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

3. Normal and Shear Stress

4. Bearing Stress

5. 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.

6. Solution:

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

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

9. 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.

10. Solution:

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

12. Example:
(a)
(b)

13. Solution
Diameter of Rod:

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

15. Example:

16. Solution
Normal Stress:

17. Bearing Stress:

18. The enD