2009+7-2InternalForces

2009+7-2InternalForces - Introduction to Solid...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Procedure for Analysis (3D) Free-Body Diagrams ± Indicate the z, y, z components of the force and couple moments and the resultant couple moments on the FBD SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Example 7.6 The uniform sign has a mass of 650kg and is supported on the fixed column. Design codes indicate that the expected maximum uniform wind loading that will occur in the area where it is located is 900Pa. Determine the internal loadings at A SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Solution ± Idealized model for the sign ± Consider FBD of a section above A since it does not involve the support reactions ± Sign has weight of W = 650(9.81) = 6.376kN ± Wind creates resultant force F w = 900N/m 2 (6m)(2.5m) = 13.5kN SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Solution ± FBD of the loadings It is easy to use vector expression SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Solution m kN k j i M k j i M W F r M M kN k i F k i F F A A w A A A A . } 5 . 40 9 . 70 1 . 19 { 0 376 . 6 0 5 . 13 25 . 5 3 0 0 ) ( ; 0 } 376 . 6 5 . 13 { 0 376 . 6 5 . 13 ; 0 r r r r r r r r r r r r r r r r r r r r + + = = + = + × + = + = = = SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Solution (F A ) z = {6.376 k }kN represents the normal force N (F A ) x = {13.5 i }kN represents the shear force
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 SJTU Introduction to Solid Mechanics -Vm211 7.1 Internal Forces Developed in Structural Members Solution ± (M A ) z = {40.5 k }kN represents the torisonal moment ± Bending moment is determined from ± where ( M A ) x = { 19.1 i }kNm and( M A ) y = { 70.9 j }kN.m 2 2 y x M M M r r r + = SJTU Introduction to Solid Mechanics -Vm211 Chapter 7: Internal Forces Internal Forces vs. External Loadings SJTU Introduction to Solid Mechanics -Vm211 ± Beams structural members designed to support loadings perpendicular to their axes ± Beams straight long bars with constant cross- sectional areas 7.2 Shear and Moment Equations and Diagrams SJTU Introduction to Solid Mechanics -Vm211 7.2 Shear and Moment Equations and Diagrams ± 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 SJTU Introduction to Solid Mechanics -Vm211 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 SJTU Introduction to Solid Mechanics -Vm211 7.2 Shear and Moment Equations and Diagrams ±
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/09/2011 for the course EE 211 taught by Professor Liuxila during the Summer '09 term at Shanghai Jiao Tong University.

Page1 / 14

2009+7-2InternalForces - Introduction to Solid...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online