Engineering Mechanics: Statics

1. Engineering Mechanics: Statics Chapter 7: Internal Forces Ahmad Faizal bin salleh 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 members (e.g. Beams) 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. Determine the internal loading here

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 • Equilibrium is maintained • Newton 3rd Law

6. 7.1 Internal Forces Developed in Structural Members • In 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 torsional or twisting moment, and Mx and Mz are the bending moment components The resultant loadings act at Geometric Centre or Centroid

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

8. 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

9. Is it possible to assume MB in CW? Solution Support Reactions 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

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

11. Solution:

12. 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

13. 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

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 Distributed Load • 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 Distributed Load • 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 Load • 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 Distributed Load 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 DiagramShear 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