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Lecture # 2

Lecture # 2. Allowable Stress. Objective:. To design simple connections and determine weather the material fails or safe, taking into consideration the computed stresses and the natural strength. Stress. ultimate. failure. x. allowable. Ultimate stress Allowable stress.

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Lecture # 2

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  1. Lecture # 2 Allowable Stress Objective: To design simple connections and determine weather the material fails or safe, taking into consideration the computed stresses and the natural strength.

  2. Stress ultimate failure x allowable Ultimate stress Allowable stress Factor of safety = F.S. = Working area Strain Allowable Stress

  3. Normal and Shear Stress

  4. Bearing Stress

  5. The two members are pinned to gather at B. Top view of the pin connections at A and B are also given in the figure. If the pin have allowable shear stress of allow = 90 MPa and the allowable tensile stress of rod CB is (t)allow = 115 MPa, determine to the nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load. Example:

  6. Solution:

  7. dA = 7 mm Ans…. dB = 10 mm Ans…. Diameter of the Pins.

  8. We will Choose dBC = 9 mm Ans….. Diameter of Rod.

  9. The control arm is subjected to the loading.\ Determine to the nearest 5 mm the required diameter of the steel pin at C if the allowable shear stress for the steel is allow = 55 MPa. Note: in the figure that the pin is subjected to double shear. Example:

  10. Solution:

  11. Use a pin having a diameter of d = 20 mm Ans…. Required Area:

  12. (a) (b) Example:

  13. Solution Diameter of Rod:

  14. Since the sectioned area A = 2(0.02 m)(t), the required thickness of the disk is Thickness of disk:

  15. Example:

  16. Solution Normal Stress:

  17. Bearing Stress:

  18. The enD

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