1. Bending Shear and Moment Diagram, Graphical method to construct shear
and moment diagram, Bending deformation of a
straight member, The flexure formula

3. Shear and moment diagram
Axial load diagram
Torque diagram
Both of these diagrams show the internal forces acting on the members. Similarly, the shear and moment diagrams show the internal shear and moment acting on the members

4. Type of Beams Simply Supported Beam Overhanging Beam Cantilever Beam
Statically Determinate
Simply Supported Beam
Overhanging Beam
Cantilever Beam

5. Propped Cantilever Beam
Type of Beams
Statically Indeterminate
Continuous Beam
Propped Cantilever Beam
Fixed Beam

7. Example 1 A B M V F Equilibrium equation for 0  x  3m:
* internal V and M should be assumed +ve

8. Shear Diagram M V F x
Sign convention: V= -9kN
Lecture 1

9. Shear Diagram F M=-9x Sign convention: M= -9x kNm X=0: M= 0
V = -9 kN F x
M=-9x
Sign convention: M= -9x kNm
X=0: M= 0
X=3: M=-27kNm

10. At cross section A-A
At section A-A
V=9kN
M=X X

11. Example 2
1) Find all the external forces

12. Moment equilibrium
Force equilibrium
Moment equilibrium
Force equilibrium

13. Boundary cond for V and M
M=5x
M=10-5x

14. Solve it
Draw the shear and moment diagrams for simply supported beam.

16. Distributed Load
For calculation purposes, distributed load can be represented as a single load acting on the center point of the distributed area.
Total force = area of distributed load (W : height and L: length)
Point of action: center point of the area

17. Example

18. Example

19. Solve it
Draw the shear and moment diagrams the beam:

20. Solving all the external loads
Distributed load will be
Solving the FBD

21. Boundary Condition
Equilibrium eq

22. Boundary Condition
Equilibrium eq

23. x=0 V= 12 kN
x=4 V=-20 kN
x=0 M= 0 kN
x=4 V=-16 kN
x=4 V= 16kN
x=6 V= 0 kN
x=4 V= -16kN
x=6 V= 0 kN

24. Graph based on equations
Straight horizontal line
y = c
y = mx + c
y=3x + 3
y=-3x + 3
y = ax2 + bx +c
y=3x2 + 3
y=-3x2 + 3

26. Graphical method Relationship between load and shear:
Relationship between shear and bending moment: 26

27. Regions of distributed load:
Dividing by Dx and taking the limit as Dxà0, the above two equations become:
Regions of distributed load:
Slope of moment diagram at each point
= shear at each point
Slope of shear diagram at each point
=  distributed load intensity at each point 27

28. Example

30. = = The previous equations become: Area under shear diagram
change in shear
Area under distributed load
change in moment = =

31. +ve area under shear diagram

34. Bending deformation of a straight member
Observation:
bottom line : longer
top line: shorter
Middle line: remain the same but rotate (neutral line)

35. Strain
Before deformation
After deformation, Dx has a radius of curvature r, with center of curvature at point O’
Similarly
Therefore

36. Maximum strain will be
-ve: compressive state
+ve: tension

37. The Flexure Formula
The location of neutral axis is when the resultant force of the tension and compression is equal to zero.
Noting

38. Since , therefore
Therefore, the neutral axis should be the centroidal axis

39. Maximum normal stress
Normal stress at y distance

40. Line NA: neutral axis
Red Line: max normal stress
c = 60 mm
Yellow Line: max compressive stress
c = 60mm
Line NA: neutral axis
Red Line: Compressive stress
y1 = 30 mm
Yellow Line: Normal stress
y2 = 50mm
Refer to Example 6.11 pp 289

41. I: moment of inertial of the cross sectional area
Find the stresses at A and B

42. I: moment of inertial of the cross sectional area
Locate the centroid (coincide with neutral axis)

43. I: moment of inertial of the cross sectional area
Profile I
I about Centroidal axis
I about Axis A-A using parallel axis theorem A A
Profile II
Total I
* Example 6-12 to 6-14 (pp )

44. Solve it
If the moment acting on the cross section of the beam is M = 6 kNm, determine the maximum bending stress on in the beam. Sketch a three dimensional of the stress distribution acting over the cross section
If M = 6 kNm, determine the resultant force the bending stress produces on the top board A of the beam

45. Total Moment of Inertia
Max Bending Stress at the top and bottom
1.45MPa
Bottom of the flange
1.14MPa
6kNm

46. Resultant F = volume of the trapezoid
300 mm
40 mm
1.45MPa
1.14MPa

47. Solve it
The shaft is supported by a smooth thrust load at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress on the shaft

48. External Forces
Draw the shear and moment diagram
Absolute Bending Stress
Mmax = 2.25kNm