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Designing Simple Connections and Determining Allowable Stresses

Learn how to design connections and assess material safety by computing stresses and applying factors of safety. This lecture covers ultimate and allowable stresses, factors of safety, and working area strain in simple connections.

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Designing Simple Connections and Determining Allowable Stresses

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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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