1. ENGR 225 Section 1.3 – 1.6

2. Internal Loadings

3. Resultant Force and Moment

4. Coplanar Loadings

5. Stress

6. Stress Derivation
A very small finite force ΔF is acting on an associated area ΔA.
Replace this force with its directional components, ΔFx, ΔFy, and ΔFz.
Take the limit as ΔA approaches zero of the quotient of the component forces and the areas
This quotient is called, stress. It represents the intensity of the force on a specified plane.

7. Normal Stress, σz
The intensity of force per unit area acting normal to ΔA.

8. Shear Stress, τzx τzy
The intensity of force per unit area acting tangent to ΔA.

9. Stress

10. Normal Stress in an Axially Loaded Bar

11. Normal Stress Distribution
dF =  dA
dF =   dA
Resultant Internal force P =   dA

12. Average Normal Stress Distribution
If we assume  is constant throughout the area or
Averaged over the cross sectional area, then
P =   dA

13. Assumptions in using relation for average normal stress
1. Uniform cross section throughout the length
2. Uniform deformation.
Applied load is along Centroidal axis
Homogenous and Isotropic material
Only axial load applied.
Weight of the bar is neglected.
Uniaxial Stress – Tensile or Compressive

15. Example 1. 7: The 80 kg lamp is supported by two rods AB and BC
Example 1.7: The 80 kg lamp is supported by two rods AB and BC. If AB has a diameter of 10 mm and BC has a diameter of 8 mm determine the average normal stress in each rod.

16. Lecture Example: Determine the stress on the floor of a 135 lb
Lecture Example: Determine the stress on the floor of a 135 lb. woman standing still wearing the following shoes that she purchased on sale. Assume that her weight is equally distributed between the heels and toes of her feet.

17. Average Shear Stress

18. Single Shear

19. Double Shear

20. Lecture Example: Determine the average normal and average shear stress developed in the wood fibers that are oriented along the section a-a.

21. Lecture Example: Determine the average compressive stress along the smooth areas of contact defined by AB and BC.

22. Lecture Example: Determine the average shear stress along the horizontal plane defined by EDB.

23. Complementary Property of Shear Stress

24. Complementary Property of Shear Stress
Two- dimensional case
zy
Force balance gives
zy = ’zy
Moment balance about x axis gives
zy = yz
Hence
zy = ’zy = yz = ’yz
yz
’yz
’zy z y

25. Allowable Stress F.S. = Ffailure / Fallowable
Failure Load : Material testing
To ensure safe working : Allowable Load < Failure Load
F.S. = Ffailure / Fallowable
Factor of Safety : accounts for
Unknown natural factors
Errors in manufacturing and assembly
Errors in load estimation
Material weathering

26. Design of Simple Connections
If  = P/A holds
F.S. = fail / allow
Hence A = P / allow

27. Rods A and B are made of steel having failure stress 510 MPa
Rods A and B are made of steel having failure stress 510 MPa. Use Factor of Safety 1.75, determine the smallest diameter to support the given load. Beam is assumed to be pin connected at A and C.

28. Determine the intensity w of the maximum distributed load that can be supported by the hanger assembly so that an allowable shear stress of 13.5 ksi is not exceeded in the 0.4 in diameter bolts at A and B and an allowable tensile stress of 22 ksi is not exceeded in the 0.5 in diameter rod AB.