Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Outline Internal Forces Developed in Structural Members Shear and Moment Equations and Diagrams Relations between Distributed Load, Shear and Moment Cables Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 normal or axial force and the shear force Couple moment M is referred as the bending moment Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.3 Determine the internal force, shear force and the bending moment acting at point B of the two-member frame. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Support Reactions FBD of the shaft Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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. Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.9 Draw the shear and moment diagrams for the overhang beam. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cables and chains used to support and transmit loads from one member to another In force analysis, weight of cables is neglected Assume cable is perfectly flexible and inextensible Due to its flexibility cables has no resistance to bending Length remains constant before and after loading Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cable Subjected to Concentrated Loads For a cable of negligible weight, it will subject to constant tensile force Known: h, L1, L2, L3 and loads P1 and P2 Form 2 equations of equilibrium Use Pythagorean Theorem to relate the three segmental lengths Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.11 Determine the tension in each segment of the cable. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.11 FBD for the entire cable. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.11 Consider leftmost section which cuts cable BC since sag yC = 12m. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cable Subjected to a Distributed Load Consider weightless cable subjected to a loading function w = w(x) measured in the x direction Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cable Subjected to a Distributed Load For FBD of the cable having length ∆ Since the tensile force changes continuously, it is denoted on the FBD by ∆T Distributed load is represented by second integration, Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.12 The cable of a suspension bridge supports half of the uniform road surface between the two columns at A and B. If this distributed loading wo, determine the maximum force developed in the cable and the cable’s required length. The span length L and, sag h are known. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Note w(x) = wo  Perform two integrations  Boundary Conditions at x = 0  Therefore, Curve becomes This is the equation of a parabola Boundary Condition at x = L/2  Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution For constant, Tension, T = FH/cosθ Slope at point B  Therefore Using triangular relationship Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution For a differential segment of cable length ds, Determine total length by integrating, Integrating yields, Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cable Subjected to its Own Weight When weight of the cable is considered, the loading function becomes a function of the arc length s rather than length x FBD of a segment of the cable is shown Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cable Subjected to its Own Weight Apply equilibrium equations to the force system Replace dy/dx by ds/dx for direct integration Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 7.4 Cables Cable Subjected to its Own Weight Therefore Separating variables and integrating Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 7.13 Determine the deflection curve, the length, and the maximum tension in the uniform cable. The cable weights wo = 5N/m. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution For symmetry, origin located at the center of the cable. Deflection curve expressed as y = f(x) Integrating term in the denominator Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Substitute So that Perform second integration or Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Evaluate constants or dy/dx = 0 at s = 0, then C1 = 0 To obtain deflection curve, solve for s Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Hence Boundary Condition y = 0 at x = 0  For deflection curve, This equations defines a catenary curve. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Boundary Condition y = h at x = L/2 Since wo = 5N/m, h = 6m and L = 20m, By trial and error, Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution For deflection curve, x = 10m, for half length of the cable Hence Maximum tension occurs when θ is maximum at s = 12.1m Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 1. In a multi-force member, the member is generally subjected to an internal _________. A) Normal force B) Shear force C) Bending moment D) All of the above. 2. In mechanics, the force component V acting tangent to, or along the face of, the section is called the _________ . A) Axial force B) Shear force C) Normal force D) Bending moment Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 3. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same? A) P, Q, and R B) P and Q C) Q and R D) None of the above. 4. A column is loaded with a horizontal 100 N force. At which section are the internal loads largest? A) P B) Q C) R D) S Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 5. Determine the magnitude of the internal loads (normal, shear, and bending moment) at point C. A) (100 N, 80 N, 80 N m) B) (100 N, 80 N, 40 N m) C) (80 N, 100 N, 40 N m) D) (80 N, 100 N, 0 N m ) 2. A column is loaded with a horizontal 100 N force. At which section are the internal loads the lowest? A) P B) Q C) R D) S Copyright © 2010 Pearson Education South Asia Pte Ltd