1. MECHANICS OF MATERIALS 7th Edition
Russell C. Hibbeler
Chapter 1: Stress

2. Introduction
Mechanics of materials is a study of the relationship between the external loads on a body and the intensity of the internal loads within the body.
This subject also involves the deformations and stability of a body when subjected to external forces.

3. Equilibrium of a Deformable Body
External Forces
Surface Forces
- caused by direct contact of other body’s surface
Body Forces
- other body exerts a force without contact

4. Equilibrium of a Deformable Body
Reactions
Surface forces developed at the supports/points of contact between bodies.

5. Equilibrium of a Deformable Body
Equations of Equilibrium
Equilibrium of a body requires a balance of forces and a balance of moments
For a body with x, y, z coordinate system with origin O,
Best way to account for these forces is to draw the body’s free-body diagram (FBD).

6. Equilibrium of a Deformable Body
Internal Resultant Loadings
Objective of FBD is to determine the resultant force and moment acting within a body.
In general, there are 4 different types of resultant loadings:
a) Normal force, N
b) Shear force, V
c) Torsional moment or torque, T
d) Bending moment, M

7. Example 1.1
Determine the resultant internal loadings acting on the cross section at C of the beam.
Solution:
Free body Diagram
Distributed loading at C is found by proportion,
Magnitude of the resultant of the distributed load,
which acts from C

8. Solution: Equations of Equilibrium
Applying the equations of equilibrium we have

9. Example 1.5
Determine the resultant internal loadings acting on the cross section at B of the pipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of 50 N and a couple moment of 70 N·m at its end A. It is fixed to the wall at C.

10. Solution Calculating the weight of each segment of pipe,
Free-Body Diagram
Applying the six scalar equations of equilibrium,

11. Stress
Distribution of internal loading is important in mechanics of materials.
We will consider the material to be continuous.
This intensity of internal force at a point is called stress.

12. Stress Normal Stress σ Force per unit area acting normal to ΔA
Shear Stress τ
Force per unit area acting tangent to ΔA

13. Average Normal Stress in an Axially Loaded Bar
When a cross-sectional area bar is subjected to axial force through the centroid, it is only subjected to normal stress.
Stress is assumed to be averaged over the area.

14. Average Normal Stress in an Axially Loaded Bar
Average Normal Stress Distribution
When a bar is subjected to a constant deformation,
Equilibrium
2 normal stress components that are equal in magnitude but opposite in direction.
σ = average normal stress P = resultant normal force A = cross sectional area of bar

15. Example 1.6
The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown.
Solution:
By inspection, different sections have different internal forces.

16. Solution: Graphically, the normal force diagram is as shown.
By inspection, the largest loading is in region BC,
Since the cross-sectional area of the bar is constant, the largest average normal stress is

17. Example 1.8
The casting is made of steel that has a specific weight of Determine the average compressive stress acting at points A and B.
Solution:
By drawing a free-body diagram of the top segment, the internal axial force P at the section is
The average compressive stress becomes

18. Average Shear Stress
The average shear stress distributed over each sectioned area that develops a shear force.
2 different types of shear:
τ = average shear stress P = internal resultant shear force A = area at that section
a) Single Shear
b) Double Shear

19. Example 1.12
The inclined member is subjected to a compressive force of 3000 N. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by EDB.
Solution:
The compressive forces acting on the areas of contact are

20. Solution:
The shear force acting on the sectioned horizontal plane EDB is
Average compressive stresses along the AB and BC planes are
Average shear stress acting on the BD plane is

21. Allowable Stress
Many unknown factors that influence the actual stress in a member.
A factor of safety is needed to obtained allowable load.
The factor of safety (F.S.) is a ratio of the failure load divided by the allowable load

22. Example 1.14
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 Note in the figure that the pin is subjected to double shear.
Solution:
For equilibrium we have

23. Solution: The pin at C resists the resultant force at C. Therefore,
The pin is subjected to double shear, a shear force of kN acts over its cross-sectional area between the arm and each supporting leaf for the pin.
The required area is
Use a pin with a diameter of d = 20 mm. (Ans)

24. Example 1.17
The rigid bar AB supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is and respectively, and the failure shear stress for each pin is , determine the largest load P that can be applied to the bar. Apply a factor of safety of F.S. = 2.
Solution:
The allowable stresses are

25. Solution:
There are three unknowns and we apply the equations of equilibrium,
We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively.
For rod AC,
Using Eq. 1,
For block B,
Using Eq. 2,

26. Solution: For pin A or C, Using Eq. 1,
When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,