. (b) that the pair of force ‘F’ forms a couple. Since couple is fully
represented by its moment we can write it as Force ‘F’ acting at ‘B’ along with
the moment Mg, as shown in Fig. (c).
Fig. (b)
4m
a
200 N
400 Nm
by
&----------->
Zz
jo
=
=
Example
Resultant:
It isa single force or pure moment or force-moment which acting
alone can produce the same effect on the body as that due to the
given system of forces.
ou
P
T
oo
M, a?
0
R=
| (2F,)? + (2F,)”
Let Ry, = XF, and Ry = XFy be the components of the resultant R in ‘x’ and ‘y I
directions.
Considering the moment of R about ‘O’
R,. 0+ Ry.x=5 Mo
Also, R.d =X Mo
=rM,
<=M,
=M,
=M,
R,
=
R,
LF
Four forces are acting on the truss as shown in fig. Determine the equivalent single
force
_ acting on the truss and the point of intersection of its line of action with a line
through
points A and G.
oe
; ie
d
iy )
240
Co$7o0
—
300
Cos40
= —!147°73N
= 147-732N
240 Sind 160-300 Sera ~180
ie. — 758: BEN
«=
758:3EN
(<)
Ct)
Z Fx
R
ZFy
Th
the
Powses
ute.
— Puoseettont -
Of
Mme
or’ San
er
thy taro
LI
ane. syn’ 27
:
Retertorn®
rhe
ys
ey
Z asTv=O 3ae pat ~- 9081 080 iN
Aee x
K
awe Seeeeee
sah va
Stnéo
Ee
oe
imesh
als
tw fd
‘2143
Replace the three forces which act on bent bar by a force-couple system at
the support point A. Then determine the x-intercept of the line of action of
the stand-alone resultant force R. All the loads are in Newtons and distances
are in Meters
Soluttor
=p
REE.
A4A00N
(—>)
seoacose26 -aaco-t2ce= ser? C4)
Loo
Sinzsco
=
3
f°
= ef it
Seer
se
SO2SS
aa
Ny Ae
The pedal-chain wheel unit of a bicycle is shown in the Fig. The left foot of the rider exerts the 40 N force,
while the use of toe clips allows the right foot to exert the nearly upward 20 N force. Determine the
equivalent force-couple system at point O. Also, determine the equation of the line of action of the system
resultant treated as a single force R. Treat the problem as two-dimensional.
Line
of
act fon
Rxed=
a:
aA:
of
Stay
Fore
CR)
=Mo
3576
\
22:28
D\
l60 mm
Equilibrium of a Body:
A body is said to be in equilibrium when it does not have any translatory or
rotatory motion in any direction. at rest or continues to be in steady linear
motion.
Mathematically,
xF = 0-—Translatory motion
x=M =0-Rotatory motion
The above equation can be re-written as
ZF,
0
Fy =0
=M=0
Types of Supports & Reaction:
Beam is defined and a horizontal structural member which has one dimension
(span
or
length)
considerably
larger
than
other
two
dimensions
(cross-
sectional). The beam is supported at their end with different types of supports.
These beams transfer the loads (including its self-weight) to the supports.
Roller Support:
In this the reaction is normal to the support (as the
(
)
rollers are frictionless). The
beam/member
is free to
move along the support i.e in horizontal direction (H=0)
and
Vv
Vs
rotate
about
the
support
(M=0).
Thus,
Roller
support has only one reaction i.e. in vertical direction
Reaction : 1(V)
M =0;
H=0
Hinged or Pinned Support:
At
H —2_
hinged
support
the
beam/member
cannot
move
or
translate in any direction as it has reactions in vertical and
horizontal directions.
about the support
But, the beam/member is free rotate
(M=0). Thus, hinged or pinned support
there are two reactions i.e. in vertical & horizontal direction.
V
Reaction : 2 (V & H)
M=0
Fixed Support:
M
At fixed support the beam/member neither translate in any
direction
nor
rotate.
The
translation
is
prevented
by
developing support reactions any direction while the rotation
is prevented by developing the support moment (M).
fixed
V
support
there
are three
reactions
horizontal directions & a support moment.
Reaction : 3 (V, H & M)
Thus,
i.e. in vertical
&
BS
how
LLL LL Lee
Type of beam:
Cantilever beam
Simply Supported beam
o
3
Fixed beam
LLL
LL LoL
LLLSL
EEL
LZZLLLLLAL
Over Hanging beams
SB
Propped Cantilever beam
Type of loading:
Ww
Ww
a
Point or Concentrated Load
W=wxl
l/2
W=wxil
po
—
/
ea
a
l
or
Uniformly Distributed Load (UDL) or Rectangular Load
W=wxl/2
w unit/length
W=wxl/2
w unit/length
or
>
1/3
l
|
|
Pp
l
21/3
Uniformly Varying Load (UVL) or Triangular Load
W,
=x!
W,
w, unit/length
=0.5
(w,-wl
w, unit/length
Trapezoidal load
|
|
Statically determinate beam/structure:
A structure is said to be statically determinate if all the unknown
components
can be determined
using the equations
reaction
of static equilibrium
alone.
i.e. No. of unknown Reactions S Equations of Equilibrium
Ex: Cantilever Beam, Beam with one hinged and another roller support
Statically indeterminate beam/structure:
A structure is said to be statically indeterminate
if all the unknown
reaction
components cannot be determined using the equations of static equilibrium
alone.
i.e. No. of unknown Reactions > Equations of Equilibrium
Ex: Propped Cantilever Beam, Fixed Beam
Determine the required mass of
the suspended cylinder if the
tension
in
the
chain
wrapped
around the freely turning gear is
to be 2 kN.
Also,
what
is the
magnitude of the resultant force
on pin A?
=Ma= 0;
— 4 (2 cos30°) + W cos45° (2 cos30°) + W sin45° (2 sin30°) = 0
ee
W = 3.586 KN
te. i,
ae
m = 3,586 (1000)/9.81
m = 366 kg
4 —3.586 cos45°— Ha = 0
«Hy= 1.464 KN
The pin A, which connects the 200-kg steel beam
column, is welded
the beam
both to the beam
by exerting a 300-N
with center of gravity at G to the vertical
and to the column. To test the weld, the 80-kg man
force on the rope which
passes through
loads
a hole in the beam
as
shown. Calculate the moment M supported by the pin.
200x495
on :
SolutiLo
;
/
/
80x48
_y
Ma
i200
o
E06
eo
Boon
,
a
————
\ ae
YA
Vv
Shia e
3
200x481x1200
Mas
+
(8 0x9.81 + 300)x 1800
F
3o0o0ox2!00
Ma = 4:94x10° NMA
4ve
a
he
Va
=
(Zoe + 80)* 4:81
Vaz 3347 KN
+ Bo0ox2
figure.
Determine the reactions at the supports at A and B shown in the
“i
120 N/m
pS
360 N/m
9m
Solution:
Jae
4m
Re
120 Nim
Determine
the support reactions at A, B and C of the loaded
connected beam as shown
below.
SOkN /m
20KN /m
—>+ve
Zu=0
Zz Ma=o
VA
5
- Pace
;
Rax5+
50=
20x4x 3+ 3x4xB0x(it 2x4)
wr Se
ts
Pk oxen
bas AB
‘the 200 okg “loaet at A» Cabculate ATh
¢ a bk
and
ne
me ” Pecks
£ atx
O42
fs th
of 4 TH
FR
24)
Suppoots
TT.
4tenstom
the e
OS
and
Sok
mar
@
hat
wf
rte
0° ®="6o™'
I
BS COSBO
$0 SOKAxX-B
165
-— Tces6o x 1-25
.
=O
aM
x
:
#1
=
ae
a
=O
‘
He= 3823:
i
fine
om
oF N
«
=
—
(t)
2s5N
a
,
EV2s~
= 5656-72N
ssaEEREEEEeeeeee
@&
ported
AC 2100 mm high (h), sup
el
pan
ar
gul
tan
rec
m
for
age door consists ofa uni
A 70 kg (W) overhead gar
by two sets of frictionless rollers at
and
r
doo
the
of
e
edg
er
upp
the
of
dle
by the cable AE attached at the mid
s
cither side of the door. The rollers A are tree
A and B shown in figure-2. Each set consists of two roller one
ition
door is held s.in pos
ded by vertical channels. If the the
Assume
ler
rol
r
fou
to move in horizontal channels, while rollers B are gui
of
h
the reaction at cac
sion in the cable AE. (b)
ten
the
(a)
e
in
rm
te
de
for BD=1050 mm,
a
1050 mm. b= 700 mm.
©
=
Sin
oso s
lo
=) G
is supported by the
AB as shown in Figure -3. which
jib
the
of
ists
cons
m
boo
ne
cra
lhe upper portion of the
mast at C. If the
cable being separately attached to the
kstay CD. each
pin at A, the guy line BC, and the bac
ine the magnitude of
ch passes over the pulley at B. determ
whi
,
line
t
hois
the
by
ted
por
sup
is
KN
load F= 5
the tension
tension in the guy line BC. and
the
um,
bri
ili
equ
for
A
at
jib
the
on
the resultant force the pin exerts
1.5 m, ¢
a radius of 0.1 mm. Consider b =
has
B
at
ey
pull
The
jib.
the
of
ght
T in the hoist line. Neglect the wei
=m.
Solution:
Two smooth cylinders A and B rest on an inclined plane and are supported by a
vertical lever CD as shown in figure shown below. The lower end C of the lever
is pinned, while, the upper end D is supported by a string DE. What is the force
exerted by cylinder B on the lever CD? Find also the reaction at hinged point C
and the tension in the string DE. Take radius of cylinder A & B as 0.1 m and 0.2
m respectively. Length of lever CD is equal to Im. Weights of cylinder A and B
are 500 N and 1200 N respectively.
7
ey
ee
oO
PEXT
aace,
B
aa
ty,
~
Soletfon:
ee
|
+on3d.= =BF
spe
/
Sphou A’
Stn Rub to Spheaa A’
Apply
500
Rae
Stw (50
S te 103-47
Rap: 265-16 N
500N
+ve
f=
es 04
R,Cosae
=
Ro
—
Ve
ZFx=O
—1200 — Rag Sf 10-53 =O
14416
N
§
Rap Cos loS3 + Re Stm3e4
Rp- 981-5N
—Re:0 as
cell with the palm
In a procedure to evaluate the strength of the triceps muscle, a person pushes down on a load
arm is 1.5 kg,
of his hand as indicated in the figure. If the load-cell reading is 160 N and mass of the lower
determine the magnitude and nature of unknown forces acting on the lower arm.
Humerus
Solution:
oO
150 mm
150 mm
| s
: - 25
1.5x9.81N
C
160 N
Free Body Diagram of Lower Arm
(
9°
N
.
Let
JT
-»
i
os
Tensfow
Comm pressive
oe eens
te
>)
1
a
T=
—P>
NM +
clin
Ree
to
a
tAfups
clear to
Hurmeut
0
7-3! X 150 — /60X 300=
/832N
Gaga
+ !60-0°
bore
The lever ABC is pin-supported at A and connected to a short link BD as shown in Fig. If the weight
of the members is negligible, determine the force of the pin on the lever at A.
The force exerted by the plunger of cylinder AB (Fig.2) on the door is 40 N directed along the line AB, and
this force tends to keep the door closed, Compute the moment of this force about the hinge O. What force
Fc normal to the plane of the door must the door stop at C exert on the door so that the combined moment
about O of the two forces is zero.
A handle for pushing a cart is shown in the following Fig. The handle has the feature that it may be easily folded
against the side of the cart when it is not needed, Determine the forces supporting handle ABC when the 50N
force is applied.
Solution:
50N
Ha
1. Asan airplane’s brakes are applied, the nose wheel exerts two forces on the end of the landing gear as
force
shown in Fig, Determine the horizontal and vertical components of reaction at the pin C and the
in strut AB.
6kKN
Solution:
ZFneeO0
5
2000
+
.
fry = 0;
6000 +
Ree
Cos40
—He=0
[A= 266r6N
Raw
Sin40 — “=O
655515
N
from a pin at endD of the
An electrical worker stands in the bucket that hangs
worker and the bucket
boom of the cherry picker as shown in figure 2. The
hs 1.1 N/m, and between
together weigh 200N. Between A and C, the boom weig
CD are uniform beams.
C and D, it weighs 0.8 N/m. Assume that AC and
B._
Determine the reactions at pins A and
+ve
8.8 CoS20x4
200 CoS2e0 XISB
—
Re Stm4ox4-
Re COoS40 x1 + 4C0S20X105 +
=O
“. Re = 753-22N
(7)
—rive
ZFx=O 3
Ha + ¥53»22 CoS4o
.
Hat
- 4 C0876 - 200COSFo- 8:8COS7O=0
504°23N
(+)
shown
with a force P that is parallel to the incline as
The man pushes the lawn mower at a steady speed
1s,
0
bag is 50 kg with mass centre at G. If
in Fig. The mass of the mower with attached grass
e
s B and C. Neglect friction. Compar
l forces Ny and N; under each pair of wheel
determine the norma
and P=0.
with the normal forces for the conditions of 0=0
Solution:
= F=0
x =
2 Fue.
d°°
5
P—
>
50x9.81
:
=O
—
Q)
Na+ Ne - 50x9.81C98
1S =0 —@)
e705 5
PCA900) ~ S0x9-81 ( Simi (215)+ cos15 (S
00))
=f1--0
+ Ne Coo)
Cas
Stmis
=O.
—
G)
|
P
Nee
127-N[¢=
214N
Net
Q26on
When
Np
Ne
=
"W
Cax@)
350
@20
N
I40N
3
Pso
The lift bar (AC) of a tow truck is modeled as a pinned beam. The hoist cable (CD) is assumed to experience no
friction as it goes over the top of the lift bar. Calculate the forces in the supporting cable (at 8) and at the hinge
of the lift bar (point 4). Assume the lift bar to be 4 m in length, ignore its mass and assume that the support cable
is attached two thirds of the way up the bar.
Solution:
C
if
=
ae
& 73 my
Ha.
Free Body Diagram of Lift Bar
z
Va
i
+ve
26, PigeO
pea
tT O?n 20x
T=
— rive
ZFy=OQ
3
+1500
267
on
X4 = —/150 0 Sind
S?m20
1O38N
1500 C0S20
|\Ha=
3B684
+7 CoS20 +4 |15p0 C0830 —HazO
N|
;
me
+ Ve
| = Fy Fux O03 :
1500 Str20
TS?nm20
+
— 1500
Sw
30+
VA-O
\A eS BT2N
=
ine
TSN (4)
Va
"|
ef
Va?
+
Ha
a
e,
Ra = B686N
A
qJan (- Be )
!
Jo... 183'] risc
Va
1.
Determine the required magnitude of force F as shown in Fig., if the resultant couple moment on the beam is
to be zero.
[F= 14.2 kN]
A skeletal diagram of the lower leg is shown in the figure. Here it can be noted that this portion of the leg is
lifted by the quadriceps muscle attached to the tip at A and the patella bone at B. This bone slides freely over
cartilage at the knee joint. The quadriceps is further extended and attached to the tibia at C. Using the
mechanical system shown to model the lower leg, determine the tension in the quadriceps at C and the
magnitude of the resultant force at the femur (pin), D, in order to hold the leg in the position shown. The
lower leg has a mass of 4.5 kg and a mass centre at G;; the foot has a mass of 2.3 kg and a mass centre at Gp.
[T = 1.01 kN; Ro = 982 kN]
75 mm
3.
A small hoist is mounted on the back of a pickup truck and is used to lift a 500 N crate as shown in Fig.
Determine (a) the force exerted on the hoist by the hydraulic cylinder BC, (b) the reaction at A.
[Rec = 5102 N; Ra = 4621N]
4.
The total weight of a wheelbarrow filled with gravel is 240 N. If the wheelbarrow is held on an 18° incline in
the position shown, determine the magnitude and the direction of (a) the force exerted by the worker on each
handle, (b) the reaction at C.
[Re =199.21 N ; F = 39.21N]
SIMPLE TRUSSES
Truss is a rigid structure consists of number of straight members connected at their
joints and designed to support loads.
Trusses are most commonly used in bridges,
roofs and towers. There should be a minimum of 3 members to form a truss. They
are classified based on the plane as follows:
Plane truss:
All the members of the truss lie in a single plane.
Ex: Roof & Bridge truss
Space truss: All the members of the truss do not lie in a single plane.
Ex: Tripod, transmission tower etc.
“Wb, 77S,
Howe
Pratt
Warren
-
-
Pratt
.
Baltimore
<
bd
K
Fink
-
-
a
=
i
Howe
~
Warren
=
Typical Bridge Truss
=
~
>,
The truss consists of straight members connected
at joints. Truss members
are connected
at their
extremities only; thus, no member is continuous
through a joint. There is no member AB; instead
there are two distinct member AD & DB (refer Fig.
A). The members of the truss are slender and can
support
little
lateral
loads;
all loads,
therefore,
must be applied to the various joints, and not to
the members themselves. The weight of the members of the truss are assumed to
be acting at the joints, half of the weight of each member being acted to each of the
two joints the member connects. The joints may be either Rigid joint (bolted or
welded
connection)
or Pin jointed. Although
the members
are actually joined
together by means of bolted or welded connections, it is assumed that the members
are pinned together, therefore, the force acting at each end of a member reduce to
single force and no moment. Thus, the only forces assumed to be applied to a truss
member are a single force at each end of the member (refer Fig. A).
Each member can then be treated as a two-force
member, and the entire truss can be considered as
a
group
of
pins
individual member
Fig. B. When
apart,
and
two-force
members.
can be acted upon as shown
An
in
the forces tend to pull the member
it is in tension.
When
the forces tend
to
Fig. B
compress the member, it is in compression.
The member of the truss can be having following sections:
FLAT BAR
ANGLE
ROUND
HEXAGON
SHEET/PLATE
CuLS O
GY
CHANNEL
WIDEFLANGE
BEAM
STANDARD/!
BEAM
SQUARE/RECT.
TUBING
PGLLGP
TEE BAR
HALF ROUND
HALFOVAL
PIPE/ROUND
TUBING
&
CHAMPFERBAR
REBAR
Classification of Pin Jointed Plane Truss:
Based on the relation between the no. of members and No. of joints the
truss can be still classified as follows:
Perfect truss:
lf No. members
are sufficient to with-stand the external load and
satisfy the following equation then the truss is said to be Perfect or
stable. They are statically determinate.
m = 2j-3
where,
’
m
=
No. of members
|.
]
j
3
=
=
No. of joints
No. of equation of equilibrium/Support reactions
Deficient truss:
If No.
members
are not sufficient to with-stand
the external
load
and satisfy the following equation then the truss is said to be
e
Imperfect or deficient & Unstable.
m < 2j-3
If No.
members
1
———! y
.
|
Redundant truss:
are more than that required
to with-stand the
external load and satisfy the following equation then the truss is
said to be Redundant& statically indeterminate.
m >2j-3
+
'
Assumptions
made
in the analysis of simple
Plane
Pin jointed
Perfect Truss:
Truss joints are frictionless pin joints. They cannot resist moments.
Load are applied only at the joints.
Each member of the truss is subjected to axial force only.
The truss is assumed perfect. (i.e. m= 2j-3)
Members of truss has negligible weight as compare to the loads applied.
Each member of the truss is two force member.
The truss is rigid and does not change in shape.
Methods of analysis of Truss
Method of Joints:
e
lf atruss is in equilibrium, then each of its joints must be in equilibrium.
¢
The method of joints consists of satisfying the equilibrium equations for
forces acting on each joint.
¢
Fx=O & 2 Fy=0 (Concurrent Force System)
Methods of joint is most suited when force in all the members of the
truss are required to be obtained.
e
Start the analysis only with a joint where there are only two unknowns
since only two equations of equilibrium are available.
B
Compute the force in each member of the loaded cantilever truss by the method of
joints.
B
5 m/
A
5m
5 m/
5m
5m
5 m
5m
30 kN
20 kN
Solution:£
T
5 m
Support Reactions:
30
60
5 m
-
*
E
m
30 kN
ME =0;
m
20 kN
T cos 30 x 4.33 + T sin 30 x 2.5
30 x 10+20 x 5 = 0
T 8 0 kN
F=0;
Fy 0;
T cos 30 = HE
HE = 69.3 kN
T sin 30 4+ VE = 30 + 20
VE 10 kN
Joint A:
2Fy 0;
FAB Sin 60 =30;
FAB 34.64 kN (T)
F=0 ;
FABCOs 60 = Fac;
Fac= 17.32 kN (C)
AB
60
AC
x
30 kN
Joint A
Joint B:
Fy=0
AB =
FBC 34.64 kN (C)
2F=0;
60
BC
Joint B
BC =
Joint C:
34.6 kN
FcD sin 60
=
CD
FBc sin 60 +20
FcD 57.73 kN (T)
F-0;
34.6 kN/60
FABSin 30+FBc cos 60 FBD
FBD 34.64 kN (T)
Fy 0;
BD
FAB COs 30 = FBc cos 30
FcD cos 60 + Fac + FBc cos 60 = FcE
FCE 63.51 kN (C)
60
60
AC ="
17.32 kN
CE
Y20 kN
Joint C
Joint E:
\DE
Fy=0;
FDE sin 60 =
10
60
FDE 11.55 kN (C)
CE
63.5 kN
Check:
E F = FoE COs 60 + FCE - HE =
69.3 kN
0
I10 kN
Joint E
Detemine the force in each member of the loaded truss.
2m
2 m
2 m
E
2
m
G
500 kg
AD
Solution:£
Support Reactions:
2m
2 m
MA 0 ;
T
2 m
G
T x 3.46 = 500 x 9.81 x 5.46 = 0
T 7.74 kN
2F=0; T=Ha
HA
HA = 7.74 kN
2Fy=0; 500 x 9.81 =Va
VA = 4.91 kN
Va
2m
500x 9.811N
FAB
Joint A:
F=0;
FaF cos 60 =
FAF
Fy 0
7.74
600
15.48 kN (C)
FAB+ 4.91 =
FAF
O-----
7.7
A
15.48 sin 60
FAB 8.49 kN (T)
4.91
Joint B:
Fy 0;
FBC COS 45 = 8.49
Fuc
F =0;
12.01 kN (T)
FBF+ 7.74
45
12.01 cos 45
7.74
FBF
B
FBF 0.756 kN (C)
8.49
Joint F:
F =0;
0.756+15.48 cos 60 = FFE cos 30
FE
FFE =9.81 kN (C)
F, 0;
Frc+9.81 sin 30
30
15.48 sin 60
0.756
60
Frc= 8.5 kN (C)
15.48
Joint D:
Fy=0;
4.91
FDE Sin
30
30°
FDE 9.81 kN (C)
F=0;
Fpc
9.81
cos
30
FoE
4.91
Fpc =8.5 kN (T)
Joint C:
Fy 0;
12.01 cos 45 +FCE Sin 30 = 8.5
8.5
300
FCE =0
45
Check:
12.01
E F = 8 . 5 - 12.01 cos 45 = 0
8.5
Determine the force in each member of the loaded truss. Make use of the symmetry
of the truss and of the loading.
D
C
B
4 m
5
A
5m
m
30 kN
60 kN
E
5m
30 kN
Solution
Support Reactions:
C
B
D
4 m
51m
5m
5 m
H
30 kN
E
m
G
60 kN
30 kN
Ra
RE
Due to symmetry in both geometry and loading
RA=RE=60 kN
Joint A:
Fy 0;
AR
FAB Sin 3 8 . 6 6 =
60
38.66
FAB 96 kN (C)
F
0;
FAH
96 cos 38.656
FAH
75 kN (T)
FAH
A
60
Joint B:
F=0;
FeC 96 cos 38.66 =60
B
38.66
FBC = 75 kN (C)
96
2Fy=0;
FBH 96 sin 38.66
BH
FBH 60 kN (T)
Joint H:
2Fy 0;
60
FHC sin 38.66 +30 = 60
38.66
Fuc = 48 kN (C)
F=0;
75
FHG
FuG= 75 + 48 cos 38.66
30
FuG = 112.5 kN (T)
FCG
Joint G:
Fy0
H
FcG = 60 kN (T)
112.5
112.5
G
60
By Symmetry,
FDE
FAB = 96 kN (C)
FFE
FAH= 75 kN (T)
FeD = FBc= 75 kN (C)
FDF = FBH = 60 kN (T)
Frc= FHc = 48 kN (C)
FGF = FHG =112.5 kN (T)
Shaan o e
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ony Poin
Sectton
ot tk Cokourpomd
Sh
For oiagam atony poin
Shaa Foru
+
Slope
SLope
J
Ogoals 76
owt
Chana benotsng
to
Imtenilty crta'butad
t
loa
ove TRe
Satfom
Sign CoDveniibm 8oR Slope
Dtheck od
Diucted
Ib
SF
SF
up to 7% ight (
dron
Valw
valn
to
nigt C
+ve
-ve
in
SFO *t
i n SFD TAn
+ve Slope
-
Ve
Slope.
Slope
BmD
6 +ve
Slope of BmD S - ve
Scanned with CamScanner
SFD and gDo Som Standand Loocding
500
Comtrlavnevar eam o?th Bit locd froo ond
Considw Soctfom *-Z
diitonu
N
& a
from free ond
B
A
L
Shaan Fozo
uniodm
CRyht e Seatlon)
SF
Indaprvcderd oz
It
Comstont
SFD
hrough our AB
B
Bencdkng Momont
Mxx W
BMD
9m
varies
MA WL
MB
)
ofth
z.
Cz:)
0
Max
kkneonly
Czo)
Benckng
Momeni
Mmay
=
WL
Cant?lavac Beam otB UdL
Shaa
unt/
Foru
lmgth
CS gp, krea)
Rnan
VA = 9L
Ve -
O
+ve
Czo)
SFD
Bendimg momont
ve
Mxx- 2
MA:-4OL
2
-
(2nddgo,
Panoboke)
MB
0
Max. Bencin9 momsnt
D
Paabolkc
Mmar: hLY | |
Scanned with CamScanner
Comt?evec
toPth
Tatongulor Losd
Shaa Foru
Vhx
un
B
h90- 92 C2nd dnga
22
Vs
O
Panabolte
WL
+ve Vy
Parabolt)
Cz-1)
SFD
mxx
5
VA
-ve
(z:o)
Cubfc
Bolin momont
xx
BMD
-
-
z - 02-C
3dagru,
6
Mg
Cz-
O
MAA
Cx0)
BeTo'n9
Maxi
Cwbre)
Som
Notes
i) A
rre
al ooys
moe*
on
0nd
3P
Max
Canttlavee
bao:
beom
ComtPuvo
Cunu
o momu nt
bendng Moont
not actimg
Tree od)
) A + vee ond
al way
e
n)
3eko
Cantove
Cunur
a
beam
Shaa FoRu
ptload
not
ading aD
ond)
Benchn g Momut and
maxim m
Shoan FoA
Sueport
o
alwuy!
Cagt?avn
beo n
Scanned with CamScanner
Stmply
Sapocted beom wth
poit load
Pontfom AC
Secl ton
L
betweon A and c
Shean FON
Cuy Steb
VA-
Ca0
CA
Bendkn9 Mormur
O
Rant
L
SFD
Cza)
mA
+e /777i
Mmes
+ve
BMD
Czo)
2
A
Pofon BC
Sactfon -
batsoeen
Pt:2etoshan
andc
Shaa Foau
Cegka sid
SFP
Ve =- ha
Ve -
A
Bendng
me o
Mc
Tb
Tb
C=o)
wab
Pmax
BMD
momnt
Cxb
Me
lood
actimg
e. a
b: l/2
Mmax
WL|
Certs o
beom Cie. Cnbad point leccd)
Scanned with CamScanner
Stmply
Shao
Sappoxted beom wttB Ud
untlung1
ore
A
Vxx = K9L
9z
VA
Czo)
mn
MAAyB
CIt dagas, knsan)
Poit 2 a o s h a n
Ve
=
0
(
tve >
VeO -
A
z2
Bemck
Ve
Ruan
SFD
Paebokt
Momunh
Mmor
Mxx
A
MA
O (a-0)
MB O
SMD
CaD
Ca/)
Stmpy Suppoted beam itk UvL CTstargulan)
unft/ ungik
Shran Eoru
V
VA
-**
C2nd dar
Cao)
6
-
aobele
Panaboke
Ca
SFD
Bendknq: momem
6
ubic
- 19.cardu)
Cubic
tve
Mmey
MA O C z 0 )
O
BMD
C-
Me:Mma =
C7)
X
Scanned with CamScanner
PROBLEMS
Dras
Bendtng
S hea Foku and
Cantflouee
beam
as
momnt
Shoon
3KN
SoutPon?
Tmoin a
dogtams
FO
bor tk
2.5EN
E
D O'Sm
m
Sect Pom
22.5 KNm-MA
D c, 8 ond4
a nd Consid Rghl Va= 75KN
Pox tfo
o TR boa m
Shean EOKu
7-5
75
CotTh out .SKn)
Wnsan
45
VE
2.5k CoTh 2.SKN)
V
5
Ve
5+
Me
KN
IX2
=
4:5ku
Sm Cwithout
VB,=
VB
25
SFE
7.5kn
125 naan
Cm s
4S+3=7:5Kw
8-25
VA 7.5 *
Paroboke
Cothout VA)
knuan
VA)
nao
22:5
Bencn4
ME O
momnt
CRyht E', No lead
Mp= -0. 5x2.5
Me
-
/:25 *wm
5 x 25 - 1x2xi : - 8.25 KMm
Mg-?Sx4
MA
BMD
/x2x 5 :15 kNm
- 5 x 5 - Ix2x 3:5 - 3 x
mA,O
-22:SKNm
wtkout M )
Coth PA)
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DRau
SED
a7md
BmD
Ove haaÉry
jor 7ko
nK
beom
4K
KNm
2m
2m
2m
Solutfon
AmE
4KN
2kN
/mn
A
A
3
72kwmn
3k
5kN
|Load Diogham
+ve
Po2
PozS
SFD
Rnuan7
Rnabolic
Potnt ontrof
BMD
max
max
2
Bendkng or Stmpy Sappostod spon 6Ewm (agmg)
ov
Benckny
hamg'ng spon CbE) kNm CHggi)
=
Poimt o ContnakAxuou
Brrcng moment 3se0 ' a '
e = 0.67m
MF0 -2 (zt)+ 5z
Scanned with CamScanner
Draw the shear force and bending moment diagrams for the beam hinged at A and roller at
D, subjected to loads as shown in the Figure.
50 kN/m
20 kN/m
im.
2 m
-2m
2 m
B
Solution
EMA 0
6RD 1[2(50)}+5[2(20))
EMp 0
6RA 5[2(50)]+1[2(20))
RA=90 kN
RD 50 kN
Segment BC:
Segment AB:
VAB= 90-50xkN
VaC 90-50(2)
VBC-10 kN
MAB 90x-50x(x/2)
MAB
90x-25x kN m
Mac 90x-2(50)(x-1)
MBc-10x+100 kN-m
Segment CD:
VcD= 90-2(50)-20(x-4)
VcD-20x+70 kN
McD 90x-2(50)(x-1)-20(x-4)x-4)/2
McD 90x-100(x-1)-10(x-4)
McD-10x+70x-60 kNm
To draw the Shear Force Diagram:
1. VAB=90-50xis linear; at x =0, Vac=90 kN; at x=2 m, Vsc =-10 kN. When
VAB 0, x= 1.8 m.
2. Vec
-10 kN along segment BC.
3. VcD-20x+70 is linear; at x = 4 m, VcD=-10 kN; at x = 6 m, Ved= -50 kN.
To draw the Bending Moment Diagram:
1. MAB 90x- 25xissecond degree, at x =0, MAB =0; at x = 1.8 m, MAB =81 kN°;
at x= 2 m, MAB = 80 kN'm.
2. Mac-10x+ 100 is linear; at x= 2 m, Msc= 80 kN m; at x = 4 m, Mec = 60 kN m.
3. McD-10x + 70x 60; at x =4 m, Mcp= 60 kN-m; at x
-
=
6 m,
Mco=0.
Scanned with CamScanner
50 kN/m
2 m
RA
90 kN
90 kN
20 kN/m
2 m
2 m
Load Diagram
Ro 50 kN
shear Diagran^
1 . 8 m>
-10 kN
-50 kN
81 kN-mi 80 kN-m
60 kN-m
Moment Diagram
Scanned with CamScanner
Draw Shear Force and Bending Moment Diagrams for the overhanging beam loaded as
shown in Fig. Also locate the salient features if any.
30 kN
20 kN
30 kN/m
AAAAAAA
AA
1.5
m
1.S m
3 m
C
Re
Solution:
Support Reaction:
Mc 0; 3Rp 20 x4.5
+
30 x 3 x
1.5-30 x 1.5
Rg 6 0 kN
2V-0;
Rc 30 x3+30+20-60
Rc
80 kN
Shear Force
Va-20 kN
Va-20+60 40 kN
Ve-20+60- 30 x 3 - 50kN
Vp-20+60-30 x3 +80 30kN
Bending Moment:
MA-0
M g - 2 0 x 1.5 =- 30 kNm
Me-20 x4.5 + 60 x 3-30 x3 x 1.5 =-45 kNm
Mp-0
Point ofZero Shear(El
VE20+60-30 xX=0
X=1.33 m
Maximum Bending Monment (Mma
Mmx ME-20 x 2.83 +60 x 1.33 -30x 1.33 x 1.33 /2
ME-3.33 kNm
Scanned with CamScanner
Shear Force Diagra
Point ob
Zeno
30
Saan
+ve
+ve
20
4/44444LdD
33m
-Ve
30
ve
20
50
Bending MomentDiagram:
-ve
-ve
3:33kNM
30
45
Scanned with CamScanner
Draw the shear force and bending moment diagrams for the overhanging beam. subjected to louds as
shown in the Fig.
61) ktu
40 ktl/m
M 120 ktN m
1 m
1 m ' 1m
3 m
R2
Solution:
60 kN
40 kN/m
M
=
120 kN-m
A
m
1
R1 =
1 m
m
3 m
R2 =
132 kN
48 kN
Load Diagram
72kN
x
=
-60 kNN
1.8 m
-48 kN
Shear Diagramm
2:3
48 kN-n
-31
4.8 kN-m
-24 kN-m
-60 kN-m
Moment
Diagram
-72 kN-m
Scanned with CamScanner
Draw Shear Force and Bending Moment Diagrams for the beam shown. Also locate the salient
features if any.
40 kN/m
15 kN/m
A
D
2m
2 m
6 m
m
IRB:108KN
RA72 KN
2
12
Paviaboke
30
+ve
3.976m
-ve
P:o ZeTo
Shoan
48
8
Linan
217.02
Cubi
1
| 96
Pauabolic
n e a
+ve
2.4
P-6 Comdoblortus
ve
Pauaboke
Scanned with CamScanner
6.
Draw S.F. and B.M. diagrams for the loaded beam as shown in the Figure below
and mark the salient features.
400 N
.oo0L
160 N/m
maAAAAAÝAAAAAAAAAAA
E
im
V dY
+oIxoàt
D
3m
5 m
5m
B= LoooNN
V
1o0ONN
3.75m-
bxD
440N
- l l xoat o480N
e - boo1+xot
ie
400N
-
PtHo
ZeroShr
E
520N
SFP
560N
T
A tn
T
1-1 o3
OS
t.
0001 ti2-Exxnal- Fxhe
t
or
io b13 T
5mlT
A126N-m
Potnt oh Cortoaaxue
Ve
conas
Ve
o-(-4eON-m,"Csx oal
BMD
720 NMn
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Raactfont
To Fend
(ZMg-o
-400x1-
Ve
t
160x
160x9x4»5
Vex 6
o0o N
4t
160xlo + 400 = VBtVe
VB
o0o N
woot
To
Draw SED
SF at
A
SF
ut B
SF
at C
SF
at
=
-
A
and
x7 +l000 -520N
-400- l60xto + 2o00= O
400-160
D
40N
and
400-160xi-560 N
480N
BMD
To Dvao
B-M at
400N
O
BM at B:A cOxI-
160xO.5
480 N-m
B-. at C: -400x 7- l 6ox 7x3.5+ 1000X6
=-720 N-m
To Fend Pot oh Zeo Shea
S.F. at F:0
2
400
160x 2 + lo00'= O
From A
3.75
B.M at F 400x a-75160x3.75+1o00x 2.75 125N-m
To Ftmd
BM. at E
Pott e
0
Covtagtoxua
A00X 2
160xG2+1000
x(Z-)
GM82
25m
=
o
Fooon A.
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CENTROID OF PLANE AREA
Centre of Mass:
.It is for the (3-Dimensional) object which has got mass.
I t is a point through which the resultant of system of parallel forces
formed due to the mass of all the particles of the entire object.
I t will not vary depending upon position and remain constant.
Centre of Gravity:
I t is for the (3-Dimensional) object which has got weight.
I t is a point through which the resultant of system of parallel forces
formed due to the weight (gravitational force) of all the particles of the
entire object.
I t will vary depending upon position (since the magnetic field vary with
position) and will not remain constant.
Centroid:
I t is for the (2-Dimensional) plane area which does not have weight.
Moment of area about any axis passing through its centroid is zero.
I t always lies on symmetric axis.
.The axis passing through Centroid or Centre of gravity is called centroidal
axis.
Centre of Gravity/Centroid of a Flat Plate:
Consider a flat plate of thickness
as shown in Fig. Let wi be weight of any (")
element portion acting at a point (x, yi). Let Wbe the total weight of the plate acting
at a point X &Y. According to the definition of .G., the point G(X, Y) is the Centre
of Gravity.
Taking moment about X-axis
WY = wI yi + w2 y2 +w3 y3 t.
=
. . .
2
wi
yi
Y = 2W
-(1)
W
Taking moment about Y-axis
WX =W1 X1 + W2 X2+ w3 X3 t....
=
2
Wi Xi
--(2)
W
Let A be the area of ith element of plate. If y is the unit weight of the material of
plate then,
Wi Y
Ait
W=2y Ait
Now, Total weight
W=y tA
W=y t2 A:
(A = 2A = Total Area)
Substituting for w and Win equations (1) & (2) we get
X A
= x
=
OR X = xaA OR =
A
Y=
A
Moment of Area abouty-axis
Total Area
A
OR
=
yda
A
OR
-Moment
of Area about x - axis
Total Area
Centroid of some standard sections:
Centroid (G) of symmetrical sections such as Circular, Rectangular and Square
lies on the exactly on the centre of the section. While, for some
unsymmetrical sections are given as follows:
Triangular section:
Let us consider a triangle with base b and height h. Consider an elemental strip
of width x at a distance y from the base and having thickness dy.
The area of elemental strip is dA
By similar triangles;
(h-y)
=x
dy
h
x= (h-)= b(1We have,
y dA
A
Sy-dy
y-
2
bh2
Ybh
2
h
Y3
Centroid lies at a distance h/3 from the base and 2h/3 from the apex
Semi-Circular section:
y
· -X
------.......-----------.x
R
Let us consider a semi-circle of radius Ras shown in Fig. Due to symmetry the
centroid lies on diametral y - axis. Let Y be the centroid a I distance from 'O'.
Let us consider an elemental angular strip at an angle 0 with x-axis and in that
angular strip further consider an element of thickness dra t a radial distance r
from 0.
The area of elem ent al strip is dA
= r d0dr
Moment of elementa I area about dia metra I x - axis = rd0 dr. r sin 0
y dA =
,,.2
sin0 d0dr
Total Moment of area about diametral x - axis
JY dA = J; JR r 2 sin8 dr d8
0
7r
= f0
R3
3 s in8 d8
=
2R 3
3
We have,
Y = f ydA
A
y
=
Y=
2R 3
-3rcR2
-2-
4R
3rr
Quarter-Circular section:
y
4R
A
''
'
X= 3rr
Y=
4R
3rr
V
X -·- ·-
._
_____ I __ __ __ __ __ __
T G(i,ji)
- -- X
I
-----+--------------> X
V
Sector of Circle:
Let us consider a sector of circle of angle 2 a as shown in Fig. Due to symm etry,
the centroid lies on x - axis. To find the distance of centroid from 'O', consider
an elemental area as shown.
y
y
I
y
Let us consider an elemental angular strip at an angle 0 with x - axis and in that
angular strip further consider an element of thickness drat a radial distance r.
The area of elemental strip is dA = r d0dr
Moment of elemental area about y - axis
=
r d0dr. r cos0
Parabolic Spandrel:
y
h
..._......__ ___..............,........,_._. X
I.
dx
a
Consider a parabolic spandrel as shown in figure. Let us consider an element
of height y = k x2 at a distance x from 0 . Let the width of the element be dx.
Area of the element - dA
= kx2 dx
The total area of parabolic spandrel- A
A
= f0a kx 2 dx
= -ka3
3
------------- (1)
Moment of area about y- axis
Jx dA = f0a kx 2 dx
x
foa kx 3dx
ka 4
--------------- -(2)
4
Moment of area about x - axis
f y dA
=
foa kx2dx
=
a
k;2
k 2x4
r -2d x
Jo
k2as
= - 10
We have,
X
= IxAdA
·
i.e.
(2) / (1)
X=
We get
We have,
Y=
I ydA
A
.
-----------------(3)
3a
4
i.e.
(3) / (1)
-
3ka 2
Y=-
We get
10
From Fig. @
h
=
ka 2
k
=
h i a2
X =
a,
y =h
Y=
3h
10
Determine the x- and y-coordinates of the centroid of the shaded area.
160-
120
Dimensions in millimeters
Solution:
160-
-80-
2
X
.
X
120
1
X
Section
Rectangular
Triangular
Area(A)
Ax
(mm)
(mm)
(mm)
(mm)
Ay
(mm3)
240 x 1200
120
60
34,56,000
17,28,000
11,94,667
9,38,667
46,50,667
26,66,667
hx 160 x80 80 +(2/3) 160 120+(1/3)80
35,200
A Xi
46,50,667=
A
35,200
2Ay26,66,667
A
35,200
=
132.12 mm
75.76
mm
Determine the x- and y-coordinates of the centroid of the shaded area.
120 mm
60 mm
40 mm
80 mm
60 mm
Solution:
120mm
60 1mm
25.46 min r=60 mm
+
60 mm
rg= 40 mnm
40 mn
80 mm
O mm |
40mm
Ta40 mm
105.46 mm
80mm
60 mm
60 mn
-20mm
Component
A, mm
Rectangle
(120)(80)
Triangle
Semicircle
Circle
x, mm
120)(60)
r(60)
=9.6
=
=
x
3.6 x
5.655
TA, mm
y, mm
yA, mm
10
60
40
+576 X 10
10
40
60
20
+144 X 10
-72 x 10
+339.3 x 10'
+596.4 X 10
-402.2 x 10'
x
-T(40) =-5.027
A
60 mm
10
x 10
13.828 x
10
60
105.46
+384 x 10
-301.6 X 10
80
A
=+757.7 x
10
A =+506.2 X
10
Location of Centroid:
XEA ETA:
X(13.828 x 10 mm) = 757.7 x 10 mm
X = 54.8 mm
EA 2yA:
Y(13.828 x 10 mm) = 506.2 x 10 mm
Y 36.6 mm
Determine the x- and y-coordinates of the centroid of the shaded area.
200
mm
200 mm
+150 mm
t
125mm
60 mm
125mm
Solution:
200 mm
2 0 0 mm
T
mm
150
1
125 mm
-
60 mm
2
125mm
X
Y
AX
Ay
(mm)
(mm)
Section
Area(A)
(mm)
(mm)
Rectangular
400 x 250
200
125
Triangular
%x 150 x 250
400+ 150/3
250/3
84,37,500
15,62,500
(T x 60)
200
125
22,60,800
-14,13,000
2,61,76,700
1,26,49,500
Circular
-
1,07,446
(mm
2,00,00,0001,25,00,000
=
4
2,61,76,700
A
F- SA
A
=
=
1,07,446
1,26,49,500
1,07,446
243.63
mm
= 117.73 mm
Determine the x- and y-coordinates of the centroid of the shaded area
T
0 mm
30 mm
100rmm
X
100 mm
>k 30 mm
100 mm
Solution:
T
50 mm
3Omm
X
100 mm
100 mm
T00
Y
mm
30 mm
x
AX
Ay
(mm)
(mm)
(mm)
115
50
26,45,0000
11,50,000
(2/8) 230
100+50/3
8,86,075
6,69,875
Triangular -(%x30 x 30)
150
100 - 30/3
-67,500
-40,500
Semi-Circular (%xTx 50)
150
4x50/3T
-5,89,048
-83,333
28,74,527
16,96,042
Area(A)
X
(mm)
(mm)
Rectangular
230 x 100
Triangular
%x 230 x 50
Section
24,373
=
A
28,74,527
28,74,34-117.94
mm
24,373
=2AY_16,96,042
A
24,373
= 69.58 mm
Locate the centroid of the shaded plane area as shown in Fig.
30 mm
r= 30 mm
--X
Solution:
30 mm
5
=30 mm
.-X
6
x
Section
Rectangular
Quarter
Circular
Area(A)
Y
AX
Ay
(mm)
(mm)
(mm)
(mm)
(mm)
30 x 60
15
90
27,000
1,62,000
30+[4x60/3T]| 60+[4 x 60/3r]| 1,56,823
% xnx602
2,41,646
Triangular
%x30x 60
(2/3) x 30
(2/3) x 60
18,090
36,180
Square
60 x 60
60
30
2,16,000
1,08,000
-[% x tx 30]
30+[4x 30/37t]
90
-60,411.5
-1,27,234.5
90-[4 x60/3Tt]
4x60/3T
1,82,469
-72,000
Semi
Circular
QuarterCircular
60]
1,75,032.5 3,48,591.5
4,886.28
F=
A
X
A
=2AY
A
1,75,032.5
= 35.82 mm
4,886.28
3,48,591.5
4,886.28
= 71.34 mm
MOMENT OF INERTIA
(SECOND MOMENT OF AREA)
Consider a plane area as shown in Fig. Let dA
be the elemental
area
with coordinates x &y.
dA
O
Moment of elemental area aboutx -
axis
= dA y
Moment of Moment of elemental area about x - axis
= dA y
i.e. Second Moment of Elemental Area about x - axis
= dA y
Second Moment of total area aboutx- axis
=
2 dA y?
OR
Moment of Inertia of plane area aboutr- axis
Ixx=Sy dA
Similarly,
Moment of Inertia of plane area abouty -
axis
yySx dA
Radius of Gyration (k):
-X
- - .
Radius of gyration of an area can be considered as the distance from the reference
(x or y) axis at which the entire area can be conceived to be concentrated without
altering the Moment of Inertia about that reference axis..
It can be calculated as follows:
k =
Aboutx - axis
k kx
About y - axis
Kyy
Theorems of Moment of Inertia
1. Perpendicular Axis Theorem
Statement: The moment of inertia of a plane are about an axis perpendicular
to its plane (i.e. Polar Moment of Inertia - I ) at any point O is equal to the
sum of moment of inertia about any two mutually perpendicular axes through
the same point and lying in the plane of are.
Proof
Let z-z axis is normal to the plane of area
According to the definition, from the figure
Moment of Inertia about z-z axis
aSr2 dA
Sx2 +y?) dA
=
z
=
Sx2 dA + Sy2 dA
1. Parallel Axis Theorem
Statement: The moment of inertia of a plane are about any axis (LA») is equal
to the sum of moment of inertia about a parallel centroidal axis (Ico) and the
product of area (4) and square of distance between the two parallel axes (y:).
Proof
dA
G
G
CG
B
Let z-z axis is normal to the plane of area
According to the definition, from the figure
Moment of Inertia about AB axis
aB
AB
AB
So+y.) dA
=
SU2+2yye + y?)
=Sy2 dA + S2 yye
dA +
Sy?
dA
dA
ABGG +2y Jy dA + Y S dA
Jy dA represent the moment of area about GG axis which is passing through the
centroid of plane area and hence the term Sy dA = 0
LAB=IoG +Ay
Moment of Inertia of Some Standard Sections
(i) Rectangular Section:
dy
Considera rectangular section of width band
depth d. Let dA be the area of the element having
d/2
thickness dy at a distance y from centroidal axis
X
X-X.
d/2
Area of the element- d A = b dy
According to the definition, from the figure
Moment of Inertia of rectangular section about centroidal axis X-X
rd/2
x
y
-d/2
dA
d/2
Ixx
bdy
-d/2
xx= b
d/2
ydy
-d/2
bd3
Ixx
12
Similarly,
db*
Iyy 12
Moment of Inertia of rectangular section about its base AB
lAB = Ixx+ A y
From Parallel axis theorem]
bd3
IAB
12
+b
bd3
TAB
3
(ii) Triangular Section:
2h/3
dy
h
X
h/3
b
Consider a triangular section of width b and height h. Let dA be the area of the
element having widthx and thickness dy at a distancey from base AB.
Area of the element- dA
=
x
dy
By similar triangles;
h
(h-y)
x =h-) =b(1Moment of Inertia of triangular section about its base AB
'AB=
x dy
lAB
las=b
4hlo
bh3
IAB 12
Moment of Inertia of triangular section about centroidal axis X-X
LAB Ixx +Ay
From Parallel axis theorem]
bh3
12
2
x
+
bh3
Ixx36
(ii) Circular Section:
dro
Kde
To Sine
X
d
Consider a circular section of radius rand diameter d. Let us consider an element of
sides ro d0and dro at a radial distance ro from centre 0.
Area of the element -dA = ro d0 dro
Perpendicular distance from the element to centroidal axis X-X - y = ro sin0
According to the definition, from the figure
Moment of Inertia of rectangular section about centroidal axis X-X Ix = Jy
Ix
0, sin
rodo dro
dA
xx
Ix
o r
sin-0 d0 dro
"r (
) do dr.
sin20127
Ix20-2dr
JO
Ixx=
Tr4
Ixx
OR
Tt d4
Ixx= 64 =
lyY
(iv) Semi Circular Section:
dt
IAB
Y
128
lx=In= 0.0068598
-X
B
(v) Quarter Circular Section:
IAB 256
Ix= Iw= 0.00343 d4
X
GE,9)
-X
The plane polygonal lamina has one circular cut of 1.6m diameter and one semi-circular
cut of 4m diameter as shown in fig. Locate the centroid of the remainder and find the
second moment of its area about centroidal axis parallel to the base.
12
6m
- l - 8 48m
16m dta
3m
-355
2m
m
-- X
2m
4m
4m
4m
AT
Atao (A)
SectPom
x
6x3
6+45
4x9
6t3
x1x6
-TIx I:6
44
Tx
12
Ix6
3
A
1+
36
8
2
378
72
243
162
44t2
-
- 2.04
804
75:39
- 5 33
2
ZA 69:7
O.5
18
2AZ51l-57 m
ZAg 247-66
59157 8.48
69.7
47* 247.66
697
3.55m
Moment
T
Inevtta
about
heal
Coortrotdas Axto
6x+yx6x3
(3.55-2)
36
Trtowguda O
9x4+ qx4 (355-2)
Ractomgal
12
+
9x +xx6 (6-3.55)
Toiomgala
36
I-61
xx -6+
o.11x21
6x
+
+
Tx6"(3.55-1)- Ctrulan
4
2
6x
(3 55-5
S l Ctulan
3 s 6 - os)|
Ractomg
2
Ixx 2 6 12+ 134 49 + 216-07 13.39
Txx
372 m
e
47:6+5631
mineTbe minlmum Pradias
Dete
the
Centofdcel
The Bgu
ayratPom abous
T h plone e o Shoom I
axe
bolos.
A
IRomm
L20mm
IX
94:6
57-5
1oomm 150mm
mm
IS0mm 1
Soludiom2
Section
A
AT
200
66/ 25xIo
495 xi0
9875
46x10
x300x240 J00
137.5
72810
xl50 x 120
175-I:35xIo-8775x1o
575x40o
6
/50
SA 50xio3
z
10.45XO
50XI03
AT
ZA
I0:45x10°
73 xIo6-A7
209 mm
6
4:73xIO =946 mm
SoXID3
2
ADOx 575 + 575x40o) C94-6- 2875)+
12
300x240 + x 30oox240 (137-5-94-6)
36
150x
120 + x150X120 (975-94-6)
36
6
Ixx=280.25xio mm
57-5x 400
400x575 (209 -20)t
12
240x
300+x240x3co C209-20o)
36
120X150 +x12ox ISo (209-150)
36
8 4 448.87xIo° rmm
Imin
=
Txx
=
80.
25
xIo o n
. . Miznimum Raciu
H*attoTnin
Tmin
280 25xid
1min
7487 mm
50xIo
L
Locate the centroid of shaded
to
area as
shown in
Fig.
Also calculate the second
moment
of
area
with
respect
vertical centroidal axis (dv).
3 0
mm
5050 mm
50 mm
Solution:
To
LocateComtaofd
x s o x 30x 2xs0+ 50x3ox 75+ Tx30 x(100+4 x30 )
3T
+
x30
x
l0o
5Ox30 +50x3o* TTx3o+x 30
8.04
m
xSOx30x
xB0
t
50x 30x1S
+
x30 ( 3 0 - +x30
3T
+ x 30x B0+ 4x30)
3
xG0x30 +50X30 + 1Xx30
TX 30
2
23.48 rom
Tmd
yy
Tyy
oresit
Tmetia
BOxS0
x3oxSo C3xso- 82.04)
36
oboul Contkoldo
-0x
t
30xs0
+
30x50 (75 - 82. 04)
12
0
0343x 60 +
Mx30 (100+ 4x 30 - 82.04)
1
| 1xx66
66+ Tx30 ( 10o 82.04)
2
128
6
3.755 xIommt
Locate the centroid of shaded plane area and determine moment of inertia of about its vertical centroidal axis.
Y
70
60
30
25 Dia
250
X-
130
92
50
100
200
Solution:
To Locate the Centroid of Shaded Area
The Section is symmetry about y-axis and hencel X = 100 mm
y = 250 x 200 x 125- x 100 x 92 x|50 +(
2;x 130 x 70 x (50+
x 2 2 0 2x70 x 70 x (35 + 180)
x 130)]
250 x 200-x 100 x 92 - T - 2 x 130 x 70 + 70 x 70
y 93.05 mm
Moment ofInertiaabout centroidalY-Y axis(l,
50x 2003 130 x 2003
Iyy
130 x
70 x 603 x
12
12
36 z *
36
130 x
70 x(100
Iy= 63.35 x 106 mm
12
25
64
92x 503
12
O
Locate the centroid of the shaded area as shown in the Figure below and determine the
Moment of Inertia with respect to centroidal Y-Axis.
+1000 mm
1000 mm
650 mnt
+400 m m -
Solution:E
Jo Lpeate T6
Cetaoid:
Secttop AlLa
500
/00OxioDo
-
500
6
500 x1o
500xIo
6
26-67xio
x 4 c0x1000
2
3
-Tix 650
6
66-67x!O
3
-
350
4x6SD
23228xio183.0a xio
3TT
-241-0SXio 383.54xmo
Todal 2A: 536 84xio
ZA
-241OSXO
3
53 6.34X (O
Z - 449 43 m
6
383-59x1o
20
536 34xI
T15:2mm
Moment_ e Inertia
about
Cendaofcla)
Y-aA7T
3
lo00 x/0Oo
t
lboox/000(449:43-500)
2
8:59x0
3
xl000xgoo 449.43t 400)
boox 4co
36
x
1300
Tix6S0 (449-43 3s0)
128
789xo
2
10
mmt
6:9H00
76710
Caleulate the second area moment with respect to both the centroidal axes for the shaded area as shown in
tigure
20
R= 100
90
R
100
All the dimensions are in mm
Solution:
Tolocate thecentroid ofshaded area:
Section
Area (A)
Squars-
100xlDo
50
QCiacle-3 77x 10D
A
4x/00
3
100-
1Dx20
TRimg-s
10Ox2
37
37
45x10
24 x1D
9.33 x1010.33x 10
=
=
B
-73-33xio66-67xi
57.98 mm
AY=TI:64 mm
ZA
5 2 xI0
10Dt
100-
zA
ZA
15x1o
o0-4x00-333
33x10
377
8900 mm
z
A
500 XID
/50
1DO-4XDO1D0-4x10D
-Cte cle-2 x0o
4
T i o nge-
AX
Y
56-0IX1o
=
635 x1o
TMomet_
Ixx
X- X
oboul
lnktio
axis
0OxIOO + 100150 -7164)
12
2
4
100-4 XID0 7164)
O 00343x200+ 7xi0D
4
71-64
0.003433 xJo0+ Tx/00 (200- 4X/0D
37T
XID
L 36
+
x20xID /0o t
764
20OX 100+x0x l00 (200-7 64)
36
Ixx
l2:63 xio mm
Momnt
lyy
e 1nextia
about
1DOXI00+ 100
Y-y
azis
C50 - 57.98)
12
O-00343x-200
t T7X10D(100-4 x100 57 98)
3n
2
4
-00342 x200 +7xloo
1Ox205
xllox
36
6
1yy
5 67x1D mm
20
4XIDO-57.98)
7
37
C100- 0
-
57.92)|
Locate the centroid of shaded region as shown in Fig. Calculate the second area moment about its vertical
centroidal axis. Take R
100mm.
Y
Y
V
X
R
X
Solution:
Y
To Locate Cartaoid
Symete
Tn Ordu
t
to
above
Contkotdal
shodd
hstamu,
z
diten mine
/00 m m
-aai
Lot es
Comtidlv
kigion
A
Arra CA
Sacttom
2
G4) a.Cikele
-
Tix1oo
333 3xIO
4x1bo
317
a . C e e l - 7xloo
lDO 4xlo0
452.06XO
377
-50ox1
Squa
EA
5 708 mm
ZAZ
zA
lDOx1D8
50mm
50
ZAT = 8 5 3 9 XiO
To
ditesmine
Momat
mactia CIy)
2
Tyy
O-00343 x200+ 11x1DD C 50- 4x100
2
O-0034 3x 0o t
77x/00
4
loo x100
12
Tyy
7. 08xo m
(50- 00 +4 X100)
317
No
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A member ABCD is subjected to point loads P1, Pi. PJ and P, as shown in Fig. Calculate
the force Pi necessary for equilibrium. if P,
= 45 lcN,
PJ
= 450 kN and P, = 130 /cN.
Detennine the net defonnation of the member, assuming the modulus of elasticity to be 210
0Pa
C
B
D
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)
625mm1
Pi
~
E
Q
PJ
....
C)
1250mm1
~
f-'-
1200 mm - --lt 600 mm
1
t--
900 mm
-----t
Solution:
1J
Valt.u
Pa.
P, + P3
IJ-5
+
0
a 'I u,. lPbf tum
nt aid~$.
=-
P~ + Pi,
J.J.So = ~ t /3o
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uod
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C-ren1~)
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Ceo~)
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=
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A bronze bar is fastened between a steel bar and an aluminium bar as shown in Fig.8. Axial loads arc
applied at the positions indicated. Find the largest value of P that will not exceed an overall dcfonnation
of J.0 mm, and also the following stresses: U0 MPa in the steel, /20 MPa in the bronze, and 80 MPa in
the aluminium. Assume that the assembly is suitably braced to prevent buckling. Use E11 = 200 GPa,
E.,
= 70 GPa. and £ 6, = 83 GPa.
+-,
P ,
Aluminum
Bronze
Stttl
A = 480 mml
A = 65IJ mml
5
A = 320 mm2
B, reor1fidt====:::::i1--•~~
1 4 - - - - - M - - - - - -........ - - - - ~
I.Sm
l.Om
2.11 m
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P◄
P◄
JJ k.,n
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=
( PL)
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\ -:JX ~tut
0•3
IP
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4sox2x,o~
= 42 · =71
4-BD
l~o
-
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80
::
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650
3.20
vaD.111,
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Both portions or the rod A /IC a~• madI,;• ot. un alum1111um lor "hic h F· -o C'., ,1, • "n1m
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p
A
r
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lO mna
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u
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d lamc -tc-r
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J.
~
80
.
Q
L)
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= O
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4
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A brass bar, having cross sectional area of 1000 mm 2, is s"!bjected to axial forces as shown in Fig. 7. Find
the net defonnation of the bar. Take E = I 05 GPa.
A
SOkN ,.
B
SOkN
t-
600 mm
C
I
..
--+-
•
I 000 mm
D
I
20~
IOkN
I·
~,
1200mm
~,
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5o ..
,. -A-:__ _ _-11 B
so ....--______ _:-l_-_-_~_.. _B_o_ ____,~l•-A
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c
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C
_B_o_:. ==~_,__jJ2_ 0_
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13
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C
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A metallic block of 300 mmx 100 mm x 40 mm size is subjected to a tensile forces of 5 kN on 100x
40 faces, 4 kNon 300 x 100 faces and 6 kN on remaining faces. Detemine the change in volume ofthe
block. Take E = 200 GPa and Poisson's ratio = 0.25. Also find the change that should be made in the
4 kN load, in order that there should be no change in the volume of the block.
Starein z, H,3 duetom
3
2 s NJmm
Solution:
DO X4D
3
6X/0
0.S Nmm
300x40D
O133
wlmm
300xlo00
Chang i
Velu
loct.
D-133) C/-2x0.25)
C:5+05+
v
300xlbox4o
v
cha
5.649 mm
v-o
in_tkN_ Lead whun
valu
e.'
E #0
C0xt y+°5) C-2p):0
-24
#FO
CAs kes ba
a-y-7*
-
7 5 75 N]mm
175
0:25-033)
N/mm
2
CCompative)
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P
300x10DO
P 525 kN Compatve)
Scanned with CamScanner
Iaboratory tests on human teeth indicate that the anea effcctive during chew ing is appruximately 0.25 cm
and that the tooth Iength is about 1.1 em. If the applicd load in the vertical direetion is 880 N and the measured
shortening
is 0.004
cm.
determine
Young's
modulus. normal
stresses and normal strains. Consider
Poisson's
ratio. 0= 0.393.
Solution:
- 880
2 5O-25 x10
- -P
-
- 35-2
Als,
F 9z -*2.
-
Oo
35-2) 53
0.393
E
-
Ma
552
880N
O.o04
2
0.25Cm
-3.64
-3.64 xio"
11
88N
3.64X1o
E 9 67Gra
Cn
.67
13 83
|
431XIo|
|143ixto
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A bar ofsteel is 60 mmx 60 mm in section and 180 mm long shown in Fig. It is subjected to a tensile load
of 300 kN along longitudinal asis and tensile loads of 750 AN and 600 kN on the lateral faces. Find the
change in dimensions of the bar and change in volume. Take E=200 GNA and a=0.3.
300 kN
150
kN
60 m
6 0
600 LN
OS
NA
600 kN
60 mm
--
300 kN
Solution:
z
6 0m m
600x10
55.55M
6oX180
G
Asoxi
69.44 MPa
60XI9D
3
300xo
Chang
=
60x60
83 33 MRa
?n Velsumu
G)
Ov
v
C-
83-33t69.44+55 55\ (1-2xD -3) x60x60x180
2
2x10S
270 mm
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Chorge in Dimmsiogt
e
to
1
Dimnsfo
55 65- 0:
Cz
-
ais
(69.44t 83.33)
85 xioS
n
Incaea
dimntion alomg
485xI6x 6o =
Dimo ito
x
mm
toy-ax
-
ey
munien
i
3 8 9 x i o x 6o = 8
along y- 0xJ
33x 1Dmm
en
'.Tncason
1
292x1ð
ime nio
alrng z-axit
x 180 O.04te?
mm
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A circle of diameter d
225 mm is scribed on an unstressed aluminium plate of thickness
Forces acting in the plane of the plate later cause normal stress a,
shown in Fig. For E 69 GPa and u = ,
82.7 Pu and a.
2) mm.
13N /Pa. as
determine the change in (a) the length of diameter AB. (b) the
length of diameter ('D. (c) the thickness of the plate. (d) the volume of the plate.
375 mm
375mm
77I7T7
Solution:
We
in
Strant
hat
note
aack
538
Dihuctfont
7 -0 -o-33x/38
-
r
e
G
Co-ordnate
xo*
-F.z
-
o 0:33C82:7+138
o 3
-055
e
x82:7-0)
-s-Pt-
/38-0.33
6 0 4 Xlo
aChar
in Dla A
OnBEzxd
AB
5 38x0x 225
0:12 mm
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)Chag in Dia CD
Jp
xd .
Oc
in 7tieKns"
eyx
O
a
225s
0:361 mm
Cho7g*
dt:
604x1o x
t
=
Pat
-/055x1ox 20
mm
0 0RI
-
Chor* in velum
e
e Plate
=e + +
v
Øv
s
38xi6"-1 055 xI6 +1604 x10)
375x375x20
3057 mm
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Elasttc Comstants (E G. K)
oungs moctulus
oR Modulus
Elastic/1y
enuan Stoes )
E
ineco Stian (e
.
Rigilty Moctulus oR
G =
3
Shsa macdulus
Shoa Stos
)
Shoan Stoatn
()
Volumetkic Sthos )
Bulk moduluk
Volumateit Stuin (E)
Ratfom betoeen
E &G
D
D
m
------.
Let u
Constder a
Thitk ns
mity and
squau
block
Subjact to
't'Whor
Stass
be
cde veloped
on
Lot
a eliago ral teniile
AC hovmq
au dagona! BD
Same omanituab
and
Shaon Stuts intenitty
o n ko as AB, Bc, cD omd DA.
PC
ABcD Srad a
Sam agnitudo
npmienu Compronan
T.
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SHrain
Tn
Diaqooa!
BD
SAran in enyth BD daue to
=
diagonal
tonstb S3oH onAc
Statn fm Lvngth BD ctu to
ie
olhagomal
Compwive Stsots
O
BD
Ao
t
Sthatn ?o BD
C) -
Faom Pg
Steatn
BD-BD
7n BD
BD
So
very Small.and
DD
D-Kt
we
can
BD
take
DBD'
honu
veey Smal.
B3E
A
BDA
BD'A -4s
Stsaln 1n BD
RBAB AD
Sthan n
D'D CoS4S
BD
ADVE
BD VADAB
BD
-
2
Equatin g
ADV
e
("
tom.g.
)
0rad
0
E
CI*H)
C: G)
E 2 6 Ctp)- (G)
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Rlatto
Let
us
betoeon F
Constd
Subjoet
oHa Co)
in
k:
cbseal elemont
o
ogual
to
ond
oliswct temut
+Aru
mutually
G
PrAponckculo eliectond.
Volumeteic StAH
b
a
Stan
Som
xamp
Bul
notking
in a l
3
Pteiity
fo
miutually
Papomc Culas
Hyobro Static
prw
a
ciuetto m
ctonwe
Volumattc tun
moculus Cr)
Volumateic S¥ros
Volummete Stva#n
Velumabtt Saln
(e) dy
C1-4)
3 C1-2P)
E3k Ci-94- (7)
Rlatton bedoem E, G emd k
Frem )
Subsutte
2G
fm Dguatfom (I)
or
E
E
3x(I-
2+) 3K (3-E
=
9kG-3KE
EG +3kE = 9kG
3kt G
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A bar of cross section 8 mm x8 mm is subjected to an axial pull of 7000 N. The lateral dimension ofthe bar
is found to be changed to 7.9985 x 7.9985 mm. ifthe modulus of rigidity of the material is 80 GPa, determine
the poisson's ratio and Young's Modulus of the material.
Solution:
8-7.9985
Lateral strain
Change inlateral dimension =
8
Original lateral dimension
O.001S
8
875 x I *
Tensile stress ( ) =
7o00 = 109:375 MR
64
P
Area
Longitudinal Strain
Lateral Strain
ux Longitudinal Strain
E
875xIo7 = H x 109.375
E
583
E
Ale hav
2G
33 x/o
C+A)
R
-)
-é)
E 9x 8 xio (CI+R)
e583 33x10
2x8x1o (ItH)
O378
From, e
weg
e
E
220 GPa
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Calculate the modulus of rigidity and bulk modulus ofa cylindrical bar of diameter 30 mm and of length 1.5 m
if the longitudinal strain in a bar during a tensile stress is four times the later strain. Find the change in volume,
when the bar is subjected to a hydrostatic pressure of 100 N/mm. Take E= 100 GPa
Solution
Vetum
ba V)
" L = x30 x1So
4
06
Laleral Suatrx 4=
M:
6
x1D m m
long/tudmal
S9ain
D.05
G, Cl+0-25)
*
xio
4 0 GPa
Bulk ocdeelug
E
3Ci-24)
1xIoS
3x kC1- 2x0s )
K
66-67 GPa
Chang in Votum
Vots retatc Staon/1Arrttca
prat
vetumatte St«aiomCa)
66.67xo :
e
Chang
4999xio
n
voluma
&V)= 14999x16 "x 0 6 xlo
Sv 159o mm
3
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