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Lecture # 23 & 24

y. y. M y. V y. z. N. M z. z. T. Compound shear stress. Compound normal stress. x. V z. x. Lecture # 23 & 24. Compound Stress. Objective:-. To find total stress (  and ) from central forces and moment. Compound normal stress:

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Lecture # 23 & 24

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  1. y y My Vy z N Mz z T Compound shear stress Compound normal stress x Vz x Lecture # 23 & 24 Compound Stress Objective:- To find total stress ( and ) from central forces and moment. Compound normal stress: Normal stress caused by different faces and moments such as (N, My, Mz) Compound shear stress: Shear stresses caused by different forces and moments such as (Vy, Vz, T)

  2. Mz My Vy Vz T N Compound Stress y My Vy N = Mz Vz x z T Select + sign wherever moment is causing tension in +ve y or +ve z- axis.

  3. Example: A force of 15000 N is applied to the edge of the member. Neglect the weight of the member and determine the state of stress at point B and C.

  4. Solution:

  5. Example: The member has a rectangular cross section. Determine the state of stress that the loading produces at point C.

  6. Solution: N = 16.45 kN V = 21.93 kN M = 32.89 kN·m

  7. Solution (cont.):

  8. Example: The solid rod has a radius of 0.75 cm. If it is subjected to the loading shown, determine the state of stress at point A.

  9. = Solution:

  10. 7000 N·cm 8000 N·cm = 500 N 800 N 11200 N·cm (c) Solution (cont.): Internal Loading.

  11. Solution (cont.): Normal Force. Shear Force.

  12. Example: The rectangular block of negligible weight is subjected to a vertical force of 40 kN, which is applied to its corner. Determine the normal-stress distribution acting on a section through ABCD. Normal Force. Bending moment.

  13. Solution:

  14. Solution (cont.):

  15. The enD

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