This document discusses various topics related to structural analysis and design including: 1. Structural mechanics enables determining forces on members based on loads on the whole structure. Stresses and deformations can then be obtained. 2. Structures can be categorized based on function, form, analysis perspective, and type of loads. Different types of loads include concentrated, distributed, uniform, and varying loads. 3. Key concepts in structural analysis are shear force, bending moment, and deflected shapes. Shear force and bending moment are calculated at cross sections and deflected shapes show beam deformation under loading.

1. CS5004
CIVIL STRUCTURAL ANALYSIS AND DESIGN

2. Structural Mechanics
• Structural Mechanics enables us to find the forces that act on individual members based on the
loads that are acting on the whole structure. Stresses, strains, internal forces and deformations in
members, then, can be obtained by using what we have already learned about the behaviour of
deformable solids.

3. categories depending on the system/mode of
classification:
(a)On the basis of its intended function/usage: Buildings, bridges, dams,
industrial sheds, cable ways,chimneys, etc.
(b)On the basis of its form/load transfer mechanism: Beams, columns, floor
slabs, arches, shells, trusses,frames,footings, etc.
(c) Considering the analysis perspective: 2-dimensional, 3-dimensional,
determinate, indeterminate, etc.

4. Various types of structures

5. categories based on various criteria for load :
(a)Based on the source/origin: Wind load, earthquake load, self weight, live load,
blast load, temperature stress, etc.
(b) Based on the direction of action: Gravity loads, lateral loads, etc.
(c) Based on time-variation: Static, dynamic, impulse, pseudo-static.
(d) Based on the mode of action/analysis point-of-view: Concentrated or point
load, distributed load, moment, pressure.

6. Load types based on direction of action
a)Lateral loads (b) Gravity loads

7. Load types based on analysis point of view
(a) Concentrated or point loads on (b)Distributed loads on a simply
a simply supported beam supported beam
(c) Pressure acting on inner surface of a cylinder

8. TYPES OF SUPPORTS
• Simple Support: If the beam rests simply on a support it is called a simple support. In such case
the reaction at the support is at right angles to the support and the beam is free to move in the
direction of its axis and also it is free to rotate about the support
• Roller Support: In this case, beam end is supported on rollers. In such cases, reaction is normal to
the support since rollers can be treated as frictionless(a). Many mechanical components are
having roller supports which roll between guides. In such cases, reaction will be normal to the
guides, in both the direction. At roller support beam is free to move along the support. It can
rotate about the support also(b).

9. TYPES OF SUPPORTS
• Hinged Support: At a hinged end, a beam cannot move in any direction. However, it can rotate
about the support . Hence the support will not develop any resisting moment, but it can develop
reaction in any direction to keep the end stationary. The reaction R can be split into its horizontal
and vertical components for the purpose of analysis.
VA = R sin θ
HA = R cos θ

10. TYPES OF SUPPORTS
• Fixed Support: At such supports, the beam end is not free to translate or rotate. Translationis
prevented by developing support reaction in any required direction. Referring to Fig the support
reaction R which is at an angle θ to x axis may be represented by HA and VA, where
VA = R sin θ
HA = R cos θ
• Rotation is prevented by developing support moment MA as shown in Fig. Thus at fixed support
A, there are three reactions HA, VA and MA.

11. TYPES OF BEAMS
• Simply Supported Beam: When both end of a beam are simply supported it is
called simply supported beam . Such a beam can support load in the direction
normal to its axis.
• Beam with One End Hinged and the Other on Rollers: If one end of a beam is
hinged and other end is on rollers, the beam can resist load in any direction.

12. TYPES OF BEAMS
• Over-hanging Beam: If a beam is projecting beyond the support. It is
called an over hanging beam. The overhang may be only on one side
as in fig or maybe on both sides as in Fig.
• Cantilever Beam: If a beam is fixed at one end and is free at the other
end, it is called cantilever beam.

13. TYPES OF BEAMS
• Propped Cantilever: It is a beam with one end fixed and the other end
simply supported.
• Both Ends Hinged: In these beams both ends will be having hinged
supports.

14. TYPES OF BEAMS
• Continuous Beam: A beam is said to be continuous, if it is supported
at more than two points.

15. TYPES OF LOADING
• Concentrated Loads: If a load is acting on a beam over a very small
length, it is approximated as acting at the mid point of that length and
is represented by an arrow as shown in Fig.

16. TYPES OF LOADING
• Uniformly Distributed Load (UDL): Over considerably long distance such load has
got uniform intensity. It is represented as shown in Fig. For finding reaction, this
load may be assumed as total load acting at the centre of gravity of the loading
(middle of the loaded length). For example, in the beam shown in Fig. the given
load may be replaced by a 20 × 4 = 80 kN concentrated load acting at a distance 2
m from the left support.

17. TYPES OF LOADING
• Uniformly Varying Load: The load shown in Fig.varies uniformly from C to D. Its
intensity is zero at C and is 20 kN/m at D. In the load diagram, the ordinate
represents the load intensity and the abscissa represents the position of load on
the beam. Hence the area of the triangle represents the total load and the
centroid of the triangle represents the centre of gravity of the load. Thus, total
load in this case is 1 / 2 × 3 × 20 = 30 kN and the centre of gravity of this loading is
at 1/ 3 × 3 =1 m from D, i.e., 1 + 3 – 1 = 3 m from A.
• For finding the reactions, we can assume that the given load is equivalent to 30
kN acting at 3 m from A.

18. Shear Force
• It is a transverse force, one part of the beam exerts on the
other part at any cross section. It is equal to the algebraic
sum of all the transverse forces (including the reaction)
either to the left or to the right of the cross-section. In the
case of horizontal beams, shear force is the unbalanced
vertical force on the left or right side of the cross section.

19. Bending Moment
• It is the couple, one part of the beam exerts on the other
part, at a cross-section. It is equal to the algebraic sum of
moments of all forces (including that of reaction) either to
the left or to the right of the cross-section about the section.
In the other words, bending moment is the unbalancing
moment of forces on any one side of the cross-section
considered.

20. SIGN CONVENTION

21. SIGN CONVENTION

22. SIGN CONVENTION

23. Cantilever beam

24. Cantilever beam

25. Cantilever beam /varying load

26. Cantilever beam

27. Simply support /point load

28. Simply support /udl

29. Simply support /uvl

30. Simply support

32. REACTIONS FROM SUPPORTS OF BEAMS
• A beam is in equilibrium under the action of the loads and the
reactions. Hence the equilibrium may be written for the system of
forces consisting of reactions and the loads. Solutions of these
equations give the unknown reactions.

33. REACTIONS FROM SUPPORTS OF BEAMS

34. REACTIONS FROM SUPPORTS OF BEAMS

35. Beam
• A structural member which is acted upon by a system of
external loads at right angles to its axis is known as beam.
Generally, a beam is a horizontal member to support floor
slabs, secondary beams, walls, stairs etc.

36. Static Equilibrium Equations
• According to the principle of statics, any structural member should satisfy the following equilibrium conditions.
1. Algebraic sum of all vertical forces should be equal to zero.
v = 0
Sum of upward vertical forces = Sum of down ward vertical forces.
()  ()
2. Algebraic sum of all horizontal forces should be equal to zero.
H = 0
Sum of forces towards right side = Sum of forces towards side
3. Algebraic sum of moments of all forces should be equal to zero.
M = 0
Sum of anticlockwise moments = Sum of clockwise moments.

37. According to static equilibrium equation
i) Statically determinate structures
ii) Statically indeterminate structures

38. Statically determinate beams
• When the reaction components of a beam can be analyzed by using
static equilibrium equations (V = 0, H = 0, M = 0) only, it is called as
statically determinate beam. In which the degree of indeterminacy is
equal to zero.
• The examples for statically determinate beams as given below
i) Cantilever beam
ii) Simply supported beam
iii) Overhanging beam

39. Statically indeterminate beam
• When the reaction components of a beam cannot be analyzed by
using static equilibrium equations (V = 0, H = 0, M = 0) only. The
beam is called as statically indeterminate beam. In which the degree
of indeterminacy is not equal to zero. (One or more than one).
• The examples for statically indeterminate beams are given below
i) Propped cantilever beam
ii) Fixed beam
iii) Continuous beam

40. Degree of Indeterminacy
• The difference between No. of unknown reaction components and
No. of known equilibrium equation is called Degree of Indeterminacy
or degree of redundancy.
• Degree of indeterminacy =(No.of unknown reaction components -
No.of known using static equilibrium equations)

41. Examples

42. Examples

43. Examples

44. Degree of Indeterminacy

45. Statically indeterminate structures

46. Static Indeterminacy

47. Indeterminacy for beams

48. Indeterminacy for Trusses

49. Example

50. Indeterminacy for Portal Frames

51. Example

52. indeterminate structures

53. Shear force (S.F) & Bending moment (B.M)
• The Shear Force at any section of a beam is the algebric sum of all the forces
acting either left or right of that section. It is denoted by F (or) SF. The symbol of
SF is F(or)V(or)SF.
• The bending moment at any section of a beam is the algebraic sum of all the
moments of the forces acting either left (or) right of that section. It is denoted by
B.M(or) M.

54. Deflected shapes of beam / Elastic line
(or) elastic curve of beam
• When a beam is subjected to transverse loads it develops shear force and bending moment at
every cross section. Due to transverse load the beam gets deflected. The deflected configuration
of the beam is known as deflected shape.
• Where, 𝜭B = Slope at B
SB = Deflection of free end(B)
• The configuration of the longitudinal axis
of the beam after bending takes place due
to loading is called elastic curve.
• The edge view of the deflected neutral surface
of a beam is known as elastic curve.

55. Deflected shapes of beam

56. Slope and Deflection
(a) Slope (𝜭 or I )
• The angle made by the tangent at a point on the elastic curve with the horizontal
is called the slope at the point. It is d enoted by 𝜭(or) i.
b)Deflection ( y)
• The vertical distance between the original axis to the elastic curve of the beam
after loading is called deflection. It is denoted by y.

57. Flexural rigidity and Stiffness of beams
(a) Flexural rigidity (EI)
• The product of values of young’s modulus and MI about Neutral axis is called flexural rigidity.
Flexural rigidity = Young’s modulus x M.I
= E x I = EI
The product of EI is expressed in N.mm2 (or) kN.m2
(b) Stiffness
• Stiffness of a beam is the property of resistance against rotation and deflection. The moment
required to produce unit rotation of slope is called stiffness of the beam. It is depends upon the
end conditions, flexural rigidity and span of the beam.
1. Stiffness factor, k = 4EI/L for fixed ends
2. Stiffness factor, k = 3EI/L for simply supported ends

58. Deflected shapes of beam
• Cantilever beam

59. Deflected shapes of beam
• Simply supported beams

60. Method of finding the slope and deflection
• The following are the various methods for slope and deflection.
1. Mohr’s area moment method
2. Double integration method
3. Macaulay’s method
4. Strain energy method
5. Conjugate beam method

61. Area moment method
It is a simple method, Mohr’s Theorem I & II are used for the determination of
slope and deflection of beams at any section with reference to the B.M.D., hence it
is called as Mohr’s area moment method.
(a) Mohr’s Theorem – I
• It states that the change of slope between any two points on an elastic curve is
equal to the area of BMD between the two points divided by flexural rigidity.
Slope =  = A / EI
(b) Mohr’s Theorem – II
• It states that the intercept taken on a vertical reference line of tangents at any
two points on an elastic curve is equal to the moment area of BMD between
these points about the reference line divided by flexural rigidity.
Y = Ax / EI = deflection.
Where, A = Net area of BMD
Ax = Net moment area of BMD.
EI = Flexural rigidity

62. A cantilever beam 2m span 200mm wide and 400mm deep. Carries a point load of 10 kN
at free end. Find the max. slope and deflection by area moment method take
E = 2.0 x 105 N/mm2.
• Given data:
By area moment method
Span  = 2m = 2000 mm , Wide b = 200mm,Depth d = 400mm, E = 2.0 x 105 N/mm2.

63. By area moment method

64. A cantilever beam of span 4m is subjected to a udl of 20 kN/m over a entire length. Find
the maximum slope and deflection. Take E = 2.1 x 105 N/mm2, I = 15 x 108mm4

65. By area moment method

66. A simply supported beam 8m long carries a point load of 90 kN at centre and udl of 5 kN/m
over at entire span. The size of beam is 200mm x 400mm E = 1.5 x 104 N/mm2. Determine
the maximum slope and deflection.

67. By area moment method

68. Highlights