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بِســــــمِ اللهِ الرَّحمَنِ الرَّحِيمِ لَا يُكَلّفُ اللهُ نَفساً إِلَّا وُسعَهَا

بِســــــمِ اللهِ الرَّحمَنِ الرَّحِيمِ لَا يُكَلّفُ اللهُ نَفساً إِلَّا وُسعَهَا صَدَقَ اللهُ العليُ العظيم. University of Kufa Faculty of Engineering Civil Department. Second Year Sub. Resistance of Materials. MSc. Ashraf A. M. R. Hiswa 2013. Outlines. Stress Concept

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بِســــــمِ اللهِ الرَّحمَنِ الرَّحِيمِ لَا يُكَلّفُ اللهُ نَفساً إِلَّا وُسعَهَا

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  1. بِســــــمِ اللهِ الرَّحمَنِ الرَّحِيمِ لَا يُكَلّفُ اللهُ نَفساً إِلَّا وُسعَهَا صَدَقَ اللهُ العليُ العظيم

  2. University of Kufa Faculty of Engineering Civil Department Second YearSub. Resistance of Materials MSc. Ashraf A. M. R. Hiswa 2013

  3. Outlines • Stress Concept • Types of Stresses • Applications on Stress • Design of Materials Due to Stress Capacities

  4. Stress Concept Stress is the distribution of a force throughout the cross-sectional area of a structural member

  5. Types of Stresses 1- Normal / Direct Stress It is product of a force normal to the area. σ = Where:- σ: Direct stress, F: force, A: cross-sectional area = h*w

  6. Types of Stresses = = = Sign Convention Units Stress = = = Pa., = = = 106 Pa. = MPa

  7. Applications on Stress Ex.1/ Calculate the stress due to the applied load if the cross section is rectangle with width of (20 mm) and height of (30 mm). Sol./ F = 60 kN (com.) A = 20*30 = 600 mm2 σ = = = - 600 MPa.

  8. Types of Stresses 2- Shear Stress It is product of a force normal to the area. = Where:- : Shear stress, F: force, A: cross-sectional area = h*w

  9. Types of Stresses : Shear stress due to force in the y-direction acts on the area normal to x-direction. = Sign Convention - τ + τ Units Stress = = = Pa., = = = 106 Pa. = MPa.

  10. Applications on Stress Ex.2/ Calculate the shear stress induced in the bolt due to the applied loads if diameter of the bolt is 10 mm. Sol./ F = 80 kN A = 2* = 2* * 102/4 = 157 mm2 τ = = = ) 509.55 MPa.)

  11. Design of Materials Due to Stress Capacities For Direct Stress ≤ For Shear Stress ≤ Where; , : Direct and Shear Stresses due to applied loads. , : Direct and Shear Resistance Stresses of Material.

  12. Applications on Stress Ex. 3/ the frame below was supposed by an engineer to support the distributed load (w=20 kN/m) over the carrier which is supported by cable from one side and a platform from the other side, calculate the cable diameter, the carrier height (h) and the required distance from the carrier to be supported by the platform (x) if you know that tension capacity of the cable is (400 MPa.) and the compression and shear resistance of the carrier are (400 MPa.), (200MPa.) respectively.

  13. Applications on Stress Sol. / According to the equilibrium law Force on platform = force in cable = 20*5/2 = 50 kN = 50000 N -Design of Cable ≤ → ≤ 400 MP. → ≤ 400 → = 125 mm2 ≤ A 125 ≤ D2/4 → D ≥ 12.6 mm → use cable of diameter of (13 mm). -Design of carrier Shear force = compression force = 50 kN. (for carrier at platform) Calculation of (x) ≤ → ≤ 400 MP. → ≤ 400 → = 125 mm2 ≤ A 125 ≤ 10*x → x ≥ 12.5 mm → use x=13 mm. Calculation of (h) ≤ → ≤ 200 MP. → ≤ 200 → = 250 mm2 ≤ A 250 ≤ 10*h → h ≥ 25 mm → use h =30 mm.

  14. References • Megson, T. H. G. (1996). STRUCTURAL AND STRESS ANALYSIS. Oxford: Butterworth-Heinemann. • Saad, M. H. (2005). ELASTICITY Theory, Applications, and Numerics. Oxford: Butterworth-Heinemann. • Hosford, W. F. (2005). Mechanical Behavior of Materials. New York: Cambridge University Press.

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