SF, BM and Truss notes for engineering mechanics

Notes on beams and truss

1. Engineering Mechanics
Shear Force, Bending Moment
&
Truss
PROF. DULAL GOLDAR.
FORMER DIRECTOR DTU
[email protected]

2. SHEAR FORCES AND BENDING MOMENTS IN DIFFERENT BEAMS.
THREE POINT LOADING
A simply supported Beam of span length, L is subjected to a
point load, P at the centre of the span. Draw S.F and B.M
diagram

3. ROLLER SUPPORT AT B
HINGE SUPPORT AT A

4. SAGGING B.M
T-Tension at Bottom
C-Compression at Top
SHEAR FORCE
BENDING MOMENT
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM
DATUM
DATUM

5. SHEAR FORCE
BENDING MOMENT
S. F. D
B. M. D

6. RESULTS
S. F. D
B. M. D

7. A POINT LOADING
A Cantilever Beam of span length, L is subjected to a point
load, P at the Free-end. Draw S.F and B.M diagram
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM
S. F. D
B. M. D
HOGGING B.M
C-Compression at
Bottom
T-Tension at Top

8. FOUR-POINT LOADING
A Simply supported Beam of span, L is subjected to Two Point
Loading symmetrically placed from supports by a distance ‘a’
from each supports. Draw SF & BM diagrams
STEP-I
STEP-II
S. F. D
B. M. D

9. STEP-III

10. PURE
BENDING
MECHANICS OF
DEFORMABLE SOLIDS
ORIGINAL
SHAPE
DEFORMED SHAPE

11. RESULTS

12. UNIFORMLY DISTRIBUTED LOAD
A simply supported beam of span ‘L’ is subjected
to uniformly distributed load (u.d.l) of ‘w’ kN/m
throughout the span length. Draw SFD & BMD
STEP-I

13. SHEAR FORCE DIAGRAM

14. BENDING MOMENT DIAGRAM

15. RESULTS

16. A U.D.L LOADING
A Cantilever Beam of span length, L is subjected to a uniformly
distributed load (u.d.l), w kN/m throughout . Draw S.F and B.M
diagram
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM

17. x

18. HOME WORK

19. EXAMPLE OF TRIANGULARLY VARYING LOADS
WATER TANK
DAM

20. TRIANGULARLY VARYING DISTRIBUTED LOAD: A Cantilever Beam AB of span length, L is
subjected to a triangularly distributed load & intensity zero at free-end B and maximum w kN/m
at A fixed support. Draw S.F.D & B.M.D.
STEP-I
STEP-II SHEAR FORCE DIAGRAM
V
x
M
x
F.B.D

21. BENDING MOMENT DIAGRAM
x/3

22. RESULTS

23. RELATION BETWEEN U.D.L. & S.F AND B.M.

24. HOME WORK

25. STEP-I
DIFFERENT POINT LOADINGS
A simply supported Beam of span length, L is subjected to Two
point loads, 2P at C & P at D as shown in Fig. . Draw S.F and
B.M diagram
STEP-II STEP-III
V
x
M
x

26. STEP-IVRESULTS
4PL/9

27. SIMPLY SUPPORTED BEAM WITH OVER-HANGING
RAILWAY SLEEPER FOUNDATION

28. CAR CHASIS

29. kN/m (u.d.l)
S.S. BEAM WITH EQUAL OVER-HANG
A simply supported beam with equal over-hang on either
side of support is subject to a u.d.l‘w’ kN/m throughout
the length of the beam as shown in Fig. . Draw SFD & BMD
C
A
a
STEP-I SHEAR FORCE DIAGRAM
STEP-II BENDING MOMENT DIAGRAM
C
A

30. STEP-III SHEAR FORCE DIAGRAM
STEP-IV BENDING MOMENT DIAGRAM
S. F. D
B. M. D

31. RESULTS
SLEEPER
UNIFORMLY DISTRIBUTED PRESSURE FROM FOUNDATION

32. Truss:
Analysis of Simple Axial Force Structures:
Structure-as an element or an assembly of elements formed into a shape to resist a set
of external forces
Structural Analysis-1. Evaluation of forces and deformations in structures
2. (i) Based on simple principles of statics
(ii) Compatibility
(iii) Properties of materials
Discrete Structures:
Composed of linear elements, inter-connected to form a
network.
Linear or Slender Element:
One dimension is large w.r.t others, load is a
function of large dimension.
Pin-Connected Frame Work: Elements are very slender or flexible wires, can not resist
any bending moment

33. Forces acting on a structure must be transferred to supports; in the
process of transferring the loads, the elements of the structure act as a
medium of communication. The element resists external forces and
generate within themselves ‘resisting forces’-(function of external
forces and their location and geometry of the structures).
Two types of axial force resisting elements
a)Wire, ropes-resists only tension.
b)Metal tubes, rods & slender wooden posts resists compressionand
tension.
Wire & Rope Structures:
Temporarystructures, tents (primarily of ropes).
Some permanents structuressuspended roof, bridges, guyed musts.
Reacting forces of the rope is always along the axis.
Bending, twisting and compression forces can not be resisted by ropes.
Analysis of resisting forces in the rope assemblies is done by the
principles of statics.
In case of indeterminate structures principles of compatibility of
deformations are used.

47. EXAMPLE:Determine the conditions for which the weight can be supported. Assume that the axles
of the pulleys are frictionless. Compute equilibrium for W1 = 100 N, W2 = 50 N and W3 = 100 N.
Solution: (Note:-Tentionin any segment of a rope is constant unless an external force is coming on the rope
in segment
C
F.B.D of point C
If the total length of the rope is unlimited, then the limiting condition
beyond which W1 & W2 can not support. W3 is obtained by the limit of
angle

49. EXANPLE:
A person weighing 80 kg is walking on a pre-tensioned rope between two poles as shown in Fig.(a) .
Determine the deflection of the rope when the person is at ¼ and ½ of the span. The tension in the rope is 1000 kg.
C
F.B.D. of the point C

51. EXAMPLE: A typical simplified suspension bridge span is shown in Fig.(a). Cable ACB supports a stiffening girder DE through
suspenders spaced very close to each other. The cable profile is a shallow parabola. Determine the relation between the
tension in the cable and load in the suspender.

55. F
SUSPENDER

56. TRUSSES:
A truss is an assembly of slender members inter-connected by idealized joints. It is called a pi-connected
frame work. Trusses are used to support (i) Roof (ii) Bridges (iii) Towers and (iv) Cranes etc. Materials used may be (a)
Timber, (b) Bamboo, (c) Steel, (d) Aluminium and some exceptional cases (e) Concrete with reinforcements (RCC).
Each element of truss capable of taking (i) Tensionor (ii) Compression depends on (a) Mechanical
property of material (b) length, (c) Shape and (d) Size etc.
Method of Analysis (A) Method of Section (B) Method of Joints (C) Graphical Method etc.
ASSUMPIONS: (1) The elements are slender and not capable of resisting Bending.
(2) The inter-connections are idealized as frictionless hinges
(3) Loads are attached at inter-connected nodes (in the plane of the frame for 2-D frame)

57. METHOD OF SECTION
C
Select a section which intersects not more
than three (3) members in a normal plane
frame. The three members should not be
collinear. It is usually possible to solve for
any of the three member forces at
intersection using the static equilibrium
conditions (i.e. ∑H = ∑V = ∑M = 0) of plane
structure
Tension Force Notation on Joints
Tension Force Notation on Members

58. EXAMPLE:
A cantilever N-Girder support three loads as shown in Fig. . Determine the forces in BD, BE and CE.
2
3

59. EXAMPLE:
Determine Forces in Members
4-Bay Simply Supported (S.S.) Truss

61. EXAMPLE:
2Bay-Warren Girder shown as loaded in Fig. . Determine forces in members
Section x
1
-y
1
Section x
2
-y
2

62. EXAMPLE:
A FINK TRUSS, LOADED WITH 600 NAT EACH PURLIN POINT
N
N
N
N

63. N

64. N

65. METHOD OF JOINTS:
The analysis of forces in members of a truss based on the free-body diagram of joints is called the method of joints.
Each joint of a truss of yhemembers connected to the joint are analysed by applying the conditions of equilibrium.
All members connected to a joint are collinear at a point with the result that the equilibrium condition of moment is
satisfied automatically. Only two independent equilibrium conditions (i.e. ∑H = 0, ∑V = 0) are effective at a joint of a
plane truss. The sequence of analysis of joints must be selected such that at any given time, only two member
forces are unknown at a joint so as to evaluate the forces in succession. The arbitrary sequence of selection of joints
may result in simultaneous equations which unnecessarily increase the number calculations. If a truss is statically
determinate then it is possible to evaluate forces of all the members by method of joints.
EXAMPLE:
THREE-BAR BRACKET TRUSS

66. EXAMPLE:
ANALYSE A BRIDGE TRUSS AS LOADED SHOWN IN FIG. .