1. Engineering Mechanics: Statics
Chapter 6:
Internal Forces
Engineering Mechanics: Statics

2. Chapter Objectives
To show how to use the method of sections for determining the internal loadings in a member.
To generalize this procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member.

3. Chapter Outline Internal Forces Developed in Structural Members
Shear and Moment Equations and Diagrams
Relations between Distributed Load, Shear and Moment

4. 7.1 Internal Forces Developed in Structural Members
The design of any structural or mechanical member requires the material to be used to be able to resist the loading acting on the member
These internal loadings can be determined by the method of sections

5. 7.1 Internal Forces Developed in Structural Members
Consider the “simply supported” beam
To determine the internal loadings acting on the cross section at C, an imaginary section is passed through the beam, cutting it into two
By doing so, the internal loadings become external on the FBD

6. 7.1 Internal Forces Developed in Structural Members
Since both segments (AC and CB) were in equilibrium before the sectioning, equilibrium of the segment is maintained by rectangular force components and a resultant couple moment
Magnitude of the loadings is determined by the equilibrium equations

7. 7.1 Internal Forces Developed in Structural Members
Force component N, acting normal to the
beam at the cut session and V, acting t
angent to the session are known as normal
or axial force
and the shear force
Couple moment M is
referred as the bending
moment

8. 7.1 Internal Forces Developed in Structural Members
For 3D, a general internal force and couple moment resultant will act at the section
Ny is the normal force, and Vx and Vz are the shear components
My is the torisonal or
twisting moment, and
Mx and Mz are the
bending moment
components

9. 7.1 Internal Forces Developed in Structural Members
For most applications, these resultant loadings will act at the geometric center or centroid (C) of the section’s cross sectional area
Although the magnitude of each loading differs at different points along the axis of the member, the method of section can be used to determine the values

10. 7.1 Internal Forces Developed in Structural Members
Free Body Diagrams
Since frames and machines are composed of multi-force members, each of these members will generally be subjected to internal shear, normal and bending loadings
Consider the frame with the blue
section passed through to
determine the internal loadings
at points H, G and F

11. 7.1 Internal Forces Developed in Structural Members
Free Body Diagrams
FBD of the sectioned frame
At each sectioned member, there is an unknown normal force, shear force and bending moment
3 equilibrium equations cannot be used
to find 9 unknowns, thus dismember
the frame and determine
reactions at each connection

12. 7.1 Internal Forces Developed in Structural Members
Free Body Diagrams
Once done, each member may be sectioned at its appropriate point and apply the 3 equilibrium equations to determine the unknowns
Example
FBD of segment DG can be used to determine
the internal loadings at G
provided the reactions of
the pins are known

13. 7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Support Reactions
Before the member is cut or sectioned, determine the member’s support reactions
Equilibrium equations are used to solve for internal loadings during sectioning of the members
If the member is part of a frame or machine, the reactions at its connections are determined by the methods used in 6.6

14. 7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Free-Body Diagrams
Keep all distributed loadings, couple moments and forces acting on the member in their exact locations, then pass an imaginary section through the member, perpendicular to its axis at the point the internal loading is to be determined
After the session is made, draw the FBD of the segment having the least loads

15. 7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Free-Body Diagrams
Indicate the z, y, z components of the force and couple moments and the resultant couple moments on the FBD
If the member is subjected to a coplanar system of forces, only N, V and M act at the section
Determine the sense by inspection; if not, assume the sense of the unknown loadings

16. 7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Equations of Equilibrium
Moments should be summed at the section about the axes passing through the centroid or geometric center of the member’s cross-sectional area in order to eliminate the unknown normal and shear forces and thereby, obtain direct solutions for the moment components
If the solution yields a negative result, the sense is opposite that assume of the unknown loadings

17. 7.1 Internal Forces Developed in Structural Members
The link on the backhoe is a two force member
It is subjected to both bending and axial load at its center
By making the member straight, only an axial force acts within the member

18. 7.1 Internal Forces Developed in Structural Members
Example 7.1
The bar is fixed at its end and is
loaded. Determine the internal normal
force at points B and C.

19. 7.1 Internal Forces Developed in Structural Members
Solution
Support Reactions
FBD of the entire bar
By inspection, only normal force Ay acts at the fixed support
Ax = 0 and Az = 0
+↑∑ Fy = 0; 8kN – NB = 0
NB = 8kN

20. 7.1 Internal Forces Developed in Structural Members
Solution
FBD of the sectioned bar
No shear or moment act on the sections since they are not required for equilibrium
Choose segment AB and DC since they contain the least number of forces

21. 7.1 Internal Forces Developed in Structural Members
Solution
Segment AB
+↑∑ Fy = 0; 8kN – NB = 0
NB = 8kN
Segment DC
+↑∑ Fy = 0; NC – 4kN= 0
NC = 4kN

22. 7.1 Internal Forces Developed in Structural Members
Example 7.2
The circular shaft is subjected to three
concentrated torques. Determine the internal
torques at points B and C.

23. 7.1 Internal Forces Developed in Structural Members
Solution
Support Reactions
Shaft subjected to only collinear torques
∑ Mx = 0;
-10N.m + 15N.m + 20N.m –TD = 0
TD = 25N.m

24. 7.1 Internal Forces Developed in Structural Members
Solution
FBD of shaft segments AB and CD

25. 7.1 Internal Forces Developed in Structural Members
Solution
Segment AB
∑ Mx = 0; -10N.m + 15N.m – TB = 0
TB = 5N.m
Segment CD
∑ Mx = 0; TC – 25N.m= 0
TC = 25N.m

26. 7.2 Shear and Moment Equations and Diagrams
Beams – structural members designed to support loadings perpendicular to their axes
Beams – straight long bars with constant cross-sectional areas
A simply supported beam is pinned at one end
and roller supported at
the other
A cantilevered beam is
fixed at one end and free
at the other

27. 7.2 Shear and Moment Equations and Diagrams
For actual design of a beam, apply
- Internal shear force V and the bending moment M analysis
- Theory of mechanics of materials
Appropriate engineering code to determine beam’s required cross-sectional area
Variations of V and M obtained by the method of sections
Graphical variations of V and M are termed as shear diagram and bending moment diagram

28. 7.2 Shear and Moment Equations and Diagrams
Internal shear and bending moment functions generally discontinuous, or their slopes will be discontinuous at points where a distributed load changes or where concentrated forces or couple moments are applied
Functions must be applied for each segment of the beam located between any two discontinuities of loadings
Internal normal force will not be considered

29. 7.2 Shear and Moment Equations and Diagrams
Load applied to a beam act perpendicular to the beam’s axis and hence produce only an internal shear force and bending moment
For design purpose, the beam’s resistance to shear, and particularly to bending, is more important than its ability to resist a normal force

30. 7.2 Shear and Moment Equations and Diagrams
Sign Convention
To define a positive and negative shear force and bending moment acting on the beam
Positive directions are denoted by an internal shear force that causes clockwise rotation of the member on which it acts and by an internal moment that causes compression or pushing on the upper part of the member

31. 7.2 Shear and Moment Equations and Diagrams
Sign Convention
A positive moment would tend to bend the member if it were elastic, concave upwards
Loadings opposite to the above are considered negative

32. 7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Support Reactions
Determine all the reactive forces and couple moments acting on the beam’
Resolve them into components acting perpendicular or parallel to the beam’s axis

33. 7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions
Specify separate coordinates x having an origin at the beam’s left end and extending to regions of the beams between concentrated force and/or couple moments or where there is no continuity of distributed loadings
Section the beam perpendicular to its axis at each distance x and draw the FBD of one of the segments

34. 7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions
V and M are shown acting in their positive sense
The shear V is obtained by summing the forces perpendicular to the beam’s axis
The moment M is obtained by summing moments about the sectioned end of the segment

35. 7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Shear and Moment Diagrams
Plot the shear diagram (V versus x) and the moment diagram (M versus x)
If computed values of the functions describing V and M are positive, the values are plotted above the x axis, whereas negative values are plotted below the x axis
Convenient to plot the shear and the bending moment diagrams below the FBD of the beam

36. 7.2 Shear and Moment Equations and Diagrams
Example 7.7
Draw the shear and bending moments
diagrams for the shaft. The support at A is a
thrust bearing and the support at C is a
journal bearing.

37. 7.2 Shear and Moment Equations and Diagrams
Solution
Support Reactions
FBD of the shaft

38. 7.2 Shear and Moment Equations and Diagrams
Solution

39. 7.2 Shear and Moment Equations and Diagrams
Solution

40. 7.2 Shear and Moment Equations and Diagrams
Solution
Shear diagram
internal shear force is always positive within the shaft AB
Just to the right of B, the shear force changes sign and remains at constant value for segment BC
Moment diagram
Starts at zero, increases linearly to B and therefore decreases to zero

41. 7.2 Shear and Moment Equations and Diagrams
Solution
Graph of shear and moment diagrams is discontinuous at points of concentrated force ie, A, B, C
All loading discontinuous are mathematical, arising from the idealization of a concentrated force and couple moment

42. 7.3 Relations between Distributed Load, Shear and Moment
Consider beam AD subjected to an arbitrary load w = w(x) and a series of concentrated forces and moments
Distributed load assumed positive when loading acts downwards

43. 7.3 Relations between Distributed Load, Shear and Moment
A FBD diagram for a small segment of the beam having a length ∆x is chosen at point x along the beam which is not subjected to a concentrated force or couple moment
Any results obtained will not apply at points of concentrated loadings

44. 7.3 Relations between Distributed Load, Shear and Moment
The internal shear force and bending moments shown on the FBD are assumed to act in the positive sense
Both the shear force and moment acting on the right-hand face must be increased by a small, finite amount in order to keep the segment in equilibrium

45. 7.3 Relations between Distributed Load, Shear and Moment
The distributed loading has been replaced by a resultant force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 < k <1

46. 7.3 Relations between Distributed Load, Shear and Moment
Slope of the = Negative of
shear diagram distributed load intensity
Slope of = Shear moment diagram

47. 7.3 Relations between Distributed Load, Shear and Moment
At a specified point in a beam, the slope of the shear diagram is equal to the intensity of the distributed load
Slope of the moment diagram = shear
If the shear is equal to zero, dM/dx = 0, a point of zero shear corresponds to a point of maximum (or possibly minimum) moment
w (x) dx and V dx represent differential area under the distributed loading and shear diagrams

48. 7.3 Relations between Distributed Load, Shear and Moment
Change in = Area under
shear shear diagram
moment shear diagram

49. 7.3 Relations between Distributed Load, Shear and Moment
Change in shear between points B and C is equal to the negative of the area under the distributed-loading curve between these points
Change in moment between B and C is equal to the area under the shear diagram within region BC
The equations so not apply at points where concentrated force or couple moment acts

50. 7.3 Relations between Distributed Load, Shear and Moment
Force
FBD of a small segment of the beam
Change in shear is negative thus the shear will jump downwards when F acts downwards on the beam

51. 7.3 Relations between Distributed Load, Shear and Moment
Force
FBD of a small segment of the beam located at the couple moment
Change in moment is positive or the moment diagram will jump upwards MO is clockwise

52. 7.3 Relations between Distributed Load, Shear and Moment
Example 7.9
Draw the shear and moment diagrams for the
beam.

53. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Support Reactions
FBD of the beam

54. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Shear Diagram
V = at x = 0
V = 0 at x = 2
Since dV/dx = -w = -500, a straight negative sloping
line connects the end points

55. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Moment Diagram
M = at x = 0
M = 0 at x = 2
dM/dx = V, positive yet linearly decreasing from
dM/dx = 1000 at x = 0 to dM/dx = 0 at x = 2