1. MAE 314 – Solid Mechanics Yun Jing
Normal Stress ( )
MAE 314 – Solid Mechanics
Yun Jing
Normal Stress

2. Pretest
This is a structure which was designed to support a 30kN load, it consists of a boom AB and of a rod BC. The boom and rod are connected by a pin at B and are supported by pins and brackets at A and C. (1) Is there a reaction moment at A? why? (2) What is the reaction force in the vertical direction at A? (3) What is the internal force in AB? (4) What is the internal force in BC?

3. Statics Review
Pins = no rxn moment
Normal Stress

4. Statics Review Solve for reactions at A & C:
Ay and Cy can not be determined from these equations.
Normal Stress

5. Statics Review Consider a free-body diagram for the boom:
substitute into the structure equilibrium equation Bx
Results:
See section 1.2 in text for complete static analysis and review of method of joints.
Normal Stress

6. Normal Stress

7. Introduction to Normal Stress
Methods of statics allow us to determine forces and moments in a structure, but how do we determine if a load can be safely supported?
Factors: material, size, etc.
Need a new concept….Stress
Normal Stress

8. Introduction to Normal Stress
Stress = Force per unit area
Normal Stress

9. Introduction to Normal Stress
If stress varies over a cross-section, the resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the axis. Thus, we can write the stress at a point as
We assume the force F is evenly distributed over the cross-section of the bar. In reality F = resultant force over the end of the bar.
Normal Stress

10. Introduction to Normal Stress
Sign convention Tensile (member is in tension)
Compressive (member is in compression)
Units (force/area)
English: lb/in2 = psi
kip/in2 = ksi
SI: N/m2 = Pa (Pascal)
kN/m2 = kPa
MPa, GPa, etc.
Tensile
Compressive
Normal Stress

11. Definitions and Assumptions
Homogenous: material is the same throughout the bar
Cross-section: section perpendicular to longitudinal axis of bar
Prismatic: cross-section does not change along axis of bar F’ F A
Prismatic
Non-Prismatic
Normal Stress

12. Definitions and Assumptions
Uniaxial bar: a bar with only one axis
Normal Stress (σ): stress acting perpendicular to the cross-section.
Deformation of the bar is uniform throughout. (Uniform Stress State)
Stress is measured far from the point of application.
Loads must act through the centroid of the cross-section.
Normal Stress

13. Definitions and Assumptions
The uniform stress state does not apply near the ends of the bar.
Assume the distribution of normal stresses in an axially loaded member is uniform, except in the immediate vicinity of the points of application of the loads (Saint-Venant’s Principle).
“Uniform” Stress
Normal Stress

14. Definitions and Assumptions
How do we know all loads must act through the centroid of the cross-section?
Let us represent P, the resultant force, by a uniform stress over the cross-section (so that they are statically equivalent).
Normal Stress

15. Definitions and Assumptions
Moments due to σ:
Set Mx = Mx and My = My
Assuming pressure is uniform, then force is only applied on one point
But the moments are the same
So M \bar = M
\bar y *P = M \bar
Equations for the centroid
Normal Stress

16. Example Problem
Can the structure we used for our statics review safely support a 30 kN load? (Assume the entire structure is made of steel with a maximum allowable stress σall=165 MPa.)
Cross-section 30 mm x 50 mm
Normal Stress

17. Example Problem
Two cylindrical rods are welded together and loaded as shown. Find the normal stress at the midsection of each rod.
Normal Stress

18. Shearing and Bearing Stress (1.6-1.8, 1.12)
MAE 314 – Solid Mechanics
Yun Jing
Shearing and Bearing Stress

19. What is Shearing Stress?
We learned about normal stress (σ), which acts perpendicular to the cross-section.
Shear stress (τ) acts tangential to the surface of a material element.
Normal stress results in a volume change.
Shear stress results in a shape change.
Shearing and Bearing Stress

20. Where Do Shearing Stresses Occur?
Shearing stresses are commonly found in bolts, pins, and rivets.
Bolt is in “single” shear
Free Body Diagram of Bolt
Force P results in shearing stress
Force F results in bearing stress (will discuss later)
Shearing and Bearing Stress

21. Shearing and Bearing Stress
Shear Stress Defined
We do not assume τ is uniform over the cross-section, because this is not the case.
τ is the average shear stress.
The maximum value of τ may be considerably greater than τave, which is important for design purposes.
Shearing and Bearing Stress

22. Free Body Diagram of Bolt Free Body Diagram of Center of Bolt
Double Shear
Bolt is in “double” shear
Free Body Diagram of Bolt
Free Body Diagram of Center of Bolt
Shearing and Bearing Stress

23. Shearing and Bearing Stress
Bearing stress is a normal stress, not a shearing stress.
Thus,
where Ab = projected area where bearing pressure is applied P = bearing force
Single shear case
Read section 1.8 in text for a detailed stress analysis of a structure.
Shearing and Bearing Stress

24. Would like to determine the stresses in the members and connections of the structure shown.
From a statics analysis:
FAB = 40 kN (compression)
FBC = 50 kN (tension)
Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection

25. Rod & Boom Normal Stresses
The rod is in tension with an axial force of 50 kN.
At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is sBC = +159 MPa.
At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,
The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.
The minimum area sections at the boom ends are unstressed since the boom is in compression.

26. Pin Shearing Stresses
The cross-sectional area for pins at A, B, and C,
The force on the pin at C is equal to the force exerted by the rod BC,
The pin at A is in double shear with a total force equal to the force exerted by the boom AB,

27. Pin Shearing Stresses
Divide the pin at B into sections to determine the section with the largest shear force,
Evaluate the corresponding average shearing stress,

28. Pin Bearing Stresses
To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,
To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,

29. Equilibrium of Shear Stresses
Consider an infinitesimal element of material. Apply a single shear stress, τ1.
Total shear force on surface is (τ1)bc.
For equilibrium in the y-direction, apply τ1 on (-) surface.
For moment equilibrium about the z-axis, apply τ2 on top and bottom surfaces.
Moment equilibrium equation about z-axis:
Thus, a shear stress must be balanced by three other stresses for the element to be in equilibrium. t2 t1 t1 t2
Shearing and Bearing Stress

30. Equilibrium of Shear Stresses
What does this tell us?
Shear stresses on opposite (parallel) faces of an element are equal in magnitude and opposite in direction.
Shear stress on adjacent (perpendicular) faces of an element are equal in magnitude and both point towards or away from each other.
Sign convention for shear stresses
Positive face – normal is in (+) x, y, or z direction
Negative face - normal is in (-) x, y, or z direction + t2 t1
Face
Direction
Shear Stress + - t1 + - t2 -
Shearing and Bearing Stress

31. Define General State of Stressy z x
σx = stress in x-direction applied in the plane normal to x-axis
σ y = stress in y-direction applied in the plane normal to y-axis
σ z = stress in z-direction applied in the plane normal to z-axis
τxy = stress in y-direction applied in the plane normal to x-axis
τ xz = stress in z-direction applied in the plane normal to x-axis
τ zy = stress in y-direction applied in the plane normal to z-axis
And so on…
Shearing and Bearing Stress

32. Define General State of Stressy z x
There are 9 components of stress:
σx, σ y, σ z, τxy, τ xz, τ yx, τ yz, τ zx, τ zy
As shown previously, in order to maintain equilibrium:
τ xy= τ yx, τ xz = τ zx, τ yz = τ zy
There are only 6 independent stress components.
Shearing and Bearing Stress

33. Shearing and Bearing Stress
Example Problem
A load P = 10 kips is applied to a rod supported as shown by a plate with a 0.6 in. diameter hole. Determine the shear stress in the rod and the plate.
Shearing and Bearing Stress

34. Shearing and Bearing Stress
Example Problem
Link AB is used to support the end of a horizontal beam. If link AB is subject to a 10 kips compressive force determine the normal and bearing stress in the link and the shear stress in each of the pins.
Shearing and Bearing Stress

35. Oblique Planes and Design Considerations (1.11, 1.13)
MAE 314 – Solid Mechanics
Yun Jing
Oblique Planes and Design Considerations

36. Stress on an Oblique Plane
What have we learned so far?
Axial forces in a two-force member cause normal stresses.
Transverse forces exerted on bolts and pins cause shearing stresses.
Oblique Planes and Design Considerations

37. Stress on an Oblique Plane
Axial forces cause both normal and shearing stresses on planes which are not perpendicular to the axis.
Consider an inclined section of a uniaxial bar.
The resultant force in the axial direction must equal P to satisfy equilibrium.
The force can be resolved into components perpendicular to the section, F, and parallel to the section, V.
The area of the section is
Oblique Planes and Design Considerations

38. Stress on an Oblique Plane
We can formulate the average normal stress on the section as
The average shear stress on the section is
Thus, a normal force applied to a bar on an inclined section produces a combination of shear and normal stresses.
Oblique Planes and Design Considerations

39. Stress on an Oblique Plane
Since σ and τ are functions of sine and cosine, we know the maximum and minimum values will occur at θ = 00, 450, and 900.
At θ=±900 σ=0
At θ=±450 σ=P/2A0
At θ=00 σ=P/A0 (max)
At θ=±900 τ=0
At θ=±450 τ=P/2A0 (max)
At θ=00 τ=0
Oblique Planes and Design Considerations

40. Stress on an Oblique Plane
What does this mean in reality?
Oblique Planes and Design Considerations

41. Design Considerations
From a design perspective, it is important to know the largest load which a material can hold before failing.
This load is called the ultimate load, Pu.
Ultimate normal stress is denoted as σu and ultimate shear stress is denoted as τu.
Oblique Planes and Design Considerations

42. Design Considerations
Often the allowable load is considerably smaller than the ultimate load.
It is a common design practice to use factor of safety.
Oblique Planes and Design Considerations

43. Oblique Planes and Design Considerations
Example Problem
Two wooden members are spliced as shown. If the maximum allowable tensile stress in the splice is 75 psi, determine the largest load that can be safely supported and the shearing stress in the splice.
Oblique Planes and Design Considerations

44. Oblique Planes and Design Considerations
Example Problem
A load is supported by a steel pin inserted into a hanging wooden piece. Given the information below, determine the load P if an overall factor of safety of 3.2 is desired.
Oblique Planes and Design Considerations