1. Engineering Mechanics: Statics
Chapter 7:
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.
To analyze the forces and study the geometry of cables supporting a load.

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

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. 7.1 Internal Forces Developed in Structural Members
Example 7.3
The beam supports the loading. Determine
the internal normal force, shear force and bending
moment acting to the left, point B and just to the
right, point C of the 6kN force.

27. 7.1 Internal Forces Developed in Structural Members
Solution
Support Reactions
9kN.m is a free vector and can be place anywhere in the FBD
+↑∑ Fy = 0; 9kN.m + (6kN)(6m) - Ay(9m) = 0
Ay = 5kN

28. 7.1 Internal Forces Developed in Structural Members
Solution
FBD of the segments AB and AC
9kN.couple moment must be kept in original position until after the section is made and appropriate body isolated

29. 7.1 Internal Forces Developed in Structural Members
Solution
Segment AB
+→∑ Fx = 0; NB = 0
+↑∑ Fy = 0; 5kN – VB = 0
VB = 5kN
∑ MB = 0; -(5kN)(3m) + MB = 0
MB = 15kN.m
Segment AC
+→∑ Fx = 0; NC = 0
+↑∑ Fy = 0; 5kN - 6kN + VC = 0
VC = 1kN
∑ MC = 0; -(5kN)(3m) + MC = 0
MC = 15kN.m

30. 7.1 Internal Forces Developed in Structural Members
Example 7.4
Determine the internal force, shear force and
the bending moment acting at point B of the
two-member frame.

31. 7.1 Internal Forces Developed in Structural Members
Solution
Support Reactions
FBD of each member

32. 7.1 Internal Forces Developed in Structural Members
Solution
Member AC
∑ MA = 0; -400kN(4m) + (3/5)FDC(8m)= 0
FDC = 333.3kN
+→∑ Fx = 0; -Ax + (4/5)(333.3kN) = 0
Ax = 266.7kN
+↑∑ Fy = 0; Ay – 400kN + 3/5(333.3kN) = 0
Ay = 200kN

33. 7.1 Internal Forces Developed in Structural Members
Solution
FBD of segments AB and BC
Important to keep distributed loading exactly as it is after the section is made

34. 7.1 Internal Forces Developed in Structural Members
Solution
Member AB
+→∑ Fx = 0; NB – 266.7kN = 0
NB = 266.7kN
+↑∑ Fy = 0; 200kN – 200kN - VB = 0
VB = 0
∑ MB = 0; MB – 200kN(4m) – 200kN(2m) = 0
MB = 400kN.m

35. 7.1 Internal Forces Developed in Structural Members
Example 7.5
Determine the normal force,
shear force and the bending
moment acting at point E of
the frame loaded.

36. 7.1 Internal Forces Developed in Structural Members
Solution
Support Reactions
Members AC and CD are two force members
+↑∑ Fy = 0;
Rsin45° – 600N = 0
R = 848.5N

37. 7.1 Internal Forces Developed in Structural Members
Solution
FBD of segment CE

38. 7.1 Internal Forces Developed in Structural Members
Solution
+→∑ Fx = 0; 848.5cos45°N - VE = 0
VE = 600 N
+↑∑ Fy = 0; sin45°N + NE = 0
NE = 600 N
∑ ME = 0; 848.5cos45°N(0.5m) - ME = 0
ME = 300 N.m
Results indicate a poor design
Member AC should be straight to eliminate bending within the member

39. 7.1 Internal Forces Developed in Structural Members
Example 7.6
The uniform sign has a mass of
650kg and is supported on the fixed
column. Design codes indicate that
the expected maximum uniform
wind loading that will occur in the
area where it is located is 900Pa.
Determine the internal loadings at A

40. 7.1 Internal Forces Developed in Structural Members
View Free Body Diagram
Solution
Idealized model for the sign
Consider FBD of a section above A since it dies not involve the support reactions
Sign has weight of
W = 650(9.81) = 6.376kN
Wind creates resultant force
Fw = 900N/m2(6m)(2.5m)
= 13.5kN

41. 7.1 Internal Forces Developed in Structural Members
Solution
FBD of the loadings

42. 7.1 Internal Forces Developed in Structural Members
Solution

43. 7.1 Internal Forces Developed in Structural Members
Solution
FAz = {6.38k}kN represents the normal force N
FAx= {13.5i}kN represents the shear force
MAz = {40.5k}kN represents the torisonal moment
Bending moment is determined from
where MAx = {-19.1i}kNm and MAy = {-70.9j}kN.m