- The document defines concepts related to forces, stress, strain and elasticity. It also discusses shear force, bending moment, and torsion. - Examples are provided for drawing shear force and bending moment diagrams for cantilever beams and simply supported beams under different loading conditions like point loads, uniformly distributed loads. - The key steps involve calculating reactions, determining shear force and bending moment values at different points, and plotting them on diagrams to understand their variations along the beam.

1. UNIT – I
• Basic Definitions of Force – Stress – Strain –
Elasticity. Shear force – Bending Moment –
• Torsion . Simple problems on Shear force
Diagram and Bending moment Diagram for
cantilever and simply supported beams.

2. • Basic Definitions of Force – Stress – Strain –
Elasticity.

3. Stress and strain
• When a body is subjected to a system of
external forces, it undergoes a deformation.
• At the same time, by virtue of its strength, it
offers a resistance against this deformation.
• This internal resistance offered by the body to
counteract the applied load is called stress.
• The resistance per unit cross sectional area is
called stress. The deformation in unit original
dimension is termed as stain.

4. • Types of stresses and strains
• Tensile stress
• When an external force produces elongation of
the body in its direction, it is termed as
• tensile force.
• P= External tensile load
• R= Resistane induced in the material of the body
• A=Cross sectional area
• Tensile stress =Tensile load / Cross sectional are
of the body

5. • Tensile strain
• Tensile strain =Increase in length/ Original
length

6. • Compressive stress
When an external force causes shortening of the body in
the direction of force, it is termed as a compressive
force. The stress developed in the body due to a
compressive force is called as compressive stress.
• P= External compressive load
• R=Resistance induced in the material of the body
• A=Cross sectional area of the body
• Compressive Stress = compressive load/ cross sectional
area of the body

7. • Compressive strain
• Compressive strain = decrease in length/ Original length
• Shear stress
• When a body is subjected to two equal and opposite forces acting
tangentially across the
• resisting section, as a result of which the body tends to shear off
across the section, then
• this tangential force is termed as shear force and the stress induced
is called shear stress.
• Shear stress = Shear force/ Shear area
• Shear strain
• Shear strain =Transverse displacement/ Distance from the fixed
base

8. • Volumetric strain
• The change in volume of an elastic body due to external force in
unit original volume is
• called as the volumetric strain
• Volumetric strain = dv/V
• Lateral strain
• When a material is subjected to uni-axial stress within the elastic
limit, it deforms not
• only longitudinally but also laterally. Under tension the lateral
dimensions diminish and
• under compression they increase. The lateral deformation per unit
original lateral
• dimension is called lateral strain.

9. • 3.4 Elasticity
• When an external force acts on a body, it
undergoes some deformation. The property
by
• which a body returns to its original shape after
the removal of external load is elasticity.

10. • 3.4.1Elastic limit
• The limiting value of stress up to and within
which the entire deformation disappears on
• removal of the external forces is called the
elastic limit of the material. The body returns
• to its original shape after the removal of the
loading when the intensity of stress is within
• a certain limit. The limit is called elastic limit.

11. • 3.4.2Hooke’s law
• Hooke’s law states that when an elastic
material is stressed within elastic limit, the
stress
• is proportional to the strain.
• K = stress/strain

12. • 3.4.3Young’s Modulus
• The ratio of the axial stress to the corresponding
axial strain, with in elastic limit is called
• the young’s modulus
• E= Axial stress/Axial strain
• 3.4.4 Shear modulus
• The ratio of the shear stress to the corresponding
shear strain is a constant, which is
• called shear modulus
• G = Shear stress/shear strain

13. • 3.4.5 Bulk Modulus
• When a body is subjected to uniform direct stress
in all the three mutually perpendicular
• direction, the ratio of the direct stress to the
corresponding volumetric strain is found to
• be a constant which is called the bulk modulus
• K = direct Stress/ Volumetric Strain
• 3.4.6 Poisson’s ratio
• Poisson’s ratio = Lateral strain/ Longitudinal Strain

14. • 3.4.7 Relation between elastic constants
• E =2G ( 1+r)
• E=3K (1-2r)
• E = 9KG/ 3 K +G
• E= Young’s modulus
• G= Shear modulus
• K=Bulk modulus
• R= poisson’s ratio
• 3.4.8 Factor of safety
• Factor safety = Ultimate stress/ Allowable stress

15. Bending moment
• Bending moment acts about the axes of the
members that are perpendicular to the
member (vertical and horizontal).
• These two illustrations explain bending
moment:

17. • The figure below illustrates all the forces,
Compression, tension, Bending moment,
torsion and shear
•

19. Torsion acts about the longitudinal axis
of the member.

20. Shear force – Bending Moment
diagram for cantilever Beam
• Shear force on cantilever beam is the sum of
vertical forces acting on a particular section of
a beam.
• While bending moment is the algebraic sum of
moments about the centroidal axis of any
selected section of all the loads acting up to
the section.

21. • Example:
• Draw shear force and bending moment
diagrams of the cantilever beam carrying
point loads. As shown in figure;

23. • SOLUTION
• Shear Force
• To draw a shear force diagram. First find value
of shear force between varying loads.
• Let start from left side.
• Shear force Between point D and C
• S.F (D-C) = -100 kg.

24. • Shear force value increases gradually as we
move towards fixed end.
• Shear force Between Point C and B
• S.F (C-B) = -(100 + 200 ) = -300 kg.

25. • Now one can see, shear force between point C
and B is the sum of point loads acting up to
that point.
• Shear force Between Point B and A
• S.F (B – A) = -(100 + 200 +300 ) = – 600 kg.

26. • One can see shear force between B and A is
the sum of all point loads acting on it.
• This shows shear force is maximum at fixed
end and minimum at free end of cantilever
beam.

28. • Bending Moment
• Bending moment at point D = B.M (D) = 0
• Bending moment at point C = B.M (C) = -(100×1) = -100 kg.m
• Bending moment at point B = B.M (B) = – (100×2 +200×1)
• B.M (B) = -400 kg.m
• Bending moment at point A = B.M (A) = (100×3 + 200×2 + 300×1)
• Total Moment at point A = B.M (A) = -1000 kg.m
• This shows that maximum bending will be at fixed end. Point loads
from free end transfer their bending strength towards fixed end. At
the end it acts as a sum of all bending moments.

29. • This shows that maximum bending will be at
fixed end. Point loads from free end transfer
their bending strength towards fixed end. At
the end it acts as a sum of all bending
moments.

31. Shear Force & Bending Moment
Diagram of Simply Supported Beam
• Shear force and bending moment diagram of
simply supported beam can be drawn by first
calculating value of shear force and bending
moment. Shear force and bending moment
values are calculated at supports and at
points where load varies.

32. • SIMPLY SUPPORTED BEAM WITH POINT LOAD
EXAMPLE
• Draw shear force and bending moment
diagram of simply supported beam carrying
point load. As shown in figure below.

34. • Solution
• First find reactions of simply supported beam.
• Both of the reactions will be equal. Since,
beam is symmetrical. i.e.,
• R1 = R2 = W/2 = 1000 kg.

35. • Now find value of shear force at point A, B and C.
• When simply supported beam is carrying point
loads. Then find shear force value in sections.
Shear force value will remain same up to point
load. Value of shear force at point load changes
and remain same until any other point load come
into action.
• Shear force between ( A – B ) = S.F (A-B) = 1000
kg

36. • Shear force between (B – C) = S.F (B -C) = 1000
– 2000
• S.F (B – C) = – 1000 kg.

37. SHEAR FORCE DIAGRAM

38. • Bending Moment
• In case of simply supported beam, bending
moment will be zero at supports. And it will be
maximum where shear force is zero.
• Bending moment at Point A and C = M(A) = M(C)
= 0
• Bending moment at point B = M(B) = R1 x
Distance of R1 from point B.
• Bending moment at point B = M (B) = 1000 x 2 =
2000 kg.m

40. SIMPLY SUPPORT BEAM WITH UDL &
POINT LOAD EXAMPLE
• Draw shear force and bending moment
diagram of simply supported beam carrying
uniform distributed load and point loads. As
shown in figure.

42. • Solution
• First find reactions R1 and R2 of simply
supported beam.
• Reactions will be equal. Since, beam is
symmetrical.
• R1 = R2 = W/2 = (600 +600 + 200 x4)/2 =
1000kg
• Hence, R1 = R2 = 1000 kg

43. • Shear Force
• Shear force between section A – B = S.F (A – B) = 1000 kg.
• Shear force at right side of point B = S.F (B) = 1000 – 600
• S. F (B) right = 400 kg.
• Now shear force at left side of point C.Because of uniform
distributed load, value of shear continuously varies from
point B to C.
• Shear force at point C (Left) = S.F (L) = 400 – (200×4)
• Shear force at point C (Left) = S.F (L) = -400 kg
• Shear force between section C – D = S.F (C-D) = -400 – 600
• Shear force between section C – D = S.F (C-D) = -1000 kg.

44. Shear Force Diagram

45. • From Shear force, one can see;
• Shear force is maximum at point A and remain same
until point load.
• At point B shear force value decreases, because of
point load.
• From B to C shear force continuously decreases,
because of udl.
• At point C shear force gradually falls, because of point
load.
• From point C to D, shear force remain same, because
no other point load is acting in this range.

46. Bending Moment
• B.M WILL BE ZERO AT SUPPORTS. I.E.,
• M(A) = M(D) = 0
• B.M at points B and C = M(B) = M(C) = 1000 x2 =
200 kg.m
• Now, how to find maximum bending moment?
• Bending moment will be maximum at point,
where shear force is zero. Hence, bending
moment will be maximum at mid point.
• M (max) = 1000×4 – 600×2 -200×2(2/2)
• M (max) = 2400 kg.m

47. Bending Moment Diagram

48. • Torsion is the shear force per unit area (shear
stress) that arises due to the application of an
external torque which is causing the structural
member to twist.
• An associated moment because of the shear
force is called twisting moment.

49. Simple problems on Shear force Diagram and

50. Bending moment Diagram for cantilever
and simply supported beams