Lecture # 2 Allowable Stress Objective:

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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.

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

Normal and Shear Stress

Bearing Stress

Example: 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.

Solution:

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

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

Example: 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.

Solution:

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

Example: (a) (b)

Solution Diameter of Rod:

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

Example:

Solution Normal Stress:

Bearing Stress:

The enD