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- 1. 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
- 2. 7.1 Internal Forces Developed in Structural MembersConsider the “simply supported” beamTo determine the internal loadings acting on the crosssection at C, an imaginary section is passed throughthe beam, cutting it into twoBy doing so, the internal loadings become external onthe FBD
- 3. 7.1 Internal Forces Developed in Structural MembersSince both segments (AC and CB) were inequilibrium before the sectioning, equilibrium ofthe segment is maintained by rectangular forcecomponents and a resultant couple momentMagnitude of the loadings is determined by theequilibrium equations
- 4. 7.1 Internal Forces Developed in Structural MembersForce component N, acting normal to thebeam at the cut session and V, acting tangent to the session are known as normalor axial forceand the shear forceCouple moment M isreferred as the bendingmoment
- 5. 7.1 Internal Forces Developed in Structural MembersFor 3D, a general internal force and couplemoment resultant will act at the sectionNy is the normal force, and Vx and Vz arethe shear componentsMy is the torisonal ortwisting moment, andMx and Mz are thebending momentcomponents
- 6. 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
- 7. 7.1 Internal Forces Developed in Structural MembersFree 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
- 8. 7.1 Internal Forces Developed in Structural MembersFree 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
- 9. 7.1 Internal Forces Developed in Structural MembersFree Body Diagrams Once done, each member may be sectioned at its appropriate point and apply the 3 equilibrium equations to determine the unknownsExample FBD of segment DG can be used to determine the internal loadings at G provided the reactions of the pins are known
- 10. 7.1 Internal Forces Developed in Structural MembersProcedure for AnalysisSupport 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
- 11. 7.1 Internal Forces Developed in Structural MembersProcedure for AnalysisFree-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
- 12. 7.1 Internal Forces Developed in Structural MembersProcedure for AnalysisFree-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
- 13. 7.1 Internal Forces Developed in Structural MembersProcedure for AnalysisEquations 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
- 14. 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
- 15. 7.1 Internal Forces Developed in Structural MembersExample 7.1The bar is fixed at its end and isloaded. Determine the internal normalforce at points B and C.
- 16. 7.1 Internal Forces Developed in Structural MembersSolutionSupport 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
- 17. 7.1 Internal Forces Developed in Structural MembersSolution 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
- 18. 7.1 Internal Forces Developed in Structural MembersSolutionSegment AB +↑∑ Fy = 0; 8kN – NB = 0 NB = 8kNSegment DC +↑∑ Fy = 0; NC – 4kN= 0 NC = 4kN
- 19. 7.1 Internal Forces Developed in Structural MembersExample 7.2The circular shaft is subjected to threeconcentrated torques. Determine the internaltorques at points B and C.
- 20. 7.1 Internal Forces Developed in Structural MembersSolutionSupport Reactions Shaft subjected to only collinear torques ∑ Mx = 0; -10N.m + 15N.m + 20N.m –TD = 0 TD = 25N.m
- 21. 7.1 Internal Forces Developed in Structural MembersSolution FBD of shaft segments AB and CD
- 22. 7.1 Internal Forces Developed in Structural MembersSolutionSegment AB ∑ Mx = 0; -10N.m + 15N.m – TB = 0 TB = 5N.mSegment CD ∑ Mx = 0; TC – 25N.m= 0 TC = 25N.m
- 23. 7.1 Internal Forces Developed in Structural MembersExample 7.3The beam supports the loading. Determinethe internal normal force, shear force and bendingmoment acting to the left, point B and just to theright, point C of the 6kN force.
- 24. 7.1 Internal Forces Developed in Structural MembersSolutionSupport 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
- 25. 7.1 Internal Forces Developed in Structural MembersSolution 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
- 26. 7.1 Internal Forces Developed in Structural MembersSolutionSegment AB +→∑ Fx = 0; NB = 0 +↑∑ Fy = 0; 5kN – VB = 0 VB = 5kN ∑ MB = 0; -(5kN)(3m) + MB = 0 MB = 15kN.mSegment AC +→∑ Fx = 0; NC = 0 +↑∑ Fy = 0; 5kN - 6kN + VC = 0 VC = 1kN ∑ MC = 0; -(5kN)(3m) + MC = 0 MC = 15kN.m
- 27. 7.1 Internal Forces Developed in Structural MembersExample 7.4Determine the internal force, shear force andthe bending moment acting at point B of thetwo-member frame.
- 28. 7.1 Internal Forces Developed in Structural MembersSolutionSupport Reactions FBD of each member
- 29. 7.1 Internal Forces Developed in Structural MembersSolutionMember 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
- 30. 7.1 Internal Forces Developed in Structural MembersSolution FBD of segments AB and BC Important to keep distributed loading exactly as it is after the section is made
- 31. 7.1 Internal Forces Developed in Structural MembersSolutionMember 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
- 32. 7.1 Internal Forces Developed in Structural MembersExample 7.5Determine the normal force,shear force and the bendingmoment acting at point E ofthe frame loaded.
- 33. 7.1 Internal Forces Developed in Structural MembersSolutionSupport Reactions Members AC and CD are two force members +↑∑ Fy = 0; Rsin45° – 600N = 0 R = 848.5N
- 34. 7.1 Internal Forces Developed in Structural MembersSolution FBD of segment CE
- 35. 7.1 Internal Forces Developed in Structural MembersSolution +→∑ Fx = 0; 848.5cos45°N - VE = 0 VE = 600 N +↑∑ Fy = 0; -848.5sin45°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
- 36. 7.1 Internal Forces Developed in Structural MembersExample 7.6The uniform sign has a mass of650kg and is supported on the fixedcolumn. Design codes indicate thatthe expected maximum uniformwind loading that will occur in thearea where it is located is 900Pa.Determine the internal loadings at A
- 37. 7.1 Internal Forces Developed in Structural MembersSolution 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
- 38. 7.1 Internal Forces Developed in Structural MembersSolution FBD of the loadings
- 39. 7.1 Internal Forces Developed in Structural MembersSolution r∑ F = 0;r r rFA − 13.5i − 6.3475k = 0r r rFA = {13.5i + 6.38k }kN r∑ M A = 0; r r r rM A + r X ( Fw + W ) = 0 r r r i j k rMA + 0 3 5.25 = 0 − 13.5 0 6.376r r r rM A = {−19.1i + 70.9 j + 40.5k }kN .m
- 40. 7.1 Internal Forces Developed in Structural MembersSolutionFAz = {6.38k}kN represents the normal force NFAx= {13.5i}kN represents the shear forceMAz = {40.5k}kN represents the torisonal momentBending moment is determined from r r2 r2 M = Mx + Mywhere MAx = {-19.1i}kNm and MAy = {-70.9j}kN.m

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