The document discusses stresses in beams, including: 1. Pure bending causes zero shear force, resulting in a constant bending moment along the beam. 2. A positively curved beam under bending moment develops compression on top and tension on bottom, with the neutral axis in between with zero stress. 3. Shear stresses in beams are highest at the center of rectangular beams and at the edge of circular beams. 4. Slope, deflection, and radius of curvature are related through beam equations involving the bending moment and moment of inertia.

1. JAHANGIRABAD INSTIUTE OF TECHNOLOGY
BARABANKI
Department of Mechanical Engineering
Stresses in Beams
RAVI VISHWAKARMA
10/06/17 RAVI VISHWAKARMA 1

2. Pure Bending
As we are aware of the fact internal reactions developed on any
cross-section of a beam may consists of a resultant normal force, a
resultant shear force and a resultant couple. In order to ensure
that the bending effects alone are investigated, we shall put a
constraint on the loading such that the resultant normal and the
resultant shear forces are zero on any cross-section perpendicular
to the longitudinal axis of the member,
That means F = 0
Thus, the zero shear force means that the bending moment is
constant or the bending is same at every cross-section of the
beam.
10/06/17 RAVI VISHWAKARMA 2

3. Normal stresses in beams
A beam subjected to a positive bending moment will tend to
develop a concave-upward curvature. Intuitively, this means the
material near the top of the beam is placed in compression along
the x direction, with the lower region in tension. At the transition
between the compressive and tensile regions, the stress becomes
zero; this is the neutral axis of the beam. If the material tends to
fail in tension, like chalk or glass, it will do so by crack initiation
and growth from the lower tensile surface. If the material is strong
in tension but weak in compression, it will fail at the top
compressive surface; this might be observed in a piece of wood by
a compressive buckling of the outer fibers.
10/06/17 RAVI VISHWAKARMA 3

4. Shearing Stresses
Shear stress in Beam cross-section (Rectangular)
d
y
10/06/17 RAVI VISHWAKARMA 4
y
meanττ
2
3
max =

5. Solid Circular Section
10/06/17 RAVI VISHWAKARMA 5
N A
R
meanττ
3
4
max =

6. Relation between Slope, Deflection and Radius
Curvature
10/06/17 RAVI VISHWAKARMA 6
X
Y
dx
dy
2
2
xd
yd
EIM =

7. Sign Convention
To find out the slope and deflection of a center line of a beam at any
point proper sign conventions will have to be taken into account, the
following sign conventions will be used:
1.X is positive when measured towards right.
2.Y is negative when measured downwards.
3.Bending moment is negative when hogging.
4.Slope is negative when the rotation is clockwise.
10/06/17 RAVI VISHWAKARMA 7

8. SLOPE AND DEFLECTION AT A SECTION
The important method used for finding out the slope
and deflection at a section in a loaded beam are
discussed as follows:
1.Double integration method
2.Moment area method
3.Macaulay method
10/06/17 RAVI VISHWAKARMA 8

9. Cantilever Beam-Point Load
A B
Downward deflection of
10/06/17 RAVI VISHWAKARMA 9
l
ymax
w
EI
Wl
B
3
3
=

10. Cantilever Beam-UDL
10/06/17 RAVI VISHWAKARMA 10
EI
Wl
8
3
l
ymax

11. Macaulay’s Method
In Macaulay’s Method a single equation is formed for all loadings on a
beam, the equation is constructed in such a way that the constants of
integration apply to all portions of the beam. This method is also called
method of singularity functions.
10/06/17 RAVI VISHWAKARMA 11
A B
W1 W2
a
b
l
( ) ( )bxWaxWR
dx
yd
EIM Ax −−−−Χ== 212
2

12. Moment Area Method
The Moment area method is partially convenient is case of beams acted
upon with point loads in which case bending moment area consists of
triangles and rectangles. In the case of distributed load the
determination of the position of centroid itself involves integration and
as such it no longer remains simpler than Macaulay method. However,
this method may be conveniently used in certain standard cases of
distributed load where the position of the centroid of the bending
moment area is known.
10/06/17 RAVI VISHWAKARMA 12

13. Fixed Beam
A fixed beam is a beam the ends of which are constrained or built in to
remain in horizontal position.
Following points are worth noting :
1.Due to the fixidity, the slope of the beam is zero at each end, and a
couple or moment will be induced at each end to satisfy this condition.
2.End moments in case of fixed beams tend to bend the beam with
convexity upwards whereas the normal downward loads tend to bend
the beam with concavity upwards.
10/06/17 RAVI VISHWAKARMA 13

14. Torsion of Shaft
To transmit energy by rotation it is necessary to apply a turning force.
In case of a shaft if the force is applied tangentially and in the plane of
transverse cross-section the torque or twisting moment may be
calculated by multiplying the force with the radius of the shaft. If the
shaft is subjected to two opposite turning moments it is said to be in
pure torsion and it will exhibit the tendency of shearing off at every
cross-section which is perpendicular to longitudinal axis.
10/06/17 RAVI VISHWAKARMA 14

15. Torsion Equation
Torsion equation is based on following assumptions:
1.The material of the shaft is uniform throughout.
2.The shaft circular section in section remains circular after loading.
3.The twist along the length of shaft is uniform throughout.
4.The distance between any two normal cross-section remains the same
after the application of torque.
5.And maximum shear stress does not exceed its elastic limit value.
Cont.…..
10/06/17 RAVI VISHWAKARMA 15

16. 10/06/17 RAVI VISHWAKARMA 16
l
C
RI
T
p
θτ ==
T=Maximum twisting torque,
D=Diameter of the shaft,
Ip=Polar moment of inertia,
τ= Shear stress
C=Modulus of rigidity
=The angle of twist (radians)Ɵ
and
l= length of shaft.

17. Power Transmitted By The Shaft
Consider a force F newton's acting tangentially on the shaft of radius R.
if the shaft due to this turning moment (F×R) starts rotating at N r.p.m.
then, Work supplied to the shaft/sec
10/06/17 RAVI VISHWAKARMA 17
100060
2
Χ
Π
=
NT
P