mechanical of solids ppt unit 2.pptx

DNR College of Engineering and
Technology
MECHANICS OF SOLIDS
by
M.THAMBI BABU
Assistant professor
Department of Mechanical Engineering
DNR College of Engineering and Technology

UNIT-II
SHEAR FORCE AND BENDING MOMENT
Definition of beam – Types of beams – Concept of shear
force and bending moment – S.F and B.M diagrams for
cantilever, simply supported and overhanging beams
subjected to point loads, u.d.l, uniformly varying loads
and combination of these loads – Point of contra flexure
– Relation between S.F., B.M and rate of loading at a
section of a beam.
Department of Mechanical Engineering DNR
College of Engineering and Technology

SHEAR FORCE AND BENDING MOMENT
Beam: It is a structural member which is subjected to transverse
load.
Transverse load: The load which is perpendicular or it has non
zero component perpendicular to longitudinal axis of beam.
Shear force: it is the unbalanced force on either side of a section
parallel to cross section.
Moment: It is the product of force and perpendicular distance
between line of action of the force and the point about which
moment is required to be calculated.
Bending moment: It is the unbalanced moment on either side of
section, in the plane of beam.

TYPES OF LOAD
Sr. No. Type of load Example Description
1. Point Load Concentrated Load acts at a point.
2.
Uniformly Distributed
Load
UDL:- uniform load distribution over
wide area. Rate of loading per unit
length.
3.
Uniformly Varying
Load (Triangular
Distributed Load)
UVL:- Intensity of load at one point
to that at the other, eg. w1/m at C to
w2/m at D
4. Couple
A beam may be subjected to a
couple.
5. Oblique Load
The effect of Horizontal
components is to cause a thrust in
the beam. Vertical components of
the load cause bending and shear
& are treated as usual vertical loads
on a beam

TYPES OF SUPPORTS
Sr. No. Type of support Example Description
1.
KnifeEdge
Support
Contact Area Insignificant.
Provides only vertical reaction
No resistance to turning or lateral
displacement
2.
Roller Support
(Horizontal
Plane)
Rollers on Horizontal Plane
Support reaction is vertical.
No resistance to turning or lateral
displacement
3.
Roller Support
(InclinedPlane)
Rollers on Inclined Plane
Support reaction is perpendicular to
inclined plane.
Allow turning or lateral
displacement.
4.
Hinged/Pin
Support
Allows turning but doesn’t allow any
lateral movement.
Support reaction could be in any
direction.
Can be determined by resolving applied
in horizontal. & Vertical. direction
5. Fixed Support
Doesn’t allow rotation or
translation.

Sr. No. Type of Beam Example Description
1. Cantilever Beam
Beams have one end rigidly
built into the support.
Large span or heavy loads
provided by additional support
are known as propel and beam
as a propped cantilever.
2.
Simply Supported
Beam
Beams with knife edge
supports or roller supports at
ends.
3.
Beams with
Overhangs
Portion of a beam that goes
beyond the support is called
overhanging, may be on one or
both ends.
4. Fixed Beams
Rigidly built-in-supports at
both ends. Beam have support
reaction and a fixing moments
at each end.
5. Continuous Beams
Beams that cover more than
one span.

Both shear force and bending moment are vector quantities
requiring a convention of signs in order that values of opposite sense
may be separated. Mathematical signs are chosen since it is in
calculation problems that it becomes necessary to use such a
conventionTABEL1. Signconvention andunits for shearingforce andbendingmoment
LOAD
EFFECT
SYMBOL
SIGN CONVENTION
UNITS
POSITIVE (+) NEGATIVE (-)
Shear
force
Q, V,S N
KN
Bending
moment
M
(bottom fibers intension) (top fibers intension)
N·mm
kN·m
BEAMS IN BENDING

BEAMS IN BENDING
The shearing force, at any transverse section in a loaded beam, is the
algebraic sum of all the forces acting on one (either) side of the
section.
The bending moment, at any transverse section in a loaded beam, is
the algebraic sum of the moments about the sections of all the forces
acting on one (either) side of the section.
RA RB
A
V1 V2 V3 V4
x
B
a4
a3
a2
a1
a b
Working to the left of x:
Qx  RA  V1 V2  V3
Mx  RA a  V1 a1  V2 a2  V3 a3
Working to the right of x:
Qx  RB V4
Mx  RB b  V4 a4

THINGS TO REMEMBER FOR DRAWING OF S.F & B.M
 Start from right hand section.
 Use Sign convention of the side which you are choosing i.e Right or Left.
 If the thing are complicated use other side of section.
 Start from zero and end to zero. B.M at the ends will be zero.
 End point of S.F. will be equal and opposite to the reaction at that point.
 Mark the points and draw the diagram considering the type of load.
 At change in nature of forces there will be two points in shear force diagram.
 At Couple there will be two points in B.M Diagram.
•Beams are assumed to be always straight, horizontal & of uniform c/s & structure,
unless otherwise specified.
•Self weight of Beam neglected unless its definite value is given.
S.F may be max. at supports or under point loads or where S.F. is zero where B.M.
may be maximum at point where S.F. is zero or where S.F. changes its nature.
•Where the B.M changes its nature is known as Point of Contra Flexure or Point of
Inflexion.

Cantilever with a Point Load at its Free End
Consider a *cantilever AB of length l and carrying a point load W at its free end B as shown fig. We know
that shear force at any section X, at a distance x from the free end, is equal to the total unbalanced vertical
force. i.e.,
Fx = – W...(Minus sign due to right downward)
and bending moment at this section,
Mx = – W . x...(Minus sign due to hogging)
Thus from the equation of shear force, we see that the shear force is constant and is equal
to – W at all sections between B and A.
And from the bending moment equation, we see that the bending moment is zero at B (where x
= 0) and increases by a straight line law to – Wl; .
at (where x = l).
Now draw the shear force and bending moment diagrams as shown in Fig.

1. Draw shear force and bending moment diagrams for a cantilever beam of
span 1.5 m carrying point loads as shown in Fig.
Given : Span (l) = 1.5 m ; Point load at B (W 1) = 1.5 kN and point load at C (W2)= 2 kN.
Shear force diagram
The shear force diagram is shown in Fig. 13.3 (b) and the values are tabulated here:
FB = – W1 = – 1.5 kN
FC = – (1.5 + W2) = – (1.5 + 2) = – 3.5 kN
FA = – 3.5 kN
Bending moment diagram
The bending moment diagram is shown in Fig. 13.3 (c) and the values are tabulated here:
MB = 0
MC = – [1.5 × 0.5] = – 0.75 kN-m
MA = – [(1.5 × 1.5) + (2 × 1)] = – 4.25 kN-m

Cantilever with a Uniformly Distributed Load
Consider a cantilever AB of length l and carrying a uniformly distributed load of w per unit length, over the
entire length of the cantilever as shown in Fig.
We know that shear force at any section X, at a distance x from B,
Fx = – w . x... (Minus sign due to right downwards)
Thus we see that shear force is zero at B (where x = 0) and increases by a straight line law to – wl
at A as shown in Fig
We also know that bending moment at X,

• 2
Thus we also see that the bending moment is zero at B (where x = 0) and increases in the form of
1.A cantilever beam AB, 2 m long carries a uniformly distributed load of 1.5
kN/m over a length of 1.6 m from the free end. Draw shear force and bending moment diagrams
for the beam.
Given :
span (l) = 2 m ; Uniformly distributed load (w) = 1.5 kN/m and length of the cantilever CB
carrying load (a) = 1.6 m.
Shear force diagram
The shear force diagram is shown in Fig. 13.5 (b) and the values are tabulated here:
FB = 0
FC = – w . a = – 1.5 × 1.6 = – 2.4 kN
FA = – 2.4 kN
Bending moment diagram
The bending moment diagram is shown in Fig. 13.5 (c) and the values are tabulated here:
MB = 0 Department of Mechanical Engineering

BENDING MOMENT DIAGRAM
CANTILEVER WITH UDL
Shear Force
S.F at B = 10 kN
S.F at C = 10 + 5 x 2 = -20kN
S.F at D = 10 + 5 x 2 + 20 = - 40kN
S.F atA = 10 + 5 x 2 + 20 + 40 x 3
= - 160 kN
Calculations
Bending Moment
BMB = 0
BMC = 10 x 2 + 5 x 2 x 2/2 = -30 kNm.
BMD = 10 x 3 + 5 x 2 x (2/2 +1)
= 50 kN.m
BME = 10 x 5 + 5 x 2 x ( 2/2 + 3) +20
x 2
= - 130 k Nm
BMA = 10 x 8 + 5 x 2 x (2/2 + 6) + 20x
5 + 40 x 3 x 3/2
= - 430 kN.m.
+ve
-ve
+ve
-ve
+ve
-ve -ve
+ve
20 kN
3m 2m
A
40 kN/m
E D C
B
2m
10 kN
5 kN/m
1m
- VE S.F.
A
E
D
C
B
20 kN
10 kN
40 kN
160 kN
430 kN.m
130 kN.m
50 kN.m
Parabolic curve
30 kN.m
- VE B.M.
Parabolic curve
A E D C B
Department of
Mechanical
Engineering
DNR College of

Simply Supported Beam with a Point Load at its Mid-point
Consider a *simply supported beam AB of span l and carrying a point load W at its mid-point C as shown
in Fig. 13.12 . Since the load is at the mid-point of the beam, therefore the reaction at the support A,
RA =R B = 0.5 W
.Simply supported beam with a point load
Thus we see that the shear force at any section between A and C (i.e., up to the point just before the load
W) is constant and is equal to the unbalanced vertical force, i.e., + 0.5 W.
Shear force at any section between C and B (i.e., just after the load W) is also constant and is equal to the
unbalanced vertical force, i.e., – 0.5 W as shown in Fig. (b).
We also see that the bending moment at A and B is zero. It increases by a straight line law and is maximum
at centre of beam, where shear force changes sign as shown in Fig. (c).

2m 2m
A
Reactions
RA +RB = 100 = 20 x 2 + 50 = 190
MA,
RB x 8 = 100 x 6 + 20 x 2 x 50 x 2
RB = 112.5 kN, RA = 77.5 kN
RB
Calculations
BMA = 0 , BMB = 0
BMC = 112.5 X 2 = 225 kN.
BMD = 112.5 X 4 – 100 X 2 – 20 X 2 X 2/2
= 210 kN.m
BME = 112.5 X 6 – 100 X 4 – 20 X 2 X 3
= 155 kN.m
BMF = 112.5 X 2.625 – 100 X 0.625 – 20 X
0.625 X 0.625/2 = 228.9 kN.m
+ve
-ve
+ve
-ve
+ve
-ve -ve
+ve
20 kN/m
EFFECT OF UNIFORMLY DISTRIBUTED LOAD
50 kN 100 kN
E D
2m
LOADED BEAM
F C
B
RA
2m
Shear Force
S.F at B = 112.5 kN (+ ve)
S.F at C = 112.5 -100 = 12.5 kN
S.F at D = 112.5 -100 – 20 x 2 = -27.5 kN
S.F at E = 112.5 – 100 – 20 X 2 – 50 =
-77.5 kN
S.F. at A = – 77.5kN
S.F. at F = 0
S.F. at F = 112.5 – 100 – 20 (x – 2)
X = 2.625 m
Bending Moment
A
E D
F C
B
E D F C
B
A
x
112.5 kN
27.5 kN
77.5 kN
225 kN.m
SHEAR FORCE DIAGRAM
228.9 kN.m
210 kN.m
115 kN.m
Department of
Mechanical Engineering
DNR College of
Engineering and

Simply Supported Beam with a Uniformly Distributed Load
Consider a simply supported beam AB of length l and carrying a uniformly
distributed load of w per unit length as shown in Fig. 13.15. Since the load is
uniformly distributed over the entire length of the beam, therefore the reactions at the
supports A,
Simply supported beam with a uniformly distributed load

a
X
X
SHEAR FORCE DIAGRAM
35
20
20
35
35
35
20
20
C
D
E
B
A
F
20
25.5
20
Draw S.F. and B.M. diagrams for the loaded Beam Point Load
20 kN
C D E
20 kN
B
1m 1.3m 1.3m 1m
LOADED BEAM
A
70 kN
RC RE
Shear Force
S.F at B = -20kN
S.F at E = 55 - 20 = 35 kN
S.F at D = -20 – 70 + 55 = -35kN
S.F at C = -20 + 55 - 70 + 55 = 20kN
Symmetrical loading
Rc = RE = 20 + 70 + 20 = 55kN
2
Calculations
Bending Moment
BME = -20 X 1 = -20 kNm.
BMD = -20 X 2.3 + 55 X 1.3 = 25.5kNm.
BMC = -20(1.3 + 1.3 +1) + 55(1.3 + 1.3)
-70 X 1.3 = -20 kNm.
BMX = - 20 (a + 1) + 55a = 35 a –20
Point of Contra flexure BMX = 0 a =
0.57
+ve
-ve
+ve
-ve
+v
e
-ve -
ve
+ve
Point of Contra flexure
Department of
Mechanical Engineering
DNR College of
Engineering and

UNIFORMLY DISTRIBUTED LOAD
WL
1/2 L
L
1/2 L
Load act centre
of UDL
W kN/m
wL
1/2 L
L
1/2 L
W kN
w = W/L kN/m
There will be Parabola in B.M. and inclined line
in S.F. diagram
Department of Mechanical
Engineering DNR College of
Engineering and Technology

EFFECT OF COUPLE
Reactions
RA + RB = 20 X 4 = 80
MA,
RB X 10 = 40 + 20 X 4 X 2
RB = 20 kN, RA = 60 kN
RB
Calculations
Bending Moment
BMA = 0, BMB = 0,
BMC = 20 x 5 = 100 kN.m (just before)
BMC = 100 – 40 = 60 (just after)
BMD = 20 x 6 – 40 = 80 kN.m
BME = 20 x 7 – 40 – 20 x 1 x 0.5
= 90 kN.m
RA
40 kN-m
4m 5m
A
20 kN/m
E D C
B
1m
X = 7m
D C
SHEAR FORCE DIAGRAM
B
A + ve S.F.
- ve
E
D C B
A
20 kN
60 kN
90 kN.m
80 kN.m
100 kN.m
60 kN.m
+ ve BM
Shear Force
S.F at B = 20 kN
S.F at C = 20 kN
S.F at A = -60kN
S.F at E = 0
S.F at E = 20 – 20 (x-6)
x = 7m
E

a = 1m
OVERHANGING BEAM WITH UDL
MA,
RB x 6 = 5 x 9 + 2 x 9 x 9/2
RB = 21 kN
RA + RB = 5 + 2 x 9 = 23, RA = 2 kN
Calculations
Reactions
+ve
-ve
+ve
-ve
+ve
-ve
RA
E
D
3m
A
2 kN/m
B
RB
C
6m
5 kN
A
D
C
11 kN
5 kN
10 kN
+ ve B
- ve
2 kN
Point of Contra
flexure
1 kNm
A
D E
B
C
24 kNm
Shear Force
S.F at C = 5 kN
S.F at B = 5 + 2 x 3 = -11 kN (justright)
S.F at B = -11 + 21 = 10 kN (justleft)
S.F at A = 2 kN (-ve)
S.F between B & A = 0
S.F at D = - 2 + 2 x a
0 = - 2 + 2 a, a = 1m
Bending Moment
BMC = 0, BMA = 0
BMB = - 5 x 3 – 2 x 3 x 3/2 = -24kN.m
b = 4m
Point of Contra flexure BME = 0
-5 (b + 3) – 2 x (b+3)
2
/ 2 + 21 b = 0
b = 4 m (+ ve value)

UNIFORMLY VARYING LOAD
W N/m
W N/m
1/3 L
L
2/3 L
Load Act the centroid of
the Triangular
Area
There will be Parabola in both S.F. and B.M

+ve
-ve
+ve
-ve
+ve
-ve
C
A
W
L D
E
x
W.x/L
B
5000 N/m
C
A
4 m
D
E
x
B
Rate of Loading
S.F. = Triangular Load area = ½ X DE X DB
W x /2L
2
W L / 2
W x /6L
3
W L / 6
2
B.M = Force X Perpendicular
= ½ X DE X DB X DB / 3 from point D to
the centroid
Cubic Parabola
13.33 kN.m
Parabola Curve
10 kN.m
DE / AC = DB / AB , DE = 5000x/4 = 1250 x
i.e rate loading at any distance x
= - 625 x2X x/3, at x = 4 ,
B.M. at A = - 13.33kN.m
S.F. at D = -1/2 X x 1250 x = - 625 x2
S.F. at B = 0 where x= 0
S.F. at A , at x = 4 , -625 X 4
2
= 10 kN
B.M. at x = -1/2 X DB X DE X DB/3
Department of Mechanical
Engineering DNR College of
Engineering and Technology

Draw S.F. and B.M. diagrams for the loaded Beam
Reactions
RA + RB = 150 + 300
MA,
RB X 6 = 150 X 5 + 300 (2/3 X 3 + 1)
RB = 275 kN, RA = 175kN
Shear Force
S.F at B = 275 kN
S.F at C = 275 – 150 = 125 kN
S.F at D = 125 kN
S.F at E = 275 – 150 – 300 = -175 kN
S.F at A = - 175kN
Calculations
+ve
-ve
+ve
-ve
+ve
-ve
RA
300 kN
1m 1m
A
E D C
B
RB
3m
150 kN
1m
F
½ WL = 300
= 200 kN/m
x
A E F D C
B
275 kN
125 kN
175 kN
+ ve S.F.
+ ve B.M.
175 kNm
442.32 kNm
400 kNm
275 kNm
A E F D C
B
Rate of Loading at distance x
w = Wx/L = w = 200 x / 3
S.F at F = -175 + ½ 200x / 3 X x
x= 2.29 m
Bending Moment
BMA = 0, BMB = 0,
BMC = 275 X 1 = 275 kN.m
BMD = 275 X 2 – 150 X 1 = 400 kN.m
BME = 175 X 1 = 175 kN.m
BMF = 175 X 3.29 – (200 X 2.29/3) ( 2.29/2 X
2.29/3)
= 442.32 kN.m
½ WL X L/3

Draw S.F. and B.M. diagrams for the loaded Beam
MA,
RB X8= 200 X 8 X 4 + ½ X 400 X 8 X 8/3
RB = 1333.33N
RA + RB = 200 X 8 + ½ X 400 X 8
RA = 1866.67 N
Rate of Loading at X-X = GH + GF
Rate of Loading at GH
DE/CD = GH/CG, GH = 400x /8 = 50x
Rate of Loading at GF = 200
Rate of Loading at X-X = GH + GF = 200 +50x
Calculations
Reactions
+ve
-ve
+ve
-ve
+ve
-ve
Bending Moment
BMF = 1333.33x – 200x X x/2 – ½ X 50x X x X
x/3…….(GH = 50x)
We have x = 4.326
Max. B.M at F = 3436.14 N/m
200 N/m
600 N/m
8m B
RB
1333.3 N
1866.6 N x = 4.325 m
3222.18 N
H
G
F x
C
200 N/m
B
A
RA
E
400 N/m
D
200 N/m
A
X
X
Shear Force at P = 0
S.F. at F =1333.33 – (load BCGF + Load CGH
= 1333.33 – (200x + ½ X 50 x X x)
x = 4.326 (quadratic equation +ve value)
F

1.A beam is a structural member which is subjected to
a.
b.
c.
d.
Axial tension or compression Transverse
loads and couples Twisting moment
No load, but its axis should be horizontal and x-section rectangular or circular
(Ans: b)
2.Which of the following are statically determinate beams?
a.
b.
c.
d.
Only simply supported beams
Cantilever, overhanging and simply supported Fixed
beams
Continuous beams
(Ans: b)
3.A cantilever is a beam whose
Both ends are supported either on rollers or hinges One end
is fixed and other end is free
Both ends are fixed
Whose both or one of the end has overhang
(Ans: b)
a.
b.
c.
d.
4.In a cantilever carrying a uniformly varying load starting from zero at the free end, the shear force dia
ram is
a.
b.
c.
d.
A horizontal line parallel to x-axis A line
inclined to x-axis
Follows a parabolic law
Follows a cubic law
(Ans: c)
5.In a cantilever carrying a uniformly varying load starting from zero at the free end, the Bending
moment diagram is
a.
b.
c.
d.
A horizontal line parallel to x-axis A line
inclined to x-axis
Follows a parabolic law
Follows a cubic law
(Ans: d)
6.In a simply supported beam, bending moment at the end
a.
b.
c.
I
s always zero if it does not carry couple at the end
Is zero, if the beam has uniformly distributed load only Is zero if
the beam has concentrated loads only

d. May or may not be zero
(Ans: a)
7.For any part of the beam, between two concentrated load Shear force diagram is a
a.
b.
c.
d.
Horizontal straight line
Vertical straight line Line
inclined to x-axis Parabola
(Ans: a)
8.For any part of a beam between two concentrated load, Bending moment diagram is a
a.
b.
c.
d.
(Ans: c)
9.For any part of a beam subjected to uniformly distributed load, Shear force diagram is
a.
b.
c.
d.
(Ans: c)
10.For any part of a beam subjected to uniformly distributed load, bending moment
diagram is
a.
b.
c.
d.
Horizontal straight line Vertical
straight line Line inclined to x-axis
Parabola
(Ans: d)
11.A sudden jump anywhere on the Bending moment diagram of a beam is
caused by
Couple acting at that point Couple acting at
some other point Concentrated load at the
point
Uniformly distributed load or Uniformly varying load on the beam (Ans: a)
a.
b.
c.
d.

16.At a point in a simply supported or overhanging beam where Shear force changes
sign and = 0, Bending moment is
a.
b.
c.
d.
Maximum Zero
Either increasing or decreasing
Infinity
(Ans: a)
17. In a cantilever subjected to a combination of concentrated load, uniformly distributed
load and uniformly varying load, Maximum bending moment is
a.
b.
c.
d.
Where shear force=0 At
the free end
At the fixed end At
the mid-point
(Ans: c)
18. Point of contra-flexure is a
a.
b.
c.
d.
Point where Shear force is maximum Point where
Bending moment is maximum Point where Bending
moment is zero
Point where Bending moment=0 but also changes sign from positive to negative
(Ans: d)

1.Define point of contra flexure. (October/November - 2018 )
Answer: The point on the beam where B.M changes sign from positive to negative or vice
versa i.e. where the B.M. is zero is known as the point of contraflexure. This is also known as
point of inflexion.
2.Define shear force and bending moment ? (October/November – 2018)
Answer:
Shear force at any section of beam is defined as the algebraic sum of all the vertical loads
acting on the beam on either side of the section under consideration.
Sign Convention
Upward loads to the left of section or downward loads to the right of it are considered as
positive.
Downward loads to the left of a section or upward loads to the right of it are considered as
negative.

The bending moment at any section of a beam is defined as the algebraic sum of moments of all
the vertical loads (acting either to the left or to the right of the section) about the section.
Sign Convention
The bending moment producing convexity downwards are known as positive bending
moment. The positive B.M. is also known as sagging bending moment. Upward loads on
either side of a section produce sagging or positive bending moments as shown in fig.
The bending moments producing convexity upwards are known as negative bending
moment. The negative B.M are also known as hogging bending moment. Downward loads
on either side of a section produce hogging or negative bending moments as shown in fig.

3.What are the types of loads? (October/November - 2018 )
1. Concentrated load or point load
2. Uniform distributed load
3. Uniform varying load

mechanical of solids ppt unit 2.pptx

Recommended

Recommended

More Related Content

Similar to mechanical of solids ppt unit 2.pptx

Similar to mechanical of solids ppt unit 2.pptx (20)

More from dnrcollege

More from dnrcollege (7)

Recently uploaded

Recently uploaded (20)

mechanical of solids ppt unit 2.pptx