Internal Forces Illustrated by Method of SectionsInternal Force. The change of interaction force among various parts of asolid body, introduced by external loadingThe Method of SectionFFnVnmII6
Internal Force • The change of interaction force among various parts of a solid body, introduced by external loading. The Method of Section F1 F2 m I F1 F2 m I R M o F1 Fn F2 m I II m Internal Forces Illustrated by Method of Sections 6
Sign Convention of Axial Forces: Coordinate: a right-handed systemMO: Centroid of the cross-sectionAx: Cross-section normal (Bar axis)F. Forces: Fn, Fsy, Fs?: Moments: T, My, M,: Axial Force (Fn)Fo Along bar axis+o Extends or contracts the bar along bar axis一F Positive for tension; Negative for compression(-)o Assuming positive for all unknown axial forces7
F1 F2 m I o x y z FN T M FSy y FSz Mz • Coordinate: a right-handed system O: Centroid of the cross-section x: Cross-section normal (Bar axis) • Forces: FN , FSy , FSz • Moments: T, My , Mz • Axial Force (FN ) o Along bar axis o Extends or contracts the bar along bar axis o Positive for tension; Negative for compression o Assuming positive for all unknown axial forces FN (+) FN (-) Sign Convention of Axial Forces 7
Procedure of Method of Sections: Sectioning the member: Taking either portion: Substituting the other portion with internal forces·EquilibratingmmFFmmFFFFNFn = FFn = -F8
• Sectioning the member • Taking either portion • Substituting the other portion with internal forces • Equilibrating Procedure of Method of Sections F F m m F FN FN = -F F F m m F FN FN = F 8
Diagram of Axial Forces: Abscissa: position of cross sections: Ordinate: axial force: Positive for tension; negative for compressionmm+X9
• Abscissa: position of cross sections • Ordinate: axial force • Positive for tension; negative for compression. F F m m FN x Diagram of Axial Forces 9
Sample Problem. Plot the diagram of axial forceB 2 kN/m CD 1 kN3kN A. Solution2 m—/-2 m—/2 m—1. Internal force in AB、CD3 kNFnFNAB = 3 kNxFNcD = -1 kN2m- x2. Internal force in BCFn(x)=3-2x(0<x<2)33. Diagram of axial forceF (kN)C4. Maximum internal force[FN,max | = 3 kN110
• Plot the diagram of axial force 3 kN A B 2 kN/m C D 1 kN 2 m 2 m 2 m 1. Internal force in AB、CD N N 3 kN 1 kN AB CD F F 2. Internal force in BC 3 kN 2 m x FN x FN,max 3 kN 3. Diagram of axial force FN kN 3 1 4. Maximum internal force Sample Problem F x x x N 3 2 0 2 • Solution 10