1. Engineering Mechanics: Statics
Chapter 7:
Internal Forces
Engineering Mechanics: Statics
Mohammad Shahril bin Salim
2012/2013

2. Chapter Objectives
Method of sections for determining the internal loadings in a member
Develop procedure by formulating equations that describe the internal shear and moment throughout a member
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
Force component N, acting normal to the beam at the cut session
V, acting tangent to the session are the shear force
Couple moment M is referred as the bending moment

6. 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

7. 7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Support Reactions
Before cut, determine the member’s support reactions
Equilibrium equations used to solve internal loadings during sectioning
Free-Body Diagrams
Keep all distributed loadings, couple moments and forces acting on the member in their exact locations
After session draw FBD of the segment having the least loads

8. 7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Free-Body Diagrams (Continue)
Indicate the z, y, z components of the force, couple moments and resultant couple moments on FBD
Only N, V and M act at the section
Determine the sense by inspection
Equations of Equilibrium
Moments should be summed at the section
If negative result, the sense is opposite

9. Example 7.3
Determine the internal force, shear force and the bending moment acting at point B of the two-member frame.

10. Solution Support Reactions FBD of each member 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

11. Solution Support Reactions Member AB +→∑ Fx = 0; NB – 266.7kN = 0
+↑∑ Fy = 0; 200kN – 200kN - VB = 0
VB = 0
∑ MB = 0; MB – 200kN(4m) + 200kN(2m) 0
MB = 400kN.m

12. Problem 1:
Determine the internal normal force, shear force, and moment at point C.

13. Solution:

14. 7.2 Shear and Moment Equations and Diagrams
Beams – structural members designed to support loadings perpendicular to their axes
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

15. 7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Support Reactions
Find all reactive forces and couple moments acting on the beam
Resolve them into components
Shear and Moment Reactions
Specify separate coordinates x
Section the beam perpendicular to its axis
V obtained by summing the forces perpendicular to the beam
M obtained by summing moments about the sectioned end

16. 7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions (Continue)
Plot (V versus x) and (M versus x)
Convenient to plot the shear and the bending moment diagrams below the FBD of the beam

17. 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.

18. Solution
Support Reactions
FBD of the shaft

19. 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.

20. 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

21. 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
The internal shear force and bending moments assumed positive sense

22. 7.3 Relations between Distributed Load, Shear and Moment
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

23. 7.3 Relations between Distributed Load, Shear and Moment
Slope of the shear diagram
Negative of distributed load intensity
Slope of shear diagram
Shear moment diagram
Area under shear diagram
Change in shear
Area under shear diagram
Change in moment

24. 7.3 Relations between Distributed Load, Shear and Moment
Force and Couple Moment
FBD of a small segment of the beam
Change in shear is negative
FBD of a small segment of the beam located at the couple moment
Change in moment is positive

25. Example 7.9
Draw the shear and moment diagrams for the overhang beam.

26. Solution The support reactions are shown.
Shear Diagram Shear of –2 kN at end A of the beam is at x = 0.
Positive jump of 10 kN at x = 4 m due to the force.
Moment Diagram