Unit 2 Shear force and bending moment.pptx

1. Mechanics of Solids
Dr. G Praveen Kumar,
Assistant Professor,
Department of Mechanical Engineering.
UNIT-II
Shear Force and Bending Moment

2. Beams
Types of beams
Types of loads
Shear force and bending moment
SFD & BMD

3. BEAMS
 Structural member which is acted upon by a
system of external loads at right angles to its axis
is known as beam.

5. TYPES OF BEAM

6. CANTILEVER BEAM
SIMPLY SUPPORTED BEAM

7. OVER HANGING BEAM
CONTINUOUS BEAM

8. TYPES OF LOADS

9. Point Load
UDL

11. SHEAR FORCE
 The shear force at any point along a loaded beam
may be defined as the algebraic sum of all vertical
forces acting on either side of the point on the beam.

12. BENDING MOMENT
 Bending moment at any point along a loaded beam
may be defined as the sum of the moments due to all
vertical forces acting on either side of the point on
the beam.

13. SHEAR FORCE AND BENDING MOMENT
DIAGRAMS
 A shear force diagram (SFD) which shows the shear
force at every section of the beam due to transverse
loading on it.
 A bending moment diagram (BMD) is a diagram which
shows the bending moment at every section of the beam
due to transverse loading on it.

15. Shear Force Calculation
SF at B = 1.5 KN
SF at C = 1.5 + 2 = 3.5 KN
SF at A = 1.5 + 2 = 3.5 KN
Bending Moment Calculation
BM at B = 0
BM at C = -1.5 x 0.5 = -0.75 KN-m
BM at A = (-1.5 x 1.5) + (-2 x 1) = -4.25 KN-m
1. Draw Shear force and Bending moment diagram for a
cantilever beam of span 1.5 m carried point load as
shown in Fig.

16. EXAMPLE

18.  Shear Force calculation
SF at B = 0
SF at C = 1x1.5 =1.5 KN
SF at A = 1.5 KN
Bending Moment calculation
BM at B = 0
BM at C = 1x1.5x(1.5/2) =1.125 KN
BM at A = 1x1.5x(1.5/2 + 0.5)
=1.875 KN
A Cantilever of length 2.0 carries a uniformly distributed load of 1 KN/m
run over a length of 1.5 m from the free end. Draw the shear force and
bending moment diagrams for the cantilever.

23. Find the reactions at A & B
1.Taking moment at A
2. Sum of forces upwards = Sum of forces downwards

24.  SF Calculation
SF at B = -W/2
SF at C = -W/2 + W
= +W/2
BM Calculation
BM at B = 0
SF at C = W/2 x L/2
= WL/4
BM atA = 0
Note:

26. SHEAR FORCE
BENDING MOMENT

27. REACTIONS AT A & B
Shear Force
SF at B = -0.5 wl
SF at C = -0.5 wl + w (l/2)
= 0
SF at A = +o.5wl
Bending Moment
BM at B = 0
SF at C = 0.5 wl (l) - w (l/2)
(l/4)
= wl2/8
SF at A = 0

29. SHEAR FORCE CALCULATION
SF at B = - 50 KN
SF at D = -50 + 40 = -10 KN
SF at C = -50 + 40 + (10 x 4) = +30 KN (Without Point Load)
SF at C = 30 + 50 = +80 KN
SF at A = +80 KN
SF at E = -RB + 40 + 10 (x-4)
0 = -50 + 40 + 10 (x-4)
X = 5 m