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

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Presentation on theme: "Engineering Mechanics: Statics Chapter 6: Internal Forces Chapter 6: Internal Forces."— Presentation transcript:

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

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 N y is the normal force, and V x and V z are the shear components M y is the torisonal or twisting moment, and M x and M z 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 A y acts at the fixed support A x = 0 and A z = 0 + ↑ ∑ F y = 0;8kN – N B = 0 N B = 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 + ↑ ∑ F y = 0;8kN – N B = 0 N B = 8kN Segment DC + ↑ ∑ F y = 0;N C – 4kN= 0 N C = 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 ∑ M x = 0; -10N.m + 15N.m + 20N.m –T D = 0 T D = 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 ∑ M x = 0;-10N.m + 15N.m – T B = 0 T B = 5N.m Segment CD ∑ M x = 0;T C – 25N.m= 0 T C = 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 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

43 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

44 7.3 Relations between Distributed Load, Shear and Moment Distributed Load 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 Distributed Load 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 Distributed Load 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 Distributed Load 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 Distributed Load Change in = Area under shearshear diagram Change in = Area under momentshear diagram

49 7.3 Relations between Distributed Load, Shear and Moment Distributed Load 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 M O 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 = +1000 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 = -1000 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


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